1 Introduction

In this article, we introduce a mathematical model for an interacting system composed of two kinds of nonrelativistic quantum particles. The number of the first kind of particles, which we call x-particles, is conserved. These particles interact by creating and annihilating bosons of a second kind, called y-particles. The interaction is supported by the set of collision configurations between at least one x-particle and one y-particle. Formally, it is given by the linear coupling \(a(\delta _x)+a^*(\delta _x)\), where \(\delta _x\) denotes the delta distribution at the position x of an x-particle and \(a^*,a\) are the creation and annihilation operators of the y-particles.

A very similar model is the Bogoliubov–Fröhlich polaron, used in physics to describe the interaction of impurities, the x-particles, with a dilute Bose–Einstein condensate [6, 7, 26]. In this context, it is natural to assume point interactions due to the dilute nature of the system. Furthermore, in the Bogoliubov approximation, the quasi-particles describing excitations of the condensate have a dispersion relation \(\omega (k)=\sqrt{a k^2 + bk^4}\) that grows quadratically for large momenta, so they behave similarly to nonrelativistic particles.

Our model is also closely related to the local, Galilean invariant, Lee model [9, 20]. In that model, there are three types of nonrelativistic particles, usually called V, \(\Theta \) and N. The V-particles can create \(\Theta \)-particles, whereupon they are transformed into N-particles. Conversely, a \(\Theta \) and an N-particle can combine to form a V-particle. These processes are such that the total mass is conserved, \(m_V=m_\Theta +m_N\). Since the N and \(\Theta \)-particles cannot create any further ones, this model is composed of invariant sectors with a bounded total number of particles. This and the constraint on the masses are the essential differences to the model we will consider.

Because of the singular interaction, the formal Hamiltonian of such a model is ultraviolet divergent and ill defined. We will define a self-adjoint and bounded-from-below Hamiltonian for our model and describe its domain using (generalised) boundary conditions that relate the (singular) behaviour of the wavefunction near the collision configurations and the wavefunction with one fewer y-particle. Such boundary conditions have appeared independently in the literature as models for nuclear reactions and simplified models of quantum field theory (see Moshinsky [15,16,17], Moshinsky–Laurrabaquio [18], Thomas [25], and Yafaev [28]). More recently, they were proposed as a general approach to the ultraviolet problem by Teufel and Tumulka [23, 24], who called them interior-boundary conditions. The result most closely related to our question is the one of [25], which treats a model with one or two x-particles and at most one y-particle, and resembles one of the sectors of the Galilean Lee model. A variant of our model, where the x-particles are not dynamical but fixed at certain locations, was treated in [28], with at most one y-particle, and by Schmidt, Teufel, Tumulka and the author [12] on the whole Fock space. Schmidt and the author [11] showed that one can define the Hamiltonian for less singular models with dynamical x-particles using this type of boundary conditions. Examples include the Nelson model and the two-dimensional variant of our problem. These results were generalised by Schmidt to general dispersion relations for the x-particles [21] and massless models [22]. We expect that our analysis can be similarly generalised.

For many models in nonrelativistic quantum field theory, the Hamiltonian may be defined by a renormalisation procedure. To our knowledge, no rigorous method was known to work for our problem, so far. It was observed numerically in [7] (for a model with the same ultraviolet behaviour as ours) that, after subtracting the expected linearly divergent quantity, there is still a logarithmic divergence in the ultraviolet cutoff. Concerning rigorous results, the less singular cases treated in [11] can be renormalised using a technique due to Nelson [8, 19]—but this does not seem to work for the three-dimensional model (the conditions of e.g. [8, Thm 3.3] are not satisfied, and the predicted divergence would be in conflict with the numerical results of [7]). Schrader [20] used a reordered resolvent expansion to renormalise the Hamiltonian for the Galilean Lee model. However, later generalisations of this method [5, 27] do not cover our model either. This seems to be related to the fact that the constraint on the masses in the Galilean Lee model alters the structure of the singularities in a specific way (see Remark 11).

In this article, we will adapt the techniques of [11] to define the Hamiltonian for our model. We explain how this result can be understood in the language of renormalisation in Remark 2. We also give an explicit characterisation of the domain, which is generally not easy to obtain by renormalisation.

2 Overview and Results

In this section, we present our main results and the reasoning behind them, while leaving the technical details for the later sections. Let us first introduce the necessary notation. We consider a fixed number M of x-particles, so the Hilbert space of our problem is

$$\begin{aligned} {\mathscr {H}}:=L^2({\mathbb {R}}^{3M})\otimes \Gamma (L^2({\mathbb {R}}^3))\,, \end{aligned}$$

where \(\Gamma (L^2({\mathbb {R}}^3))\) is the bosonic Fock space over \(L^2({\mathbb {R}}^3)\). We will denote the sector of \({\mathscr {H}}\) with ny-particles by \({\mathscr {H}}^{(n)}\), by \(\psi ^{(n)}\) the component of \(\psi \in {\mathscr {H}}\) in this sector and by \(N=\mathrm {d}\Gamma (1)\) the number operator. The Hamiltonian for the non-interacting model is

$$\begin{aligned} L=-\frac{1}{2m}\sum _{\mu =1}^M \Delta _{x_\mu } + \mathrm {d}\Gamma (-\Delta _y+1), \end{aligned}$$
(1)

where m is the mass of the x-particles, and we have set the mass of the y-particles to \(\frac{1}{2}\) and their rest energy to one. Its domain D(L) is given by

$$\begin{aligned} D(L)=\left\{ \psi \in D(N) \Big \vert \forall n\in {\mathbb {N}}: \psi ^{(n)}\in H^2({\mathbb {R}}^{3(M+n)}) \right\} . \end{aligned}$$

The interaction operator is formally given by

$$\begin{aligned} \sum _{\mu =1}^M \left( a^*(\delta _{x_\mu })+a(\delta _{x_\mu }) \right) . \end{aligned}$$

The obvious problem with this operator is that the creation operator is not a densely defined operator on \({\mathscr {H}}\), as

$$\begin{aligned} a^*(\delta _{x_\mu })\psi ^{(n)}= \frac{1}{\sqrt{n+1}} \sum _{i=1}^n \delta _{x_\mu }(y_i)\psi ^{(n)}(X, {\hat{Y}}_i) \end{aligned}$$

(where \(X=(x_1, \dots , x_M)\), \( {\hat{Y}}_i=(y_1,\dots ,y_{i-1}, y_{i+1},\dots , y_{n+1})\in {\mathbb {R}}^{3n}\)) is not an element of \({\mathscr {H}}^{(n+1)}\) for any nonzero \(\psi ^{(n)}\in {\mathscr {H}}^{(n)}\). The annihilation operator is less problematic, since

$$\begin{aligned} a(\delta _{x_\mu })\psi ^{(n)}= \frac{1}{\sqrt{n}} \sum _{i=1}^n \psi ^{(n)}(X, Y)\vert _{y_i=x_\mu }= \sqrt{n} \psi ^{(n)}(X, Y)\vert _{y_n=x_\mu }, \end{aligned}$$
(2)

is a well-defined element of \({\mathscr {H}}\) for \(\psi \in D(L)\). Note, however, that the creation operator is a well-defined operator from \({\mathscr {H}}\) to a space of distributions. In this sense, it is the adjoint of the annihilation operator, \((a(\delta _x))^*:{\mathscr {H}}^{(n)}\rightarrow D(L)'\).

Our approach is not to give a meaning to the interaction operator directly, but rather to implement an interaction using boundary conditions on the sets

$$\begin{aligned} {\mathscr {C}}^n=\left\{ (X,Y)\in {\mathbb {R}}^{3M}\times {\mathbb {R}}^{3n} \Bigg \vert \prod _{\mu =1}^M\prod _{i=1}^n |x_\mu -y_i|=0 \right\} \end{aligned}$$
(3)

of collision configurations between the x-particles and the y-particles. We will then see that the resulting operator includes an interaction of the desired form when interpreted in the sense of distributions. This is similar to the representation of point interactions, with fixed particle number, by boundary conditions (see e.g. [2,3,4, 13, 14]).

In order to impose boundary conditions for L, we first restrict L to the domain

$$\begin{aligned} D(L_0)=\left\{ \psi \in D(N) \Big \vert \forall n\in {\mathbb {N}}: \psi ^{(n)}\in H^2_0({\mathbb {R}}^{3(M+n)}{\setminus } {\mathscr {C}}^n) \right\} \end{aligned}$$
(4)

of functions that vanish on \({\mathscr {C}}^n\). The adjoint \(L_0^*\) of \(L_0=L\vert _{D(L_0)}\) is then an extension of L. Its domain contains functions of the form

$$\begin{aligned} G\varphi ^{(n)}:= - \sum _{\mu =1}^M L^{-1} a^*(\delta _{x_\mu }) \varphi ^{(n)}\in {\mathscr {H}}^{(n+1)}, \end{aligned}$$
(5)

with \(\varphi \in {\mathscr {H}}\). These functions diverge like \(|x_\mu -y_i|^{-1}\) near the set where \(|x_\mu -y_i|=0\) (see Eq. (14)). We interpret

$$\begin{aligned} \big (B_\mu \psi ^{(n+1)}\big )(X,Y)= \beta \lim _{ y_{n+1}\rightarrow x_\mu } \left|x_\mu -y_{n+1} \right| \psi ^{(n+1)}(X,Y,y_{n+1}), \end{aligned}$$

with some constant \(\beta \) (depending on m and n), as a singular boundary value of such functions (the limit exists almost everywhere and defines an element of \({\mathscr {H}}^{(n)}\); see Lemma 7). Note that, by symmetry, the limit is the same for \(y_j\rightarrow x\), for any \(j=1,\dots , n+1\). We set \(B=\sum _{\mu =1}^M B_\mu \) and choose \(\beta \) so that \(B G\varphi =M \varphi \). The free operator L then corresponds to the operator with the Dirichlet-type boundary condition \( B\psi =0\).

There is a second relevant boundary value operator, given by the finite part of \(\psi ^{(n+1)}\) at \(y_{n+1}=x_\mu \). More precisely

$$\begin{aligned}&\big (A_\mu \psi ^{(n+1)}\big )(X,Y)\\&\quad = \lim _{r \rightarrow 0}\frac{\sqrt{n+1}}{4\pi } \int \limits _{S^2} \left( \psi ^{(n+1)}(X,Y,x_\mu +r \omega )- \frac{\alpha }{r}(B\psi ^{(n+1)})(X,Y)\right) \mathrm {d}\omega , \end{aligned}$$

where \(\alpha =(M\beta )^{-1}\) and the limit is taken in the distributional sense (see Lemma 9). Note that this is a local boundary value operator extending the evaluation at \(y_{n+1}=x_\mu \). We set \(A=\sum _{\mu =1}^M A_\mu \), which then extends the annihilation operator.

It has been shown for variants of our model (with fixed x-particles in [12], and in two dimensions in [11]) that the operator \(L_0^*+A\) is self-adjoint on a domain characterised by the boundary conditions \(B \psi ^{(n+1)}=M \psi ^{(n)}\), for all \(n\in {\mathbb {N}}\). These conditions should be viewed as relations on the space \(D(L)\oplus G {\mathscr {H}}\subset D(L_0^*)\). Since \(G {\mathscr {H}}\cap D(L)=\{0\}\), any vector in that space can be written as \(\psi = \varphi + G \xi \) with unique \(\varphi \in D(L)\), \(\xi \in {\mathscr {H}}\). Because \(B\varphi =0\) and \(BG \xi = M \xi \), the condition \(B\psi = M\psi \) is then equivalent to \(\psi =\xi \), or \(\varphi =\psi -G\psi \in D(L)\). To see how this boundary condition is related to the creation and annihilation operators, first observe that \(G\psi \) is in the kernel of \(L_0^*\), for every \(\psi \in {\mathscr {H}}\). That is, for any \(\varphi \in D(L_0)\), which vanishes on the set \({\mathscr {C}}^n\) of collision configurations by definition, we have

$$\begin{aligned} \langle L_0^* G\psi , \varphi \rangle _{\mathscr {H}}&= \langle \psi ,G^* L_0 \varphi \rangle _{\mathscr {H}}\nonumber \\&=-\sum _{n=0}^\infty \sum _{\mu =1}^M \langle \psi ^{(n)},a(\delta _{x_\mu })\varphi ^{(n+1)} \rangle _{{\mathscr {H}}^{(n)}}=0. \end{aligned}$$
(6)

Then, taking any \(\psi \in {\mathscr {H}}\) such that \(\psi -G\psi \in D(L)\) gives

$$\begin{aligned} L_0^*\psi + A\psi = L(\psi - G\psi ) + A \psi = L\psi + \sum _{\mu =1}^M a^*(\delta _{x_\mu })\psi +A\psi , \end{aligned}$$
(7)

in the sense of distributions. Thus, \(L_0^*+A\) represents an operator with creation and annihilation operators up to a choice of domain, including the boundary condition, and the choice of the extension A. The latter choice is made in a natural way such that the operator is local.

In the proof of self-adjointess for the two-dimensional model [11], a key step is then to use again that \(G^*L=-\sum _{\mu =1}^M a(\delta _{x_\mu })\) and that A extends the annihilation operator, to rewrite the Hamiltonian as

$$\begin{aligned} L_0^*\psi + A\psi&=L(1 - G)\psi + A \psi \nonumber \\&= (1-G)^*L(1-G)\psi + A \psi - \sum _{\mu =1}^M a(\delta _{x_\mu })(1-G)\psi \nonumber \\&=(1-G)^*L(1-G)\psi + AG\psi . \end{aligned}$$
(8)

The operator \(T=AG\) is a symmetric operator on a domain \(D(T)\supset D(L)\), and the proof is then essentially reduced to proving appropriate bounds for this operator on the domain of \(L_0^*\) with the boundary condition \(\psi -G\psi \in D(L)\). This operator also appears in the theory of point interactions, where it is known as the Skornyakov–Ter-Martyrosyan operator.

In the three-dimensional case, the range of G is not contained in the domain \(D(T)=D(L^{1/2})\) of T. Since a vector \(\psi \) satisfying \(\psi -G\psi \in D(L)\) is in \(D(L^{1/2})\) if and only if \(G\psi \) is, T and thus also A do not map all vectors satisfying \((1-G)\psi \in D(L)\) to \({\mathscr {H}}\) (see also Lemma 8 and Remark 10). Consequently, this is not a good domain for \(L_0^* +A\). This forces us to modify our approach, by treating T as a perturbation to the free part L. We let

$$\begin{aligned} K=L+T \end{aligned}$$
(9)

with \(D(K)=D(L)\). This operator is self-adjoint and bounded from below (see Sect. 3.3). We may restrict it to the kernel of \(a(\delta _x)\), as above, and define

$$\begin{aligned} K_0= K\vert _{D(L_0)}. \end{aligned}$$

The domain of the adjoint \(K_0^*\) contains functions of the form

$$\begin{aligned} G_T\varphi ^{(n)}:= - \sum _{\mu =1}^M (K+c_0)^{-1} a^*(\delta _{x_\mu }) \varphi ^{(n)} \in {\mathscr {H}}^{(n+1)}, \end{aligned}$$

where \(c_0>0\) is chosen appropriately. These functions are in the kernel of \(K_0^*+c_0\), by the reasoning of Eq. (6). Their main divergence is still proportional to \(|x_\mu -y_i|^{-1}\), but their asymptotic expansion also contains a logarithmically divergent term, as we will show. Thus, the boundary operator B will still be defined on such functions and we can extend \(a(\delta _{x_\mu })\), by taking the finite part in this expansion, which yields an operator \(A_T\).

$$\begin{aligned}&\big (A_T\psi ^{(n+1)}\big )(X,Y)\\&\quad = \lim _{r\rightarrow 0}\sum _{\mu =1}^M \frac{1}{4\pi }\int \limits _{S^2} \left( \sqrt{n+1} \psi ^{(n+1)}(X,Y,x_\mu +r\omega )+ f(r) (B\psi ^{(n+1)})(X,Y)\right) \mathrm {d}\omega , \end{aligned}$$

where \(f(r)=\alpha r^{-1}+\gamma \log r\) is a fixed function, with explicit constants \(\alpha \), as for A, and \(\gamma \) (depending on m, see Eq. (32)). This local boundary value operator then extends the sum of the \(a(\delta _{x_\mu })\) to elements of the range of \(G_T\) (see Proposition 13). Its action there is given by

$$\begin{aligned} A_T G_T =T +S, \end{aligned}$$

with an operator S that is symmetric on \(D(S)=D(L^\varepsilon )\) for any \(\varepsilon >0\).

For any \(\psi \in D(K_0^*)\) such that \(\psi -G_T\psi \in D(L)\), we then have, as in (7),

$$\begin{aligned} L\psi + \sum _{\mu =1}^M a^*(\delta _{x_\mu })+A_T\psi&= (K+c_0)(1-G_T)\psi -c_0\psi - T\psi + A_T\psi \nonumber \\&= K_0^*\psi + (A_T-T)\psi . \end{aligned}$$
(10)

The left-hand side is similar to (7), but we have chosen a different domain \(D(K_0^*)\) and thus also a different extension \(A_T\). This is the expression that will replace \(L_0^*+A\) in our case of the three-dimensional model. Note that \(A_T-T=A_T(1-G_T) + S\), so this operator is better behaved than \(A_T\) alone and it will map vectors satisfying \((1-G_T)\psi \in D(L)=D(K)\) to \({\mathscr {H}}\).

Using \(G_T^*(K+c_0)=-\sum _{\mu =1}^M a(\delta _{x_\mu })\), we can further rewrite the Hamiltonian, as in (8), and obtain

$$\begin{aligned} K_0^* + (A_T-T)&=(K+c_0)(1-G_T)+\sum _{\mu =1}^M a(\delta _{x_\mu }) (1-G_T) +S- c_0\nonumber \\&=(1-G_T)^*(K+c_0)(1-G_T)+S -c_0. \end{aligned}$$
(11)

Our main result is:

Theorem 1

Let A and \(A_T\), be the extensions of \(\sum _{\mu =1}^M a(\delta _{x_\mu })\) defined above, \(AG=T\) and \(A_TG_T-T=S\). Let

$$\begin{aligned} D(H)=\left\{ \psi \in D(N) \Big \vert \forall n\in {\mathbb {N}}: \psi ^{(n+1)} - G_T\psi ^{(n)}\in H^2\left( {\mathbb {R}}^{3(M+n+1)}\right) \right\} , \end{aligned}$$

then the operator

$$\begin{aligned} H&=K_0^* +(A_T-T)\\&=(1-G_T)^*(K+c_0)(1-G_T)+S -c_0 \end{aligned}$$

is self-adjoint on D(H) and bounded from below. Furthermore, for any \(n\in {\mathbb {N}}\) and \(\psi \in D(H)\), we have (setting \(\psi ^{(-1)}=0\))

$$\begin{aligned} H\psi ^{(n)} = L\psi ^{(n)} + \sum _{\mu =1}^M a^*(\delta _{x_\mu })\psi ^{(n-1)} +A_T\psi ^{(n+1)} \end{aligned}$$

as elements of \(H^{-2}({\mathbb {R}}^{3(M+n)})\).

The condition \(\psi ^{(n+1)} - G_T\psi ^{(n)}\in H^2({\mathbb {R}}^{3(M+n+1)})\) means that functions in D(H) satisfy the boundary condition \(B\psi =M\psi \), among all functions of the form \(\varphi + G_T\xi \) with \(\varphi \in D(L)\) and \(\xi \in {\mathscr {H}}\).

Remark 2

The transformation \((1-G_T)\) achieves something similar to the Gross transformation in the Nelson model, in that it provides a transformation relating the interacting operator and a perturbation of the free operator. Applying the inverse of \((1-G_T)\) and its adjoint (which exist by Proposition 12) to expression (11), we have explicitly

$$\begin{aligned} (1-G_T^*)^{-1}H (1-G_T)^{-1}= & {} L+T+c_0 - c_0 (1-G_T^*)^{-1}(1-G_T)^{-1}\\&+(1-G_T^*)^{-1}S(1-G_T)^{-1}. \end{aligned}$$

Our proof of Theorem 1 implies that this is indeed a self-adjoint operator on D(L). The key difference is that the Gross transformation is a Weyl operator, and thus constructed starting from a one-particle function, while \(G_T\) contains in T an n-body interaction that cannot be expressed in such a way. Note also that \((1-G_T)\) is not unitary.

Applying the corresponding transformation to the model with an ultraviolet cutoff \(\Lambda >0\) allows for a reformulation of our result in the language of renormalisation, as in [11]. More precisely, one introduces the cutoff interaction \(v_\Lambda \) and the associated objects (here for \(M=1\))

$$\begin{aligned} G_\Lambda&= - L^{-1} a^*(v_\Lambda ),\\ T_\Lambda&= - a(v_\Lambda )L^{-1} a^*(v_\Lambda )+ \frac{m\Lambda }{\pi ^2 (2m+1)}, \\ K_\Lambda&= L + T_\Lambda ,\\ G_{T_\Lambda }&=- (K_\Lambda + c_0)^{-1}a^*(v_\Lambda ). \end{aligned}$$

Then, by the same algebra that leads to Eqs. (10), (11),

$$\begin{aligned} L+ a^*(v_\Lambda ) + a(v_\Lambda )= (1-G_{T_\Lambda })^* K_\Lambda (1-G_{T_\Lambda }) + S_\Lambda - c_0 - \frac{m \Lambda }{\pi ^2 (2m+1)}, \end{aligned}$$

with

$$\begin{aligned} S_\Lambda =- a(v_\Lambda )( K_\Lambda ^{-1}-L^{-1} ) a^*(v_\Lambda )=a(v_\Lambda )K_\Lambda ^{-1} T_\Lambda L^{-1} a^*(v_\Lambda ). \end{aligned}$$

Along the lines of [11, Sect. 3.4], one can then obtain a renormalisation procedure as follows. >From our estimates on T and S (given in Lemma 8 and Lemma 19), one deduces that, as \(\Lambda \rightarrow \infty \), \(T_\Lambda \) converges to T, and \(S_\Lambda - c \gamma _m \log \Lambda \) converges to S (for some \(c\in {\mathbb {R}}\), \(\gamma _m\) as in (30)), strongly as operators from D(T), respectively D(S), to \({\mathscr {H}}\). Together with similar convergence results for \(G_\Lambda \), \(G_{T_\Lambda }\), one then obtains strong resolvent convergence of

$$\begin{aligned} L+ a^*(v_\Lambda ) + a(v_\Lambda )+\frac{m \Lambda }{\pi ^2 (2m+1)} - c \gamma _m \log \Lambda \end{aligned}$$

to the operator H. This shows exactly the divergence in \(\Lambda \), with a linear and a logarithmic term, observed numerically in [7] (confirming this observation in view of earlier work [26], where convergence without the logarithmic term was claimed). Additionally, our analysis for \(M>1\) shows that the logarithmic term is proportional to M, even though one might naively expect it to be of order \(M^2\), given the form of \(S_\Lambda \). This point of view is developed in more detail for the Bogoliubov–Fröhlich Hamiltonian in [10].

The rest of the article is devoted to the proof of Theorem 1. We start by discussing in detail the properties of the maps G and \(G_T\) as well as the domain of H. We then define the operator \(A_T\) as a (distribution-valued) extension of the annihilation operator to \(D(L)\oplus G_T {\mathscr {H}}\) and give some estimates on its regularity and growth in the particle number in Sect. 4. These results are put together in the proof of Theorem 1 in Sect. 5.

3 The Domain

Before discussing the domain of H, let us recall some properties of extensions of the Laplacian \(L_0\), the restriction of L to vectors vanishing on the collision configurations \({\mathscr {C}}^n\) (cf. Eq. (4)). Spelling out the definition of the operator G, Eq. (5), we have

$$\begin{aligned} G\psi ^{(n)}=&- \left( -\frac{1}{2m}\Delta _X - \Delta _Y+n+1\right) ^{-1}\frac{1}{\sqrt{n+1}}\sum _{\mu =1}^M\sum _{i=1}^{n+1} \delta _{x_\mu }(y_i)\psi ^{(n)}(X,{\hat{Y}}_i), \end{aligned}$$

where we recall that \(X=(x_1,\dots ,x_M)\in {\mathbb {R}}^{3M}\) stands for the positions of the x-particles, \( Y=(y_1,\dots , y_{n+1})\in {\mathbb {R}}^{3(n+1)}\) for those of the y-particles, and \({\hat{Y}}_i\in {\mathbb {R}}^{3n}\) is Y without the ith entry. We view G both as an operator on \({\mathscr {H}}\) and an operator from \({\mathscr {H}}^{(n)}\) to \({\mathscr {H}}^{(n+1)}\), without distinguishing these by the notation.

The functions in the range of G have a specific singular behaviour on the set \({\mathscr {C}}^{n+1}\) (defined in Eq. (3)) of collision configurations. Since

$$\begin{aligned} (-\Delta _x + \lambda ^2)^{-1}\delta = \frac{\mathrm {e}^{- \lambda |x|}}{4\pi |x|}, \end{aligned}$$

the function \(G\psi ^{(n)}\) will diverge like \(|x_\mu -y_i|^{-1}\) on the plane \(\{x_\mu =y_i\}\). The singular set \({\mathscr {C}}^{n+1}\) is just the union of these planes, and due to symmetry, the divergence on each of these is exactly the same. To be more explicit, first observe that the Fourier transform of \(\delta _{x_\mu }(y_i)\psi ^{(n)}(X,{\hat{Y}}_i)\) equals

$$\begin{aligned} \frac{1}{(2\pi )^{3/2}}{\widehat{\psi }}^{(n)}(P+e_\mu k_i, {\hat{K}}_i), \end{aligned}$$
(12)

where \(e_\mu k_i\) is the vector in \({\mathbb {R}}^{3M}=\bigoplus _{\mu =1}^M {\mathbb {R}}^3\) with the \(\mu \)th component equal to \(k_i\) and all other components equal to zero, and \(p_\mu \), \(x_\mu \); \(k_i\), \(y_i\) are conjugate Fourier variables. Now consider, for simplicity, the case \(M=1\). Choosing the centre of mass coordinate \(s=\frac{2m}{2m+1}x+ \frac{1}{2m+1}y_{i}\) and the (weighted) relative coordinate \(r=\sqrt{\frac{2m}{2m+1}}(y_{i}-x)\) for the pair x, \(y_i\) gives

$$\begin{aligned}&\left( -\frac{1}{2m}\Delta _x - \Delta _Y+n+1\right) ^{-1}\delta _{x}(y_{i})\psi ^{(n)}(x,{\hat{Y}}_{i})\nonumber \\&\quad =\frac{1}{(2\pi )^{3(n+1)/2+3}}\int \limits _{{\mathbb {R}}^{3(n+2)}} \frac{\mathrm {e}^{\mathrm {i}px+\mathrm {i}KY}{\widehat{\psi }}^{(n)}(p+ k_{i}, {\hat{K}}_{i})}{\frac{1}{2m}p^2 + K^2+n+1}\mathrm {d}p \mathrm {d}K \end{aligned}$$
(13)
$$\begin{aligned}&\quad =\frac{1}{(2\pi )^{3(n+1)/2}}\left( \frac{2m}{2m+1}\right) ^{3/2} \frac{1}{4\pi |r|} \int \limits _{{\mathbb {R}}^{3(n+1)}} \mathrm {e}^{-|r|\sqrt{n+1+\frac{1}{2m+1}\xi ^2 + {\hat{K}}_{i}^2}} \nonumber \\&\qquad \times \mathrm {e}^{\mathrm {i}{\hat{K}}_{i} {\hat{Y}}_{i}+\mathrm {i}\xi s}{\widehat{\psi }}^{(n)}(\xi , {\hat{K}}_{i})\mathrm {d}\xi \mathrm {d}{\hat{K}}_{i}. \end{aligned}$$
(14)

This function diverges like \(\frac{m}{2\pi (2m+1)\left|x-y_i \right|}\psi ^{(n)}\left( \frac{2m}{2m+1}x+ \frac{1}{2m+1}y_{i},{\hat{Y}}_i\right) \) as \(r\rightarrow 0\), i.e. \(\left|x-y_i \right|\rightarrow 0\) (note that r is the weighted relative coordinate). Observe also that (14) is a smooth function of s and \({\hat{Y}}_{i}\) for \(|r|>0\) because of the exponential decay of its Fourier transform.

3.1 Properties of G

We will now discuss the mapping properties of the operator G. As already noted in Eq. (6), G maps elements of \({\mathscr {H}}\) to the kernel of \(L_0^*\). Furthermore, G is bounded from \({\mathscr {H}}\) to \(D(L^s)\), \(s<\tfrac{1}{4}\):

Lemma 3

Let \(0\le s < \tfrac{1}{4}\). There exists a constant C such that for all \(\psi \in {\mathscr {H}}\)

$$\begin{aligned} \left\| L^s G\psi \right\| _{\mathscr {H}}\le C \Vert N^{s-\frac{1}{4}} \psi \Vert _{\mathscr {H}}. \end{aligned}$$

Proof

Since we will not aim for optimal estimates w.r.t. the number M of x-particles, we can just treat the case \(M=1\) and then bound the sum over the x-particles by the Cauchy–Schwarz inequality.

For \(M=1\) and arbitrary \(n\in {\mathbb {N}}\), we start from expression (13) and use the simple inequality

$$\begin{aligned}&{Re}\left( \overline{{\hat{\psi }}^{(n)}}(p+k_i,{\hat{K}}_i){\hat{\psi }}^{(n)}(p+k_j,{\hat{K}}_j)\right) \\&\quad \le \frac{1}{2} \left( \frac{|{\hat{\psi }}^{(n)}(p+k_i,{\hat{K}}_i)|^2 k_j^2 }{k_i^2} + \frac{|{\hat{\psi }}^{(n)}(p+k_j,{\hat{K}}_j)|^2 k_i^2 }{k_j^2}\right) . \end{aligned}$$

This implies (writing L(p, K) for the Fourier representation of L)

$$\begin{aligned}&\left\| L^s G\psi ^{(n)} \right\| ^2_{{\mathscr {H}}^{(n+1)}} \le \frac{1}{(2\pi )^3 (n+1)} \sum _{i,j=1}^{n+1} \int \frac{|{\hat{\psi }}^{(n)}(p+k_i,{\hat{K}}_i)|^2 k_j^2 }{L(p,K)^{2-2s}k_i^2} \mathrm {d}p \mathrm {d}K \\&\quad \le \frac{\left\| \psi ^{(n)} \right\| ^2_{{\mathscr {H}}^{(n)}}}{(2\pi )^3} \sup _{(p,{\hat{K}}_{n+1})\in {\mathbb {R}}^{3(n+1)}}\sum _{j=1}^{n+1} \int \frac{k_j^2 \mathrm {d}k_{n+1}}{\left( \frac{1}{2m}(p-k_{n+1})^2 + K^2 + n+1\right) ^{2-2s} k_{n+1}^2}. \end{aligned}$$

By splitting the summand with \(j=n+1\) from the rest and changing variables \(k_{n+1}\mapsto k_{n+1}/\sqrt{n+1}\), respectively \(k_{n+1}\mapsto k_{n+1}/\sqrt{n+1+{\hat{K}}_{n+1}^2}\), we can further bound this by

$$\begin{aligned}&\left\| L^s G\psi ^{(n)} \right\| ^2_{{\mathscr {H}}^{(n+1)}} \nonumber \\&\quad \le \frac{\left\| \psi ^{(n)} \right\| ^2_{{\mathscr {H}}^{(n)}}}{(2\pi )^3} \left( \int \frac{ \mathrm {d}k_{n+1}}{(k_{n+1}^2 + n+1)^{2-2s}} + \sup _{{\hat{K}}_{n+1}} {\hat{K}}_{n+1}^2 \int \frac{\mathrm {d}k_{n+1}}{(K^2 +n+1)^{2-2s}k_{n+1}^2} \right) \nonumber \\&\quad \le C^2 (n+1)^{2s-\frac{1}{2}}\left\| \psi ^{(n)} \right\| ^2_{{\mathscr {H}}^{(n)}}. \end{aligned}$$
(15)

This proves the claim. \(\square \)

As an immediate Corollary, we have that \(\sum _{\mu =1}^M a(\delta _{x_\mu })\) maps D(L) to \({\mathscr {H}}\), continuously.

Corollary 4

The operator \(-G^*=\sum _{\mu =1}^M a(\delta _{x_\mu })L^{-1}\) is bounded on \({\mathscr {H}}\).

Remark 5

(The domain of \(L_0^*\)). The operator G can be used to parameterise the domain of \(L_0^*\). From the proof of Lemma 3, one easily infers that \(G:{\mathscr {H}}\rightarrow \ker L_0^*\) can be extended to \(D(L^{-1/4}):=D(L^{1/4})'\). We then have a characterisation of \(D(L_0^*)\) by

$$\begin{aligned} D(L_0^*) = D(L)\oplus G D(L^{-1/4}). \end{aligned}$$

Since we only work on the subspace of \(D(L_0^*)\) where G acts on \({\mathscr {H}}\), we do not need this parameterisation and we will not give a detailed proof. The central argument to obtain this parameterisation is to show that the norm \(\left\| G\psi \right\| _{{\mathscr {H}}^{(n+1)}}\) is equivalent to the norm on \(H^{-1/2}({\mathbb {R}}^{3(M+n)})\) by a generalisation of [3, Lem B.2] to an arbitrary number of particles. This implies the parameterisation above by [1, Prop. 2.9].

On this domain, one could consider models with more general boundary conditions of the form

$$\begin{aligned} a A\psi + b B\psi = \psi . \end{aligned}$$

These correspond to additional direct point interactions between the x- and y-particles (see [12, Sect. 4]) for the case with static x-particles. We will not pursue this here.

As an operator on \({\mathscr {H}}\), G has another important property:

Proposition 6

The operator \(1-G\) is invertible on \({\mathscr {H}}\), and both \(1-G\) and \((1-G)^{-1}\) map the domain D(N) of the number operator to itself.

Proof

We claim that

$$\begin{aligned} (1-G)^{-1}=\sum _{j=0}^\infty G^j \end{aligned}$$

is a bounded operator on \({\mathscr {H}}\). To prove this, we first note that, since G maps \({\mathscr {H}}^{(j)}\) to \({\mathscr {H}}^{(j+1)}\), the sum on \({\mathscr {H}}^{(n)}\) is actually finite,

$$\begin{aligned} \left( (1-G)^{-1}\psi \right) ^{(n)}=\sum _{j=0}^n G^{j} \psi ^{(n-j)}. \end{aligned}$$

The operator is thus well defined and one easily checks that it is an inverse to \(1-G\). To show boundedness, we use Lemma 3 to obtain

$$\begin{aligned} \left\| G^{j} \psi ^{(n-j)} \right\| _{{\mathscr {H}}^{(n)}} \le C^j \left( {\prod _{i=0}^{j-1} (n-i)}\right) ^{-\frac{1}{4}}\left\| \psi ^{(n-j)} \right\| \le \frac{C^j}{(j!)^{\frac{1}{4}}} \left\| \psi ^{(n-j)} \right\| _{{\mathscr {H}}^{(n-j)}}. \end{aligned}$$
(16)

This gives a bound on \((1-G)^{-1}\) by

$$\begin{aligned} \left\| (1-G)^{-1}\psi \right\| _{\mathscr {H}}\le \sum _{j=0}^\infty \left\| G^{j} \psi \right\| \le \left\| \psi \right\| _{\mathscr {H}}\left( \sum _{j=0}^\infty \frac{C^j}{(j!)^{\frac{1}{4}}} \right) . \end{aligned}$$

The operator \(1-G\) maps D(N) to itself if G does. From the estimate of Eq. (15), we see that

$$\begin{aligned} \left\| n (G\psi )^{(n)} \right\| \le C \left\| n^{\frac{3}{4}} \psi ^{(n-1)} \right\| \le 2 C \left\| (n-1) \psi ^{(n-1)} \right\| , \end{aligned}$$

which proves that G leaves D(N) invariant.

To prove the same for \((1-G)^{-1}\), observe that \(G^j\) maps D(N) to itself for any finite j, because G does. It is thus sufficient to prove the claim for \(\sum _{j=j_0}^\infty G^j\), for some \(j_0\in {\mathbb {N}}\). From Lemma 3, we obtain for \(j\ge 4\), as in Eq. (16),

$$\begin{aligned} \left\| n G^j \psi ^{(n-j)} \right\| _{{\mathscr {H}}^{(n)}}&\le \frac{n}{((n-3)(n-2)(n-1)n)^{\frac{1}{4}}}\frac{C^j}{((j-4)!)^{\frac{1}{4}}} \left\| \psi ^{(n-j)} \right\| _{{\mathscr {H}}^{(n-j)}}\\&\le {\tilde{C}} \frac{C^j}{((j-4)!)^{\frac{1}{4}}} \left\| \psi ^{(n-j)} \right\| _{{\mathscr {H}}^{(n-j)}}. \end{aligned}$$

This shows that \(\sum _{j=4}^\infty G^j\) maps \({\mathscr {H}}\) to D(N) and completes the proof. \(\square \)

3.2 Singular Boundary Values

The asymptotic behaviour on \({\mathscr {C}}^{n+1}\) can be used to define local boundary operators on \(D(L_0^*)\). We will define these only on the range of G (acting on \({\mathscr {H}}\)). These boundary values will be defined by certain limits as \(x_\mu -y_{n+1}\rightarrow 0\). These limits exist only almost everywhere or in the sense of distributions. In the following, an expression such as \(\lim _{\left|x_\mu -y_{n+1} \right|\rightarrow 0} \Phi (X,Y,y_{n+1})\) should thus be read as the limit for \(\varepsilon \rightarrow 0\) of the functions \(\Phi _\varepsilon (X,Y)\) given by restricting \(\Phi (X,Y,y_{n+1})\) to the plane where \(\varepsilon =y_{n+1}-x_\mu \in {\mathbb {R}}^3\) is fixed. The boundary values we define can be extended to \(D(L_0^*)\) (cf. Remark 5) in the sense of distributions on the boundary without the singular set \({\mathscr {C}}^{n}\) (see [12, Lem. 6] for the case of a fixed source), but we will not need this here. Set, for suitable \(\varphi \in {\mathscr {H}}^{(n+1)}\),

$$\begin{aligned} \left( B\varphi \right) (X,Y): =-\frac{2\pi (2m+1)}{m}\sqrt{n+1}\sum _{\mu =1}^M \lim _{|x_\mu -y_{n+1}|\rightarrow 0} |x_\mu -y_{n+1}|\varphi (X,Y,y_{n+1}). \end{aligned}$$
(17)

We then have:

Lemma 7

Let \(n\in {\mathbb {N}}\) and \(\psi \in {\mathscr {H}}^{(n)}\). The element \(G\psi \in {\mathscr {H}}^{(n+1)}\) has a representative such that the limit in Eq. (17) exists for almost every (X, Y) and the equality

$$\begin{aligned} \left( BG\psi \right) (X,Y) = M \psi (X, Y) \end{aligned}$$

holds.

Proof

The fact that \(G\psi \in {\mathscr {H}}^{(n+1)}\) was proved in Lemma 3. It follows from the analogue of Eq. (14) for arbitrary M that \(L^{-1}\delta _{x_\nu }(y_i)\psi (X,{\hat{Y}}_i)\) has a smooth representative outside of the plane \(\{x_\nu =y_i\}\). After multiplying by \(|x_\mu -y_{n+1}|\), the limit as \(|x_\mu -y_{n+1}|\rightarrow 0\) is thus zero a.e., except if \(\mu =\nu \) and \(i=n+1\). In the latter case, the limit equals \(\psi (X, Y)\), in \(L^2\) and thus almost everywhere, by strong continuity of the semigroup \(\mathrm {e}^{-t\sqrt{-\Delta + \lambda ^2}}\). There is one such contribution for every \(\mu \), giving the prefactor M in the statement. \(\square \)

The limit in (17) defining B vanishes on D(L), because \(\psi ^{(n)}\in H^2({\mathbb {R}}^{3(M+n)})\) has a well-defined evaluation in the \(L^2\)-sense, the Sobolev trace, on the set \(\{x_\mu =y_{n+1}\}\). We can thus define an operator

$$\begin{aligned} B:D(L)\oplus G{\mathscr {H}}\rightarrow {\mathscr {H}}, \qquad \psi + G\varphi \mapsto M\varphi , \end{aligned}$$

which can be calculated using the local expression (17) outside of the singular sets.

The annihilation operator

$$\begin{aligned} a(\delta _{x_\mu })\psi ^{(n+1)} = \sqrt{n+1} \psi ^{(n)}(X, Y)\vert _{y_{n+1}=x_\mu }, \end{aligned}$$

is not defined on the singular functions in the range of G. It can, however, be extended by considering only the finite part at \(y_{n+1}=x_\mu \), i.e. subtracting the explicit divergence before evaluation:

$$\begin{aligned}&\left( A\varphi \right) (X,Y) \nonumber \\&\quad :=\lim _{r\rightarrow 0}\sum _{\mu =1}^M \frac{1}{4\pi }\int \limits _{S^2}\left( \sqrt{n+1} \varphi (X,Y,x_\mu +r\omega )+ \frac{m}{2\pi (2m+1)} \frac{1}{r} \frac{(B\varphi )(X,Y)}{M}\right) \mathrm {d}\omega . \end{aligned}$$
(18)

At least formally, \(T\psi =AG\psi \) then defines an operator which preserves the number of particles (we will define an operator T below, and then prove equality in Lemma 9). This operator also appears in the theory of point interactions, where it is known as the Skornyakov–Ter-Martyrosyan operator. For fixed \(n\in {\mathbb {N}}\), T is composed of two contributions. The first is the finite part of the term \(L^{-1}\delta _{x_\mu }(y_{n+1})\psi (X,Y)\), which we call the diagonal part \(T_\mathrm {d}\). The second contribution is just the evaluation of terms of the form \(L^{-1}\delta _{x_\nu }(y_i)\psi (X,{\hat{Y}}_{i})\) with either \(\nu \ne \mu \) or \(i\ne n+1\). These diverge on different planes and are smooth on \(\{x_\mu =y_{n+1}\}\) outside of the lower-dimensional set \({\mathscr {C}}^{n}\). We call this the off-diagonal part \(T_\mathrm {od}\). The action of T can be read off from Eqs. (13), (14) and is conveniently expressed in Fourier variables. For an n-particle wavefunction \(\psi ^{(n)}\), we have

$$\begin{aligned} \widehat{T_\mathrm {d} \psi ^{(n)}}(P,K) =&\frac{1}{4\pi }\left( \frac{2m}{2m+1}\right) ^{\tfrac{3}{2}} \sum _{\mu =1}^M \sqrt{n+1 + \tfrac{1}{2m+1} p_\mu ^2 + \tfrac{1}{2m}{\hat{P}}_\mu ^2 + K^2} {\hat{\psi }}^{(n)}(P,K), \end{aligned}$$
(19)

which is obtained from (14) by developing the exponential, and

$$\begin{aligned} \widehat{T_\mathrm {od} \psi ^{(n)}}(P,K)=&-\frac{1}{(2\pi )^{3}}\bigg ( \sum _{\mu ,\nu =1}^M \sum _{i=1}^n \int \frac{{\hat{\psi }}^{(n)}(P-e_\mu \xi + e_\nu k_i,{\hat{K}}_i,\xi )}{n+1+\frac{1}{2m} (P-e_\mu \xi )^2 +K^2 + \xi ^2} \mathrm {d}\xi \nonumber \\&+ \sum _{\mu \ne \nu =1}^M \int \frac{{\hat{\psi }}^{(n)}(P-e_\mu \xi + e_\nu \xi ),K)}{n+1+\frac{1}{2m} (P-e_\mu \xi )^2 +K^2 + \xi ^2}\mathrm {d}\xi \bigg ), \end{aligned}$$
(20)

which corresponds to the evaluation of (13) at \(y_{n+1}=x\).

It is a well-known result in the theory of point interactions (see e.g. [3, 13]) that, for a fixed number n of particles, \(T=T_\mathrm {d}+T_\mathrm {od}\) is bounded from \(H^{1}({\mathbb {R}}^{3(M+n)})\) to \({\mathscr {H}}^{(n)}\), and symmetric. Since, in our problem, n is not fixed, the dependence of the bound on n is important. A bound on \(T_\mathrm {od}\) which is independent of n, even though the number of terms grows linearly in n, was proved for the norm of \(T_\mathrm {od}\) as an operator form \(H^{1/2}\) to \(H^{-1/2}\) by Moser and Seiringer [13] (with \(M=1\)). We adapt their method to prove the same for the operator from \(H^1\) to \({\mathscr {H}}^{(n)}\), and arbitrary M, in Lemma 17. This gives:

Lemma 8

For all \(n\in {\mathbb {N}}\), define the operator \(T=T_\mathrm {d}+T_\mathrm {od}\) by (19), (20) on the domain \(D(T)^{(n)}= {\mathscr {H}}^{(n)}\cap H^1({\mathbb {R}}^{3(M+n)})\) and denote the induced operator on \({\mathscr {H}}\) by the same symbol. The operator T is symmetric on \(D(T)=D(L^{1/2})\), and there exists a constant C such that for all \(\psi \in D(L^{1/2})\)

$$\begin{aligned} \left\| T\psi \right\| _{{\mathscr {H}}} \le C \Vert (1+L^{1/2}) \psi \Vert _{{\mathscr {H}}}. \end{aligned}$$

Proof

The statement is trivial for \(T_\mathrm {d}\). The bound on \(T_\mathrm {od}\) is proved in Lemma 17. To show symmetry of \(T_\mathrm {od}\) on its domain, one may use the representation

$$\begin{aligned} T_\mathrm {od} \psi ^{(n)}(X,{\hat{Y}}_{n+1}) = -\sum _{(\mu ,n+1)\ne (\nu , i)} \tau _{x_\mu }(y_{n+1})L^{-1}\delta _{x_\nu }(y_i) \psi ^{(n)}(X,{\hat{Y}}_i), \end{aligned}$$
(21)

where \(\tau _{x_\mu }(y_{n+1})\) denotes evaluation at \(y_{n+1} = x_\mu \) (outside of \({\mathscr {C}}^{n}\)). This proves the claim, because of the bounds on \(T_\mathrm {od}\) obtained before and because \(\tau ^*_x=\delta _x\), as a map from smooth functions to distributions. \(\square \)

For the boundary operator A, we can now prove:

Lemma 9

Let \(\psi \in {\mathscr {H}}^{(n)}\). The limit in Eq. (18) with \(\varphi =G\psi \) exists in \(H^{-1}({\mathbb {R}}^{3(M+n)})\) and \(AG\psi =T\psi \).

Proof

It is sufficient to prove the claim for a fixed \(\mu \in \{1,\dots , M\}\). We start with the diagonal part \(T_\mathrm {d}\), which is clearly an operator from \({\mathscr {H}}^{(n)}\) to \(H^{-1}({\mathbb {R}}^{3(M+n)})\). Consider the representation of G in Eq. (14). This shows, for \(M=1\), that the limit for \(|x-y_{n+1}|\rightarrow 0\) of

$$\begin{aligned} \left( L^{-1}\delta _x(y_{n+1})\psi \right) (x,Y) - \frac{m}{2\pi (m+1)} \frac{1}{|x-y_{n+1}|} \psi \left( \tfrac{2m}{2m+1}x+\tfrac{1}{2m+1}y_{n+1},Y\right) \end{aligned}$$

exists in \(H^{-1}\) for \(\psi \in L^2\) and equals \(-(2\pi )^3T_\mathrm {d}\psi \) because this limit is (up to a prefactor) just the derivative at \(r=0\) of the semigroup \(\mathrm {e}^{-r T_\mathrm {d}}\). To complete the proof for \(M=1\), we need to show that the error made by replacing \(\psi (x,Y)\) with \(\psi (\tfrac{2m}{2m+1}x+\tfrac{1}{2m+1}y_{n+1},Y)\) converges to zero as \(r\rightarrow 0\). This follows, by duality, from the fact that for \(f\in H^1({\mathbb {R}}^{3(1+n)})\)

$$\begin{aligned} \lim _{r\rightarrow 0}&\,\frac{1}{r}\int \limits _{S^2} \left( f(x-\tfrac{1}{2m+1}r\omega ,Y)- f(x,Y)\right) \mathrm {d}\omega = -\frac{1}{2m+1}\int \limits _{S^2} \omega \cdot \nabla f(x,Y) \mathrm {d}\omega =0 \end{aligned}$$
(22)

in \({\mathscr {H}}^{(n)}\). The generalisation to arbitrary M is straightforward, completing the argument for \(T_\mathrm {d}\).

The off-diagonal part \(T_\mathrm {od}\) is a bounded operator from \({\mathscr {H}}^{(n)}\) to \(H^{-1}({\mathbb {R}}^{3(M+n)})\) by Lemma 8 and duality. The estimates of Lemma 17 also yield a uniform bound for the operator obtained by evaluation at \(y_{n+1}-x_\mu =\varepsilon \), whose integral kernel differs from that of \(T_\mathrm {od}\) by a factor \(\mathrm {e}^{\mathrm {i}\varepsilon \xi }\). This implies dominated convergence for \(\varepsilon \rightarrow 0\). \(\square \)

We have thus defined a second boundary value operator

$$\begin{aligned} A:D(L)\oplus G{\mathscr {H}}\rightarrow D(L^{-1/2}), \qquad \psi + G\varphi \mapsto \sum _{\mu =1}^M a(\delta _{x_\mu })\psi + T\varphi , \end{aligned}$$

where T is extended to \(T:{\mathscr {H}}^{(n)}\rightarrow H^{-1}({\mathbb {R}}^{3(M+n)})\) by duality. This is an extension of the annihilation operator, originally defined on D(L). There are of course many such extensions, e.g. the map \(G\varphi \mapsto 0\) provides an example. The extension A is special (though still not unique) in that it is also a sum of local operators, in the sense that \((A\psi )^{(n)}(X,Y)\) is a sum over \(\mu \) in which each term is completely determined by \(\psi \) restricted to any neighbourhood of the point \((X,Y,x_\mu )\in {\mathscr {C}}^{n+1}\).

Remark 10

(The model with one y-particle). The results we have obtained so far are sufficient to discuss the model with at most one y-particle. For the cases \(M=1,2\) this is essentially the model introduced in [25]. Certain sectors of the Galilean Lee model [9, 20] can also be described in a very similar way.

Consider the subspace D of \({\mathscr {H}}^{(0)}\oplus {\mathscr {H}}^{(1)}\) formed by elements \(\psi =(\psi ^{(0)},\psi ^{(1)})\) with \(\psi ^{(0)}\in D(L)=H^2({\mathbb {R}}^{3M})\), \(\psi ^{(1)}\in D(L)\oplus G{\mathscr {H}}^{(0)}\). On this space, both \(B\psi ^{(1)}\) and \(A\psi ^{(1)}\) are well defined. Since \(BG\varphi =M\varphi \), we also have

$$\begin{aligned} \psi ^{(1)}- GB\psi ^{(1)}/M\in D(L)=H^2({\mathbb {R}}^{3(M+1)}) . \end{aligned}$$

Using first that \(L_0^*G=0\) and then the symmetry of (L, D(L)), we find the identity for \(\psi , \varphi \in D\)

$$\begin{aligned}&\langle L_0^* \psi ^{(1)},\varphi ^{(1)} \rangle - \langle \psi ^{(1)},L_0^*\varphi ^{(1)} \rangle \\&\quad =\langle L_0^* (\psi ^{(1)}-GB\psi ^{(1)}/M),\varphi ^{(1)} \rangle - \langle \psi ^{(1)},L_0^*(\varphi ^{(1)}-GB\varphi ^{(1)}/M) \rangle \\&\quad =\langle L (\psi ^{(1)}-GB\psi ^{(1)}/M),GB\varphi ^{(1)}/M \rangle {-} \langle GB \psi ^{(1)}/M,L(\varphi ^{(1)}-GB\varphi ^{(1)}/M)\rangle . \end{aligned}$$

Since \(LG=-\sum _{\mu =1}^M a^*(\delta _{x_\mu })\), this equals

$$\begin{aligned}&\frac{1}{M}\sum _{\mu =1}^M \bigg (-\langle a(\delta _{x_\mu })(\psi ^{(1)}-GB\psi ^{(1)}/M), B\varphi ^{(1)} \rangle \\&\quad + \langle B \psi ^{(1)},a(\delta _{x_\mu })(\varphi ^{(1)}-GB\varphi ^{(1)}/M) \rangle \bigg ). \end{aligned}$$

If \(B\psi ^{(1)}\) and \(B\varphi ^{(1)}\) are elements of the domain \(D(T)^{(0)}= H^1({\mathbb {R}}^{3M})\) of the symmetric operator T, we can add the term \((\langle T B\psi ^{(1)}, B\varphi ^{(1)}\rangle - \langle B\psi ^{(1)},T B\varphi ^{(1)}\rangle )/M^2=0\) to this equation. Since \(AG=T\), it then becomes

$$\begin{aligned} \langle L_0^* \psi ^{(1)},\varphi ^{(1)} \rangle - \langle \psi ^{(1)},L_0^*\varphi ^{(1)} \rangle =\langle B\psi ^{(1)}/M,A\varphi ^{(1)}\rangle -\langle A\psi ^{(1)},B\varphi ^{(1)}/M\rangle . \end{aligned}$$
(23)

This implies that the operator

$$\begin{aligned} H^{(1)}(\psi ^{(0)},\psi ^{(1)}) = (L\psi ^{(0)} + A\psi ^{(1)}, L_0^* \psi ^{(1)}) \end{aligned}$$

is symmetric if we impose the boundary condition \(B\psi ^{(1)}=M\psi ^{(0)}\), i.e. on the domain

$$\begin{aligned} D(H^{(1)})=\Big \{ \psi \in {\mathscr {H}}^{(0)}\oplus {\mathscr {H}}^{(1)} \Big \vert \psi ^{(0)}\in D(L),\, \psi ^{(1)}\in D(L)\oplus G{\mathscr {H}}^{(0)},\, B\psi ^{(1)}=M\psi ^{(0)}\Big \}. \end{aligned}$$

One can prove that \(H^{(1)}\) is self-adjoint, for example by constructing its resolvent along the lines of [9, 25] or by adapting our proof in Sect. 5. For \(\psi \in D(H^{(1)})\), we have

$$\begin{aligned} L_0^* \psi ^{(1)}=L(\psi ^{(1)}-G\psi ^{(0)})=L\psi ^{(1)} + \sum _{\mu =1}^M a^*(\delta _{x_\mu })\psi ^{(0)}, \end{aligned}$$
(24)

where the right-hand side is a sum in \(H^{-2}({\mathbb {R}}^{3(M+1)})\). We can thus also write

$$\begin{aligned} H^{(1)}\psi =L\psi + A\psi ^{(1)}+ \sum _{\mu =1}^M a^*(\delta _{x_\mu })\psi ^{(0)}. \end{aligned}$$

It is important to note that we have used the fact that \(\psi ^{(0)}\in H^2({\mathbb {R}}^{3M})\subset D(T)^{(0)}\). This does not carry over to cases with more y-particles, since \(B\psi ^{(1)}=\psi ^{(0)}\) implies that \(\psi ^{(1)}\) is not in \(D(T)^{(1)}=H^1({\mathbb {R}}^{3(M+1)})\) (or even \(H^{1/2}({\mathbb {R}}^{3(M+1)})\)), if \(\psi ^{(0)}\ne 0\). Consequently, \(A\psi ^{(2)}=a(\delta _x)\left( \psi ^{(2)} - G\psi ^{(1)}\right) + T \psi ^{(1)}\) does not define an element of \({\mathscr {H}}^{(1)}\), and the definition \(H^{(2)}\) by analogy with \(H^{(1)}\) does not yield a well-defined operator from \(D(H^{(2)})\) to \({\mathscr {H}}\).

3.3 The Operator K and Its Extension

Up to now, we have developed the theory very much in parallel to [11], where we treated in particular the two-dimensional variant of our model. There, we proved that \(L_0^*+A\) is self-adjoint of the domain \(D(L)\oplus G{\mathscr {H}}\) with the boundary condition \(B\psi =M\psi \). For the three-dimensional model, this cannot be true as such. In this case, \(AG=T\) is defined on \(D(L^{1/2})\) but if \(B\psi ^{(n)}=M\psi ^{(n-1)}\ne 0\), then \(\psi ^{(n)}\notin H^1({\mathbb {R}}^{3(M+n)})\cap {\mathscr {H}}^{(n)}=D(T)^{(n)}\), since such a function must diverge like \(\psi ^{(n-1)}(X, {\hat{Y}}_n)|x_\mu -y_n|^{-1}\) as \(y_n\rightarrow x_\mu \). Consequently, we should not expect \(L_0^*+A\) to map this domain to \({\mathscr {H}}\). This problem cannot be remedied by simply interpreting the operators as quadratic forms, since one can also show that \(G{\mathscr {H}}^{(n-1)}\cap H^{1/2} ({\mathbb {R}}^{3(M+n)}) =\{0\}\) (see [11, Prop. 4.2]).

Remark 11

Of course, the fact that \(G\psi \) is not in the domain \(D(T)=D(L^{1/2})\) of T does not immediately imply that \(TG\psi \notin {\mathscr {H}}\), but only that \(T_\mathrm {d}G\psi \notin {\mathscr {H}}\) and \(T_\mathrm {od}G\psi \notin {\mathscr {H}}\), separately. Cancellations between these two terms can make TG well defined on \(D(L)\oplus G{\mathscr {H}}\). In this case, one can proceed with the operator \(L_0^*+A\). This happens in the variant of our model with fixed x-particles [12], which formally corresponds to taking \(m=\infty \) (see Eq. (32)). Similar cancellations occur in the Galilean Lee model [9, 20], when formulated in our language, due to the constraint on the masses of the different particles. Proposition 13 shows that, in our model, there are no such cancellations.

Our solution to the problem of defining AG is to change the regularity of elements of the domain in such a way that the singularities of A and \(L_0^*\) cancel, similar to the cancellation in \(L+a^*\) on \(D(H^{(1)})\subset D(L_0^*)\) in the case of one y-particle (see Eq. (24)). Let K be the operator

$$\begin{aligned} K=L+T \end{aligned}$$

with \(D(K)=D(L)\) and T given by (19), (20). By Lemma 8 and the Kato–Rellich theorem, K is self-adjoint and bounded from below. Let \(K_0\) be the restriction of K to \(D(L_0)\), the vectors \(\psi \in D(L)\) with \(\psi ^{(n)}\in H^2_0({\mathbb {R}}^{3(M+n)}{\setminus }{\mathscr {C}}^n)\). In analogy with G, we define a map

$$\begin{aligned} G_T:{\mathscr {H}}\rightarrow \ker (K_0^* + c_0), \qquad G_T\psi := - \sum _{\mu =1}^M (K+c_0)^{-1} a^*(\delta _{x_\mu }) \psi , \end{aligned}$$
(25)

where \(c_0>-\min \sigma (K)\) is a fixed constant. The important properties of G carry over to \(G_T\) by perturbation theory.

Proposition 12

Let \(c_0>-\min \sigma (K)\) and \(G_T\) be defined by (25). For \(0\le s < \tfrac{1}{4}\), there exists a constant C such that

$$\begin{aligned} \left\| L^s G_T\psi \right\| _{\mathscr {H}}\le C \Vert N^{s-\frac{1}{4}} \psi \Vert _{\mathscr {H}}. \end{aligned}$$

Moreover, \(1-G_T\) is invertible on \({\mathscr {H}}\) and \(1-G_T\) as well as \((1-G_T)^{-1}\) map the domain of the number operator D(N) to itself.

Proof

By the resolvent formula, we have

$$\begin{aligned} G_T\psi = G\psi - L^{-1}(T+c_0) G_T \psi . \end{aligned}$$
(26)

The bound on \(L^sG_T\) then follows from the bound on \(L^s G\), Lemma 3, and the fact that \(L^{-1/2}T\) is bounded on \({\mathscr {H}}\), by Lemma 8. The continuity and the mapping properties of

$$\begin{aligned} (1-G_T)^{-1}=\sum _{j=0}^\infty G^j_T \end{aligned}$$

follow from this by exactly the same proof as in Proposition 6. \(\square \)

This proposition implies that \((1-G_T)\psi \in D(N)\) if and only if \(\psi \in D(N)\), so we have

$$\begin{aligned} D(H)&= \{ \psi \in D(N): (1-G_T)\psi \in D(L) \}\nonumber \\&= \{ \psi \in {\mathscr {H}}: (1-G_T)\psi \in D(L) \}\nonumber \\&= (1-G_T)^{-1} D(L). \end{aligned}$$
(27)

This also shows that D(H) is dense in \({\mathscr {H}}\), by continuity and surjectivity of \((1-G_T)^{-1}\). Concerning the regularity of vectors in the range of \(G_T\), applying identity (26) twice, we find for any \(n\ge 1\)

$$\begin{aligned} G_T\psi ^{(n)}=G\psi ^{(n)} - L^{-1}(T+c_0) G \psi ^{(n)} + (L^{-1}(T+c_0))^2 G_T \psi ^{(n)}. \end{aligned}$$
(28)

The first term is an element of \(H^{s}({\mathbb {R}}^{3(M+n+1)})\), \(s<\frac{1}{2}\), by Lemma 3 and diverges like \(|x_\mu -y_i|^{-1}\) near \({\mathscr {C}}^{n+1}\) by Lemma 7. The operator \(TL^{-1}T\) is bounded on \({\mathscr {H}}^{(n)}\) by Lemma 8, so the last term above is an element of \(H^2({\mathbb {R}}^{3(M+n+1)})\). It can thus be evaluated on the co-dimension-three set \({\mathscr {C}}^{n+1}\) of collision configurations. The term \(L^{-1}(T+c_0) G \psi ^{(n)}\) is in \(H^1({\mathbb {R}}^{3(M+n+1)})\) (even in \(H^s\), \(s<3/2\)), so it should have a less pronounced divergence on \({\mathscr {C}}^{n+1}\) than the \(|x_\mu -y_i|^{-1}\)-divergence of \(G\psi ^{(n)}\). In fact, we will show in Proposition 13 that this term diverges logarithmically. We can thus define \(BG_T \psi =M\psi \) by the same limit (17) as for G. We will define a modification \(A_T\) of the operator A on the range of \(G_T\) in the next section (see Proposition 13).

4 Extension of the Annihilation Operator

In this section, we will analyse the divergence of \(G_T\psi \) on the sets \({\mathscr {C}}^n\) in order to define a local boundary operator \(A_T\) that extends \(\sum _{\mu =1}^M a(\delta _{x_\mu })\) to \(D(L)\oplus G_T{\mathscr {H}}\). One should think of functions in the range of \(G_T\) as having an expansion of the form

$$\begin{aligned} (G_T\psi )^{(n+1)}(X,Y) \sim \psi ^{(n)}(X,{\hat{Y}}_{n+1}) \left( \frac{b_1}{|x_\mu -y_{n+1}|} + b_2 \log |x_\mu -y_{n+1}| \right) + F(X,Y) \end{aligned}$$

near \(x_\mu =y_{n+1}\). Here, the constant \(b_1\) comes from the expansion of G given in Eq. (14), \(b_2\) is determined by the second term in Eq. (28), and F(X, Y) has an appropriate limit as \(\left|x_\mu -y_{n+1} \right|\rightarrow 0\). As in Sect. 3.2, we would then define boundary value operators B, \(A_T\) such that any \(\varphi \in D(L)\oplus G_T {\mathscr {H}}\) has the expansion for \(\left|x_\mu -y_{n+1} \right|\rightarrow 0\)

$$\begin{aligned} \varphi ^{(n+1)}(X,Y)\sim \frac{(B\varphi ^{(n+1)})(X,{\hat{Y}}_{n+1})}{M} f(\left|x_\mu -y_{n+1} \right|) + (A_T \varphi ^{(n+1)})(X,{\hat{Y}}_{n+1}) + o(1), \end{aligned}$$

where \(f(\left|x_\mu -y_{n+1} \right|)\) is the divergent function in the expansion above. We will justify this intuition by defining the operators \(B, A_T\) and showing that they are given by appropriate limits, in the sense of distributions.

We define the map

$$\begin{aligned} B: D(L)\oplus G_T {\mathscr {H}}\rightarrow {\mathscr {H}},\qquad \varphi + G_T\psi \mapsto M \psi . \end{aligned}$$

As a consequence of Proposition 13, B is a sum of local boundary operators given by the same expression, Eq. (17) (where the limit is taken in \(H^{-1}\)), as the corresponding operator on \( D(L)\oplus G{\mathscr {H}}\). To define \(A_T\), we set for appropriate \(\varphi \in {\mathscr {H}}^{(n+1)}\)

$$\begin{aligned}&\big (A_T\varphi \big )(X,Y)\nonumber \\&\quad := \lim _{r\rightarrow 0}\sum _{\mu =1}^M \frac{1}{4\pi }\int \limits _{S^2} \left( \sqrt{n+1} \varphi (X,Y,x_\mu +r\omega )+ f_m(r) (B\varphi )(X,Y)\right) \mathrm {d}\omega , \end{aligned}$$
(29)

where \(r>0\),

$$\begin{aligned} f_m(r)=\frac{1}{M} \left( \frac{m}{2\pi (2m+1)} \frac{1}{r} + \gamma _m\log (r)\right) , \end{aligned}$$

and \(\gamma _m\) is the constant

$$\begin{aligned} \gamma _m=\frac{1}{(2\pi )^3}\left( \frac{2m}{2m+1}\right) ^3 \left( \frac{2\sqrt{m(m+1)}}{2m+1} - (2m+1) \tan ^{-1}\left( \frac{1}{2\sqrt{m(m+1)}}\right) \right) .\nonumber \\ \end{aligned}$$
(30)

Note that for \(\varphi \in H^2({\mathbb {R}}^{3(M+n+1)})\), \(A_T\varphi \) equals the usual annihilation operator. We will show that the formula for \(A_T\) defines a map

$$\begin{aligned} A_T:D(L)\oplus G_T {\mathscr {H}}\rightarrow D(L^{-1/2})\,,\qquad \varphi +G_T\psi \mapsto \sum _{\mu =1}^M a(\delta _{x_\mu }) \varphi + T\psi +S\psi ,\nonumber \\ \end{aligned}$$
(31)

where \(T=AG\) as before and \(S:D(L^\varepsilon )\rightarrow {\mathscr {H}}\) is symmetric, for any \(\varepsilon >0\).

Observe also that

$$\begin{aligned} \lim _{m\rightarrow \infty } \gamma _m=\frac{1}{(2\pi )^3}\left( 1-(\tan ^{-1})'(0)\right) =0. \end{aligned}$$
(32)

which explains why there is no logarithmically divergent term in the case of fixed x-particles, treated in [12]. The result of this section is:

Proposition 13

Let \(\psi \in {\mathscr {H}}^{(n)}\). Then for \(\varphi =G_T\psi \) the limit in Eq. (29) exists in \(H^{-1}({\mathbb {R}}^{3(M+n)})\) and \(S\psi :=(A_TG_T-T)\psi \) defines a symmetric operator on \(D(S)={\mathscr {H}}^{(n)} \cap H^\varepsilon ({\mathbb {R}}^{3(M+n)})\), for any \(\varepsilon >0\).

We will give an outline of the proof here and provide some of the more technical points as separate lemmas in the appendix.

Proof

Let \(R\psi :=- L^{-1}T G \psi \). Then, in view of Eq. (28), we have

$$\begin{aligned} G_T\psi = G\psi + R\psi + \left( (L^{-1}(T+c_0))^2G_T -c_0 L^{-1} G\right) \psi . \end{aligned}$$
(33)

The sum of \(G\psi \) and the 1 / r-term in \(f_m(r)\) converges to \(AG\psi =T\psi \in H^{-1}\) by Lemma 9. Since the last term in Eq. (33) is an element of \(H^2({\mathbb {R}}^{3(M+n+1)})\), it has a Sobolev trace on \(\{x_\mu =y_{n+1}\}\) and the usual annihilation operator is well defined on this term. We denote this evaluation by

$$\begin{aligned} S_\mathrm {reg}\psi := \sum _{\mu =1}^M a(\delta _{x_\mu }) \left( (L^{-1}(T+c_0))^2 G_T -c_0 L^{-1} G\right) \psi . \end{aligned}$$
(34)

It then remains to show the convergence of the sum of \(R\psi \) and the logarithmic term in \(f_m(r)\). It is sufficient to prove convergence in \({\mathscr {H}}^{(n)}\) for \(\psi \in H^\varepsilon ({\mathbb {R}}^{3(M+n)})\), \(\varepsilon >0\), convergence in \(H^{-\varepsilon }\) for \(\psi \in {\mathscr {H}}^{(n)}\) then follows by duality.

We will focus on the calculation of the asymptotic behaviour at \(r=0\) in the case \(M=1\), \(n=0\); here, the full argument is provided in Lemma 18. We set \(R_\mathrm {d}=- L^{-1}T_\mathrm {d} G\), \(R_\mathrm {od}=- L^{-1}T_\mathrm {od} G\) and start by analysing \((R_\mathrm {d}\psi )(x,y)\) at \(x=y\). By Eqs. (13), (19), we have, with a change of variables \(\sigma =p+k\), \(\rho =k -\frac{1}{(2m+1)} \sigma \),

$$\begin{aligned}&(R_\mathrm {d}\psi )(x,y)\nonumber \\&\quad =\frac{1}{(2\pi )^{9/2}}\frac{1}{4\pi }\left( \frac{2m}{2m+1}\right) ^{3/2} \int \frac{\mathrm {e}^{\mathrm {i}px+ \mathrm {i}ky}\sqrt{1+\frac{1}{2m+1}p+ k^2} }{L(p,k)^2}{\hat{\psi }}(p + k) \mathrm {d}p \mathrm {d}k\nonumber \\&\quad = \frac{1}{2(2\pi )^{4}}\left( \frac{2m}{2m+1}\right) ^{3/2} \nonumber \\&\qquad \times \frac{1}{(2\pi )^{3/2}} \int \frac{\mathrm {e}^{\mathrm {i}\sigma s + \mathrm {i}\rho r}\sqrt{1+\frac{2m+2}{2m+1}\rho ^2 + b_1 \sigma ^2 + b_2 \rho \sigma } }{(1+\frac{1}{2m+1} \sigma ^2 + \frac{2m+1}{2m}\rho ^2 )^2}{\hat{\psi }}(\sigma ) \mathrm {d}\sigma \mathrm {d}\rho , \end{aligned}$$
(35)

where s, r are the centre of mass and relative coordinate and \(b_1=\tfrac{4m^2+2m+1}{(2m+1)^3}\), \(b_2=\tfrac{2}{(2m+1)^2}\). This acts on \(\psi \) as a Fourier multiplier with the function given by the \(\rho \)-integral. The singularity of this integral depends only on the behaviour of the integrand at infinity, so we replace the square root in the numerator by \(\sqrt{(2m+2)/(2m+1)}|\rho |\). The error we make by this replacement is integrable in \(\rho \), and the evaluation at \(x=y\) gives rise to a Fourier multiplier with a function that grows no faster than \(|\sigma |^\varepsilon \) (see Lemma 19). We thus have to calculate the asymptotic behaviour as \(r\rightarrow 0\) of

$$\begin{aligned} \frac{1}{2(2\pi )^4}\left( \frac{2m}{2m+1}\right) ^{3/2}\sqrt{\frac{2m+2}{2m+1}} \int \frac{\mathrm {e}^{\mathrm {i}\rho r}|\rho |}{(1 +\frac{1}{2m+1}\sigma ^2+\frac{2m+1}{2m} \rho ^2)^2}\mathrm {d}\rho . \end{aligned}$$
(36)

The integral

$$\begin{aligned} \int _{{\mathbb {R}}^3} \frac{\mathrm {e}^{\mathrm {i}\lambda \xi }|\xi |}{(1+\xi ^2)^2} \mathrm {d}\xi = 4\pi \int _0^\infty \frac{\sin (t)t^2}{(\lambda ^2+t^2)^2} \mathrm {d}t= 2\pi \int _0^\infty \frac{\sin (t)+t\cos (t)}{\lambda ^2+t^2} \mathrm {d}t \end{aligned}$$

has an expansion given by \(-4\pi \log (\lambda ) + {\mathcal {O}}(1)\) as \(\lambda \rightarrow 0\). We thus find that expression (36) behaves like

$$\begin{aligned} -\frac{1}{(2\pi )^3}\left( \frac{2m}{2m+1}\right) ^{7/2} \sqrt{\frac{2m+2}{2m+1}}\log \left( r\sqrt{\tfrac{2m}{2m+1} (1+\tfrac{1}{2m+1}\sigma ^2)}\right) , \end{aligned}$$

up to remainders that are uniformly bounded in \(\sigma \) as \(r\rightarrow 0\).

We now turn to \(R_\mathrm {od}=L^{-1}T_\mathrm {od} L^{-1}\delta _x\psi \). We have (cf. (20) and note that \(T_\mathrm {od}\) here is the operator on the one-particle space)

$$\begin{aligned}&\left( \widehat{T_\mathrm {od}L^{-1}\delta _x\psi }\right) (p,k)\\&\quad =-\frac{1}{(2\pi )^{9/2}} \int \frac{1}{2+\frac{1}{2m}(p-\xi )^2+ \xi ^2+k^2}\frac{{\hat{\psi }}(p+k)}{1+\frac{1}{2m}(p+k-\xi )^2+\xi ^2} \mathrm {d}\xi . \end{aligned}$$

Using the same variables \(\rho , \sigma , s,r\) as before, this gives

$$\begin{aligned}&\left( R_\mathrm {od}\psi \right) (s,r)\nonumber \\&\quad = -\frac{1}{(2\pi )^{6+3/2}} \int \frac{\mathrm {e}^{\mathrm {i}\sigma s}\mathrm {e}^{\mathrm {i}\rho r}}{(1+\tfrac{1}{2m+1}\sigma ^2+\tfrac{2m+1}{2m}\rho ^2)(1 + \frac{1}{2m}(\sigma -\xi )^2 +\xi ^2)}\nonumber \\&\qquad \frac{{\widehat{\psi }}(\sigma )}{2+\xi ^2 + \tfrac{1}{2m+1}(\sigma -\xi )^2+\tfrac{2m+1}{2m}(\rho +\tfrac{1}{2m+1}\xi )^2} \mathrm {d}\xi \mathrm {d}\sigma \mathrm {d}\rho . \end{aligned}$$
(37)

Similar to the case of \(R_\mathrm {d}\), this acts as a Fourier multiplier by the function given by the integral over \(\xi \) and \(\rho \). To simplify the calculation of this integral, we replace \((\sigma -\xi )^2\) in the denominator by \(\sigma ^2+\xi ^2\) (the error again has better decay in \(\xi \) and \(\rho \) (see Eq. (46))). For the expression resulting from the denominator of the last line, we then gather the terms

$$\begin{aligned} \xi ^2+ \tfrac{1}{2m+1} \xi ^2 + \tfrac{2m+1}{2m}(\rho +\tfrac{1}{2m+1}\xi )^2&=\tfrac{2m+1}{2m}(\xi +\tfrac{1}{2m+1}\rho )^2 + \tfrac{2m+2}{2m+1} \rho ^2, \end{aligned}$$

making apparent that the \(\xi \)-integral is now a convolution. This can be evaluated using the Fourier transform

$$\begin{aligned} \int _{{\mathbb {R}}^3} \frac{e^{-\mathrm {i}kx} e^{-\lambda |x|}}{|x|^2}\mathrm {d}x =4\pi \int _0^\infty \frac{\sin (|k|t)}{|k|t} e^{-\lambda t}\mathrm {d}t =\frac{4\pi }{|k|} \tan ^{-1}\left( \frac{|k|}{\lambda }\right) . \end{aligned}$$

The result is

$$\begin{aligned}&\int \frac{1}{(\beta +\frac{2m+1}{2m}\xi ^2)(\gamma +\frac{2m+1}{2m}(\xi +\frac{1}{2m+1}\rho )^2)}\mathrm {d}\xi \nonumber \\&\quad =2\pi ^2 \left( \frac{2m}{2m+1}\right) ^2 \frac{2m+1}{|\rho |}\tan ^{-1}\left( \frac{|\rho |}{\sqrt{2m(2m+1)}(\sqrt{\beta }+\sqrt{\gamma })}\right) , \end{aligned}$$
(38)

where \(\beta =1+\tfrac{1}{2m}\sigma ^2\), \(\gamma =2+\tfrac{1}{2m+1}\sigma ^2 + \tfrac{2m+2}{2m+1} \rho ^2\). >From this, we can see that the divergence of \(R_\mathrm {od}\psi \) stems from the insufficient (cubic) decay of the integrand for large \(|\rho |\), as for \(R_\mathrm {d}\). For the analysis of this divergence, we can thus replace the \(\tan ^{-1}\) by its limit as \(\rho \rightarrow \infty \), which equals (note the \(\rho \)-dependence of \(\gamma \))

$$\begin{aligned} \lim _{\rho \rightarrow \infty }\tan ^{-1}\left( \frac{|\rho |}{\sqrt{2m(2m+1)}(\sqrt{\beta }+\sqrt{\gamma })}\right) = \tan ^{-1}\left( \frac{1}{2\sqrt{m(m+1)}}\right) . \end{aligned}$$

The asymptotics of the remaining \(\mathrm {d}\rho \)-integral can be evaluated as for \(R_\mathrm {d}\). One finds that \((R_\mathrm {od}\psi )(s,r)\) has the asymptotic behaviour

$$\begin{aligned} \frac{1}{(2\pi )^3} \left( \frac{2m}{2m+1}\right) ^2 2m \tan ^{-1}\left( \frac{1}{2\sqrt{m(m+1)}}\right) \log (r)\psi (s), \end{aligned}$$

with a convergent remainder, as \(r\rightarrow 0\). Bounds on the convergent part as an operator on \(H^\varepsilon \) are provided in Lemma 19. Consequently,

$$\begin{aligned} (R\psi )(s,r)= -\gamma _m\psi (s)\log (r) + {\mathcal {O}}(1), \end{aligned}$$

with a remainder that converges in \({\mathscr {H}}^{(n)}\) as \(r\rightarrow 0\).

This remains true with \(\psi (s)\) replaced by \(\psi (x)=\psi (s+\tfrac{1}{2m+1} r)\) after averaging \(\omega =(y-x)/\left|y-x \right|\), as in Eq. (22). This completes the proof for \(M=1\), \(n=0\).

For arbitrary M and n, the key observation is that although R is given in terms of sums corresponding to different combinations of creating and annihilating a particle on the planes \(\{x_\nu =y_i\}\), \(\nu \in \{1,\dots M\}\), \(i\in \{1,\dots , n\}\), only some of the contributions are actually singular. These behave in a similar way as for \(M=1\), \(n=0\) (see Lemma 18 for details).

We then have

$$\begin{aligned} A_T G_T\psi = T\psi +S\psi . \end{aligned}$$

Similarly to T, the operator S is a sum of real Fourier multipliers (in this case of logarithmic growth) and integral operators. Its symmetry on D(S) is shown in Lemma 20. \(\square \)

5 Proof of Theorem 1

We will now prove the self-adjointness of the operator

$$\begin{aligned} H =(1-G_T)^*(K+c_0) (1-G_T) -c_0 + S \end{aligned}$$

on the domain

$$\begin{aligned} D(H)=\left\{ \psi \in {\mathscr {H}} \Big \vert \psi - G_T\psi \in D(L) \right\} . \end{aligned}$$

Equality of this domain and the one given in Theorem 1 is shown in Eq. (27). The remaining statements of the theorem follow from (11) and (10) in view of the results of Sect. 3.

Lemma 14

The operator \(H_0=(1-G_T)^*K(1-G_T)\) is self-adjoint on D(H) and bounded from below.

Proof

This follows directly from the invertibility of \((1-G_T)\). \(\square \)

Lemma 15

Let (S, D(S)) be the symmetric operator on \({\mathscr {H}}^{(n)}\) defined in Proposition 13 and denote its extension to \({\mathscr {H}}\) by the same symbol. Then (S, D(S)) is infinitesimally \(H_0\)-bounded.

Proof

We decompose \(S\psi =SG_T\psi + S(1-G_T)\psi \) and estimate both terms separately. For the action of S on the singular part, \(G_T\psi \), we have by Lemma 19 and Proposition 12

$$\begin{aligned} \left\| SG_T \psi \right\| _{\mathscr {H}}\le C_\varepsilon \left\| L^\varepsilon G_T \psi \right\| _{\mathscr {H}}\le C'_\varepsilon \left\| \psi \right\| _{\mathscr {H}}, \end{aligned}$$

for any \(0<\varepsilon <\tfrac{1}{4}\). On the regular part, we have, again by Lemma 19,

$$\begin{aligned} \left\| S(1-G_T) \psi \right\| _{\mathscr {H}}&\le C_\varepsilon \left\| L^\varepsilon (1-G_T) \psi \right\| _{\mathscr {H}}\\&\le \delta \left\| L (1-G_T) \psi \right\| _{\mathscr {H}}+ C_{\delta } \left\| (1-G_T) \psi \right\| \\&\le \delta \Vert (1-G_T)^{-1} \Vert \left\| H_0 \psi \right\| _{\mathscr {H}}+ C_\delta \left\| (1-G_T) \right\| \left\| \psi \right\| _{\mathscr {H}}, \end{aligned}$$

for arbitrary \(\delta >0\). This proves the claim. \(\square \)

In view of Eq. (11), this proves that H is self-adjoint on D(H), by the Kato–Rellich theorem. It is also immediate that H is bounded from below.