Keywords

1 Introduction

The Wigner–Eckart theorem [1, 2] has a wide application when considering collective effects associated with spin ordering in systems of identical particles that possess angular moments of arbitrary magnitude. This includes the effects of spin order in ferro- and antiferromagnetic electronic systems, as well as magnetic effects that occur in normal atomic Bose and Fermi condensates. To describe spin ordering in systems of spin 1/2 particles, the Hamiltonian is traditionally used, written in the spin representation in the form obtained by Heisenberg, Dirac, and Van Vleck [3,4,5]. A system of particles with spin different from 1/2, “high” spin, cannot be described in the spin representation using the Heisenberg–Dirac–Van Vleck Hamiltonian, since this Hamiltonian implies taking into account the symmetry properties of the total wave function for a system of particles with spin 1/2.

Systems of particles with spins other than 1/2, or with so-called “high” spins, have a number of properties that are fundamentally different from the properties of similarly arranged systems of localized 1/2 spins. The structure of the ground state, the spectrum of excitations, behavior at critical points—everything has a character that is completely different from the behavior of magnetic systems with half spin. Intensive theoretical studies of anisotropic antiferromagnets with a particle spin of 1 [6,7,8] were stimulated by experiments on a Bose–Einstein condensate confined by optical traps that do not destroy the degrees of freedom due to the spin of atoms, which made it possible to observe macroscopic quantum phenomena associated with spin ordering [6]. Phenomena such as fragmentation of the ground state were discovered, when not one, but several ground states with different magnetic orderings of the phase are observed [6, 7]. Thus, for systems with spins s = 1, the ground state is a singlet state, although magnetic fluctuations are very strong with the appearance of the so-called magnetic or cyclic phase [8]. For systems with spins s = 2, the ground state is ferromagnetic [9]. After successful cooling of atomic Bose systems and their theoretical description [10, 11], experiments were carried out to cool the Fermi system, in particular K40, when 7 × 105 atoms were cooled to a degeneracy temperature below 300 nK [12].

New phenomena have been discovered in the Fermi system, such as a scaly structure of spatial distribution [13, 14], suppression of elastic and inelastic collisions [15, 16], and the existence of zero sound at low temperatures [17,18,19,20,21,22,23].

The model Hamiltonian is the cornerstone of the mathematical description of high-spin systems when the exchange interaction is taken into account. It describes the entire system of pairwise interacting particles and usually contains terms introduced phenomenologically, which contain the sum of scalar products of spin operators in high powers (biquadratic, bicubic, and so on) and depend on the total spin of each pair of atoms. It is thanks to these terms that it is possible to explain the presence in the system of atoms of nonlinear effects associated with magnetic fluctuations of the ground state.

In the theory of condensed matter, studies of the magnetic properties of systems of identical particles with high spin began long before the experimental production of the Bose–Einstein condensate in 1995 by the work of Haldane [24, 25]. To describe the antiferromagnetic chain, Haldane and later his followers Affleck, Kennedy, Lieb, and Tasaki [26, 27] used a similar Hamiltonian within the extended S = 1 model, taking into account the biquadratic term in the exchange interaction with indefinite coefficients, with the help of which the intermediate phases were analyzed. Various phase diagrams of transitions near the critical point with different possible scenarios for the development of events, which directly depend on the ratio of the coefficients in the original Hamiltonian, were obtained numerically. In the semiclassical approximation, a soliton solution for excitations was found, and intermediate phases were also analyzed by studying the extended S = 1 model of resonating valence bonds taking into account the biquadratic term in the exchange interaction.

$$\hat{H}_{AKLT} = J\sum\limits_{i} {\left\{ {\left( {{\hat{\mathbf{S}}}_{i} \cdot {\hat{\mathbf{S}}}_{i + 1} } \right) + \beta \left( {{\hat{\mathbf{S}}}_{i} \cdot {\hat{\mathbf{S}}}_{i + 1} } \right)^{2} } \right\}} .$$
(1)

For example, it was shown in [21] that such a Hamiltonian admits the existence of solutions for the coefficient β = 1/3; in this case, the ground state is a stable simple valence bond (VBS), which is separated from the excitations by a gap. Since the ground state at β = 0 (as shown in Haldane's work [24, 25] exhibits a linear long-range order correlation and is adiabatically related to the β = 1/3 state, Schollwöck [28] concludes that the Haldane phase carries VBS character.

All the variety of phases predicted theoretically for the spin-1 chain is directly related to the values ​​of the coefficients at the bilinear and biquadratic terms in the exchange interaction Hamiltonian written in the spin representation. The problem with the description of the spin-1 system of identical particles by the mentioned Hamiltonian Eq. (1) is that it gives the splitting of energy of each couple of particles on three sublevels, despite the fact that in this case energy splitting should be only on two sublevels—symmetric and antisymmetric if remember the reason of this splitting.

$$\begin{gathered} E_{AKLT} = \left\{ \begin{gathered} J(1 + \beta ),\,(S = 2), \hfill \\ J( - 1 + \beta ),\,(S = 1), \hfill \\ J( - 2 + 4\beta ),\,(S = 0), \hfill \\ \end{gathered} \right. \hfill \\ \beta = 1/3 \hfill \\ E_{Sch} = \left\{ \begin{gathered} \frac{4}{3}J,(S = 2),\,symmetric \hfill \\ - \frac{2}{3}J,(S = 1),\,antisymmetric \hfill \\ - \frac{2}{3}J,(S = 0),\,symmetric \hfill \\ \end{gathered} \right. \hfill \\ \beta = 0 \hfill \\ E_{VBS} = \left\{ \begin{gathered} J,(S = 2),\,\,symmetric \hfill \\ - J,(S = 1),\,antisymmetric \hfill \\ - 2J,(S = 0),\,symmetric. \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$$

In all these cases, the symmetric state has two different sublevels, though in the coordinate representation, the same energy correction for two identical particles has two sublevels, symmetric and antisymmetric: \(\varepsilon^{(1)} = K \pm A\).

Therefore, knowing the exact form of the Hamiltonian describing the behavior of a spin-1 system is a fundamental point for the statistical description of systems with a “high” spin and analysis of critical phenomena occurring in such systems. In [24,25,26], the Hamiltonian of a system of particles with a spin other than 1/2 was obtained from first principles; following these works, in this chapter, we will derive in detail the Hamiltonian of the exchange interaction in the spin representation and consider some effects of magnetic ordering in systems of identical particles with arbitrary spin.

2 General Prerequisites and Provisions of the Theory of Angular Momentum

Before proceeding to the description of spin interactions in systems of identical particles, let us remind the reader of some general relations of the theory of angular momenta. We have already noted above that angular momentum in classical mechanics is introduced as a conserved quantity (integral of motion) in the presence of an additional symmetry of space with respect to rotation associated with the property of isotropy of space.

2.1 Turn Operator

Consider two coordinate systems with a common origin at some point Q, S, and S′. Rotation, as an isometric transformation between two systems that do not change the position of the origin, can be defined in terms of two alternative approximations [2]. The first (“passive” turn) is that the physical body itself remains motionless. In this case, the transformation S → S′ means the rotation of the coordinate system. In this case, the physical vector or tensor at some point Q in the body does not change; however, the direction of the basis vectors in both coordinate systems is different. Consequently, the components of vectors and tensors have different values ​​in both coordinate systems, and these values ​​must be related by a linear rotation transformation. The Cartesian coordinates of the point Q, x, y, z and x′, y′, z′, are defined relative to the corresponding systems, S and S′.

Within the framework of the second description (“active” rotation), a physical body undergoes a real rotation. The initial coordinate system, S, is called “laboratory.” The coordinate system rigidly connected to the body, S′, rotates relative to the laboratory one. Rotation of the body means that, for example, the vector r at some point undergoes a rotation with the body and receives a new value r . Cartesian coordinates of the point Q, x, y, z and x′, y′, z′, are determined relative to the laboratory coordinate system.

Both methods of description are widely used in various applications. Mathematically, they are related as “forward” and “inverse” transformations.

Isometry means that regardless of the method of describing rotations, these coordinate transformations leave the lengths of the vectors unchanged. If l is a vector connecting two points in three-dimensional space, P (x1, y1, z1) and Q (x2, y2, z2), then

$$\begin{gathered} \left| {\mathbf{l}} \right| = \left| {P\left( {x_{1} ,y_{1} ,z_{1} } \right) - Q\left( {x_{2} ,y_{2} ,z_{2} } \right)} \right| = \left| {P\left( {x^{\prime}_{1} ,y^{\prime}_{1} ,z^{\prime}_{1} } \right) - Q\left( {x^{\prime}_{2} ,y^{\prime}_{2} ,z^{\prime}_{2} } \right)} \right| = \hfill \\ = \sqrt {\left( {x_{1} - x_{2} } \right)^{2} + \left( {y_{1} - y_{2} } \right)^{2} + \left( {z_{1} - z_{2} } \right)^{2} } \hfill \\ = \sqrt {\left( {x^{\prime}_{1} - x^{\prime}_{2} } \right)^{2} + \left( {y^{\prime}_{1} - y^{\prime}_{2} } \right)^{2} + \left( {z^{\prime}_{1} - z^{\prime}_{2} } \right)^{2} } . \hfill \\ \end{gathered}$$
(2)

The rotations of the coordinate system do not violate the mutual orientation of the coordinate axes, that is, the right coordinate system remains right after the rotations, and the left one remains left. To demonstrate the derivation of an explicit expression for the rotation operator, it is convenient first to consider active rotation. In this case, the rotation occurs at an angle φ around the axis determined by the direction of the unit vector n. We are considering isotropic space. The physical properties of the body do not depend on rotations in such a space; however, the functional description of the body depends on them. As mentioned above, active rotation SS′ changes the direction of vector r. The new vector is r (Fig. 1).

Fig. 1
A diagram depicts the turn description in a coordinate system. The rotation occurs at an angle phi around the axis of the unit vector n.

Turn description

The change in the radius vector r during rotation can be described as follows:

$$\begin{gathered} \left| {\delta {\mathbf{r}}} \right| = r\sin \theta \delta \varphi , \hfill \\ \delta {\mathbf{\varphi }}\parallel {\mathbf{z}}, \Rightarrow \hfill \\ \delta {\mathbf{r}} = \delta {\mathbf{\varphi }} \times {\mathbf{r}}. \hfill \\ \end{gathered}$$
(3)

Similarly, the change in the velocity vector when turning will be as follows:

$$\begin{gathered} \left| {\delta {\mathbf{v}}} \right| = v\sin \theta \delta \varphi , \hfill \\ \delta {\mathbf{\varphi }}\parallel {\mathbf{z}}, \Rightarrow \hfill \\ \delta {\mathbf{v}} = \delta {\mathbf{\varphi }} \times {\mathbf{v}}. \hfill \\ \end{gathered}$$
(4)

To demonstrate the derivation of an explicit expression for the rotation operator, it is convenient to first consider active rotation. In this case, the rotation occurs at an angle φ around the axis determined by the direction of the unit vector n. We are considering isotropic space. The physical properties of the body do not depend on rotations in such a space; however, the functional description of the body depends on them. Let there be some scalar coordinate function that describes the body in the laboratory coordinate system S (x, y, z). Here, we do not specify the properties of this function, except that it must belong to the class of differentiable functions. As stated above, active rotation changes the direction of the vector r. The new vector is r. The function \(\Psi \left( {\mathbf{r}} \right)\) is converted to a function \(\Psi \left( {{\mathbf{r}}^{\prime}} \right)\) to be linked to and from the initial function by linear transformations

$$\begin{gathered} \Psi \left( {{\mathbf{r}}^{\prime}} \right) = \hat{\mathscr{R}}_{a} \left( {{\mathbf{n}},\varphi } \right)\Psi \left( {\mathbf{r}} \right), \hfill \\ \Psi \left( {\mathbf{r}} \right) = \hat{\mathscr{R}}_{a}^{ - 1} \left( {{\mathbf{n}},\varphi } \right)\Psi \left( {{\mathbf{r}}^{\prime}} \right), \hfill \\ \end{gathered}$$
(5)

where \(\hat{\mathscr{R}}_{a}\) is the operator of active rotation, and \(\hat{\mathscr{R}}_{a}^{ - 1}\) is the operator of inverse active rotation.

$$\begin{gathered} \delta {\mathbf{r}} = \delta {\mathbf{\varphi }} \times {\mathbf{r}} \equiv \delta \varphi {\mathbf{n}} \times {\mathbf{r}}, \hfill \\ \Psi \left( {{\mathbf{r}}^{\prime}} \right) = \hat{\mathscr{R}}_{a} \left( {\delta \varphi ,{\mathbf{n}}} \right)\Psi \left( {\mathbf{r}} \right), \hfill \\ {\mathbf{r}}^{\prime} = {\mathbf{r}} + \delta {\mathbf{r}}, \hfill \\ \Psi \left( {{\mathbf{r}}^{\prime}} \right) = \Psi \left( {{\mathbf{r}} + \delta {\mathbf{r}}} \right) = \sum\limits_{\nu = 0}^{\infty } {\frac{1}{\nu !}} \left( {\delta {\mathbf{r}} \cdot \frac{\partial }{{\partial {\mathbf{r}}}}} \right)^{\nu } \Psi \left( {\mathbf{r}} \right) = \hfill \\ = \sum\limits_{\nu = 0}^{\infty } {\frac{1}{\nu !}} \left( {\delta \varphi {\mathbf{n}} \times {\mathbf{r}} \cdot \frac{\partial }{{\partial {\mathbf{r}}}}} \right)^{\nu } \Psi \left( {\mathbf{r}} \right) = \sum\limits_{\nu = 0}^{\infty } {\frac{1}{\nu !}} \left( {\delta \varphi {\mathbf{n}} \times {\mathbf{r}} \cdot \nabla } \right)^{\nu } \Psi \left( {\mathbf{r}} \right) = \hfill \\ = \sum\limits_{\nu = 0}^{\infty } {\frac{1}{\nu !}} \left( {{\mathbf{r}} \times \nabla \cdot {\mathbf{n}}\delta \varphi } \right)^{\nu } \Psi \left( {\mathbf{r}} \right) = \sum\limits_{\nu = 0}^{\infty } {\frac{1}{\nu !}} \left( {\delta \varphi {\mathbf{n}} \cdot {\mathbf{r}} \times \nabla } \right)^{\nu } \Psi \left( {\mathbf{r}} \right) = e^{{\delta \varphi {\mathbf{n}} \cdot {\mathbf{r}} \times \nabla }} \Psi \left( {\mathbf{r}} \right). \hfill \\ \end{gathered}$$
(6)

Thus, we have a finite rotation operator in the form:

$$\hat{\mathscr{R}}_{a} = e^{{\delta \varphi {\mathbf{n}} \cdot {\mathbf{r}} \times \nabla }} .$$
(7)

2.2 Angular Momentum Conservation

In the presence of isotropy of space, the Lagrange function describing a physical system should not change with any rotation (active or passive), which means

$$\begin{gathered} L\left( {{\mathbf{r}} + \delta {\mathbf{r}},{\mathbf{v}} + \delta {\mathbf{v}}} \right) = L\left( {{\mathbf{r}},{\mathbf{v}}} \right) \hfill \\ \delta {\mathbf{r}} = \delta {\mathbf{\varphi }} \times {\mathbf{r}} \hfill \\ \delta {\mathbf{v}} = \delta {\mathbf{\varphi }} \times {\mathbf{v}}. \hfill \\ \end{gathered}$$
(8)

Let us expand the Lagrange function in a series in a small parameter, taking into account the smallness of the angle of rotation, we can restrict ourselves to the first term of the expansion:

$$\begin{gathered} L\left( {{\mathbf{r}},{\mathbf{v}}} \right) = L\left( {{\mathbf{r}} + \delta {\mathbf{r}},{\mathbf{v}} + \delta {\mathbf{v}}} \right) = L\left( {{\mathbf{r}},{\mathbf{v}}} \right) + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{r}}}} \cdot \delta {\mathbf{r}} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}}} \cdot \delta {\mathbf{v}} + ..., \Rightarrow \hfill \\ \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{r}}}} \cdot \delta {\mathbf{r}} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}}} \cdot \delta {\mathbf{v}} = 0, \hfill \\ \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{r}}}} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}}} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{v}} = \left( {\frac{d}{dt}\frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}}}} \right) \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}} \hfill \\ + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}}} \cdot \frac{d}{dt}\left( {\delta {\mathbf{\varphi }} \times {\mathbf{r}}} \right) = \left( {\frac{d}{dt}{\mathbf{p}}} \right) \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}} + {\mathbf{p}} \cdot \frac{d}{dt}\left( {\delta {\mathbf{\varphi }} \times {\mathbf{r}}} \right) = \frac{d}{dt}\left( {{\mathbf{p}} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}}} \right) \hfill \\ = \frac{d}{dt}\left( {\delta {\mathbf{\varphi }} \cdot {\mathbf{r}} \times {\mathbf{p}}} \right) = \delta {\mathbf{\varphi }} \cdot \frac{d}{dt}\left( {{\mathbf{r}} \times {\mathbf{p}}} \right) = 0 \hfill \\ \Rightarrow \left( {{\mathbf{r}} \times {\mathbf{p}}} \right) = \mathscr{I}_{\varphi } . \hfill \\ \end{gathered}$$
(9)

This conserved quantity (integral of motion) \(\mathscr{I}_{\varphi }\) is the angular momentum of the particle:

$$\left( {{\mathbf{r}} \times {\mathbf{p}}} \right) = {\mathbf{m}}.$$
(10)

Consider now a system consisting of a large number of particles, then the requirement for the Lagrange function, due to the isotropy of space, leads to the following relations:

$$\begin{gathered} L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right) = \hfill \\ = L\left( {{\mathbf{r}}_{1} + \delta {\mathbf{r}}_{1} ,{\mathbf{r}}_{2} + \delta {\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} + \delta {\mathbf{r}}_{N} ,{\mathbf{v}}_{1} + \delta {\mathbf{v}}_{1} ,{\mathbf{v}}_{2} + \delta {\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} + \delta {\mathbf{v}}_{N} } \right) = \hfill \\ = L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right) + \hfill \\ + \sum\limits_{i = 1}^{N} {\left( {\frac{{\partial L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right)}}{{\partial {\mathbf{r}}_{i} }} \cdot \delta {\mathbf{r}}_{i} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}_{i} }} \cdot \delta {\mathbf{v}}_{i} + ...} \right)} \Rightarrow \hfill \\ \sum\limits_{i = 1}^{N} {\left( {\frac{{\partial L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right)}}{{\partial {\mathbf{r}}_{i} }} \cdot \delta {\mathbf{r}}_{i} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}_{i} }} \cdot \delta {\mathbf{v}}_{i} } \right)} = 0, \hfill \\ \sum\limits_{i = 1}^{N} {\left( {\frac{{\partial L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right)}}{{\partial {\mathbf{r}}_{i} }} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}_{i} }} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{v}}_{i} } \right)} = \hfill \\ = \sum\limits_{i = 1}^{N} {\left( {\frac{d}{dt}\left( {\frac{{\partial L\left( {{\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ,...{\mathbf{r}}_{N} ,{\mathbf{v}}_{1} ,{\mathbf{v}}_{2} ,...{\mathbf{v}}_{N} } \right)}}{{\partial {\mathbf{v}}_{i} }}} \right) \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} + \frac{{\partial L\left( {{\mathbf{r}},{\mathbf{v}}} \right)}}{{\partial {\mathbf{v}}_{i} }} \cdot \frac{d}{dt}\left( {\delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} } \right)} \right)} = \hfill \\ = \sum\limits_{i = 1}^{N} {\left( {\frac{d}{dt}\left( {{\mathbf{p}}_{i} } \right) \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} + {\mathbf{p}}_{i} \cdot \frac{d}{dt}\left( {\delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} } \right)} \right)} = \frac{d}{dt}\sum\limits_{i = 1}^{N} {{\mathbf{p}}_{i} \cdot \delta {\mathbf{\varphi }} \times {\mathbf{r}}_{i} = } \delta {\mathbf{\varphi }}\frac{d}{dt}\sum\limits_{i = 1}^{N} {{\mathbf{r}}_{i} \times {\mathbf{p}}_{i} } = 0, \hfill \\ \Rightarrow \sum\limits_{i = 1}^{N} {{\mathbf{r}}_{i} \times {\mathbf{p}}_{i} } = \sum\limits_{i = 1}^{N} {{\mathbf{m}}_{i} } = \mathscr{I}_{\varphi } \equiv {\mathbf{M}}. \hfill \\ \end{gathered}$$
(11)

Thus, the integral of motion in the presence of space isotropy is the total angular momentum of the system.

2.3 Angular Momentum in Quantum Mechanics

As is known, the correspondence principle remains valid in the formalism of quantum mechanics. That is, operators of these physical quantities correspond to all physical quantities, and the relations between these operators retain the same form as they had in classical mechanics; the same applies to dynamic relations. Thus, the operator of the angular momentum of one particle will have the form Eq. (9), with the only difference that the momentum of the particle p and the radius vector r that determines its location in space will be represented by the corresponding operators.

$$\begin{gathered} {\hat{\mathbf{m}}} = \hbar {\hat{\mathbf{j}}} = {\hat{\mathbf{r}}} \times {\hat{\mathbf{p}}}, \Rightarrow \hfill \\ {\hat{\mathbf{j}}} = \frac{1}{\hbar }{\hat{\mathbf{r}}} \times {\hat{\mathbf{p}}}, \hfill \\ \hat{j}_{x} ,\hat{j}_{y} ,\hat{j}_{z} , \hfill \\ {\hat{\mathbf{j}}} = \hat{j}_{x} {\mathbf{e}}_{x} + \hat{j}_{y} {\mathbf{e}}_{y} + \hat{j}_{z} {\mathbf{e}}_{z} , \hfill \\ \end{gathered}$$
(1)

where the unit for moment of momentum is the same as the unit for “action” ([m][r] [r]/[t]]) in classical mechanics. In the quantum-mechanical expression for the moment of momentum operator, the dimension of “action” is taken into account by Planck’s constant, which expresses the quantum of action.

The operators of the projections of the angular momentum on the Cartesian axes do not commute with each other, which means that it is impossible to simultaneously exact measure any two projections on the corresponding axes. (see the Robertson–Schrödinger uncertainty [29,30,31].)

$$\begin{gathered} \left[ {\hat{j}_{x} \hat{j}_{y} } \right] = i\hat{j}_{z} ,\left[ {\hat{j}_{y} \hat{j}_{z} } \right] = i\hat{j}_{x} ,\left[ {\hat{j}_{z} \hat{j}_{x} } \right] = i\hat{j}_{y} \hfill \\ \left[ {\hat{j}_{\alpha } \hat{j}_{\beta } } \right] = i\varepsilon_{xyz} \hat{j}_{\gamma } , \hfill \\ \Delta A\Delta B \ge \left| {\left\langle { - \frac{i}{2}\left[ {\hat{A}\hat{B}} \right]} \right\rangle } \right|, \Rightarrow \Delta j_{\alpha } \Delta j_{\beta } \ge \left| {\left\langle { - \frac{i}{2}\left[ {\hat{j}_{\alpha } \hat{j}_{\beta } } \right]} \right\rangle } \right| = \left| {\left\langle { - \frac{i}{2}i\varepsilon_{xyz} \hat{j}_{\gamma } } \right\rangle } \right| = \frac{1}{2}\left\langle {j_{\gamma } } \right\rangle . \hfill \\ \end{gathered}$$
(13)

The operator of the square of the angular momentum is introduced, as in classical mechanics, as the scalar product:

$$\begin{gathered} \widehat{{j^{2} }} = {\hat{\mathbf{j}}} \cdot {\hat{\mathbf{j}}} = \hat{j}_{x}^{2} + \hat{j}_{y}^{2} + \hat{j}_{z}^{2} , \hfill \\ \left[ {\widehat{{j^{2} }}\hat{j}_{x} } \right] = \left[ {\widehat{{j^{2} }}\hat{j}_{y} } \right] = \left[ {\widehat{{j^{2} }}\hat{j}_{z} } \right] = 0. \hfill \\ \end{gathered}$$
(14)

It is clearly seen that the operator of the square of the angular momentum commutes with the operator of the projection of the moment on any axis. This means that the square of the angular momentum and any of its projections are simultaneously precisely measurable quantities. Let us set one of the projections, let it be the operator of the projection of the moment on the z-axis, then both of these operators \(\hat{j}^{2} ,\hat{j}_{z}\) have a common set \(\left\{ {\left| {\lambda m} \right\rangle } \right\}\) of eigenvectors of the state, where λ–is an eigenvalue of the operator \(\hat{j}^{2}\) and m is an eigenvalue of the angular momentum projection operator \(\hat{j}_{z}\):

$$\begin{gathered} 0 = \left[ {\hat{j}^{2} \hat{j}_{z} } \right], \Rightarrow \hfill \\ \hat{j}_{z} \left| {\lambda m} \right\rangle = m\left| {\lambda m} \right\rangle \hfill \\ \hat{j}^{2} \left| {\lambda m} \right\rangle = \lambda \left| {\lambda m} \right\rangle . \hfill \\ \end{gathered}$$
(15)

The set of eigenvectors \(\left\{ {\left| {\lambda m} \right\rangle } \right\}\) is orthogonal and complete since Eq. (15) corresponds to the properties of the Sturm–Liouville problem:

$$\begin{gathered} \left\langle {\lambda m} \right|\left. {\lambda ^{\prime}m^{\prime}} \right\rangle = \delta_{\lambda \lambda ^{\prime}} \delta_{mm^{\prime}} , \hfill \\ \sum\limits_{{m_{\min } }}^{{m_{\max } }} {\left| {\lambda m} \right\rangle } \left\langle {\lambda m} \right| = \hat{1}. \hfill \\ \end{gathered}$$
(16)

Let us present the operators \(\hat{j}^{ + }\) and \(\hat{j}^{ - }\) as follows:

$$\begin{gathered} \hat{j}^{ + } = \hat{j}_{x} + i\hat{j}_{y} , \hfill \\ \hat{j}^{ - } = \hat{j}_{x} - i\hat{j}_{y} , \hfill \\ \hat{j}^{ - } \equiv \hat{j}^{ + \dag } = \hat{j}^{\dag }_{x} - i\hat{j}^{\dag }_{y} \equiv \hat{j}_{x} - i\hat{j}_{y} , \hfill \\ \hat{j}^{ - \dag } = \hat{j}^{\dag }_{x} + i\hat{j}^{\dag }_{y} = \hat{j}_{x} + i\hat{j}_{y} = \hat{j}^{ + } . \hfill \\ \end{gathered}$$
(16)

These operators are Hermitian conjugate one to another. A direct product of these operators gives the following expression:

$$\begin{gathered} \hat{j}^{ + } \cdot \hat{j}^{ - } = \hat{j}^{2}_{x} + \hat{j}^{2}_{y} - i\hat{j}_{x} \hat{j}_{y} + i\hat{j}_{y} \hat{j}_{x} = \hat{j}^{2}_{x} + \hat{j}^{2}_{y} - i\left( {\hat{j}_{x} \hat{j}_{y} - \hat{j}_{y} \hat{j}_{x} } \right) = \hat{j}^{2}_{x} + \hat{j}^{2}_{y} + \hat{j}_{z} = \hfill \\ = \hat{j}^{2} - \hat{j}_{z}^{2} + \hat{j}_{z} , \hfill \\ \hat{j}^{ - } \cdot \hat{j}^{ + } = \hat{j}^{ + } \cdot \hat{j}^{ - } - 2\hat{j}_{z} = \hat{j}^{2} - \hat{j}_{z}^{2} - \hat{j}_{z} , \hfill \\ \end{gathered}$$
(17)

and a commutator of these operators is

$$\begin{gathered} \left[ {\hat{j}^{ + } \hat{j}^{ - } } \right] = \left[ {\left( {\hat{j}_{x} + i\hat{j}_{y} } \right)\left( {\hat{j}_{x} - i\hat{j}_{y} } \right)} \right] = - ii\hat{j}_{z} + i( - i)\hat{j}_{z} = 2\hat{j}_{z} , \hfill \\ \hat{j}^{ + } \hat{j}^{ - } - \hat{j}^{ - } \hat{j}^{ + } = 2\hat{j}_{z} . \hfill \\ \end{gathered}$$
(18)

Let us prove the following statement:

Lemma_I.

$$\underline{{m^{2} \le \lambda }} \left\{ { - \sqrt \lambda \le m \le \sqrt \lambda } \right\}.$$

Proof:

$$\begin{gathered} \hat{j}^{2} = \hat{j}_{x}^{2} + \hat{j}_{y}^{2} + \hat{j}_{z}^{2} , \hfill \\ \hat{j}_{x} = \hat{j}_{x}^{\dag } , \Rightarrow \left\langle {\lambda m^{\prime}} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle = \left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda m^{\prime}} \right\rangle^{*} , \hfill \\ \hat{j}_{y} = \hat{j}_{y}^{\dag } , \Rightarrow \left\langle {\lambda m^{\prime}} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle = \left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda m^{\prime}} \right\rangle^{*} , \hfill \\ \left\langle {\lambda m} \right|\hat{j}^{2} \left| {\lambda m} \right\rangle = \left\langle {\lambda m} \right|\hat{j}_{z}^{2} \left| {\lambda m} \right\rangle + \left\langle {\lambda m} \right|\hat{j}_{x} \hat{j}_{x} \left| {\lambda m} \right\rangle + \left\langle {\lambda m} \right|\hat{j}_{y}^{2} \left| {\lambda m} \right\rangle , \hfill \\ \lambda = m^{2} + \left\langle {\lambda m} \right|\hat{j}_{x} \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left| {\lambda m^{\prime}} \right\rangle } \left\langle {\lambda m^{\prime}} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle + \left\langle {\lambda m} \right|\hat{j}_{y} \sum\limits_{{m_{\min } }}^{{m_{\max } }} {\left| {\lambda m^{\prime}} \right\rangle } \left\langle {\lambda m^{\prime}} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle = \hfill \\ = m^{2} + \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda m^{\prime}} \right\rangle } \left\langle {\lambda m^{\prime}} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle + \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda m^{\prime}} \right\rangle } \left\langle {\lambda m^{\prime}} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle = \hfill \\ = m^{2} + \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left| {\left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda m^{\prime}} \right\rangle } \right|}^{2} + \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left| {\left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda m^{\prime}} \right\rangle } \right|}^{2} , \hfill \\ \sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left| {\left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda m^{\prime}} \right\rangle } \right|}^{2} \ge 0,\sum\limits_{{m^{\prime} = m_{\min } }}^{{m_{\max } }} {\left| {\left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda m^{\prime}} \right\rangle } \right|}^{2} \ge 0, \Rightarrow \hfill \\ \lambda \ge m^{2} . \hfill \\ \end{gathered}$$
(19)

Let us prove the next statement:

Lemma_II.

$$\begin{gathered} \hat{j}^{ + } \left| {\lambda m} \right\rangle = const\left| {\lambda (m + 1)} \right\rangle , \hfill \\ \hat{j}^{ - } \left| {\lambda m} \right\rangle = const\left| {\lambda (m - 1)} \right\rangle . \hfill \\ \end{gathered}$$

Proof:

$$\begin{gathered} ]\left| \chi \right\rangle = \hat{j}^{ + } \left| {\lambda m} \right\rangle , \hfill \\ \left[ {\hat{j}_{z} \hat{j}_{x} } \right] = i\hat{j}_{y} \Rightarrow \hat{j}_{z} \hat{j}_{x} = \hat{j}_{x} \hat{j}_{z} + i\hat{j}_{y} \hfill \\ \left[ {\hat{j}_{y} \hat{j}_{z} } \right] = i\hat{j}_{x} \Rightarrow - i\hat{j}_{x} + \hat{j}_{y} \hat{j}_{z} = \hat{j}_{z} \hat{j}_{y} . \hfill \\ \hat{j}_{z} \hat{j}_{x} - \hat{j}_{x} \hat{j}_{z} = i\hat{j}_{y} , \Rightarrow \hat{j}_{z} \hat{j}_{x} = i\hat{j}_{y} + \hat{j}_{x} \hat{j}_{z} , \hfill \\ \hat{j}_{y} \hat{j}_{z} - \hat{j}_{z} \hat{j}_{y} = i\hat{j}_{x} \Rightarrow \hat{j}_{y} \hat{j}_{z} - i\hat{j}_{x} = \hat{j}_{z} \hat{j}_{y} , \hfill \\ \hat{j}_{z} \left| \chi \right\rangle = \hat{j}_{z} \hat{j}^{ + } \left| {\lambda m} \right\rangle = \hat{j}_{z} \left( {\hat{j}_{x} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle = \hat{j}_{z} \hat{j}_{x} \left| {\lambda m} \right\rangle + i\hat{j}_{z} \hat{j}_{y} \left| {\lambda m} \right\rangle = \hfill \\ = \left( {\hat{j}_{x} \hat{j}_{z} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle + \left( {\hat{j}_{x} + i\hat{j}_{y} \hat{j}_{z} } \right)\left| {\lambda m} \right\rangle \hfill \\ = \left( {m\hat{j}_{x} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle + \left( {\hat{j}_{x} + i\hat{j}_{y} m} \right)\left| {\lambda m} \right\rangle = \hfill \\ = \left( {m(\hat{j}_{x} + i\hat{j}_{y} ) + \hat{j}_{x} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle = (m + 1)\hat{j}^{ + } \left| {\lambda m} \right\rangle = (m + 1)\left| \chi \right\rangle , \hfill \\ \left| \chi \right\rangle = const\left| {\lambda (m + 1)} \right\rangle = \hat{j}^{ + } \left| {\lambda m} \right\rangle . \hfill \\ ]\left| \eta \right\rangle = \hat{j}^{ - } \left| {\lambda m} \right\rangle \hfill \\ \hat{j}_{z} \left| \eta \right\rangle = \hat{j}_{z} \hat{j}^{ - } \left| {\lambda m} \right\rangle = \hat{j}_{z} \left( {\hat{j}_{x} - i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle \hfill \\ = \left( {\hat{j}_{x} \hat{j}_{z} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle - \left( {\hat{j}_{x} + i\hat{j}_{y} \hat{j}_{z} } \right)\left| {\lambda m} \right\rangle = \hfill \\ = \left( {m\hat{j}_{x} + i\hat{j}_{y} } \right)\left| {\lambda m} \right\rangle - \left( {\hat{j}_{x} + i\hat{j}_{y} m} \right)\left| {\lambda m} \right\rangle = \left( {m\hat{j}_{x} + i\hat{j}_{y} - \left( {\hat{j}_{x} + i\hat{j}_{y} m} \right)} \right)\left| {\lambda m} \right\rangle = \hfill \\ = \left( {m(\hat{j}_{x} - i\hat{j}_{y} ) - \left( {\hat{j}_{x} - i\hat{j}_{y} } \right)} \right)\left| {\lambda m} \right\rangle = (m - 1)\hat{j}^{ - } \left| {\lambda m} \right\rangle = (m - 1)\left| \eta \right\rangle , \hfill \\ \left| \eta \right\rangle = \hat{j}^{ - } \left| {\lambda m} \right\rangle = const\left| {\lambda (m - 1)} \right\rangle . \hfill \\ \end{gathered}$$
(20)

Consequently, the operators \(\hat{j}^{ - } ,\hat{j}^{ + }\) introduced by us can be considered as operators of decreasing and increasing by 1 the projection of the angular momentum on the z-axis, respectively.

Theorem 1: The eigenvalue of the operator of the square of the angular momentum \(\hat{j}^{2}\) is the number j(j + 1), where j is the maximal eigenvalue of the angular momentum projection onto the z-axis.

Proof:

$$\begin{gathered} ]m_{\max } = j \hfill \\ \hat{j}^{ + } \left| {\lambda m_{\max } } \right\rangle = \hat{j}^{ + } \left| {\lambda j} \right\rangle = 0, \hfill \\ \hat{j}^{2} \left| {\lambda m_{\max } } \right\rangle = \lambda \left| {\lambda m_{\max } } \right\rangle ,\hat{j}^{2} = \hat{j}^{ - } \cdot \hat{j}^{ + } + \hat{j}_{z}^{2} + \hat{j}_{z} \hfill \\ \hat{j}^{2} \left| {\lambda j} \right\rangle = (\hat{j}^{ - } \cdot \hat{j}^{ + } + \hat{j}_{z}^{2} + \hat{j}_{z} )\left| {\lambda j} \right\rangle = (j^{2} + j)\left| {\lambda j} \right\rangle = j(j + 1)\left| {\lambda j} \right\rangle . \hfill \\ \end{gathered}$$
(21)

Theorem 2: The minimum eigenvalue of the operator of the angular momentum projection on the z-axis is mmin = −j.

Proof:

(22)

The second solution contradicts Lemma1.

Thus, we come to the fundamental statement that the projection of the angular momentum on the quantization axis can change from the minimum value −j to the maximum value +j, running all possible values through one:

j,j + 1, −j + 2, … j−2, j−1, j.

The latter means that in nature there are only two types of angular momenta that satisfy this condition: integer, 0, 1, 2, … etc. and half-integers, 1/2, 3/2, …

Matrix elements of the operators of projections on the x- and y-axes in the j z -representation

$$\begin{gathered} \left[ {\hat{j}^{ + } \hat{j}^{ - } } \right] = 2\hat{j}_{z} , \hfill \\ \hat{j}^{ - } \cdot \hat{j}^{ + } = \hat{j}^{ + } \cdot \hat{j}^{ - } - 2\hat{j}_{z} = \hat{j}^{2} - \hat{j}_{z}^{2} - \hat{j}_{z} . \hfill \\ \hat{j}^{ + } \cdot \hat{j}^{ - } = \hat{j}^{2} - \hat{j}_{z}^{2} + \hat{j}_{z} ; \hfill \\ \hat{j}_{z} \left| {\lambda m} \right\rangle = m\left| {\lambda m} \right\rangle \hfill \\ \left\langle {\lambda m} \right|\hat{j}^{ + } \cdot \hat{j}^{ - } \left| {\lambda m} \right\rangle = \left\langle {\lambda m} \right|\hat{j}^{2} - \hat{j}_{z}^{2} + \hat{j}_{z} \left| {\lambda m} \right\rangle \hfill \\ = \left\langle {\lambda m} \right|j(j + 1) - m_{{}}^{2} + m\left| {\lambda m} \right\rangle = j(j + 1) - m\left( {m - 1} \right), \hfill \\ \left\langle {\lambda m^{\prime}} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle = const\left\langle {\lambda m^{\prime}} \right|\left| {\lambda (m - 1)} \right\rangle = const\delta_{{m^{\prime}\left( {m - 1} \right)}} \hfill \\ \left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda m^{\prime}} \right\rangle = const^{\prime}\left\langle {\lambda m} \right|\left| {\lambda (m^{\prime} + 1)} \right\rangle = const^{\prime}\delta_{{m\left( {m^{\prime} + 1} \right)}} \hfill \\ \hat{j}^{ + } = \hat{j}^{ - \dag } \_\_\_\hat{j}^{ - } = \hat{j}^{ + \dag } , \Rightarrow \hfill \\ \sum\limits_{m^{\prime} = - j}^{j} {\left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda m^{\prime}} \right\rangle \cdot \left\langle {\lambda m^{\prime}} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle } = \left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle \cdot \left\langle {\lambda (m - 1)} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle = \hfill \\ \equiv \left\langle {\lambda (m - 1)} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle^{*} \cdot \left\langle {\lambda (m - 1)} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle = \left| {\left\langle {\lambda (m - 1)} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle } \right|^{2} = \hfill \\ \equiv \left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle \cdot \left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle^{*} = \left| {\left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle } \right|^{2} \hfill \\ = j(j + 1) - m\left( {m - 1} \right), \Rightarrow \hfill \\ \left\langle {\lambda m} \right|\hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle = \left\langle {\lambda (m - 1)} \right|\hat{j}^{ - } \left| {\lambda m} \right\rangle = \sqrt {j(j + 1) - m\left( {m - 1} \right)} . \hfill \\ \hat{j}^{ + } \left| {\lambda (m - 1)} \right\rangle = \sqrt {j(j + 1) - m\left( {m - 1} \right)} \left| {\lambda m} \right\rangle , \hfill \\ \hat{j}^{ - } \left| {\lambda m} \right\rangle = \sqrt {j(j + 1) - m\left( {m - 1} \right)} \left| {\lambda (m - 1)} \right\rangle . \hfill \\ \end{gathered}$$
(23)

Then,

$$\begin{gathered} \hat{j}_{x} = \frac{1}{2}\left( {\hat{j}^{ + } + \hat{j}^{ - } } \right), \hfill \\ \left\langle {\lambda m} \right|\hat{j}^{ - } \left| {\lambda (m - 1)} \right\rangle = \sqrt {j(j + 1) - (m - 2)\left( {m - 1} \right)} \left\langle {\lambda m} \right|\left| {\lambda (m - 2)} \right\rangle = 0 \hfill \\ \left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda (m - 1)} \right\rangle = \frac{1}{2}\sqrt {j(j + 1) - m\left( {m - 1} \right)} = \left\langle {\lambda (m - 1)} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle , \hfill \\ \left\langle {\lambda (m - 1)} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle = \frac{1}{2}\sqrt {j(j + 1) - m\left( {m - 1} \right)} , \hfill \\ \hat{j}_{y} = \frac{1}{2i}\left( {\hat{j}^{ + } - \hat{j}^{ - } } \right), \hfill \\ \left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda (m - 1)} \right\rangle = \frac{1}{2i}\sqrt {j(j + 1) - m\left( {m - 1} \right)} = - \left\langle {\lambda (m - 1)} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle . \hfill \\ \end{gathered}$$
(24)

Using the general expressions Eq. (24) obtained for the matrix elements of the operators of the projections of the angular momentum on the x-, y-, z-axes, we write out the matrices of these operators on the Cartesian basis in an explicit form.

  1. (i)

    j = 1/2

    $$\begin{gathered} \hat{j}_{z} = \hat{s}_{z} = \left( {\begin{array}{*{20}c} {1/2} & 0 \\ 0 & { - 1/2} \\ \end{array} } \right),\,\hat{j}_{x} = \hat{s}_{x} = \left( {\begin{array}{*{20}c} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \\ \end{array} } \right) = \frac{1}{2}\left( {\begin{array}{*{20}c} 0 & 1 \\ 1 & 0 \\ \end{array} } \right), \hfill \\ \hat{j}_{y} = \left( {\begin{array}{*{20}c} 0 & {\frac{ - i}{2}} \\ \frac{i}{2} & 0 \\ \end{array} } \right) = \frac{1}{2}\left( {\begin{array}{*{20}c} 0 & { - i} \\ i & 0 \\ \end{array} } \right). \hfill \\ \end{gathered}$$
    $$\begin{gathered} \left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda (m - 1)} \right\rangle = \left\langle {\lambda (m - 1)} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle = \hfill \\ = \frac{1}{2}\sqrt {j(j + 1) - m\left( {m - 1} \right)} , \hfill \\ \left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda (m - 1)} \right\rangle = - \left\langle {\lambda (m - 1)} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle = \hfill \\ = \frac{1}{2i}\sqrt {j(j + 1) - m\left( {m - 1} \right)} , \hfill \\ \left\langle {\lambda 1/2} \right|\hat{j}_{x} \left| {\lambda ( - 1/2)} \right\rangle = \left\langle {\lambda ( - 1/2)} \right|\hat{j}_{x} \left| {\lambda 1/2} \right\rangle = \hfill \\ = \frac{1}{2}\sqrt {1/2(1/2 + 1) - 1/2\left( { - 1/2} \right)} = \frac{1}{2}, \hfill \\ \left\langle {\lambda 1/2} \right|\hat{j}_{y} \left| {\lambda ( - 1/2)} \right\rangle = - \left\langle {\lambda (1/2 - 1)} \right|\hat{j}_{y} \left| {\lambda 1/2} \right\rangle = \hfill \\ = \frac{1}{2i}\sqrt {1/2(1/2 + 1) - 1/2\left( {1/2 - 1} \right)} = \frac{ - i}{2}. \hfill \\ \end{gathered}$$
    (25)
  2. (ii)

    j = 1

    $$\begin{gathered} \left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda (m - 1)} \right\rangle = \frac{1}{2}\sqrt {j(j + 1) - m\left( {m - 1} \right)} = \left\langle {\lambda (m - 1)} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle , \hfill \\ \left\langle {\lambda m = 1} \right|\hat{j}_{x} \left| {\lambda m = (1 - 1)} \right\rangle = \left\langle {\lambda ,(1 - 1)} \right|\hat{j}_{x} \left| {\lambda ,1} \right\rangle = \hfill \\ = \left\langle {\lambda ,1} \right|\hat{j}_{x} \left| {\lambda ,0} \right\rangle = \frac{1}{2}\sqrt {1(1 + 1) - 1\left( {1 - 1} \right)} = \frac{\sqrt 2 }{2} = \frac{1}{\sqrt 2 }, \hfill \\ \left\langle {\lambda m = 0} \right|\hat{j}_{x} \left| {\lambda (0 - 1)} \right\rangle = \left\langle {\lambda (0 - 1)} \right|\hat{j}_{x} \left| {\lambda 0} \right\rangle = \hfill \\ = \frac{1}{2}\sqrt {1(1 + 1) - 0\left( {0 - 1} \right)} = \frac{1}{\sqrt 2 }, \hfill \\ \left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda (m - 1)} \right\rangle = - \left\langle {\lambda (m - 1)} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle = \hfill \\ = \frac{1}{2i}\sqrt {j(j + 1) - m\left( {m - 1} \right)} , \hfill \\ m = 1,\,0, - 1. \hfill \\ \hat{j}_{z} = \left( {\begin{array}{*{20}c} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & { - 1} \\ \end{array} } \right),\,\hat{j}_{x} = \left( {\begin{array}{*{20}c} 0 & {\frac{1}{\sqrt 2 }} & 0 \\ {\frac{1}{\sqrt 2 }} & 0 & {\frac{1}{\sqrt 2 }} \\ 0 & {\frac{1}{\sqrt 2 }} & 0 \\ \end{array} } \right) = \frac{1}{\sqrt 2 }\left( {\begin{array}{*{20}c} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} } \right),\,\hat{j}_{y} = \frac{1}{\sqrt 2 }\left( {\begin{array}{*{20}c} 0 & { - i} & 0 \\ i & 0 & { - i} \\ 0 & i & 0 \\ \end{array} } \right). \hfill \\ \end{gathered}$$
    (26)
  3. (iii)

    j = 3/2

    $$\hat{j}_{z} = \left( {\begin{array}{*{20}c} {3/2} & 0 & 0 & 0 \\ 0 & {1/2} & 0 & 0 \\ 0 & 0 & { - 1/2} & 0 \\ 0 & 0 & 0 & { - 3/2} \\ \end{array} } \right),$$
    $$\begin{gathered} \left\langle {\lambda m} \right|\hat{j}_{x} \left| {\lambda (m - 1)} \right\rangle = \frac{1}{2}\sqrt {j(j + 1) - m\left( {m - 1} \right)} = \left\langle {\lambda (m - 1)} \right|\hat{j}_{x} \left| {\lambda m} \right\rangle , \hfill \\ \left\langle {\lambda - 3/2} \right|\hat{j}_{x} \left| {\lambda - 1/2} \right\rangle = \frac{1}{2}\sqrt {3/2 \cdot 5/2 - 1/2 \cdot 3/2} = \frac{{\sqrt 3 }}{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left\langle {\lambda - 1/2} \right|\hat{j}_{x} \left| {\lambda - 3/2} \right\rangle , \hfill \\ \left\langle {\lambda 1/2} \right|\hat{j}_{x} \left| {\lambda ( - 1/2)} \right\rangle = \frac{1}{2}\sqrt {15/4 + 1/4} = \left\langle {\lambda ( - 1/2)} \right|\hat{j}_{x} \left| {\lambda 1/2} \right\rangle = 1, \hfill \\ \left\langle {\lambda 1/2} \right|\hat{j}_{x} \left| {\lambda 3/2)} \right\rangle = \left\langle {\lambda (3/2)} \right|\hat{j}_{x} \left| {\lambda 1/2} \right\rangle = \frac{1}{2}\sqrt {15/4 - 3/2\left( {1/2} \right)} = \frac{{\sqrt 3 }}{2}, \hfill \\ \hat{j}_{x} = \left( {\begin{array}{*{20}c} 0 & {\sqrt 3 /2} & 0 & 0 \\ {\sqrt 3 /2} & 0 & 1 & 0 \\ 0 & 1 & 0 & {\sqrt 3 /2} \\ 0 & 0 & {\sqrt 3 /2} & 0 \\ \end{array} } \right), \hfill \\ \hat{j}_{y} = \left( {\begin{array}{*{20}c} 0 & { - i\sqrt 3 /2} & 0 & 0 \\ {i\sqrt 3 /2} & 0 & { - i} & 0 \\ 0 & i & 0 & { - i\sqrt 3 /2} \\ 0 & 0 & {i\sqrt 3 /2} & 0 \\ \end{array} } \right), \hfill \\ \left\langle {\lambda m} \right|\hat{j}_{y} \left| {\lambda (m - 1)} \right\rangle = \frac{1}{{2i}}\sqrt {j(j + 1) - m\left( {m - 1} \right)} = - \left\langle {\lambda (m - 1)} \right|\hat{j}_{y} \left| {\lambda m} \right\rangle , \hfill \\ \left\langle {\lambda 3/2} \right|\hat{j}_{y} \left| {\lambda (1/2)} \right\rangle = \frac{1}{{2i}}\sqrt {15/4 - 3/4} = - i\frac{{\sqrt 3 }}{2} = - \left\langle {\lambda (1/2)} \right|\hat{j}_{y} \left| {\lambda 3/2} \right\rangle , \hfill \\ \left\langle {\lambda 1/2} \right|\hat{j}_{y} \left| {\lambda ( - 1/2)} \right\rangle = \frac{1}{{2i}}\sqrt {15/4 + 1/4} = - i = - \left\langle {\lambda ( - 1/2)} \right|\hat{j}_{y} \left| {\lambda 3/2} \right\rangle , \hfill \\ \left\langle {\lambda ( - 1/2)} \right|\hat{j}_{y} \left| {\lambda - 3/2} \right\rangle = \frac{1}{{2i}}\sqrt {15/4 - 3/4} = - i\frac{{\sqrt 3 }}{2} \hfill \\ \quad \quad \quad \quad = - \left\langle {\lambda - 3/2} \right|\hat{j}_{y} \left| {\lambda ( - 1/2)} \right\rangle . \hfill \\ \end{gathered} .$$
    (27)

2.4 Addition of Angular Momenta in Quantum Mechanics. The Wigner–Eckart Theorem

Consider two rotations. There can be two rotations in which the particle participates. For example, an electron in an atom has orbital rotation and proper rotation (spin), or, in another case, we have two electrons in an atom in certain orbital states l1 and l2, forming a complete orbital state L. In any case, let us have two rotations j1 and j2, where we do not care about the physical nature of these rotations, but we will be interested in the connection of these rotations with the formation of the total angular momentum J. The operator of the total angular momentum \({\hat{\mathbf{J}}} = {\mathbf{e}}_{x} \widehat{{J_{x} }} + {\mathbf{e}}_{y} \widehat{{J_{y} }} + {\mathbf{e}}_{z} \widehat{{J_{z} }}\) connects with the operators \(\widehat{{{\mathbf{j}}_{1} }} = {\mathbf{e}}_{x} \widehat{{j_{1x} }} + {\mathbf{e}}_{y} \widehat{{j_{1y} }} + {\mathbf{e}}_{z} \widehat{{j_{1z} }}\) and \(\widehat{{{\mathbf{j}}_{2} }} = {\mathbf{e}}_{x} \widehat{{j_{2x} }} + {\mathbf{e}}_{y} \widehat{{j_{2y} }} + {\mathbf{e}}_{z} \widehat{{j_{2z} }}\) as follows:

$$\begin{gathered} {\hat{\mathbf{J}}} = {\mathbf{e}}_{x} \widehat{{J_{x} }} + {\mathbf{e}}_{y} \widehat{{J_{y} }} + {\mathbf{e}}_{z} \widehat{{J_{z} }} = \hfill \\ = \widehat{{{\mathbf{j}}_{1} }} + \widehat{{{\mathbf{j}}_{2} }} = {\mathbf{e}}_{x} \left( {\widehat{{j_{1x} }} + \widehat{{j_{2x} }}} \right) + {\mathbf{e}}_{y} \left( {\widehat{{j_{1y} }} + \widehat{{j_{2y} }}} \right) + {\mathbf{e}}_{z} \left( {\widehat{{j_{1z} }} + \widehat{{j_{2z} }}} \right). \hfill \\ \end{gathered}$$
(28)

And

$$\begin{gathered} {\hat{\mathbf{J}}} \cdot {\hat{\mathbf{J}}} = \widehat{{J^{2} }} = \widehat{{J_{x}^{2} }} + \widehat{{J_{y}^{2} }} + \widehat{{J_{z}^{2} }} = \hfill \\ = \left( {\widehat{{{\mathbf{j}}_{1} }} + \widehat{{{\mathbf{j}}_{2} }}} \right) \cdot \left( {\widehat{{{\mathbf{j}}_{1} }} + \widehat{{{\mathbf{j}}_{2} }}} \right) = \widehat{{j_{1}^{2} }} + \widehat{{j_{2}^{2} }} + 2\left( {\widehat{{{\mathbf{j}}_{1} }} \cdot \widehat{{{\mathbf{j}}_{2} }}} \right). \hfill \\ \end{gathered}$$
(29)

It is easy to prove that

$$\begin{gathered} \left[ {\widehat{{J^{2} }}\widehat{{j_{1}^{2} }}} \right] = \left[ {\widehat{{J^{2} }}\widehat{{j_{2}^{2} }}} \right] = 0, \hfill \\ \left[ {\widehat{{J^{2} }}\widehat{{J_{z} }}} \right] = 0, \hfill \\ \left[ {\widehat{{J_{z} }}\widehat{{j_{1z} }}} \right] = \left[ {\widehat{{J_{z} }}\widehat{{j_{2z} }}} \right] = 0, \hfill \\ \left[ {\widehat{{J^{2} }}\widehat{{j_{1z} }}} \right] = - 2i\widehat{{j_{1y} }}\widehat{{j_{2x} }} + 2i\widehat{{j_{1x} }}\widehat{{j_{2y} }} \ne 0, \hfill \\ \left[ {\widehat{{J^{2} }}\widehat{{j_{2z} }}} \right] = - 2i\widehat{{j_{2y} }}\widehat{{j_{1x} }} + 2i\widehat{{j_{2x} }}\widehat{{j_{1y} }}, \ne 0. \hfill \\ \end{gathered}$$
(30)

And

$$\begin{gathered} \left[ {\widehat{{J^{2} }}\widehat{{j_{1}^{2} }}} \right] = \left[ {\widehat{{J^{2} }}\widehat{{j_{2}^{2} }}} \right] = 0, \hfill \\ \left[ {\widehat{{j_{1}^{2} }}\widehat{{j_{1z} }}} \right] = \left[ {\widehat{{j_{2}^{2} }}\widehat{{j_{2z} }}} \right] = 0, \hfill \\ \left[ {\widehat{{J_{z} }}\widehat{{j_{1z} }}} \right] = \left[ {\widehat{{J_{z} }}\widehat{{j_{2z} }}} \right] = 0, \hfill \\ \left[ {\widehat{{J^{2} }}\widehat{{j_{1z} }}} \right] = - 2i\widehat{{j_{1y} }}\widehat{{j_{2x} }} + 2i\widehat{{j_{1x} }}\widehat{{j_{2y} }} \ne 0, \hfill \\ \left[ {\widehat{{J^{2} }}\widehat{{j_{2z} }}} \right] = - 2i\widehat{{j_{2y} }}\widehat{{j_{1x} }} + 2i\widehat{{j_{2x} }}\widehat{{j_{1y} }}, \ne 0. \hfill \\ \end{gathered}$$
(31)

That means we have two different orthogonal and complete basis. The first one is the basis of the following eigenstates:

$$\begin{gathered} \left\{ {\left| {JMj_{1} j_{2} } \right\rangle } \right\},\left\langle {JMj_{1} j_{2} } \right|\left. {J^{\prime}M^{\prime}j_{1} j_{2} } \right\rangle = \delta_{JJ^{\prime}} \delta_{MM^{\prime}} , \hfill \\ \sum\limits_{{J = \left| {j_{1} - j_{2} } \right|}}^{{j_{1} + j_{2} }} {\sum\limits_{M = - J}^{ + j} {\left| {JMj_{1} j_{2} } \right\rangle \left\langle {JMj_{1} j_{2} } \right|} } = \hat{1}. \hfill \\ \widehat{{J^{2} }}\left| {JMj_{1} j_{2} } \right\rangle = J\left( {J + 1} \right)\left| {JMj_{1} j_{2} } \right\rangle , \hfill \\ \widehat{{J_{z} }}\left| {JMj_{1} j_{2} } \right\rangle = M\left| {JMj_{1} j_{2} } \right\rangle , \hfill \\ \widehat{{j_{1}^{2} }}\left| {JMj_{1} j_{2} } \right\rangle = j_{1} \left( {j_{1} + 1} \right), \hfill \\ \widehat{{j_{2}^{2} }}\left| {JMj_{1} j_{2} } \right\rangle = j_{2} \left( {j_{2} + 1} \right). \hfill \\ \end{gathered}$$
(32)

The second is

$$\begin{gathered} \left\{ {\left| {j_{1} j_{2} m_{1} m_{2} } \right\rangle = \left| {j_{1} m_{1} } \right\rangle \left| {j_{2} m_{2} } \right\rangle } \right\}, \hfill \\ \left\langle {j_{1} m_{1} j_{2} m_{2} } \right.\left| {j_{1} m^{\prime}_{1} j_{2} m^{\prime}_{2} } \right\rangle = \delta_{{m_{1} m_{1} ^{\prime}}} \delta_{{m_{2} m_{2} ^{\prime}}} , \hfill \\ \sum\limits_{{m_{1} = - j_{1} }}^{{j_{1} }} {\sum\limits_{{m_{2} = - j_{2} }}^{{j_{2} }} {\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle \left\langle {j_{1} m_{1} j_{2} m_{2} } \right. = \hat{1},} } \hfill \\ \widehat{{j_{1}^{2} }}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle = j_{1} \left( {j_{1} + 1} \right)\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle , \hfill \\ \widehat{{j_{2}^{2} }}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle = j_{2} \left( {j_{2} + 1} \right)\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle , \hfill \\ \widehat{{j_{1z} }}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle = m_{1} \left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle , \hfill \\ \widehat{{j_{2z} }}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle = m_{2} \left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle , \hfill \\ \widehat{{J_{z} }}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle = M\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle . \hfill \\ m_{1} + m_{2} = M. \hfill \\ \end{gathered}$$
(33)

These bases are related by the unitary transformation:

$$\left| {JMj_{1} j_{2} } \right\rangle = \sum\limits_{\begin{subarray}{l} m_{1} = - j_{1} \\ m_{2} = - j_{2} . \\ m_{1} + m_{2} = M \end{subarray} }^{\begin{subarray}{l} j_{1} \\ j_{2} \end{subarray} } {C_{{j_{1} m_{1} j_{2} m_{2} }}^{JM} \left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle } .$$
(34)

Here, the coefficients \(C_{j_1 m_1 j_2 m_2 }^{JM}\) are the Clebsch–Gordan coefficients. This is the Wigner–Eckart theorem. These coefficients could be calculated by the exact algebraic method in the following way:

Acting by the following operator on the wave vector (34) with respect to notations (23), we have

$$\begin{gathered} \widehat{{J^{2} }} = \widehat{{J_{x}^{2} }} + \widehat{{J_{y}^{2} }} + \widehat{{J_{z}^{2} }} = = \widehat{{j_{1}^{2} }} + \widehat{{j_{2}^{2} }} + 2\left( {\widehat{{{\mathbf{j}}_{1} }} \cdot \widehat{{{\mathbf{j}}_{2} }}} \right), \Rightarrow \hfill \\ \widehat{{J^{2} }} - \widehat{{j_{1}^{2} }} - \widehat{{j_{2}^{2} }} = 2\left( {\widehat{{{\mathbf{j}}_{1} }} \cdot \widehat{{{\mathbf{j}}_{2} }}} \right) = \widehat{{j_{1}^{ + } }}\widehat{{j_{2}^{ - } }} + \widehat{{j_{1}^{ - } }}\widehat{{j_{2}^{ + } }} + 2\widehat{{j_{1z} }}\widehat{{j_{2z} }}, \hfill \\ \end{gathered}$$
(35)
$$\begin{gathered} \left( {\widehat{{J^{2} }} - \widehat{{j_{1}^{2} }} - \widehat{{j_{2}^{2} }}} \right)\left| {JMj_{1} j_{2} } \right\rangle = \sum\limits_{\begin{subarray}{l} m_{1} = - j_{1} \\ m_{2} = - j_{2} . \\ m_{1} + m_{2} = M \end{subarray} }^{\begin{subarray}{l} j_{1} \\ j_{2} \end{subarray} } C{_{{j_{1} m_{1} j_{2} m_{2} }} ^{JM} \left\{ {2\widehat{{j_{1z} }}\widehat{{j_{2z} }} + \widehat{{j_{1}^{ - } }}\widehat{{j_{2}^{ + } }} + \widehat{{j_{1}^{ + } }}\widehat{{j_{2}^{ - } }}} \right\}\left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle } , \hfill \\ \left( {J(J + 1) - j_{1} (j_{1} + 1) - j_{2} (j_{2} + 1)} \right)\sum\limits_{\begin{subarray}{l} m_{1} = - j_{1} \\ m_{2} = - j_{2} . \\ m_{1} + m_{2} = M \end{subarray} }^{\begin{subarray}{l} j_{1} \\ j_{2} \end{subarray} } C{_{{j_{1} m_{1} j_{2} m_{2} }}^{JM} \left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle } = \hfill \\ = \sum\limits_{\begin{subarray}{l} m_{1} = - j_{1} \\ m_{2} = - j_{2} . \\ m_{1} + m_{2} = M \end{subarray} }^{\begin{subarray}{l} j_{1} \\ j_{2} \end{subarray} } C{_{{j_{1} m_{1} j_{2} m_{2} }}^{JM} \left\{ {2m_{1} m_{2} \left| {j_{1} m_{1} j_{2} m_{2} } \right\rangle + } \right.} \hfill \\ + \sqrt {j_{2} (j_{2} + 1) - (m_{2} + 1)m_{2} } \sqrt {j_{1} (j_{1} + 1) - m_{1} (m_{1} - 1)} \left| {j_{1} \left( {m_{1} - 1} \right)j_{2} \left( {m_{2} + 1} \right)} \right\rangle + \hfill \\ \left. { + \sqrt {j_{1} (j_{1} + 1) - (m_{1} + 1)m_{1} } \sqrt {j_{2} (j_{2} + 1) - m_{2} (m_{2} - 1)} \left| {j_{1} \left( {m_{1} + 1} \right)j_{2} \left( {m_{2} - 1} \right)} \right\rangle } \right\}, \hfill \\ \end{gathered}$$
(36)

from which we have a system of equations determined by the coefficients \(C_{{j_{1} m_{1} j_{2} m_{2} }}^{JM}\)

$$\begin{gathered} \left( {J(J + 1) - j_{1} (j_{1} + 1) - j_{2} (j_{2} + 1) - 2m_{1} m_{2} } \right)C_{{j_{1} m_{1} j_{2} m_{2} }}^{JM} = \hfill \\ =C_{{j_{1} m_{1} + 1j_{2} m_{2} - 1}}^{JM} \sqrt {j_{2} (j_{2} + 1) - (m_{2} - 1)m_{2} } \sqrt {j_{1} (j_{1} + 1) - m_{1} (m_{1} + 1)} + \hfill \\ +C_{{j_{1} m_{1} - 1j_{2} m_{2} + 1}}^{JM} \sqrt {j_{1} (j_{1} + 1) - (m_{1} - 1)m_{1} } \sqrt {j_{2} (j_{2} + 1) - m_{2} (m_{2} + 1)} , \hfill \\ \sum\limits_{\begin{subarray}{l} m_{1} = - j_{1} \\ m_{2} = - j_{2} . \\ m_{1} + m_{2} = M \end{subarray} }^{\begin{subarray}{l} j_{1} \\ j_{2} \end{subarray} } {\left( {C_{{j_{1} m_{1} j_{2} m_{2} }}^{JM} } \right)^{2} } = 1. \hfill \\ \end{gathered}$$
(37)

As an example, we consider the Clebsch–Gordan coefficients for the addition of two spins s = 1 with a formation of the two different states |J = 2, M = 0, j1 = 1, j2 = 1 > and

|J = 0, M = 0, j1 = 1, j2 = 1 > :

$$\begin{gathered} \left| {J = 2,M = 0j_{1} = 1,j_{2} = 1} \right\rangle =C_{111 - 1}^{20} \left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ +C_{1 - 111}^{20} \left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle +C_{1010}^{20} \left| {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle , \hfill \\ \left| {J = 0,M = 0j_{1} = 1,j_{2} = 1} \right\rangle =C_{111 - 1}^{00} \left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ +C_{1 - 111}^{00} \left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle +C_{1010}^{00} \left| {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle . \hfill \\ \end{gathered}$$
(38)

For the \(\left| {J = 2,M = 0j_{1} = 1,j_{2} = 1} \right\rangle\) state, we have the system of the following equations determined by the Clebsch–Gordan coefficients:

$$\begin{gathered} \left| {J = 2,M = 0j_{1} = 1,j_{2} = 1} \right\rangle = \left. {C_{{111 - 1}}^{{20}} } \right|\left. {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{20}} } \right|\left. {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle \hfill \\ + \left. {C_{{1010}}^{{20}} } \right|\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle , \hfill \\ \left( {2(2 + 1) - 1(1 + 1) - 1(1 + 1)} \right)\left\{ {\left. {C_{{111 - 1}}^{{20}} } \right|} \right.\left. {\underline{{j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1}} } \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{20}} } \right|\left. {\underline{\underline{{j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1}}} } \right\rangle \hfill \\ + \left. {C_{{1010}}^{{20}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\} = \hfill \\ \left\{ {\widehat{{j_{1}^{ - } }}\widehat{{j_{2}^{ + } }} + \widehat{{j_{1}^{ + } }}\widehat{{j_{2}^{ - } }} + 2\widehat{{j_{{1z}} }}\widehat{{j_{{2z}} }}} \right\}\left\{ {\left. {C_{{111 - 1}}^{{20}} } \right|} \right.\left. {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{20}} } \right|\left. {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle \hfill \\ + \left. {C_{{1010}}^{{20}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\} = \hfill \\ = \left[ {C_{{111 - 1}}^{{20}} } \right.\left. {\sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} } \right|\left. {j_{1} = 1,m_{1} = 0j_{2} = 1m_{2} = 0} \right\rangle \hfill \\ + C_{{1 - 111}}^{{20}} \cdot 0 + \hfill \\ + C_{{1010}}^{{20}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \left. {\left. {\underline{\underline{{\left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = + 1} \right\rangle }}} } \right\rangle } \right] + \hfill \\ + \left[ {C_{{111 - 1}}^{{20}} } \right. \cdot 0 + \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \hfill \\ \,\left. {C_{{1 - 111}}^{{20}} } \right|\left. {j_{1} = 1,m_{1} = 0j_{2} = 1m_{2} = 0} \right\rangle + \hfill \\ + C_{{1010}}^{{20}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \left. {\underline{{\left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle }} } \right] + \hfill \\ + \left\{ {2( - 1) \cdot 1\,} \right.\left. {C_{{111 - 1}}^{{20}} } \right|\left. {\underline{{j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1}} } \right\rangle + \hfill \\ + 2 \cdot 1 \cdot ( - 1)\left. {C_{{1 - 111}}^{{20}} } \right|\left. {\underline{\underline{{j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1}}} } \right\rangle \hfill \\ + 0 \cdot \left. {C_{{1010}}^{{20}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\}, \hfill \\ \left\{ {\begin{array}{*{20}l} \begin{gathered} \left( {2(2 + 1) - 1(1 + 1) - 1(1 + 1)} \right)C_{{111 - 1}}^{{20}} \hfill \\ = C_{{1010}}^{{20}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} + 2\left( { - 1} \right) \cdot 1C_{{111 - 1}}^{{20}} , \hfill \\ \end{gathered} \\ \begin{gathered} \left( {2(2 + 1) - 1(1 + 1) - 1(1 + 1)} \right)C_{{1 - 111}}^{{20}} \hfill \\ = C_{{1010}}^{{20}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} + 2 \cdot 1 \cdot ( - 1)C_{{1 - 111}}^{{20}} , \hfill \\ \end{gathered} \\ {\left( {2(2 + 1) - 1(1 + 1) - 1(1 + 1)} \right)C_{{1010}}^{{20}} = } \\ \begin{gathered} = \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} C_{{1 - 111}}^{{20}} \hfill \\ + C_{{111 - 1}}^{{20}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} . \hfill \\ \end{gathered} \\ \end{array} } \right. \hfill \\ \end{gathered}$$
(39)
$$\begin{gathered} \left\{ {\begin{array}{*{20}l} {2_{111 - 1}^{20} = C_{1010}^{20} } \\ {2_{1 - 111}^{20} = C_{1010}^{20} } \\ {C_{1010}^{20} = C_{1 - 111}^{20} +C_{111 - 1}^{20} } \\ \end{array} } \right., \hfill \\ \left( {C_{111 - 1}^{20} } \right)^{2} + \left( {C_{1 - 111}^{20} } \right)^{2} + \left( {C_{1010}^{20} } \right)^{2} = 1, \hfill \\C_{111 - 1}^{20} = \frac{1}{\sqrt 6 } = C_{1 - 111}^{20} , \hfill \\C_{1010}^{20} = \frac{2}{\sqrt 6 } = \sqrt{\frac{2}{3}} . \hfill \\ \end{gathered}$$
(40)
$$\begin{gathered} \left| {J = 2,M = 0j_{1} = 1,j_{2} = 1} \right\rangle = \frac{1}{\sqrt 6 }\left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ + \frac{1}{\sqrt 6 }\left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle + \frac{2}{\sqrt 6 }\left| {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle = \hfill \\ = \frac{1}{\sqrt 6 }\left\{ {\left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle } \right. + \left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle + \hfill \\ + 2\left. {\left| {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\}. \hfill \\ \end{gathered}$$
(41)

For the state

$$\begin{gathered} \left| {J = 0,M = 0j_{1} = 1,j_{2} = 1} \right\rangle = C_{111 - 1}^{00} \left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ +C_{1 - 111}^{00} \left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle \hfill \\ +C_{1010}^{00} \left| {j_{1} = 1,\,m_{1} = 0,j_{2} = 1,\,m_{2} = 0} \right\rangle , \hfill \\ \end{gathered}$$

we have

$$\begin{gathered} \left| {J = 0,M = 0j_{1} = 1,j_{2} = 1} \right\rangle = \left. {C_{{111 - 1}}^{{00}} } \right|\left. {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{00}} } \right|\left. {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle \hfill \\ + \left. {C_{{1010}}^{{00}} } \right|\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle , \hfill \\ \left( { - 1(1 + 1) - 1(1 + 1)} \right)\left\{ {\left. {C_{{111 - 1}}^{{00}} } \right|} \right.\left. {\underline{{j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1}} } \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{00}} } \right|\left. {\underline{\underline{{j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1}}} } \right\rangle \hfill \\ + \left. {C_{{1010}}^{{00}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\} = \hfill \\ \left\{ {\widehat{{j_{1}^{ - } }}\widehat{{j_{2}^{ + } }} + \widehat{{j_{1}^{ + } }}\widehat{{j_{2}^{ - } }} + 2\widehat{{j_{{1z}} }}\widehat{{j_{{2z}} }}} \right\}\left\{ {\left. {C_{{111 - 1}}^{{00}} } \right|} \right.\left. {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle + \hfill \\ + \left. {C_{{1 - 111}}^{{00}} } \right|\left. {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle \hfill \\ + \left. {C_{{1010}}^{{00}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\} = \hfill \\ = \left[ {C_{{111 - 1}}^{{00}} } \right.\left. {\sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} } \right|\left. {j_{1} = 1,m_{1} = 0j_{2} = 1,\,m_{2} = 0} \right\rangle \hfill \\ + C_{{1 - 111}}^{{00}} \cdot 0 + \hfill \\ + C_{{1010}}^{{00}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \hfill \\ \left. {\left. {\underline{\underline{{\left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = + 1} \right\rangle }}} } \right\rangle } \right] + \hfill \\ + \left[ {C_{{111 - 1}}^{{00}} } \right. \cdot 0 + \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \hfill \\ \,\left. {C_{{1 - 111}}^{{00}} } \right|\left. {j_{1} = 1,m_{1} = 0j_{2} = 1,m_{2} = 0} \right\rangle + \hfill \\ + C_{{1010}}^{{00}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} \hfill \\ \left. {\underline{{\left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle }} } \right] + \hfill \\ + \left\{ {2( - 1) \cdot 1\,} \right.\left. {C_{{111 - 1}}^{{00}} } \right|\left. {\underline{{j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1}} } \right\rangle + \hfill \\ + 2 \cdot 1 \cdot ( - 1)\left. {C_{{1 - 111}}^{{00}} } \right|\left. {\underline{\underline{{j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1}}} } \right\rangle \hfill \\ + 0 \cdot \left. {C_{{1010}}^{{00}} } \right|\left. {\left. {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\}, \hfill \\ \left\{ {\begin{array}{*{20}l} \begin{gathered} \left( { - 1(1 + 1) - 1(1 + 1)} \right)C_{{111 - 1}}^{{00}} \hfill \\ = C_{{1010}}^{{00}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} + 2\left( { - 1} \right) \cdot 1C_{{111 - 1}}^{{00}} , \hfill \\ \end{gathered} \\ \begin{gathered} \left( { - 1(1 + 1) - 1(1 + 1)} \right)C_{{1 - 111}}^{{00}} \hfill \\ = C_{{1010}}^{{00}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} + 2 \cdot 1 \cdot ( - 1)C_{{1 - 111}}^{{00}} , \hfill \\ \end{gathered} \\ {\left( { - 1(1 + 1) - 1(1 + 1)} \right)C_{{1010}}^{{00}} = } \\ \begin{gathered} = \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} C_{{1 - 111}}^{{00}} \hfill \\ + C_{{111 - 1}}^{{00}} \sqrt {1(1 + 1) - 0 \cdot 1} \sqrt {1(1 + 1) - 1 \cdot 0} . \hfill \\ \end{gathered} \\ \end{array} } \right. \hfill \\ \end{gathered}$$
(42)
$$\begin{gathered} \left\{ {\begin{array}{*{20}l} { -C_{111 - 1}^{00} = C_{1010}^{00} } \\ { -C_{1 - 111}^{00} =C_{1010}^{00} } \\ { - 2C_{1010}^{00} = C_{1 - 111}^{00} +C_{111 - 1}^{00} } \\ \end{array} } \right., \hfill \\ \left( {C_{111 - 1}^{00} } \right)^{2} + \left( {C_{1 - 111}^{00} } \right)^{2} + \left( {C_{1010}^{00} } \right)^{2} = 1, \hfill \\C_{111 - 1}^{00} = C_{1 - 111}^{00} = -C_{1010}^{00} = \frac{1}{\sqrt 3 }. \hfill \\ \end{gathered}$$
(43)

Finally, we have for the state \(\left| {J = 0,M = 0j_{1} = 1,j_{2} = 1} \right\rangle\)

$$\begin{gathered} \left| {J = 0,M = 0j_{1} = 1,j_{2} = 1} \right\rangle = \frac{1}{\sqrt 3 }\left\{ {\left| {j_{1} = 1,m_{1} = 1,j_{2} = 1,m_{2} = - 1} \right\rangle } \right. + \hfill \\ + \left| {j_{1} = 1,m_{1} = - 1,j_{2} = 1,m_{2} = 1} \right\rangle - \left. {\left| {j_{1} = 1,m_{1} = 0,j_{2} = 1,m_{2} = 0} \right\rangle } \right\}. \hfill \\ \end{gathered}$$
(44)

3 Spin Hamiltonian of Identical Particles with Arbitrary Spin

3.1 Interaction in a System of Two Identical Particles with Spin

Consider a Hamiltonian describing pair interaction in the coordinate representation of two nonrelativistic identical particles. We represented the complete Hamiltonian as the sum of the Hamiltonians of two independent particles \(\hat{H}^{0} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) = \hat{h}_{I} ({\mathbf{r}}_{1} ) + \hat{h}_{II} ({\mathbf{r}}_{2} )\) and the interaction operator \(\hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\):

$$\hat{H}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) = \hat{H}^{0} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) + \hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ).$$
(45)

In the nonrelativistic case, the total Hamiltonian Eq. (45), in the absence of spin–orbit interaction, is invariant with respect to the permutation of particles described by the operator \(\hat{P}_{1,2}\) and also commutes with the operator \({\hat{\mathbf{S}}}\) of the total spin of particles included in the system:

$$\begin{gathered} \left[ {\hat{H}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\hat{P}_{1,2} } \right] = 0, \hfill \\ \left[ {\hat{H}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ){\hat{\mathbf{S}}}} \right] = 0. \hfill \\ \end{gathered}$$
(46)

This means that in the stationary case the total energy E of the system, the magnitude of the total spin S, its projection Sz, and the parity g of the permutation are conserved. These quantities are simultaneously precisely measurable and define the complete set of parameters that characterize the system. The eigenfunctions of the Hamiltonians \(\hat{h}_{I} ({\mathbf{r}}_{1} )\) and \(\hat{h}_{II} ({\mathbf{r}}_{2} )\), describing the motions of particles 1 and 2, corresponding to states I k, II l with energies \(\varepsilon_{I}\) and \(\varepsilon_{II}\), respectively, are the products of the coordinate and spin parts of the first and second particles, respectively:

$$\begin{gathered} \hat{h}_{I} ({\mathbf{r}}_{1} )\psi_{I} ({\mathbf{r}}_{1} ,\xi_{1} ) = \varepsilon_{I} \psi_{I} ({\mathbf{r}}_{1} ,\xi_{1} ), \hfill \\ \hat{h}_{II} (\vec{r}_{2} )\psi_{II} ({\mathbf{r}}_{2} ,\xi_{2} ) = \varepsilon_{Il} \psi_{II} ({\mathbf{r}}_{2} ,\xi_{2} ), \hfill \\ \psi_{I} ({\mathbf{r}}_{1} ,\xi_{1} ) = \phi_{I} ({\mathbf{r}}_{1} )\chi_{1} (\xi_{1} ), \hfill \\ \psi_{II} ({\mathbf{r}}_{2} ,\xi_{2} ) = \phi_{II} ({\mathbf{r}}_{2} )\chi_{2} (\xi_{2} ). \hfill \\ \end{gathered}$$
(47)

The solution of the stationary Schrödinger equation for two identical particles, in the absence of interaction,

$$\hat{H}^{0} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ) = E_{I,II}^{0} \Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ),$$

in accordance with Eq. (46), is (2 s + 1) times degenerate in the value of the total spin S, where s is the value of the spin of the particle. The total wave function \(\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} )\), taking into account the above-indicated degeneracy, corresponding to the energy value \(E_{I,II}^{0} = \varepsilon_{I} + \varepsilon_{II}\), is an eigenfunction of this Hamiltonian:

$$\hat{H}^{0} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ) = \left( {\varepsilon_{I} + \varepsilon_{II} } \right)\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ),$$
(48)

and meets the properties of symmetry with respect to the permutation operation.

$$\hat{P}_{1,2} \Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ) = \left( { - 1} \right)^{2s} \Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ).$$
(49)

Here, the parity of the complete permutation \(g = \left( { - 1} \right)^{2s}\). The operation of a total permutation of indices 1 and 2 includes a permutation of the spatial coordinates of particles 1 and 2 and the values of the projections of the spins on the Z-axis of individual particles:

$$\hat{P}_{1,2} = \hat{P}_{{{\mathbf{r}}_{1} {\mathbf{r}}_{2} }} \hat{P}_{{s_{z1} ,s_{z2} }} .$$
(50)

In the nonrelativistic case, a two-particle function can always be represented as a simple product of its coordinate and spin parts:

$$\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ) = \Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) \cdot {\rm X}(\xi_{1} ,\xi_{2} ).$$
(51)

The symmetry properties of each part of the wave function with respect to the permutation operation are determined as follows:

  1. (1)

    The symmetry of the spin part of a system of two identical particles is determined by the symmetry of the Clebsch–Gordan coefficients [2]

    $${\rm X}(\xi_{1} ,\xi_{2} ) = \left| {S,S_{z} } \right\rangle = \sum\limits_{{s_{1z} s_{2z} }} {C_{{s_{1} s_{1z} s_{2} s_{2z} }}^{{SS_{z} }} } \left| {s_{1} ,s_{1z} } \right\rangle \left| {s_{2} ,s_{2z} } \right\rangle ,$$

    where

    $$C_{{ss_{1z} ss_{2z} }}^{{SS_{z} }} = \left( { - 1} \right)^{2s - S} C_{{ss_{2z} ss_{1z} }}^{{SS_{z} }} ,$$

    and

    $$\hat{P}_{{s_{z1} ,s_{z2} }} {\rm X}(\xi_{1} ,\xi_{2} ) = \left( { - 1} \right)^{2s - S} {\rm X}(\xi_{1} ,\xi_{2} ).$$
    (52)

    Here, S is the total spin of the couple of particles.

  2. (2)

    The symmetry of the coordinate part of the two-particle function Eq. (51) is determined as the result of the action of the permutation operator \(\hat{P}_{{{\mathbf{r}}_{1} {\mathbf{r}}_{2} }}\) on the function Eq. (51), taking into account Eqs. (49) and  (52):

    $$\begin{gathered} \hat{P}_{1,2} \Psi_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ) = \hat{P}_{{{\mathbf{r}}_{{\mathbf{1}}} {\mathbf{,r}}_{{\mathbf{2}}} }} \Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\hat{P}_{{s_{z1} ,s_{z2} }} {\rm X}(\xi_{1} ,\xi_{2} ) = \hfill \\ = \left( { - 1} \right)^{2s} \Psi_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ), \hfill \\ \hat{P}_{{{\mathbf{r}}_{{\mathbf{1}}} {\mathbf{,r}}_{{\mathbf{2}}} }} \Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\left( { - 1} \right)^{2s - S} {\rm X}(\xi_{1} ,\xi_{2} ) = \left( { - 1} \right)^{2s} \Psi_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} ), \hfill \\ \hat{P}_{{{\mathbf{r}}_{{\mathbf{1}}} {\mathbf{,r}}_{{\mathbf{2}}} }} \Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) = \left( { - 1} \right)^{S} \Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ). \hfill \\ \end{gathered}$$
    (53)

Taking into account the interaction in the first order of the stationary perturbation theory gives the first correction to energy Eq. (48) in the following form:

$$\begin{gathered} E^{(1)} = \left\langle {\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} )} \right|\hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\left| {\Psi^{0}_{I,II} ({\mathbf{r}}_{1} ,\xi_{1} ,{\mathbf{r}}_{2} ,\xi_{2} )} \right\rangle = \hfill \\ = \left\langle {\Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )} \right|\hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\left| {\Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )} \right\rangle \left\langle {{\rm X}(\xi_{1} ,\xi_{2} )} \right.\left| {{\rm X}(\xi_{1} ,\xi_{2} )} \right\rangle = K \pm A, \hfill \\ K = \left\langle {\phi_{I} ({\mathbf{r}}_{1} )\phi_{II} ({\mathbf{r}}_{2} )} \right|\hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\left| {\phi_{I} ({\mathbf{r}}_{2} )\phi_{II} ({\mathbf{r}}_{1} )} \right\rangle , \hfill \\ A = \left\langle {\phi_{I} ({\mathbf{r}}_{1} )\phi_{II} ({\mathbf{r}}_{2} )} \right|\hat{V}({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} )\left| {\phi_{I} ({\mathbf{r}}_{2} )\phi_{II} ({\mathbf{r}}_{1} )} \right\rangle , \hfill \\ \end{gathered}$$
(54)

where K is the “direct” Coulomb contribution and A is the exchange contribution. This type of solution is due to the symmetry of the coordinate part of the wave function (see, for example, [31]) which has the following form:

$$\Phi_{I,II} ({\mathbf{r}}_{1} ,{\mathbf{r}}_{2} ) = {\rm N}\left( {\phi_{I} ({\mathbf{r}}_{1} )\phi_{II} ({\mathbf{r}}_{2} ) + \left( { - 1} \right)^{S} \phi_{I} ({\mathbf{r}}_{2} )\phi_{II} ({\mathbf{r}}_{1} )} \right).$$
(55)

Thus, the correction to the energy due to the interaction in the coordinate representation is determined according to Eq. (55) by the value of the total spin:

$$E^{(1)} = K + \left( { - 1} \right)^{S} A.$$
(56)

It should be noted here that if we are dealing with a system of identical spin-less particles (s = 0), then in expressions (45)–(48), the spin part should be omitted (spin function \(\chi_{1} (\xi_{1} ) = \chi_{2} (\xi_{2} ) = {\rm X}(\xi_{1} \xi_{2} ) = 1\)). Nevertheless, spin-less particles belong to the bosonic class (s = 0—is the integer number), then the condition Eq. (49) must be satisfied with the factor \(g = \left( { - 1} \right)^{2s} = 1\), which means the symmetry of the total wave function of the system of two spin-less particles. The degeneracy of the two-particle state with the energy \(E_{I,II}^{0} = \varepsilon_{I} + \varepsilon_{II}\) is equal to (2 s + 1) = 1, i.e., this state is non-degenerated. That is why the energy correction Eq. (53) has only one solution corresponding to the symmetric form of the wave function \(E^{(1)} = K + A,\) that is in an agreement with the general condition Eq. (56) for the total spin S = 0.

3.2 Spin Hamiltonian of a System of Identical Particles

We showed above (see Eqs. 49, 50, 52, and 53) that the symmetries of the coordinate and spin parts of the wave function are uniquely related. This means that one can introduce operators acting in the spin space and directly acting on the spin variables of the wave function, in such a way that the eigenvalues of these operators will be numbers or combinations of numbers that are directly related to the symmetry of the coordinate part.

Let us set ourselves the goal of constructing a Hamiltonian in the spin representation for a system of identical particles, which describes the interaction defined in the coordinate representation, and its eigenvalue will be a solution in the form Eq. (56) [32,33,34,35]. For this, it is necessary to obtain an explicit form of the parity operator of the permutation of the coordinate part of the wave function, denote it by \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\), which, however, acts on the spin variables of the wave function in the spin representation. The eigenvalue of this operator must be the parity value \(\Lambda_{r} = \left( { - 1} \right)^{S}\) of the permutation of the coordinate part of the wave function Eqs. (53), (55):

$$\begin{gathered} {\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} {\rm X}(\xi_{1} ,\xi_{2} ) = \left( { - 1} \right)^{S} {\rm X}(\xi_{1} ,\xi_{2} ), \Leftrightarrow {\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} \left| {S,S_{z} ;s1s2} \right\rangle , \hfill \\ {\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} \left| {S,S_{z} ;s1s2} \right\rangle = \left( { - 1} \right)^{S} \left| {S,S_{z} ;s1s2} \right\rangle , \Rightarrow \hfill \\ {\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} \sum\limits_{{s_{1z} s_{2z} }} {C_{{s_{1} s_{1z} s_{2} s_{2z} }}^{{SS_{z} }} } \left| {s_{1} ,s_{1z} } \right\rangle \left| {s_{2} ,s_{2z} } \right\rangle = \left( { - 1} \right)^{S} \sum\limits_{{s_{1z} s_{2z} }} {C_{{s_{1} s_{1z} s_{2} s_{2z} }}^{{SS_{z} }} } \left| {s_{1} ,s_{1z} } \right\rangle \left| {s_{2} ,s_{2z} } \right\rangle . \hfill \\ \end{gathered}$$
(57)

We will look for the parity operator \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\) in the form of a series expansion in powers of the scalar products of the spin operators of interacting particles:

$${\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} = c_{2s} \left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2s} + c_{2s - 1} \left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2s - 1} + ... + c_{1} \left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) + c_{0} .$$
(58)

Substituting (41) into (40), and taking into account

$${\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} = \frac{1}{2}\left( {\widehat{{S^{2} }} - \widehat{{s_{1}^{2} }} - \widehat{{s_{2}^{2} }}} \right) = \frac{1}{2}\left( {\widehat{{S^{2} }} - 2\widehat{{s^{2} }}} \right),$$
(59)

we have a system of algebraic equations that uniquely determines the sought coefficients in expansion Eq. (58):

$$\left\{ \begin{gathered} c_{2s} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s} + c_{2s - 1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s - 1} + ... \hfill \\ \left. { + c_{1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right) + c_{0} } \right|_{S = 0} = \left( { - 1} \right)^{0} , \hfill \\ c_{2s} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s} + c_{2s - 1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s - 1} + ... \hfill \\ \left. { + c_{1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right) + c_{0} } \right|_{S = 1} = \left( { - 1} \right)^{1} , \hfill \\ ... \hfill \\ c_{2s} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s} + c_{2s - 1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s - 1} + ... \hfill \\ \left. { + c_{1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right) + c_{0} } \right|_{S} = \left( { - 1} \right)^{S} , \hfill \\ ... \hfill \\ c_{2s} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s} + c_{2s - 1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right)^{2s - 1} + ... \hfill \\ \left. { + c_{1} \left( {\frac{S(S + 1) - 2s(s + 1)}{2}} \right) + c_{0} } \right|_{S = 2s} = \left( { - 1} \right)^{2s} . \hfill \\ \end{gathered} \right.$$
(60)

Thus, in the first approximation of perturbation theory, Hamiltonian Eq. (45) can be replaced by an effective Hamiltonian acting in the spin space, for which the commutation relations Eq. (45) remain valid.

$$\hat{H} = E^{0} + K + A{\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} ,$$
(61)

where the parity operator \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\) is uniquely determined by Eqs. (58) and (60).

Generalizing to the case of a system of particles with an arbitrary spin s, and dropping the constant corresponding to the energy of the state \(E^{0}\) of non-interacting particles, we obtain a Hamiltonian that takes into account the pair interaction in the form:

$$\hat{H}_{{\text{int}}} = \sum\limits_{k < l} {K_{kl} } + \sum\limits_{k < l} {A_{kl} {\hat{\mathscr{R}}}_{{s_{k} ,s_{l} }} } .$$
(62)

Let us give several examples for the Hamiltonian of identical particles with allowance for the pair interaction, written in the spin representation.

(i) s= 1/2

The system of equations Eq. (60) is a system of two equations for determining the coefficients \(c_{1}\) and \(c_{0}\):

$$\left\{ \begin{gathered} c_{1} \frac{1}{2}\left( {1(1 + 1) - 2\frac{1}{2}\left( {\frac{1}{2} + 1} \right)} \right) + c_{0} = \left( { - 1} \right)^{1} \hfill \\ c_{1} \frac{1}{2}\left( {0 - 2\frac{1}{2}\left( {\frac{1}{2} + 1} \right)} \right) + c_{0} = \left( { - 1} \right)^{0} , \hfill \\ \end{gathered} \right.$$
(63)

whence we determine the coefficients \(c_{1} = - 2\) and \(c_{0} = - \frac{1}{2}\), then the parity operator of the pairwise permutation has the form: \({\hat{\mathscr{R}}}_{{s_{k} \cdot s_{l} }} = - \frac{1}{2}\left( {1 + 4{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right).\)[3], and Hamiltonian Eq. (62) takes the well-known form of the Heisenberg–Dirac–Van Vleck (HDV) Hamiltonian:

$$\begin{gathered} \hat{H}_{{{\text{int}} 1/2}} = \sum\limits_{k < l} {K_{kl} } - \frac{1}{2}\sum\limits_{k < l} {A_{kl} (1 + 4{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} ) = } \hfill \\ = \sum\limits_{k < l} {(K_{kl} } - \frac{{A_{kl} }}{2}) - \sum\limits_{k < l} {2A_{kl} {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } = \hfill \\ = \sum\limits_{k < l} {(K_{kl} } - \frac{{A_{kl} }}{2}) - \sum\limits_{k < l} {J_{kl} {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } , \hfill \\ \end{gathered}$$
(64)

where \(J_{kl}\) is a Heisenberg parameter, \(J_{kl} = 2A_{kl}\).

In the case when the constant of pair exchange interaction, or, which is the same, the Heisenberg parameter is the same for all pairs of particles, the Schrödinger equation with Hamiltonian Eq. (64) admits an exact solution. Indeed, in the indicated case, Hamiltonian Eq. (64) can be rewritten as follows:

$$\begin{gathered} \hat{H}_{{{\text{int}} 1/2}} = \sum\limits_{k < l} {(K_{kl} } - \frac{{A_{kl} }}{2}) - J\sum\limits_{k < l} {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } = const - \frac{J}{2}\left( {\widehat{{\Sigma^{2} }} - \sum\limits_{k = 1}^{\rm N} {\widehat{{s_{k}^{2} }}} } \right) = \hfill \\ = const - \frac{J}{2}\left( {\widehat{{\Sigma^{2} }} - {\rm N}\widehat{{s^{2} }}} \right), \hfill \\ \end{gathered}$$
(65)

where \(\widehat{{\Sigma^{2} }}\) is the operator of the total spin quadrat of the particles system, \({\rm N}\) is the total number of particles. The eigenvector \(\left| {\Sigma ,{\rm M}} \right\rangle\) of this Hamiltonian is the vector with the determined value of the total spin \(\Sigma\) of the system and its projection \({\rm M}\) onto the quantization axis z.

$$\begin{gathered} \left\{ {\frac{{{\rm N}({\rm N} - 1)}}{2}\left( {K - \frac{A}{2}} \right) - \frac{J}{2}\left( {\widehat{{\Sigma^{2} }} - {\rm N}\widehat{{s^{2} }}} \right)} \right\}\left| {\Sigma ,{\rm M}} \right\rangle = \hfill \\ = \left\{ {\frac{{{\rm N}({\rm N} - 1)}}{2}\left( {K - \frac{A}{2}} \right) - \frac{J}{2}\left( {\Sigma \left( {\Sigma + 1} \right) - {\rm N}s\left( {s + 1} \right)} \right)} \right\}\left| {\Sigma ,{\rm M}} \right\rangle . \hfill \\ \end{gathered}$$
(66)

The eigenvalue of the interaction energy of the spin system will have the following form:

$$\begin{gathered} E_{{\text{int}}} = \frac{{{\rm N}({\rm N} - 1)}}{2}\left( {K - \frac{A}{2}} \right) - \frac{J}{2}\left( {\Sigma \left( {\Sigma + 1} \right) - {\rm N}\frac{1}{2}\left( {\frac{1}{2} + 1} \right)} \right) = \hfill \\ = \frac{{{\rm N}({\rm N} - 1)}}{2}\left( {K - \frac{A}{2}} \right) - A\left( {\Sigma \left( {\Sigma + 1} \right) - {\rm N}\frac{3}{4}} \right). \hfill \\ \end{gathered}$$
(67)

If the exchange interaction constant A > 0 (J > 0) is positive, then the most favorable state is the state of the system with the maximum possible value of the total spin \(\Sigma = \frac{N}{2}\). That is, ferromagnetic ordering is established in the system with energy \(E_{{\text{int}}} = \frac{{{\rm N}\left( {{\rm N} - 1} \right)}}{2}\left( {K - A} \right)\). In the case of a negative value A < 0 (J < 0)—the most favorable energetically state with the minimum total spin \(\Sigma \to 0\), that is, antiferromagnetic ordering of spins with the energy \(E_{{\text{int}}} = \frac{\rm N}{4}\left\{ {2({\rm N} - 1)K - A\left( {{\rm N} - 4} \right)} \right\}\).

(ii) s= 1

For particles with spins s = 1, expression (43) gives a system of three equations:

$$\left\{ \begin{gathered} c_{2} \left( {\frac{2(2 + 1) - 2 \cdot 1(1 + 1)}{2}} \right)^{2} + c_{1} \left( {\frac{2(2 + 1) - 2 \cdot 1 \cdot (1 + 1)}{2}} \right) + c_{0} = \left( { - 1} \right)^{2} \hfill \\ c_{2} \left( {\frac{1(1 + 1) - 2 \cdot 1 \cdot (1 + 1)}{2}} \right)^{2} + c_{1} \left( {\frac{1(1 + 1) - 2 \cdot (1 + 1)}{2}} \right) + c_{0} = \left( { - 1} \right)^{1} \hfill \\ c_{2} \left( {\frac{0 \cdot (0 + 1) - 2 \cdot 1 \cdot (1 + 1}{2}} \right)^{2} + c_{1} \left( {\frac{0 \cdot (0 + 1) - 2 \cdot 1 \cdot (1 + 1)}{2}} \right) \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{0} = \left( { - 1} \right)^{0} , \hfill \\ \end{gathered} \right.$$
(68)

which uniquely determines the coefficients: \(c_{2} = c_{1} = - c_{0} = 1.\) The parity operator \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\) for a pair of particles in this case is

$${\hat{\mathscr{R}}}_{{s_{k} \cdot s_{l} }} = - 1 + \left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) + \left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{2} .$$
(69)

Hamiltonian Eq. (62) includes, in addition to the bilinear term, also the biquadratic contribution:

$$\hat{H}_{{{\text{int}} 1}} = \sum\limits_{k < l} {K_{kl} } + \sum\limits_{k < l} {A_{kl} \left( {\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{2} + \left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) - 1} \right).}$$
(70)

Interaction energy for a couple of particles k, l:

$$\begin{gathered} {\text{E}}_{{\text{k,l}}}^{{{\text{exc}}}} = A_{k,l} \left\{ {\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{2} + \frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right) - 1} \right\} = \hfill \\ = A_{kl} \left\{ {\left( {\frac{1}{2}\left( {S(S + 1) - 4} \right)} \right)^{2} + \frac{1}{2}\left( {S(S + 1) - 4} \right) - 1} \right\} = \hfill \\ = A_{kl} \left\{ {\left( {\frac{S(S + 1)}{2}} \right)^{2} - \frac{3S(S + 1)}{2} + 1} \right\}. \hfill \\ \end{gathered}$$
(71)

If the exchange interaction constant for a couple of particles A > 0, then the favorable state is the antisymmetric state with the total spin S = 1 (, Sz =  ± 1.0) and the correction to the energy, the degeneracy multiplicity of which is gS = 3. In the case A < 0, an advantageous state will be a symmetric total spin degenerate state with S = 0 (Sz = 0) and S = 2 (Sz =  ± 2, ± 1, 0) and a correction to the energy, the degeneracy multiplicity of which taking into account total spin projections gS = 6.

(iii) s= 3/2

For a system of particles with spins s = 3/2, the parity operator \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\) has the following form:

$${\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} 3/2}} = - \frac{2}{9}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{3} - \frac{11}{{18}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2} + \frac{9}{8}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) + \frac{67}{{32}}.$$
(72)

Hamiltonian Eq. (62) takes a bicubic form:

$$\hat{H}_{{{\text{int}} 1}} = \sum\limits_{k < l} {K_{kl} } + \sum\limits_{k < l} {A_{kl} \left( { - \frac{2}{9}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{3} - \frac{11}{{18}}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{2} + \frac{9}{8}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) + \frac{67}{{32}}} \right).}$$
(73)

Interaction energy of a couple of particles k, l:

$$\begin{gathered} E_{k,l}^{exc} = A\left\{ { - \frac{2}{9}\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{3} - \frac{11}{{18}}\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{2} } \right. + \hfill \\ \left. { + \frac{9}{8}\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right) + \frac{67}{{32}}} \right\} = \hfill \\ = A\left\{ { - \frac{2}{9}\left( {\frac{1}{2}\left( {S(S + 1) - 3 \cdot \frac{5}{2}} \right)} \right)^{3} - \frac{11}{{18}}\left( {S(S + 1) - \frac{15}{2}} \right)^{2} } \right. + \hfill \\ \left. { + \frac{9}{8}\left( {S(S + 1) - \frac{15}{2}} \right) + \frac{67}{{32}}} \right\}. \hfill \\ \end{gathered}$$
(74)

If the exchange interaction constant A > 0, then an advantageous state is a degenerate symmetric state with a total spin S = 1 and S = 3. The correction to the energy in this case \({\text{E}}_{{\text{k,l}}}^{{{\text{exc}}}} = - A_{k,l}\), the degeneracy rate of which is determined by the entire set of projections on the z-axis of the indicated values of the total spin: for S = 1, Sz =  ± 1.0 and for S = 3, Sz =  ± 3, ± 2, ± 1.0, that is, gS = 10. In the case A < 0, an antisymmetric degenerate state with S = 0 (Sz = 0) and S = 2 (Sz =  ± 2, ± 1.0) and a correction to the energy, which is characterized by the degeneracy multiplicity gS = 6, will be an advantageous state.

(iv) s  = 2

For a system of particles with spins s = 2, the parity operator \({\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }}\) has the following form:

$${\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} 2}} = \frac{1}{36}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{4} + \frac{1}{6}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{3} - \frac{13}{{36}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2} - \frac{5}{2}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) - 1.$$
(75)

The Hamiltonian Eq. (62) takes a bi-tetras form:

$$\begin{gathered} \hat{H}_{{{\text{int}} 2}} = \sum\limits_{k < l} {K_{kl} } + \sum\limits_{k < l} {A_{kl} \,\left( {\frac{1}{36}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{4} + \frac{1}{6}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{3} } \right.} \, \hfill \\ \left. {\quad \quad \quad \quad \quad \quad \quad - \frac{13}{{36}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2} - \frac{5}{2}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) - 1} \right). \hfill \\ \end{gathered}$$
(76)

The eigenvalue of the interaction energy of a couple of particles is

$$\begin{gathered} E_{1,2}^{exc} = \frac{A}{6}\left\{ {\frac{1}{6}\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{4} + \left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{3} } \right. - \hfill \\ \left. { - \frac{13}{6}\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{2} - 15\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right) - 6} \right\} = \hfill \\ = \frac{A}{6}\left\{ {\frac{1}{6}\left( {\left( {\frac{S(S + 1)}{2} - 2 \cdot 3} \right)} \right)^{4} + \left( {\frac{S(S + 1)}{2} - 6} \right)^{3} } \right. - \hfill \\ \left. { - \frac{13}{6}\left( {\frac{S(S + 1)}{2} - 6} \right)^{2} - 15\left( {\frac{S(S + 1)}{2} - 6} \right) - 6} \right\}. \hfill \\ \end{gathered}$$
(77)

If the exchange interaction constant A > 0, then the favorable state is the degenerate antisymmetric state with the total spin S = 3, (Sz =  ± 3, ± 2, ± 1, 0) and S = 1 (Sz =  ± 1, 0) and the correction to energy, the degeneracy multiplicity of which is gS = 10. In the case A < 0, a symmetric degenerate state with S = 0 (Sz = 0), S = 2 ((Sz =  ± 2, ± 1, 0), S = 4 (Sz =  ± 4, ± 3, ± 2, ± 1, 0) and a correction to the energy, the degeneracy multiplicity of which is gS = 15.

(v) s  = 5/2

The system of equations for determining the coefficients Eq. (60) written for the case of particles with spin s = 5/2 leads to a system of six equations for determining six coefficients

$$\left\{ \begin{gathered} c_{5} \left( {15 - \frac{35}{4}} \right)^{5} + c_{4} \left( {15 - \frac{35}{4}} \right)^{4} + c_{3} \left( {15 - \frac{35}{4}} \right)^{3} + c_{2} \left( {15 - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( {15 - \frac{35}{4}} \right) + c_{0} = - 1 \hfill \\ c_{5} \left( {10 - \frac{35}{4}} \right)^{5} + c_{4} \left( {10 - \frac{35}{4}} \right)^{4} + c_{3} \left( {10 - \frac{35}{4}} \right)^{3} + c_{2} \left( {10 - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( {10 - \frac{35}{4}} \right) + c_{0} = 1 \hfill \\ c_{5} \left( {6 - \frac{35}{4}} \right)^{5} + c_{4} \left( {6 - \frac{35}{4}} \right)^{4} + c_{3} \left( {6 - \frac{35}{4}} \right)^{3} + c_{2} \left( {6 - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( {6 - \frac{35}{4}} \right) + c_{0} = - 1 \hfill \\ c_{5} \left( {3 - \frac{35}{4}} \right)^{5} + c_{4} \left( {3 - \frac{35}{4}} \right)^{4} + c_{3} \left( {3 - \frac{35}{4}} \right)^{3} + c_{2} \left( {3 - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( {3 - \frac{35}{4}} \right) + c_{0} = 1 \hfill \\ c_{5} \left( {1 - \frac{35}{4}} \right)^{5} + c_{4} \left( {1 - \frac{35}{4}} \right)^{4} + c_{3} \left( {1 - \frac{35}{4}} \right)^{3} + c_{2} \left( {1 - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( {1 - \frac{35}{4}} \right) + c_{0} = - 1 \hfill \\ c_{5} \left( { - \frac{35}{4}} \right)^{5} + c_{4} \left( { - \frac{35}{4}} \right)^{4} + c_{3} \left( { - \frac{35}{4}} \right)^{3} + c_{2} \left( { - \frac{35}{4}} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - \frac{35}{4}} \right) + c_{0} = 1, \hfill \\ \end{gathered} \right.$$
(78)

whence we find the following coefficients:

$$c_{5} = - \frac{1}{450},c_{4} = - \frac{1}{40},c_{3} = \frac{53}{{1200}},c_{2} = \frac{{{14911}}}{{{14400}}},c_{1} = \frac{{{35687}}}{{{23040}}},c_{0} = - \frac{{{47417}}}{{{18432}}}.$$
(79)

We obtain the form of the Hamiltonian of the exchange interaction of a system of identical particles with spins s = 3, in the spin representation

$$\begin{gathered} \hat{H}_{{{\text{int}} 5/2}} = \sum\limits_{k < l} {K_{kl} } + \hfill \\ + \sum\limits_{k < l} {A_{kl} \left( { - \frac{1}{450}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{5} - \frac{1}{40}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{4} + \frac{53}{{1200}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{3} + } \right.} \hfill \\ + \left. {\frac{14911}{{14400}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2} + \frac{35687}{{23040}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) - \frac{{{47417}}}{{{18432}}}} \right). \hfill \\ \end{gathered}$$
(80)

(vi) s  = 3

For particles with spin s = 3 from Eq. (60), we have a system of seven equations that completely determine the coefficients:

$$\left\{ \begin{gathered} c_{6} \left( {21 - 12} \right)^{6} + c_{5} \left( 9 \right)^{5} + c_{4} \left( 9 \right)^{4} + c_{3} \left( 9 \right)^{3} + c_{2} \left( 9 \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( 9 \right) + c_{0} = 1 \hfill \\ c_{6} \left( {15 - 12} \right)^{6} + c_{5} \left( 3 \right)^{5} + c_{4} \left( 3 \right)^{4} + c_{3} \left( 3 \right)^{3} + c_{2} \left( 3 \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( 3 \right) + c_{0} = - 1 \hfill \\ c_{6} \left( {10 - 12} \right)^{6} + c_{5} \left( { - 2} \right)^{5} + c_{4} \left( { - 2} \right)^{4} + c_{3} \left( { - 2} \right)^{3} + c_{2} \left( { - 2} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - 2} \right) + c_{0} = 1 \hfill \\ c_{6} \left( {6 - 12} \right)^{6} + c_{5} \left( { - 6} \right)^{5} + c_{4} \left( { - 6} \right)^{4} + c_{3} \left( { - 6} \right)^{3} + c_{2} \left( { - 6} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - 6} \right) + c_{0} = - 1 \hfill \\ c_{6} \left( {3 - 12} \right)^{6} + c_{5} \left( { - 9} \right)^{5} + c_{4} \left( { - 9} \right)^{4} + c_{3} \left( { - 9} \right)^{3} + c_{2} \left( { - 9} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - 9} \right) + c_{0} = 1 \hfill \\ c_{6} \left( {1 - 12} \right)^{6} + c_{5} \left( { - 11} \right)^{5} + c_{4} \left( { - 11} \right)^{4} + c_{3} \left( { - 11} \right)^{3} + c_{2} \left( { - 11} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - 11} \right) + c_{0} = - 1 \hfill \\ c_{6} \left( { - 12} \right)^{6} + c_{5} \left( { - 12} \right)^{5} + c_{4} \left( { - 12} \right)^{4} + c_{3} \left( { - 12} \right)^{3} + c_{2} \left( { - 12} \right)^{2} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + c_{1} \left( { - 12} \right) + c_{0} = 1. \hfill \\ \end{gathered} \right.$$
(81)

Whence we obtain the coefficients and the form of the Hamiltonian of the exchange interaction of a system of identical particles with spins s = 3, in the spin representation:

$$\begin{gathered} c_{6} = \frac{1}{{{8100}}},c_{5} = \frac{19}{{{8100}}},c_{4} = - \frac{2}{{{2025}}},c_{3} = - \frac{{{571}}}{{{2700}}}, \hfill \\ c_{2} = - \frac{{{713}}}{{{900}}},c_{1} = \frac{{{87}}}{{{50}}},c_{0} = \frac{151}{{25}}. \hfill \\ \end{gathered}$$
(82)
$$\begin{gathered} \hat{H}_{{{\text{int}} 3}} = \sum\limits_{k < l} {K_{kl} } + \hfill \\ + \sum\limits_{k < l} {A_{kl} \left( {\frac{1}{8100}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{6} + \frac{19}{{8100}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{5} - \frac{2}{2025}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{4} - \frac{571}{{2700}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{3} - } \right.} \hfill \\ - \left. {\frac{713}{{900}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)^{2} + \frac{87}{{50}}\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right) + \frac{{{151}}}{{{25}}}} \right). \hfill \\ \end{gathered}$$
(83)

4 Applications of Spin Hamiltonians

Interest in one-dimensional chains of ions with spins S = 1 arose long ago, beginning with the well-known works of Haldane [24, 25]. In these works, the authors suggested that the total spin of the entire chain manifests itself in the ground state as an exponentially decreasing spin–spin correlation, and unlike a chain of ½-spins, spin-1 chains have an excitation spectrum separated from the ground state by an energy gap, which causes a fundamentally different behavior at the critical point.

Single-ion anisotropy with a corresponding manifestation of magnetic properties exists in many compounds, for example, in CsNiCl3 (weak axial anisotropy), NENP [Ni (C2 H8N 2) 2 NO2) ClO4] (weak axial anisotropy), CsFeBr3, NENC [Ni (C2H8 N2) 2 Ni (CN) 4], and DTN [NiCl2–4SC (NH2) 2] (strong planar anisotropy [27]), which could become objects for experimental verification of the results obtained from the theory of critical phenomena in systems with “large” spin.

4.1 Antiferromagnetic Ordering in a Chain of Identical Particles with Spin s = 1

Here, we consider collective effects associated with magnetic ordering in a chain of ions whose spin is equal to one. The exchange interaction of ions with the nearest neighbors gives rise to a pair potential, which rapidly decreases with the distance between particles, so that the interaction with ions of the second coordination sphere can be considered negligible, and in our consideration, equal to zero.

In the case of a chain of ions with a pairwise interaction of the nearest neighbors, the Hamiltonian, taking into account Eq. (70), can be written in the following form:

$$\hat{H}_{{{\text{int}} 1}} = \sum\limits_{k} {\left\{ {K_{k,k + 1} + A_{k,k + 1} \left[ {\left( {{\hat{\mathbf{s}}}_{i} \cdot {\hat{\mathbf{s}}}_{i + 1} } \right)^{2} + {\hat{\mathbf{s}}}_{i} \cdot {\hat{\mathbf{s}}}_{i + 1} - 1} \right]} \right\}} ,$$
(84)

which, in contrast to the Heisenberg Hamiltonian, contains, along with the bilinear, also the biquadratic term of the scalar product of the spin operators. Hamiltonian Eq. (84), as was done above, when deriving expression Eq. (71), can be rewritten in the form where the operators of the scalar products of the spins of neighboring ions will be replaced by the operators of the square of the total spin of the same pair:

$${\hat{\mathbf{s}}}_{i} \cdot {\hat{\mathbf{s}}}_{i + 1} = \frac{1}{2}\left( {\widehat{{S^{2} }} - \widehat{{s_{i}^{2} }} - \widehat{{s_{i + 1}^{2} }}} \right).$$

The exchange contribution to the energy \(E_{k,k + 1}^{exc}\) from each pair is written as follows:

$$E_{k,k + 1}^{exc} = A_{k,k + 1} \left\{ {\left( {\frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right)} \right)^{2} + \frac{1}{2}\left( {S(S + 1) - 2s(s + 1)} \right) - 1} \right\}.$$
(85)

In the case of a positive value of the exchange integral, the energy has a minimum value for the total spin of a pair of particles S = 1, which corresponds to an antisymmetric state for each pair of spins of the nearest neighbors.

To calculate the magnetic state of the entire chain, we will use the following model of dividing it into segments: first, we select segments containing two interacting ions along the entire length of the chain, and then consider the pairwise interaction of two segments that already form quartile segments, etc.

Hamiltonian of exchange interaction Eq. (84) for a chain of spins can be represented as a series

$$\hat{H}_{{{\text{int}} 1}} = K\frac{N}{2} + \sum\limits_{k = 1,N/2} {A_{k,k + 1} } \left\langle { \bullet_{k} \bullet_{k + 1} } \right\rangle + \sum\limits_{q = 1,N/4} {A_{q,q + 1}^{(1)} } \left\langle {\left\langle { \bullet \bullet } \right\rangle_{q} \left\langle { \bullet \bullet } \right\rangle_{q + 1} } \right\rangle + ...,$$
(86)

where the exchange interaction operators of a couple of neighboring ions are symbolically denoted as \(A_{k,k + 1} \left[ {\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{k + 1} } \right)^{2} + {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{k + 1} - 1} \right] = A_{k,k + 1} \left\langle { \bullet_{k} \bullet_{k + 1} } \right\rangle\). For a pair of couples is the following denotation with renormalized constant \(A_{q,q + 1}^{(1)}\) and total spins of the pair \({\hat{\mathbf{S}}}_{q} = \left( {{\hat{\mathbf{s}}}_{k} + {\hat{\mathbf{s}}}_{k + 1} } \right),\)\({\hat{\mathbf{S}}}_{q + 1} = \left( {{\hat{\mathbf{s}}}_{k + 2} + {\hat{\mathbf{s}}}_{k + 3} } \right)\)

$$A_{q,q + 1}^{(1)} \left[ {\left( {{\hat{\mathbf{S}}}_{q} \cdot {\hat{\mathbf{S}}}_{q + 1} } \right)^{2} + {\hat{\mathbf{S}}}_{q} \cdot {\hat{\mathbf{S}}}_{q + 1} - 1} \right] = A_{q,q + 1}^{(1)} \left\langle {\left\langle { \bullet \bullet } \right\rangle_{q} \left\langle { \bullet \bullet } \right\rangle_{q + 1} } \right\rangle .$$

We search the ground state energy by using the method of successive approximations taking into account the block structure of the Hamiltonian. In this case, the lowest state corresponding to the minimum energy in a pair of blocks is the antisymmetric state with the total spin of the pair S = 1. The energy of these couples is

$$\begin{gathered} E_{k,k + 1}^{exc} = A_{k,k + 1} \left\{ {\left( {\frac{1}{2}\left( {1(1 + 1) - 2 \cdot 1(1 + 1)} \right)} \right)^{2} } \right. \hfill \\ \left. { + \frac{1}{2}\left( {1 \cdot (1 + 1) - 2 \cdot 1 \cdot (1 + 1)} \right) - 1} \right\} = - A_{k,k + 1} = - A^{(0)} . \hfill \\ \end{gathered}$$
(87)

Thus, the value of the total spin Σ of the entire chain consisting of N ions will be equal to Σ = 1, which corresponds to the specific magnetization of the system per particle \(\frac{M}{N} = \frac{\gamma \Sigma }{N} = \frac{\gamma \cdot 1}{N} \to 0\) and belongs to the antiferromagnetic state of the chain. In this case, the total energy will have the following form:

$$E_{{\text{int}}} \approx K\frac{N}{2} - \frac{N}{2}\left( {A^{(0)} + \frac{{A^{(1)} }}{2} + ... + \frac{{A^{(\nu )} }}{{2^{\nu } }}} \right).$$
(88)

Assuming that the exchange interaction constant A (ν) at each subsequent ν-th renormalization step is related to the constant A (ν−1) of the previous (ν−1)-th step as

$$A^{(\nu )} = A^{(\nu - 1)} \xi ,$$
(89)

where ξ is some constant characterizing the decrease in the exchange interaction during the transition to the next coordination sphere in the crystal, and for a chain of ions, to the next step of enlarging the segment. Then the energy per ion of the ground state of the antiferromagnetic chain of infinite length will tend to the value

$$\frac{{E_{{\text{int}}} }}{N} \approx \frac{K}{2} - \frac{1}{2}\sum\limits_{\nu = 0}^{\infty } {\frac{{A^{(\nu )} }}{{2^{\nu } }}} = \frac{1}{2}\left( {K - A^{(0)} \sum\limits_{\nu } {\frac{{\xi^{\nu } }}{{2^{\nu } }}} } \right) = \left( {\frac{1}{2}K - A^{(0)} \frac{1}{2 - \xi )}} \right).$$
(90)

Thus, the contribution of the exchange interaction in the chain has a singularity when the condition \(\xi \to 2\) is satisfied, which is naturally associated with an increase in the order parameter in the system, which leads to a disorder–order phase transition with the antiferromagnetic ordering. The renormalization coefficient can be obtained in the following way [35]. Let's write the statistical sum as follows:

$$\begin{gathered} Z = \hfill \\ = \sum\limits_{\begin{subarray}{l} S_{1,2} = 0, \\ \left( {M_{1,2} } \right) \end{subarray} }^{2} {\sum\limits_{\begin{subarray}{l} S_{3,4} = 0, \\ \left( {M_{3,4} } \right) \end{subarray} }^{2} {...\sum\limits_{\begin{subarray}{l} S_{N - 1,N} = 0, \\ \left( {M_{1N - 1,N} } \right) \end{subarray} }^{2} {\left\langle {S_{1,2} M_{1,2} } \right|\exp } } } \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} } \right)\left| {S_{1,2} M_{1,2} } \right\rangle \times \hfill \\ \left\langle {S_{3,4} M_{3,4} } \right|\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{3} \cdot s_{4} }} } \right)\left| {S_{3,4} M_{3,4} } \right\rangle ... \times \hfill \\ ...\left\langle {S_{N - 1,N} M_{N - 1,N} } \right|\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{N - 1} \cdot s_{N} }} } \right)\left| {S_{N - 1,N} M_{N - 1,N} } \right\rangle = \hfill \\ = \sum\limits_{\begin{subarray}{l} S_{1,2,3,4} , \\ \left( {M_{1,2,3,4} } \right) \end{subarray} } {\sum\limits_{\begin{subarray}{l} S_{5,6,7,8,} \\ \left( {M_{5,6,7,8} } \right) \end{subarray} } {...\sum\limits_{\begin{subarray}{l} S_{N - 3,N - 2,N - 1,N} \\ \left( {M_{N - 3,N - 2,N - 1,N} } \right) \end{subarray} } {f\left( {\frac{{A^{(1)} }}{T}} \right)\left\langle {S_{1,2,3,4} M_{1,2,3,4} } \right|} } } \hfill \\ \quad \exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{1,2} \cdot S_{3,4} }} } \right)\left| {S_{1,2,3,4} M_{1,2,3,4} } \right\rangle \times \hfill \\ f\left( {\frac{{A^{(1)} }}{T}} \right)\left\langle {S_{5,6,7,8} M_{5,6,7,8} } \right|\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{5,6} \cdot S_{7,8} }} } \right)\left| {S_{5,6,7,8,2} M_{5,6,7,8} } \right\rangle ... \times \hfill \\ f\left( {\frac{{A^{(1)} }}{T}} \right)\left\langle {S_{N - 3,N - 2,N - 1,N} M_{N - 3,N - 2,N - 1,N} } \right|\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{N - 3,N - 2} \cdot S_{N - 1,N} }} } \right) \hfill \\ \quad \left| {S_{N - 3,N - 2,N - 1,N} M_{N - 3,N - 2,N - 1,N} } \right\rangle , \hfill \\ {\hat{\mathscr{R}}}_{{s_{k} \cdot s_{l} }} = - 1 + \left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) + \left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{2} . \hfill \\ \end{gathered}$$
(91)

Here is the parity operator \({\hat{\mathscr{R}}}_{{s_{k} \cdot s_{l} }}\) in the form Eq. (69), \({\hat{\mathbf{S}}}_{1,2}\) and \({\hat{\mathbf{S}}}_{3,4}\) are the operators of the total spins of pairs of particles 1, 2 and 3, 4, respectively, the function \(f\left( {\frac{{A^{(1)} }}{T}} \right)\) should not depend on the values of the spins \(S_{1,2}\) and \(S_{3,4}\). Taking into account the summation over the values of the spin projections onto an arbitrary quantization axis and the use of expressions for the reduced matrix elements of the statistical operator, we will have an expression for the partition function in the form

$$\begin{gathered} Z = \hfill \\ = \sum\limits_{{S_{1,2} = 0,}}^{2} {\sum\limits_{{S_{3,4} = 0,}}^{2} {...\sum\limits_{{S_{N - 1,N} = 0,}}^{2} {g_{1,2} \left\langle {S_{1,2} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} } \right)} \right|} } } \left| {S_{1,2} } \right\rangle \times \hfill \\ g_{3,4} \left\langle {S_{3,4} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{3} \cdot s_{4} }} } \right)} \right|\left| {S_{3,4} } \right\rangle ... \times \hfill \\ ...g_{N - 1,N} \left\langle {S_{N - 1,N} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{N - 1} \cdot s_{N} }} } \right)} \right|\left| {S_{N - 1,N} } \right\rangle = \hfill \\ = \sum\limits_{{S_{1,2,3,4} ,}} {\sum\limits_{{S_{5,6,7,8,} }} {...\sum\limits_{{S_{N - 3,N - 2,N - 1,N} }} {f\left( {\frac{{A^{(1)} }}{T}} \right)g_{1,2,3,4} \left\langle {S_{1,2,3,4} } \right|} } } \hfill \\ \quad \left| {\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{1,2} \cdot S_{3,4} }} } \right)} \right|\left| {S_{1,2,3,4} } \right\rangle \times \hfill \\ f\left( {\frac{{A^{(1)} }}{T}} \right)g_{5,6,7,8} \left\langle {S_{5,6,7,8} } \right|\left| {\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{5,6} \cdot S_{7,8} }} } \right)} \right|\left| {S_{5,6,7,8,2} } \right\rangle ... \times \hfill \\ f\left( {\frac{{A^{(1)} }}{T}} \right)g_{N - 3,N - 2,N - 1,N} \left\langle {S_{N - 3,N - 2,N - 1,N} } \right| \hfill \\ \quad \left| {\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{N - 3,N - 2} \cdot S_{N - 1,N} }} } \right)} \right|\left| {S_{N - 3,N - 2,N - 1,N} } \right\rangle , \hfill \\ \end{gathered}$$
(92)

where gklmn are degeneration factors of the spin projections onto an arbitrary quantization axis mentioned above. A matrix element of the operator exponent can be calculated in the following way:

$$\begin{gathered} \left\langle {S_{N - 1,N} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{N - 1} \cdot s_{N} }} } \right)} \right|\left| {S_{N - 1,N} } \right\rangle = \hfill \\ \left\langle {S_{N - 1,N} } \right|\left| {\sum\limits_{\eta = 0}^{\infty } {\frac{1}{\eta !}\left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{N - 1} \cdot s_{N} }} } \right)^{\eta } } } \right|\left| {S_{N - 1,N} } \right\rangle = \hfill \\ = \left\langle {S_{N - 1,N} } \right|\left| {\sum\limits_{\eta = 0}^{\infty } {\frac{1}{\eta !}\left( { - \frac{{A^{(0)} }}{T}} \right)^{\eta } \left( { - 1} \right)^{{\eta S_{N - 1,N} }} } } \right|\left| {S_{N - 1,N} } \right\rangle = \hfill \\ = \left\langle {S_{N - 1,N} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}\left( { - 1} \right)^{{S_{N - 1,N} }} } \right)} \right|\left| {S_{N - 1,N} } \right\rangle = \hfill \\ = \exp \left( { - \left( { - 1} \right)^{{S_{N - 1,N} }} \frac{{A^{(0)} }}{T}} \right). \hfill \\ \end{gathered}$$
(93)

In this case, the following condition must be satisfied for all possible values of the spins \(S_{3,4}\):

$$\begin{gathered} g_{1,2} \left\langle {S_{1,2} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{1} \cdot s_{2} }} } \right)} \right|\left| {S_{1,2} } \right\rangle \cdot g_{3,4} \left\langle {S_{3,4} } \right|\left| {\exp \left( { - \frac{{A^{(0)} }}{T}{\hat{\mathscr{R}}}_{{s_{3} \cdot s_{4} }} } \right)} \right|\left| {S_{3,4} } \right\rangle = \hfill \\ = \sum\limits_{{S_{1,2,3,4} ,}} {\left( {\frac{{A^{(1)} }}{T}} \right)g_{1,2,3,4} \left\langle {S_{1,2,3,4} } \right|\left| {\exp \left( { - \frac{{A^{(1)} }}{T}{\hat{\mathscr{R}}}_{{S_{1,2} \cdot S_{3,4} }} } \right)} \right|\left| {S_{1,2,3,4} } \right\rangle } , \Rightarrow \hfill \\ g_{1,2} \exp \left( { - \left( { - 1} \right)^{{S_{1,2} }} \frac{{A^{(0)} }}{T}} \right) \cdot g_{3,4} \exp \left( { - \left( { - 1} \right)^{{S_{3,4} }} \frac{{A^{(0)} }}{T}} \right) = \hfill \\ = \sum\limits_{{S_{1,2,3,4} ,}} {f\left( {\frac{{A^{(1)} }}{T}} \right)g_{1,2,3,4} \exp \left( { - \left( { - 1} \right)^{{S_{1,2,3,4} }} \frac{{A^{(1)} }}{T}} \right)} . \hfill \\ \end{gathered}$$
(94)

Then condition Eq. (94) can be rewritten for different spin configurations:

  1. (1)

    If \(S_{1,2} = S_{3,4} = 1\), then we have the condition Eq. (94) in the form

    $$\begin{gathered} 3\exp \left( { - \left( { - 1} \right)\frac{{A^{(0)} }}{T}} \right) \cdot 3\exp \left( { - \left( { - 1} \right)\frac{{A^{(0)} }}{T}} \right) = \hfill \\ = \sum\limits_{{S_{1,2,3,4} ,}} {f\left( {\frac{{A^{(1)} }}{T}} \right)g_{1,2,3,4} \exp \left( { - \left( { - 1} \right)^{{S_{1,2,3,4} }} \frac{{A^{(1)} }}{T}} \right)} = \hfill \\ = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {1 \cdot \exp \left( { - \frac{{A^{(1)} }}{T}} \right) + 3 \cdot \exp \left( {\frac{{A^{(1)} }}{T}} \right) + 5\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\} = \hfill \\ = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {3 \cdot \exp \left( {\frac{{A^{(1)} }}{T}} \right) + 6\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}, \Rightarrow \hfill \\ \boxed{3\exp \left( {\frac{{2A^{(0)} }}{T}} \right) = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 3\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}}. \hfill \\ \end{gathered}$$
    (95)
  2. (2)

    If \(S_{1,2} \ne S_{3,4}\), then

  3. (3)

    \(S_{1,2} = 1,\,S_{3,4} = 0,\, \Rightarrow S_{1,2,3,4} = 1.\)

    $$\begin{gathered} 3\exp \left( { - \left( { - 1} \right)\frac{{A^{(0)} }}{T}} \right) \cdot \exp \left( { - \frac{{A^{(0)} }}{T}} \right) = f\left( {\frac{{A^{(1)} }}{T}} \right)\exp \left( { - \frac{{A^{(1)} }}{T}} \right), \Rightarrow \hfill \\ \boxed{3 = f\left( {\frac{{A^{(1)} }}{T}} \right)\exp \left( { - \frac{{A^{(1)} }}{T}} \right)}. \hfill \\ \end{gathered}$$
    (96)

\(S_{1,2} = 1,S_{3,4} = 2, \Rightarrow S_{1,2,3,4} = 1,2,3\).

$$\begin{gathered} 3\exp \left( { - \left( { - 1} \right)\frac{{A^{(0)} }}{T}} \right) \cdot 5\exp \left( { - \frac{{A^{(0)} }}{T}} \right) = \hfill \\ = \sum\limits_{{S_{1,2,3,4} ,}} {f\left( {\frac{{A^{(1)} }}{T}} \right)g_{1,2,3,4} \exp \left( { - \left( { - 1} \right)^{{S_{1,2,3,4} }} \frac{{A^{(1)} }}{T}} \right)} = \hfill \\ = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {3\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 5\exp \left( { - \frac{{A^{(1)} }}{T}} \right) + 7\exp \left( {\frac{{A^{(1)} }}{T}} \right)} \right\}, \hfill \\ 3\exp \left( {\frac{{A^{(0)} }}{T}} \right) \cdot 5\exp \left( { - \frac{{A^{(0)} }}{T}} \right) = 15 = f\left( {\frac{{A^{(1)} }}{T}} \right) \hfill \\ \left\{ {10\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 5\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}, \Rightarrow \hfill \\ \boxed{3 = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {2\exp \left( {\frac{{A^{(1)} }}{T}} \right) + \exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}}, \hfill \\ \end{gathered}$$
(97)

Then, summarizing (a) and (b) for case 2) we have the following condition:

$$\begin{gathered} 3 = f\left( {\frac{{A^{(1)} }}{T}} \right)\exp \left( { - \frac{{A^{(1)} }}{T}} \right) \hfill \\ + \hfill \\ 3 = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {2\exp \left( {\frac{{A^{(1)} }}{T}} \right) + \exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\} \hfill \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\hfill \\ 6 = f\left( {\frac{{A^{(1)} }}{T}} \right)\left\{ {2\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 2\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}, \Rightarrow \hfill \\ \boxed{3 = f\left( {\frac{{A^{(1)} }}{T}} \right) \cdot 2Cosh\left( {\frac{{A^{(1)} }}{T}} \right)}. \hfill \\ \end{gathered}$$
(98)

From Eq. (81), we obtain the explicit form of

$$f\left( {\frac{{A^{(1)} }}{T}} \right) = \frac{3}{{2Cosh\left( {\frac{{A^{(1)} }}{T}} \right)}} \cdot$$
(99)

Substituting Eq. (99) into Eq. (95), we find

$$\begin{gathered} 3\exp \left( {\frac{{2A^{(0)} }}{T}} \right) = \frac{3}{{2Cosh\left( {\frac{{A^{(1)} }}{T}} \right)}}\left\{ {\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 3\exp \left( { - \frac{{A^{(1)} }}{T}} \right)} \right\}, \Rightarrow \hfill \\ \exp \left( {\frac{{2A^{(0)} }}{T}} \right) = \frac{{\exp \left( {\frac{{A^{(1)} }}{T}} \right) + 3\exp \left( { - \frac{{A^{(1)} }}{T}} \right)}}{{\exp \left( {\frac{{A^{(1)} }}{T}} \right) + \exp \left( { - \frac{{A^{(1)} }}{T}} \right)}}, \Rightarrow \hfill \\ \exp \left( {\frac{{2A^{(1)} }}{T}} \right) = \frac{{3 - \exp \left( {\frac{{2A^{(0)} }}{T}} \right)}}{{\exp \left( {\frac{{2A^{(0)} }}{T}} \right) - 1}} \hfill \\ \end{gathered}$$
$$\begin{gathered} \exp \left( {\frac{{2A^{(1)} }}{T}} \right) = \frac{{3 - \exp \left( {\frac{{2A^{(0)} }}{T}} \right)}}{{\exp \left( {\frac{{2A^{(0)} }}{T}} \right) - 1}} = \frac{{2 - \exp \left( {\frac{{2A^{(0)} }}{T}} \right) + 1}}{{\exp \left( {\frac{{2A^{(0)} }}{T}} \right) - 1}} \hfill \\ = \frac{{\exp \left( { - \frac{{A^{(0)} }}{T}} \right) - Sinh\left( {\frac{{A^{(0)} }}{T}} \right)}}{{Sinh\left( {\frac{{A^{(0)} }}{T}} \right)}} = \hfill \\ = \frac{{\exp \left( { - \frac{{A^{(0)} }}{T}} \right)}}{{Sinh\left( {\frac{{A^{(0)} }}{T}} \right)}} - 1 \hfill \\ \frac{{A^{(1)} }}{T} = \frac{1}{2}\ln \left( {\frac{{\exp \left( { - \frac{{A^{(0)} }}{T}} \right)}}{{Sinh\left( {\frac{{A^{(0)} }}{T}} \right)}} - 1} \right). \hfill \\ \end{gathered}$$
(100)

Figure 2 shows that at low temperatures \(\frac{{A^{(0)} }}{T} \ge 0.35\), \(A^{(1)}\) increases very quickly, which indicates the appearance of long-range order in the system of spins, and at a certain temperature \(\frac{{A^{(0)} }}{0.55} = T*\), the ratio of \(A^{(1)} /A^{(0)} = \xi\) reaches a critical value \(\xi = 2\), which leads to a phase transition from a paramagnetic state to an antiferromagnetic state.

Fig. 2
Two graphs plot the dependence A power of 1 over T as a function of A power of 0 over T. Each value are represented in the graph.

Dependence \(\frac{{A^{(1)} }}{T}\) as a function of \(\frac{{A^{(0)} }}{T}\)

4.2 Spin Waves and Spin Dark–Bright Soliton in the Spin-1 Chain

The ground state of this system is antiferromagnetic, then in the approximation of interaction with nearest neighbors, we will consider the spin of a pair of atoms \(S_{1,2} = 1\), and the eigenvalue of the operator \(\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)\) is equal to \(\overline{{\left( {{\hat{\mathbf{s}}}_{1} \cdot {\hat{\mathbf{s}}}_{2} } \right)}} = \overline{{\frac{1}{2}\left( {\hat{S}_{1,2}^{2} - 2\widehat{{s^{2} }}} \right)}} = \frac{1}{2}\left( {S_{1,2} \left( {S_{1,2} + 1} \right) - 2s(s + 1)} \right) = - 1\).

Let one spin numbered k from a chain of atoms be “flipped”. Then the operator of the excitation energy can be written in the form

$$\begin{gathered} \hat{V} = \hat{H}_{{\text{int}}} - E_{0} = \\ A\left\{ {{\hat{\mathbf{s}}}_{k - 1} \cdot {\hat{\mathbf{s}}}_{k} + {\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{k + 1} + ({\hat{\mathbf{s}}}_{k - 1} \cdot {\hat{\mathbf{s}}}_{k} )^{2} + ({\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{k + 1} )^{2} \,\, - 2\left( {( - 1)s_{k}^{2} + ( - 1)^{2} s_{k}^{4} } \right)} \right\}, \\ \end{gathered}$$
(101)

where Ecorresponds to the ground state energy in the zero (pairwise) approximation \(E_{0} = A\sum\limits_{i} {(( - 1)^{2} s_{i}^{4} + ( - 1)s_{i}^{2} - 1)}\).

In the semiclassical continual approximation, the eigenvalues of the mechanical moments of ions can be represented as a continuous function of the distance from the ion. Then the site spin vector k ± 1 can be written as

$${\mathbf{s}}_{k \pm 1} = {\mathbf{s}}_{k} \pm b\frac{{\partial {\mathbf{s}}_{k} }}{\partial x} + \frac{{b^{2} }}{2}\frac{{\partial^{2} {\mathbf{s}}_{k} }}{{\partial x^{2} }}.$$
(102)

In this expression, b is the lattice constant. After substituting expression Eq. (102) into Eq. (101) in the semiclassical approximation, taking into account the replacement of the spin operators by their vector functions, we obtain the expression for the excitation energy V.

$$V = 2A\left\{ {s_{k}^{2} \left( {2 + b^{2} \left( {\frac{{\partial {\mathbf{s}}_{k} }}{\partial x}} \right)^{2} + \frac{{b^{4} }}{4}\left( {\frac{{\partial^{2} {\mathbf{s}}_{k} }}{{\partial x^{2} }}} \right)^{2} } \right) + \frac{{b^{2} }}{2}(1 + 2s_{k}^{2} ){\mathbf{s}}_{k} \cdot \frac{{\partial^{2} {\mathbf{s}}_{k} }}{{\partial x^{2} }}} \right\}.$$
(103)

On the other hand, in the approximation of the average effective field H * formed by the entire set of magnetic moments of flipped spins, the excitation energy will be written as the energy of interaction of each proper magnetic moment with this field:

$$V = - g{\mathbf{s}}_{k} \cdot {\mathscr{H}}^{*} .$$
(104)

Here g is the Lande factor. Then, in accordance with Bloch’s theorem, the intrinsic magnetic moment of each ion will precess in the effective magnetic field as

$$\begin{gathered} \frac{{\partial {\mathscr{M}}}}{\partial t} = \gamma ({\mathscr{H}}^{*} \times {\mathscr{M}}), \hfill \\ \gamma = \frac{e}{mc}. \hfill \\ \end{gathered}$$
(105)

An explicit expression for the field strength is determined using expressions Eqs. (103) and (104) and has the following form:

$${{\mathscr{H}}}^{*} = \frac{ - A}{g}\left\{ {{\mathbf{s}}_{k} \left( {2 + b^{2} \left( {\frac{{\partial {\mathbf{s}}_{k} }}{\partial x}} \right)^{2} + \frac{{b^{4} }}{4}\left( {\frac{{\partial^{2} {\mathbf{s}}_{k} }}{{\partial x^{2} }}} \right)^{2} } \right) + \frac{{b^{2} }}{2}(1 + 2s_{k}^{2} ) \cdot \frac{{\partial^{2} {\mathbf{s}}_{k} }}{{\partial x^{2} }}} \right\}.$$
(106)

Writing out Eq. (105) for each component of the magnetic moment vector \({\mathscr{M}} = g{\mathbf{s}}\) and taking into account the obvious equality \({\mathbf{s}} \times {\mathbf{s}} = 0\), we can rewrite Eq. (106) in the linear approximation for cyclic coordinates

$${\dot{\mathscr{M}}}^{{\mathbf{ + }}} = i\frac{\gamma As}{g}\left\{ { - 2{\mathscr{M}}^{{\mathbf{ + }}} + \frac{{b^{2} }}{2}\left( {1 + 2s(s + 1)} \right)\frac{{\partial^{2} {\mathscr{M}}^{{\mathbf{ + }}} }}{{\partial x^{2} }}} \right\}.$$
(107)

Then, if we look for a solution for \({\mathscr{M}}^{{\mathbf{ + }}}\) in the form of a monochromatic plane wave \({\mathscr{M}}^{{\mathbf{ + }}} = \Delta {\mathscr{M}}\exp (ikx - i\omega t)\), we obtain a dispersion relation for spin waves in an S = 1 antiferromagnet, which is different from that known for antiferromagnets with 1/2 spin [36].

$$\omega = \frac{\gamma As}{g}\left( { - 2 + \frac{5}{2}b^{2} k^{2} } \right).$$
(108)

Thus, for small deviations from the equilibrium antiferromagnetic state, spin excitations in the form of transverse spin waves with a dispersion dependence resembling the dispersion relation for spin waves in s = 1/2 ferromagnet will propagate along the system of integer spins s = 1. However, it can be seen from relation Eq. (108) that the excitation spectrum is separated from the ground state by an energy gap Δ = ℏω = 2 ℏγAs/g which can be associated with the existence of a minimum limit value for the wave vector \(k_{\min } = \sqrt {\frac{4}{{5b^{2} }}}\) and the corresponding limitation on the maximum wavelength.

If, however, the first non-vanishing nonlinear corrections are retained in expression Eqs. (106) and (107), then we obtain a system of equations for the transverse components of the intrinsic magnetic moments:

$$\begin{gathered} \dot{\mathscr{M}}^{ + } = - i\frac{{Ab^{2} }}{\hbar }\left[ {\frac{{\partial ^{2} \mathscr{M}^{ + } }}{{\partial x^{2} }} + \left( {\frac{{\mathscr{M}^{ +2 } }}{{b^{2} g^{2} }} - \frac{{4\mathscr{M}^{ -2 } }}{{b^{2} g^{2} }}} \right) \mathscr{M}^{ + } } \right], \hfill \\ \dot{\mathscr{M}}^{ - } = - i\frac{{Ab^{2} }}{\hbar }\left[ {\frac{{\partial ^{2} \mathscr{M}^{ - } }}{{\partial x^{2} }} + \left( {\frac{{\mathscr{M}^{ -2 } }}{{b^{2} g^{2} }} - \frac{{4\mathscr{M}^{ +2 } }}{{b^{2} g^{2} }} - \Xi } \right) \mathscr{M}^{ - } } \right], \hfill \\ \end{gathered}$$
(109)

Here is

$$\begin{gathered} \mathscr{M}^{ \pm } = \mathscr{M}_{x} \pm i\mathscr{M}_{y} , \hfill \\ \Xi = \frac{A}{{g^{2} }}b. \hfill \\ \end{gathered}$$
(110)

This system of equations is similar to the equations for the Dark–Bright soliton [37, 38]. The solution is written in the form

$$\begin{gathered} \mathscr{M}^{ + } = \sqrt {Aa^{3} } (i\sin \alpha + \cos \alpha \tanh \{ k[x - q(t)]\} ) \hfill \\ \mathscr{M}^{ - } = \sqrt {\frac{{m_{0}^{2} ka}}{2}} e^{i\Omega t} e^{ixktg\alpha } \sec h\{ k(x - q(t))\} . \hfill \\ \end{gathered}$$
(111)

Precession frequency taking into account the shift and soliton coordinate is

$$\begin{gathered} \Omega = \frac{A}{\hbar }\left( {1 - \frac{{tg^{2} \alpha }}{2}} \right)\left( {kb} \right)^{2} - \Xi , \hfill \\ q(t) = q(0) + kt\frac{{Aa^{2} }}{\hbar }tg\alpha . \hfill \\ \end{gathered}$$
(112)

Such integrable systems are known as the Manakov equations [39]. These systems store free energy. However, since the free energy decreases with increasing soliton velocity, the soliton is formally unstable (it is accelerated). If we add an inhomogeneous potential, then the system becomes non-integrable and the soliton can interact in a nontrivial way with the environment. Nevertheless, if the interaction potential changes slowly on the scale of the soliton length, then this leads to a coordinated motion (change) of the soliton and the potential, and the free energy in this case is an adiabatic invariant.

4.3 Ground States of an Isotropic Magnetic System of Particles with Spin s = 3/2

The Hamiltonian of an isotropic magnet with magnetic ion spin s = 3/2 has the form Eq. (73) with the omitted contribution of the direct Coulomb interaction that does not include an explicit dependence of spin operators.

$$\hat{H}_{{{\text{int}} 1}} = - \sum\limits_{k < l} {A_{kl} \left( {\frac{2}{9}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{3} + \frac{11}{{18}}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right)^{2} - \frac{9}{8}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) - \frac{67}{{32}}} \right).}$$
(113)

We express this Hamiltonian in terms of irreducible combinations of the Steven spin operators [40, 41] describing multipole moments. These operators for spin 3/2 are described in detail in [4041], here is the form of these operators:

Quadrupole:

$$\begin{gathered} \hat{Q}_{2}^{0} = 3\hat{s}_{z}^{2} - \hat{s}^{2} \hfill \\ \hat{Q}_{2}^{1} = \left\{ {\hat{s}_{z} \hat{s}_{x} } \right\} \hfill \\ \tilde{\hat{Q}}_{2}^{1} = \left\{ {\hat{s}_{z} \hat{s}_{y} } \right\} \hfill \\ \hat{Q}_{2}^{2} = \frac{1}{2}\left( {\widehat{{s^{ + } }}^{2} + \widehat{{s^{ - } }}^{2} } \right) \hfill \\ \tilde{\hat{Q}}_{2}^{2} = \frac{1}{2i}\left( {\widehat{{s^{ + } }}^{2} - \widehat{{s^{ - } }}^{2} } \right). \hfill \\ \end{gathered}$$
(114)

Here, the curly brackets stand for anticommutators of the operators, for example, \(\left\{ {\hat{j}_{z} ,\hat{j}_{x} } \right\} = \hat{j}_{z} \hat{j}_{x} + \hat{j}_{x} \hat{j}_{z}\).

Octupole:

$$\begin{gathered} \hat{Q}_{3}^{0} = 5\hat{s}_{z}^{3} - 3\widehat{{s^{2} }}\hat{s}_{z} + \hat{s}_{z} , \hfill \\ \hat{Q}_{3}^{1} = \frac{1}{2}\left\{ {\left( {5\widehat{{s_{z} }}^{2} - \widehat{{s^{2} }} - \frac{1}{2}} \right),\hat{s}_{x} } \right\}, \hfill \\ \tilde{\hat{Q}}_{3}^{1} = \frac{1}{2}\left\{ {\left( {5\widehat{{s_{z} }}^{2} - \widehat{{s^{2} }} - \frac{1}{2}} \right),\hat{s}_{y} } \right\}, \hfill \\ \hat{Q}_{3}^{2} = \frac{1}{4}\left\{ {\hat{s}_{z} ,\left( {\widehat{{s^{ + } }}^{2} + \widehat{{s^{ - } }}^{2} } \right)} \right\}, \hfill \\ \tilde{\hat{Q}}_{3}^{2} = \frac{1}{4i}\left\{ {\hat{s}_{z} ,\left( {\widehat{{s^{ + } }}^{2} - \widehat{{s^{ - } }}^{2} } \right)} \right\}, \hfill \\ \hat{Q}_{3}^{3} = \frac{1}{2}\left( {\widehat{{s^{ + } }}^{3} + \widehat{{s^{ - } }}^{3} } \right), \hfill \\ \tilde{\hat{Q}}_{3}^{3} = \frac{1}{2i}\left( {\widehat{{s^{ + } }}^{3} - \widehat{{s^{ - } }}^{3} } \right). \hfill \\ \end{gathered}$$
(115)

So we have the Hamiltonian Eq. (113) in the form

$$\begin{gathered} \hat{H}_{{{\text{int}} 1}} = - \frac{1}{2}\sum\limits_{k < l} {A_{kl} \left\{ { - \frac{2}{9}} \right.\left( {\frac{1}{10}\hat{Q}_{3k}^{0} \cdot \hat{Q}_{3l}^{0} + \frac{3}{20}\left( {\hat{Q}_{3k}^{1} \cdot \hat{Q}_{3l}^{1} + \hat{\tilde{Q}}_{3k}^{1} \cdot \hat{\tilde{Q}}_{3l}^{1} } \right)} \right. + } \hfill \\ + \frac{3}{2}\left( {\hat{Q}_{3k}^{2} \cdot \hat{Q}_{3l}^{2} + \hat{\tilde{Q}}_{3k}^{2} \cdot \hat{\tilde{Q}}_{3l}^{2} } \right) + \left. {\frac{1}{4}\left( {\hat{Q}_{3k}^{3} \cdot \hat{Q}_{3l}^{3} + \hat{\tilde{Q}}_{3k}^{3} \cdot \hat{\tilde{Q}}_{3l}^{3} } \right)} \right) + \hfill \\ + \frac{1}{12}\left( {\frac{1}{3}\hat{Q}_{3k}^{0} \cdot \hat{Q}_{3l}^{0} + (\hat{Q}_{3k}^{1} \cdot \hat{Q}_{3l}^{1} + \hat{\tilde{Q}}_{3k}^{1} \cdot \hat{\tilde{Q}}_{3l}^{1} )} \right. + \hfill \\ + \left( {\hat{Q}_{3k}^{2} \cdot \hat{Q}_{3l}^{2} + \hat{\tilde{Q}}_{3k}^{2} \cdot \hat{\tilde{Q}}_{3l}^{2} } \right) + \left. {\frac{1}{4}\left( {\hat{Q}_{3k}^{3} \cdot \hat{Q}_{3l}^{3} + \hat{\tilde{Q}}_{3k}^{3} \cdot \hat{\tilde{Q}}_{3l}^{3} } \right)} \right) + \hfill \\ + \left. {\frac{1}{5}\left( {{\hat{\mathbf{s}}}_{k} \cdot {\hat{\mathbf{s}}}_{l} } \right) + \frac{67}{{32}}} \right\}. \hfill \\ \end{gathered}$$
(116)

Separating the mean fields per unit note in Hamiltonian (21) and performing diagonalization on the basis of eigen spin states of the spin moment projection operators \(\hat{s}_{z}\), we find the single-node Hamiltonian in the form

$$\hat{H}_{0} = - A\eta \left( {\frac{1}{5}\hat{s}_{z} + \frac{1}{36}\hat{Q}_{2}^{0} + \frac{1}{45}\hat{Q}_{3}^{0} + \frac{1}{18}\hat{Q}_{3}^{3} } \right),$$
(117)

where η is the coordination number (the number of nearest neighbors), A is the exchange integral for the nearest neighbors, the operator of the quadrupole moment is given by Eq. (114), the operator of the octupole moment is rewritten in the form \(\hat{Q}_{3}^{0} = 5\hat{s}_{z}^{3} - 3\widehat{{s^{2} }}\hat{s}_{z} + \hat{s}_{z} = 5\hat{s}_{z}^{3} - 3\frac{3}{2} \cdot \frac{5}{2}\hat{s}_{z} + \hat{s}_{z} = 5\hat{s}_{z}^{3} - \frac{41}{4}\hat{s}_{z}\), and the operator of the octupole transition is \(\hat{Q}_{3}^{3} = \frac{1}{2}\left( {\widehat{{s^{ + } }}^{3} + \widehat{{s^{ - } }}^{3} } \right)\).

Using the Bogoliubov transformations for the set of one-body eigenstates \(\left\{ {\left| {3/2,j_{z} } \right\rangle } \right\}\)

$$\begin{gathered} \left| {\chi_{ + } } \right\rangle = u\left| {3/2,3/2} \right\rangle + v\left| {3/2, - 3/2} \right\rangle , \hfill \\ \left| {\chi_{ - } } \right\rangle = - v\left| {3/2,3/2} \right\rangle + u\left| {3/2, - 3/2} \right\rangle , \hfill \\ \left| {\chi_{1/2} } \right\rangle = \left| {3/2,1/2} \right\rangle , \hfill \\ \left| {\chi_{ - 1/2} } \right\rangle = \left| {3/2, - 1/2} \right\rangle , \hfill \\ u^{2} + v^{2} = 1. \hfill \\ \end{gathered}$$
(118)

we diagonalize Hamiltonian Eq. (117) and set the parameters \(u = \cos \phi ,\)\(v = \sin \phi ,\) where \(\tan \phi = v/u\), find the eigenvalues of energy corresponding to states Eq. (118)

$$\begin{gathered} E_{ + } = E_{\min } = - \frac{\eta A}{6}\left( {\frac{1}{2}\left\langle {Q_{2}^{0} } \right\rangle + \frac{1}{5}\left( {9\left\langle {s_{z} } \right\rangle + \left\langle {Q_{3}^{0} } \right\rangle } \right)(u^{2} - v^{2} ) + \left\langle {Q_{3}^{3} } \right\rangle 2uv} \right) = \hfill \\ = - \frac{\eta A}{6}\left( {\frac{1}{2}\left\langle {Q_{2}^{0} } \right\rangle + \frac{1}{5}\left( {9\left\langle {s_{z} } \right\rangle + \left\langle {Q_{3}^{0} } \right\rangle } \right)\cos (2\phi ) + \left\langle {Q_{3}^{3} } \right\rangle \sin (2\phi )} \right), \hfill \\ E_{ - } = E_{\max } = - \frac{\eta A}{6}\left( {\frac{1}{2}\left\langle {Q_{2}^{0} } \right\rangle - \frac{1}{5}\left( {9\left\langle {s_{z} } \right\rangle + \left\langle {Q_{3}^{0} } \right\rangle } \right)(u^{2} - v^{2} ) - \left\langle {Q_{3}^{3} } \right\rangle 2uv} \right) = \hfill \\ - \frac{\eta A}{6}\left( {\frac{1}{2}\left\langle {Q_{2}^{0} } \right\rangle - \frac{1}{5}\left( {9\left\langle {s_{z} } \right\rangle + \left\langle {Q_{3}^{0} } \right\rangle } \right)\cos (2\phi ) - \left\langle {Q_{3}^{3} } \right\rangle \sin (2\phi )} \right), \hfill \\ \end{gathered}$$
$$\begin{gathered} E_{1/2} = \frac{\eta A}{2}\left( {\frac{1}{6}\left\langle {Q_{2}^{0} } \right\rangle - \frac{1}{5}\left( {\left\langle {s_{z} } \right\rangle - \left\langle {Q_{3}^{0} } \right\rangle } \right)} \right), \hfill \\ E_{ - 1/2} = \frac{\eta A}{2}\left( {\frac{1}{6}\left\langle {Q_{2}^{0} } \right\rangle + \frac{1}{5}\left( {\left\langle {s_{z} } \right\rangle - \left\langle {Q_{3}^{0} } \right\rangle } \right)} \right). \hfill \\ \end{gathered}$$
(119)

We chose the parameters u = sinφ and v = cosφ of the Bogoliubov transformations such that the state |χ + 〉 corresponds to the normal sublevel with the minimum energy, which holds under the condition

$$- \frac{3}{5}\eta A\left( {\left( {\left\langle {s_{z} } \right\rangle + \frac{1}{9}\left\langle {Q_{3}^{0} } \right\rangle } \right) \cdot \sin (2\phi ) - \frac{5}{9}\left\langle {Q_{3}^{3} } \right\rangle \cos (2\phi )} \right) = 0.$$
(120)

Bearing in mind relations Eqs. (114) and (115), we find the values of the multipole moments averaged over the above states

$$\begin{gathered} \left\langle {Q_{3}^{0} } \right\rangle = 5\left\langle {\left( {s_{z} } \right)^{3} } \right\rangle - \frac{41}{4}\left\langle {s_{z} } \right\rangle \hfill \\ \left\langle {Q_{2}^{0} } \right\rangle = 3\left\langle {\left( {s_{z} } \right)^{2} } \right\rangle - \frac{15}{4} \hfill \\ \left\langle {Q_{3}^{3} } \right\rangle = \frac{1}{2}\left( {\left\langle {\left( {s^{ + } } \right)^{3} } \right\rangle + \left\langle {\left( {s^{ - } } \right)^{3} } \right\rangle } \right) \hfill \\ \end{gathered}$$
(121)

which take the following values for the state |χ + 〉

$$\left\langle {s_{z} } \right\rangle = \frac{3}{2}\left( {u^{2} - v^{2} } \right) = \frac{3}{2}\cos 2\phi ,$$
$$\left\langle {Q_{2}^{0} } \right\rangle = 3\left\langle {\left( {s_{z} } \right)^{2} } \right\rangle - \frac{15}{4} = 3\left( {u^{2} + v^{2} } \right) = 3,$$
(122)
$$\left\langle {Q_{3}^{0} } \right\rangle = 5\left\langle {\left( {s_{z} } \right)^{3} } \right\rangle - \frac{41}{4}\left\langle {s_{z} } \right\rangle = \frac{3}{2}\left( {u^{2} - v^{2} } \right) = \frac{3}{2}\cos 2\phi ,$$
$$\begin{gathered} \left\langle {Q_{3}^{3} } \right\rangle = \frac{uv}{2}\left( {\left\langle {3/2,3/2} \right|\left( {s^{ + } } \right)^{3} \left| {3/2, - 3/2} \right\rangle + \left\langle {3/2, - 3/2} \right|\left( {s^{ - } } \right)^{3} \left| {3/2,3/2} \right\rangle } \right) = \hfill \\ = 3uv = 3\sin 2\phi , \hfill \\ \end{gathered}$$

where 〈sz〉 is the average value of the spin projection to the z-axis per one particle, \(\left\langle {Q_{2}^{0} } \right\rangle\) is the average quadrupole moment, \(\left\langle {Q_{3}^{0} } \right\rangle\) is the average octupole moment, and \(\left\langle {Q_{3}^{3} } \right\rangle\) is the average value of the octupole transition parameter.

Substituting Eqs. (122) into Eq. (119), we find the expressions for the energies of the respective states

$$\begin{gathered} E_{ + } = E_{\min } = - \frac{3\eta A}{4} \hfill \\ E_{ - } = E_{\max } = \frac{\eta A}{4} \hfill \\ E_{{ \pm \frac{1}{2}}} = \frac{\eta A}{4} \hfill \\ \end{gathered}$$
(123)

which vary according to the Bogoliubov transformation (the rotation transformation with the tangent of the rotation angle defined as tanφ = v/u).

Taking into account Eq. (123), condition Eq. (120) of the extreme of the state |χ + 〉 with the minimum energy E is transformed to

$$E^{\prime}_{ + } = \frac{\eta }{5}A\cos (2\phi )\sin (2\phi ) = 0$$
(124)

from which the following possible solutions arise, which we will analyze for two fundamentally different cases of the values of the exchange integral, A > 0 and A < 0:

Case I. A > 0

  1. (1)

    The first solution of Eq. (124), \(\cos (2\phi_{1} ) = 0, \Rightarrow \phi_{1} = \frac{\pi }{4}.\) This state is characterized by the average spin projection \(\left\langle {s_{z} } \right\rangle = \frac{3}{2}\cos 2\phi_{1} = 0\), by the average quadrupole moment \(\left\langle {Q_{2}^{0} } \right\rangle = 3\left\langle {\left( {s_{z} } \right)^{2} } \right\rangle - \frac{15}{4} = 3\), by the average octupole moment \(\left\langle {Q_{3}^{0} } \right\rangle = 5\left\langle {\left( {s_{z} } \right)^{3} } \right\rangle - \frac{41}{4}\left\langle {s_{z} } \right\rangle = \frac{3}{2}\cos 2\phi_{1} = 0\), and by the average value of the octupole transition parameter \(\left\langle {Q_{3}^{3} } \right\rangle = 3\sin 2\phi_{1} = 3\). Thus, the first solution corresponds to a nematic state.

    To clarify the stability of the said state, we check the sign of the second derivative of E+ in Eq. (123) with respect to the angular parameter.

    : \(E_{ + }^{^{\prime\prime}} = \frac{2\eta }{5}A\cos 4\phi \left| {_{{\phi_{1} }} } \right. = - \frac{2\eta }{5}A < 0\). Thus, the above nematic state is unstable. Following the method [41], the excitation spectrum in the nematic phase has the form

    $$\begin{gathered} \varepsilon_{1,2} (k) = \frac{3\eta }{{2\sqrt 2 }} \times \hfill \\ \sqrt {\left( {A - A\left( {\mathbf{k}} \right)} \right)\left( {\left( {\frac{11}{{18}}} \right) - \frac{5}{4}\left( \frac{2}{9} \right)} \right)\left\{ {A\left( {\left( {\frac{11}{{18}}} \right) - \frac{5}{4}\left( \frac{2}{9} \right)} \right)} \right.} \hfill \\ \sqrt {\left. { - \frac{1}{2}A\left( {\mathbf{k}} \right)\left( {\left( { - \frac{9}{8}} \right) + \frac{3}{2}\left( {\frac{11}{{18}}} \right) + \frac{63}{{16}}\left( \frac{2}{9} \right)} \right)} \right\}} = \hfill \\ = \frac{{\eta \left( {A - A\left( {\mathbf{k}} \right)} \right)}}{2\sqrt 2 } = \frac{\eta A}{{2\sqrt 2 }}\left( {1 - {\text{Re}} \sum\limits_{\nu } {e^{{i{\mathbf{k}} \cdot {\mathbf{b}}_{\nu } }} } } \right), \hfill \\ \varepsilon_{3} (k) = \frac{9\eta }{4}\sqrt {\left( \frac{2}{9} \right)\left( {A - A\left( {\mathbf{k}} \right)} \right)\left\{ {\left( \frac{2}{9} \right)\left( {A - A\left( {\mathbf{k}} \right)} \right) - A\left( {\mathbf{k}} \right)\Delta_{1} } \right\}} = \hfill \\ = \frac{{\eta \left( {A - A\left( {\mathbf{k}} \right)} \right)}}{6} = \frac{\eta }{6}A\left( {1 - {\text{Re}} \sum\limits_{\nu } {e^{{i{\mathbf{k}} \cdot {\mathbf{b}}_{\nu } }} } } \right), \hfill \\ \Delta_{1} = \left( { - \frac{9}{8}} \right) - \frac{1}{2}\left( {\frac{11}{{18}}} \right) + \frac{103}{{16}} \cdot \left( \frac{2}{9} \right) = 0. \hfill \\ \end{gathered}$$
    (125)

    Here, we used the pre-factors from Eq. (113): the parameter c3 = \(\left( \frac{2}{9} \right)\) in front of the bicubic term in Eq. (113), the parameter c2 = \(\left( {\frac{11}{{18}}} \right)\) in front of the biquadratic term, and the parameter c1 = \(\left( { - \frac{9}{8}} \right)\)– in front of the bilinear (Heisenberg) term. Here, k is the magnon wave vector, bv is the lattice vector in the direction toward the vth nearest neighbor, A(k) is the Fourier transform of the exchange integral, and the gap parameter is Δ1 = 0, which corresponds to the state of the phase transition to a magnetic phase. At the corner point of the Brillouin zone, i.e., at k = 0, where A(k) = –A, we have

    $$\begin{gathered} \varepsilon_{3} (k) = \frac{9\eta }{4}A\sqrt {c_{3} (c_{1} - \frac{{c_{2} }}{2} + \frac{135}{{16}}c_{3} )} = \frac{9\eta }{4}A\sqrt {c_{3} \Delta_{2} } ; \hfill \\ \Delta_{2} = (c_{1} - \frac{{c_{2} }}{2} + \frac{135}{{16}}c_{3} ) = \frac{2}{9} \hfill \\ \end{gathered}$$
    (126)

    and the condition \(c_{3} \Delta_{2} = \left( \frac{2}{9} \right)^{2} > 0\) ensures the stability of the nematic phase with respect to the phase transition to the antiferromagnetic state. This condition is always satisfied for any sign of the exchange integral A. In other words, the phase transition from the nematic phase to the antiferromagnetic state is prohibited.

  2. (2)

    The second solution \(\sin 2\phi_{2} = 0, \Rightarrow \phi_{2} = 0\), then

  3. the average spin projection value is \(\left\langle {s_{z} } \right\rangle = \frac{3}{2}\cos 2\phi_{2} = \frac{3}{2}\),

  4. the average quadrupole moment is \(\left\langle {Q_{2}^{0} } \right\rangle = 3\),

  5. the average octupole moment is \(\left\langle {Q_{3}^{0} } \right\rangle = \frac{3}{2}\cos 2\phi_{2} = \frac{3}{2}\),

  6. the average value of the octupole transition parameter is

    $$\left\langle {Q_{3}^{3} } \right\rangle = 3\sin \left( {2\phi_{2} } \right) = 0.$$

Thus, the second solution corresponds to a ferromagnetic state.

STABILITY of THE SOLUTION.

  • We check the sign of the second derivative of \({E}_{+}\) with respect to the angular parameter: \(E_{ + }^{^{\prime\prime}} = \frac{2\eta }{5}A\cos 4\phi \left| {_{\phi 2} } \right. = \frac{2\eta }{5}A > 0\). Thus, the ferromagnetic state is stable.

The Excitation Spectrum in the Ferromagnetic Phase (A>0)

In the nearest-neighbor approximation in the case of the isotropic magnet, three modes merge into a single gapless Goldstone magnon mode:

$$\begin{gathered} \varepsilon_{1} (k) = \frac{3\eta }{4}\left( {A - A\left( {\mathbf{k}} \right)} \right)\left( {\left( { - \frac{9}{8}} \right) + \frac{3}{2}\left( {\frac{11}{{18}}} \right) + \frac{63}{{16}}\left( \frac{2}{9} \right)} \right) = \hfill \\ = \frac{{\eta \left( {A - A\left( {\mathbf{k}} \right)} \right)}}{2}, \hfill \\ \varepsilon_{2} (k) = \frac{3\eta }{2}\left( {A - A\left( {\mathbf{k}} \right)} \right)\left( {\frac{11}{{18}} - \frac{5}{4}\left( \frac{2}{9} \right) + A\Delta_{1} } \right) = \hfill \\ = \frac{\eta }{2}\left( {A - A\left( {\mathbf{k}} \right)} \right), \hfill \\ \varepsilon_{3} (k) = \frac{9\eta }{2}\left\{ {\left( \frac{2}{9} \right)\left( {A - A\left( {\mathbf{k}} \right)} \right) - A\left( {\mathbf{k}} \right)\Delta_{1} } \right\} = \hfill \\ = \frac{{\eta \left( {A - A\left( {\mathbf{k}} \right)} \right)}}{2} = \frac{\eta }{2}A\left( {1 - {\text{Re}} \sum\limits_{\nu } {e^{{i{\mathbf{k}} \cdot {\mathbf{b}}_{\nu } }} } } \right), \hfill \\ \Delta_{1} = \left( { - \frac{9}{8}} \right) - \frac{1}{2}\left( {\frac{11}{{18}}} \right) + \frac{103}{{16}} \cdot \left( \frac{2}{9} \right) = 0. \hfill \\ \end{gathered}$$
(127)

A gap parameter \({\Delta }_{1}=0\) corresponds to the state of the phase transition from a ferromagnetic phase to a nematic state.

At the corner point of the Brillouin zone, i.e., at k = 0, where A(k) = –A, the condition \(c_{3} \Delta_{2} = \left( \frac{2}{9} \right)^{2} > 0\) ensures the stability of the ferromagnetic phase with respect to the phase transition to an antiferromagnetic phase. A transition to an antinematic phase is forbidden since under weakening of correlations, i.e., at A → 0, the system passes to a paramagnetic disordered phase. The total angular momentum remains maximal and spin waves with the Goldstone, the gapless spectrum can propagate on the background of ferromagnetic ordering (See Fig. 3).

Fig. 3
A schematic diagram of a 3 over 2 spin system with possible phase transitions nematic, ferromagnet, antiferromagnet, and antinematic.

Schematic diagram of the possible phase transitions for the 3/2-spin system

Case II, A<0, of the negative Exchange Interaction Parameter.

There is a case of two sublattices.

(1) The first solution is given by the equation \(\cos (2\phi_{1} ) = 0,\) yielding the parameter \(\phi_{1} = \frac{\pi }{4}\), which determines for two sublevels the following values:

  • the average spin projection value, \(\begin{gathered} \left\langle {s_{s} } \right\rangle_{1} = \frac{3}{2}\cos 2\phi_{1} = 0, \hfill \\ \left\langle {s_{z} } \right\rangle_{2} = - \frac{3}{2}\cos 2\phi_{1} = 0, \hfill \\ \end{gathered}\)

  • the average quadrupole moment, \(\begin{gathered} \left\langle {Q_{2}^{0} } \right\rangle_{1} = 3, \hfill \\ \left\langle {Q_{2}^{0} } \right\rangle_{2} = 3, \hfill \\ \end{gathered}\)

  • the average octupole moment, \(\begin{gathered} \left\langle {Q_{3}^{0} } \right\rangle_{1} = \frac{3}{2}\cos 2\phi_{1} = 0, \hfill \\ \left\langle {Q_{3}^{0} } \right\rangle_{2} = - \frac{3}{2}\cos 2\phi_{1} = 0, \hfill \\ \end{gathered}\)

  • the average value of the octupole transition parameter, \(\begin{gathered} \left\langle {Q_{3}^{3} } \right\rangle_{1} = 3\sin \left( {2\phi_{1} } \right) = 3, \hfill \\ \left\langle {Q_{3}^{3} } \right\rangle_{2} = - 3\sin \left( {2\phi_{1} } \right) = - 3. \hfill \\ \end{gathered}\)

Thus, the first solution corresponds to the antinematic state with the inclusion of two sublattices.

STABILITY of THE SOLUTION. We check the sign of the second derivative of \({E}_{+}\) with respect to the angular parameter: \(E_{ + }^{^{\prime\prime}} = \frac{2\eta }{5}A\cos 4\phi \left| {_{{\phi_{1} }} } \right. = - \frac{2\eta }{5}A > 0\) Thus, the antinematic state is stable.

The Excitation Spectrum in the Antinematic Phase (A < 0) should consist of three modes, but the first two modes disappear and only one Goldstone gapless mode remains.

$$\begin{gathered} \varepsilon_{1,2} (k) = \frac{3\eta }{{2\sqrt 2 }} \cdot \sqrt {\left( {A + A\left( {\mathbf{k}} \right)} \right)} \times \hfill \\ \sqrt {\left( {\left( {\frac{11}{{18}}} \right) - \frac{11}{4}\left( \frac{2}{9} \right)} \right)\left\{ {A\left( {\left( {\frac{11}{{18}}} \right) - \frac{11}{4}\left( \frac{2}{9} \right)} \right)} \right.} \hfill \\ \sqrt {\left. { - \frac{1}{2}A\left( {\mathbf{k}} \right)\left( {\left( { - \frac{9}{8}} \right) - \frac{3}{2}\left( {\frac{11}{{18}}} \right) + \frac{191}{{16}}\left( \frac{2}{9} \right)} \right)} \right\}} = 0, \hfill \\ \varepsilon_{3} (k) = \frac{9\eta }{4}\sqrt {\left( \frac{2}{9} \right)\left( {A - A\left( {\mathbf{k}} \right)} \right)} \times \hfill \\ \sqrt {\left\{ {\left( \frac{2}{9} \right)\left( {A - A\left( {\mathbf{k}} \right)} \right) - A\left( {\mathbf{k}} \right)\left( { - \frac{9}{8} - \frac{1}{2}\left( {\frac{11}{{18}}} \right) + \frac{119}{{16}}\left( \frac{2}{9} \right)} \right)\Delta_{1} } \right\}} = \hfill \\ = \frac{{\eta \left( {A - A\left( {\mathbf{k}} \right)} \right)}}{\sqrt 2 } = \frac{\eta }{\sqrt 2 }A\left( {1 - {\text{Re}} \sum\limits_{\nu } {e^{{i{\mathbf{k}} \cdot {\mathbf{b}}_{\nu } }} } } \right). \hfill \\ \end{gathered}$$
(128)

This mode is the same as in the case of the ferromagnetic phase considered above. The stability of the antinematic phase requires the fulfillment of two conditions: AΔ1 > 0 and AΔ2 < 0. The first condition is violated in principle, as Δ1 = 0, which corresponds to a phase transition to the antiferromagnetic state. The second condition does hold, since AΔ2 =\(\frac{2}{9}\) A < 0 and vanishes only under weakening of correlations, i.e., when the exchange integral vanishes. Thus, the antinematic (stable) state is in neutral equilibrium with respect to the transition to the antiferromagnetic state in the parameter AΔ1 =  0 and in absolutely stable equilibrium with respect to the transition to the forbidden ferromagnetic state in accordance with the parameter AΔ2 <  0 (See Fig. 3).

(2) The second solution is given by the equation sin(2ϕ2) = 0 yielding the parameter \(\phi_{2} = 0\), which determines the following parameters of two sublattices:

  • the average spin projection value, \(\begin{gathered} \left\langle {s_{s} } \right\rangle_{1} = \frac{3}{2}\cos 2\phi_{2} = \frac{3}{2}, \hfill \\ \left\langle {s_{z} } \right\rangle_{2} = - \frac{3}{2}\cos 2\phi_{2} = - \frac{3}{2}, \hfill \\ \end{gathered}\)

  • the average quadrupole moment, \(\begin{gathered} \left\langle {Q_{2}^{0} } \right\rangle_{1} = 3, \hfill \\ \left\langle {Q_{2}^{0} } \right\rangle_{2} = 3, \hfill \\ \end{gathered}\)

  • the average octupole moment, \(\begin{gathered} \left\langle {Q_{3}^{0} } \right\rangle_{1} = \frac{3}{2}\cos 2\phi_{2} = \frac{3}{2}, \hfill \\ \left\langle {Q_{3}^{0} } \right\rangle_{2} = - \frac{3}{2}\cos 2\phi_{2} = - \frac{3}{2}, \hfill \\ \end{gathered}\)

  • the average value of the octupole transition parameter,

    \(\begin{gathered} \left\langle {Q_{3}^{3} } \right\rangle_{1} = 3\sin \left( {2\phi_{2} } \right) = 0, \hfill \\ \left\langle {Q_{3}^{3} } \right\rangle_{2} = - 3\sin \left( {2\phi_{2} } \right) = 0. \hfill \\ \end{gathered}\)

STABILITY of THE SOLUTION we check the sign of the second derivative of \({E}_{+}\) with respect to the angular parameter: \(E_{ + }^{^{\prime\prime}} = \frac{2\eta }{5}A\cos 4\phi \left| {_{\phi 2} } \right. = \frac{2\eta }{5}A < 0\).

Thus, the antiferromagnetic state is unstable.

The excitation spectrum of the antiferromagnetic phase (A < 0) should consist of three modes. However, the first two modes disappear, leaving only one gapless Goldstone mode.

$$\begin{gathered} {\text{c1 = - 9/8,\_c2 = 11/18, c3 = 2/9}} \hfill \\ \varepsilon_{1} (k) = \frac{3\eta }{4}\sqrt {\left( {A^{2} - A^{2} \left( {\mathbf{k}} \right)} \right)} \left( { - \left( { - \frac{9}{8}} \right) + \frac{5}{2}\left( {\frac{11}{{18}}} \right) - \frac{191}{{16}}\left( \frac{2}{9} \right)} \right) = 0, \hfill \\ \varepsilon_{2} (k) = \frac{3\eta }{2}\sqrt {\left( {A^{2} - A^{2} \left( {\mathbf{k}} \right)} \right)} \left( {\frac{3}{2}\left( {\frac{11}{{18}}} \right) - \left( { - \frac{9}{8}} \right) - \frac{147}{{16}}\left( \frac{2}{9} \right)} \right) = 0, \hfill \\ \varepsilon_{3} (k) = \frac{9\eta }{4}\sqrt {\left( \frac{2}{9} \right)^{2} A^{2} \left( {\mathbf{k}} \right) - A^{2} \left( {\frac{1}{2}\left( {\frac{11}{{18}}} \right) - \left( { - \frac{9}{8}} \right) - \frac{119}{{16}}\left( \frac{2}{9} \right)} \right)^{2} } = \hfill \\ = \frac{9\eta }{4}\left( \frac{2}{9} \right)\sqrt {A^{2} \left( {\mathbf{k}} \right) - A^{2} } = \frac{\eta }{2}\sqrt {A^{2} \left( {\mathbf{k}} \right) - A^{2} } . \hfill \\ \end{gathered}$$
(129)

The condition for the stability of the antiferromagnetic state \(A\Delta_{1} < 0\),\(A\Delta_{2} < 0\) is satisfied only halfway, since \(A\Delta_{1} = 0\) which corresponds to a phase transition to the antinematic phase and \(A\Delta_{2} < 0\), which indicates the stability of the system with respect to the transition to the nematic state.

Thus, the antiferromagnetic (unstable) state corresponds to a neutral equilibrium with respect to a transition to the antinematic state, according to the parameter AΔ1 = 0, and is absolutely stable with respect to a transition to the nematic state, which is forbidden, in accordance with the parameter AΔ2 < 0. (See Fig. 3.)