1 Introduction

Uncertainty relations are fundamental in quantum mechanics, underlying many conceptual differences between classical and quantum theories. The Heisenberg–Robertson uncertainty relations [1] are expressed in terms of the product \(V_{\rho }(A)V_{\rho }(B)\) of the variances of the measurement results of the observables A and B, and the product can be null even when one of the two variances is different from zero. Here, we provide a different uncertainty relation, based on the sum \(V_{\rho }(A)+V_{\rho }(B)\), that is guaranteed to be nontrivial whenever the observables are incompatible on the state. Previous uncertainty relations that provide a bound to the sum of the variances comprise a lower bound in terms of the variance of the sum of observables [2], a lower bound based on the entropic uncertainty relations [3], a sum uncertainty relation for angular momentum observables [4], sum uncertainty relations for arbitrary N observables [5], a series of uncertainty inequalities in the qubit system and a state-independent bound for the sum of variances [6], a unified and exact framework for the variance-based uncertainty relations [7], a lower bound based on the Wigner–Yanase skew information or Wigner–Yanase–Dyson skew information uncertainty relations [8,9,10,11,12]. Uncertainty relations are useful in many areas related or even unrelated to quantum mechanics: entanglement detection [13, 14], quantum cryptography [15], signal processing [16], etc. Owing back to entanglement measure, Zidan’s model for quantum computing [17,18,19,20] was discovered. This model was used to solve an extended version of the Deutsch–Jozsa algorithm. This extension was intractable for more than 27 years using the quantum circuit model [18]. So, developing uncertainty relations could be helpful to produce new quantum technologies.

The quantum mechanical uncertainties associated with observables A and B in the state \(|\psi \rangle \) are defined via \(\varDelta A^2=\langle \psi |A^2|\psi \rangle -\langle \psi |A|\psi \rangle ^2\) and \(\varDelta B^2=\langle \psi |B^2|\psi \rangle -\langle \psi |B|\psi \rangle ^2.\) Similarly, we can define the uncertainty in the sum of two observables as \(\varDelta (A+B)^2=\langle \psi |(A+B)^2|\psi \rangle -\langle \psi |(A+B)|\psi \rangle ^2.\) The expectation value of A is defined as \(\langle A\rangle =\langle \psi |A|\psi \rangle .\) The Heisenberg–Robertson uncertainty relation [1] bounds the product of the variances through the expectation value of the commutator

$$\begin{aligned} \varDelta A^2\varDelta B^2\ge \left| \frac{1}{2}\langle [A,B]\rangle \right| ^2. \end{aligned}$$
(1)

It was strengthened by Schr\(\ddot{\mathrm {o}}\)dinger [21], obtaining

$$\begin{aligned} \varDelta A^2\varDelta B^2\ge \left| \frac{1}{2}\langle [A,B]\rangle \right| ^2+\left| \frac{1}{2}\langle \{A,B\}\rangle -\langle A\rangle \langle B\rangle \right| ^2. \end{aligned}$$
(2)

In [22], Maccone and Pati gave two stronger uncertainty relations, relating to the sum of variances with respect to pure state, whose lower bound is guaranteed to be nontrivial whenever the two observables are incompatible on the state of the system. The following two inequalities have lower bounds which are nontrivial. The first inequality is

$$\begin{aligned} \varDelta A^2+\varDelta B^2\ge \pm \mathrm {i}\langle [A,B]\rangle +\left| \langle \psi |A\pm \mathrm {i}B|\psi ^{\bot }\rangle \right| ^2, \end{aligned}$$
(3)

which is valid for arbitrary states \(|\psi ^{\bot }\rangle \) orthogonal to the state of the system \(|\psi \rangle \).

A second inequality with nontrivial bound even if \(|\psi \rangle \) is an eigenstate either of A or of B is

$$\begin{aligned} \varDelta A^2+\varDelta B^2\ge \frac{1}{2}\left| \langle \psi _{A+B}^{\bot }|A+B|\psi \rangle \right| ^2, \end{aligned}$$
(4)

where \(|\psi _{A+B}^{\bot }\rangle =(A+B-\langle A+B\rangle )|\psi \rangle /N\) is a state orthogonal to the state of the system \(|\psi \rangle \). One can also obtain an amended Heisenberg–Robertson inequality

$$\begin{aligned} \varDelta A\varDelta B\ge \pm \mathrm {i}\langle [A,B]\rangle \Bigg /\left( 1-\frac{1}{2}\left| \left\langle \psi \left| \frac{A}{\varDelta A}\pm \mathrm {i} \frac{B}{\varDelta B}\right| \psi ^\bot \right\rangle \right| ^2\right) , \end{aligned}$$
(5)

which reduces to (1) when minimizing the lower bound over \(|\psi ^\bot \rangle \) and becomes an equality when maximizing it.

In [25], the authors prove that the following stronger inequality exists:

$$\begin{aligned}&\varDelta A^2+\varDelta B^2+\varDelta C^2\ge \frac{1}{3}\varDelta (A+B+C)^2\pm \mathrm {i}\frac{\sqrt{3}}{3}\left\langle [A,B,C]\right\rangle \nonumber \\&\quad +\frac{2}{3}\left| \left\langle \psi \left| \left( A+ e^{\pm \mathrm {i}\frac{2\pi }{3}}B+e^{\pm \mathrm {i}\frac{4\pi }{3}}C\right) \right| \psi ^{\bot }\right\rangle \right| ^2, \end{aligned}$$
(6)

which is valid for arbitrary states \(\left| \psi ^{\bot }\right\rangle \) orthogonal to the state of the system \(|\psi \rangle ,\) where \(\langle [A,B,C]\rangle =\langle [A,B]\rangle +\langle [B,C]\rangle +\langle [C,A]\rangle \) and the sign should be chosen so that \(\pm \mathrm {i}\frac{\sqrt{3}}{3}\langle [A,B,C]\rangle \) (a real quantity) is positive.

For a quantum state \(\rho \) and observables A and B, the Heisenberg uncertainty relation was expressed as follows:

$$\begin{aligned} V_\rho (A)V_\rho (B)\ge \frac{1}{4}\mid \mathrm {Tr}(\rho [A,B])\mid ^2. \end{aligned}$$
(7)

The further stronger result was given by Schrödinger:

$$\begin{aligned} V_\rho (A)V_\rho (B)-|\mathrm {Re}\{\mathrm {Cov}_\rho (A,B)\}|^2\ge \frac{1}{4}\mid \mathrm {Tr}(\rho [A,B])\mid ^2, \end{aligned}$$
(8)

where the covariance is defined by \(\mathrm {Cov}_\rho (A,B)=\mathrm {Tr}\big [\rho (A-(\mathrm {Tr}\rho A)I)(B-(\mathrm {Tr}\rho B)I)\big ],\) and \(V_\rho (A)=\mathrm {Cov}_\rho (A,A).\)

In this paper, we will give two stronger uncertainty relations, relating to the sum of variances with respect to density matrix, whose lower bound is guaranteed to be nontrivial whenever the two observables are incompatible on the state of the system; moreover, two stronger uncertainty relations in terms of the product of the variances of two observables will be established in Sect. 2. Also, several stronger uncertainty relations for three observables will be established in Sect. 3.

2 Stronger uncertainty relations for two observables

Let H be a separable complex Hilbert space with the inner product \((\cdot ,\cdot )\), and B(H) the algebra of all bounded linear operators on H. The set of all trace-class operators on H is denoted by \(L^1(H)\). Recall that an operator \(A\in B(H)\) is said to be a Hilbert–Schmidt operator if

$$\begin{aligned} \Vert A\Vert _2:=\left( \sum _{n\in I}\langle e_n|A^*A|e_n\rangle \right) ^{1/2}<\infty \end{aligned}$$

for some orthonormal basis \(\{|e_n\rangle \}_{n\in I}\) for H. The set of all Hilbert–Schmidt operators on H is denoted by \(L^2(H)\). It is well known that both \(L^1(H)\) and \(L^2(H)\) are self-ideals of the \(C^*\)-algebra B(H), and the product of two Hilbert–Schmidt operators on H is a trace class on H. The Hilbert–Schmidt inner product \(\langle A,B\rangle :=\mathrm {Tr} A^{*}B\).

In what follows, for an operator \(A\in B(H)\), the adjoint of A is denoted by \(A^*\). An operator \(A\in B(H)\) is said to be self-adjoint if \(A=A^*\). The set of all self-adjoint operators on H is denoted by S(H). A state is given by a positive operator \(\rho \) of trace 1, called a density operator. The set of all states is denoted by D(H).

For a mixed state \(\rho \) with the spectral decomposition \(\rho =\sum _i\lambda _i|\psi _i\rangle \langle \psi _i|,\) denote \(\varDelta A_i^2=\langle \psi _i|A^2|\psi _i\rangle -\langle \psi _i|A|\psi _i\rangle ^2\), \(\varDelta B_i^2=\langle \psi _i|B^2|\psi _i\rangle -\langle \psi _i|B|\psi _i\rangle ^2\). As is well known, the variance \(V_{\rho }(A)\) is a concave function with respect to \(\rho ,\) which is the following lemma.

Lemma 1

[23] Let \(\rho _i\in D(H), \forall i, A\in S(H).\) Then,

$$\begin{aligned} V_{\sum _ip_i\rho _i}(A)\ge \sum _ip_iV_{\rho _i}(A), \end{aligned}$$

where \(\sum _ip_i=1, p_i\ge 0,\forall i.\)

Theorem 1

Let \(\rho \in D(H),\) and \(\rho =\sum _i\lambda _i|\psi _i\rangle \langle \psi _i|\). If there exists \(|\psi ^\bot \rangle \) such that \(\langle \psi ^\bot |\psi _i\rangle =0,\forall i\), then

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \pm \mathrm {i}\mathrm {Tr}(\rho [A,B])+\left\| \rho ^\frac{1}{2}(A\pm \mathrm {i}B)|\psi ^{\bot }\rangle \right\| _2^2. \end{aligned}$$

Proof

We see from (3) that

$$\begin{aligned} \varDelta A_i^2+\varDelta B_i^2\ge & {} \pm \mathrm {i}\langle [A,B]\rangle +\left| \langle \psi _i|A\pm \mathrm {i}B|\psi ^{\bot }\rangle \right| ^2\\= & {} \pm \mathrm {i}\mathrm {Tr}([A,B]|\psi _i\rangle \langle \psi _i|)+\mathrm {Tr}\left( (A\pm \mathrm {i}B)|\psi ^{\bot }\rangle \langle \psi ^{\bot }|(A\mp \mathrm {i}B)|\psi _i\rangle \langle \psi _i|\right) . \end{aligned}$$

By multiplying both members by \(\lambda _i\) and summing over i, we obtain the mixed-state extension of (3):

$$\begin{aligned} \sum _i\lambda _i\varDelta A_i^2+\sum _i\lambda _i\varDelta B_i^2\ge \pm \mathrm {i}\mathrm {Tr}([A,B]\rho )+\mathrm {Tr}\left( (A\pm \mathrm {i}B)|\psi ^{\bot }\rangle \langle \psi ^{\bot }|(A\mp \mathrm {i}B)\rho \right) . \end{aligned}$$

By the concavity of the variance, we get \(V_\rho (A)\ge \sum _i\lambda _i\varDelta A_i^2,V_\rho (B)\ge \sum _i\lambda _i\varDelta B_i^2.\) Hence, we obtain the conclusion. \(\square \)

Write \(O(H)=\{\rho \in D(H)|\exists \rho _{\bot }\in D(H), \mathrm {s.t.}\ \rho \rho _{\bot }=0\}.\)

Theorem 2

Let \(\rho \in O(H), A,B\in S(H),\) and \(\exists \ \rho _{\bot }\in D(H),\mathrm {s.t.}\ \rho \rho _{\bot }=0.\) Then,

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \pm \mathrm {i}\mathrm {Tr}(\rho [A,B])+\left\| \rho ^\frac{1}{2}(A\pm \mathrm {i}B)\rho _{\bot }^\frac{1}{2}\right\| _2^2. \end{aligned}$$

Proof

Denote \(C=A-\mathrm {Tr}(\rho A), D=B-\mathrm {Tr}(\rho B).\) So \(\Vert C\rho ^{\frac{1}{2}}\Vert _2^2=\mathrm {Tr}(\rho C^2)=V_{\rho }(A), \) and \(\Vert \mathrm {i}D\rho ^{\frac{1}{2}}\Vert _2^2=V_{\rho }(B).\) Thus,

$$\begin{aligned} \left\| (C\mp \mathrm {i} D)\rho ^{\frac{1}{2}}\right\| _2^2= & {} \left\langle C \rho ^{\frac{1}{2}}\mp \mathrm {i} D\rho ^{\frac{1}{2}},C \rho ^{\frac{1}{2}}\mp \mathrm {i} D\rho ^{\frac{1}{2}}\right\rangle \\= & {} \left\| C\rho ^{\frac{1}{2}}\right\| _2^2+\left\| \mathrm {i}D\rho ^{\frac{1}{2}}\right\| _2^2\mp \mathrm {i}\left\langle C \rho ^{\frac{1}{2}},D \rho ^{\frac{1}{2}}\right\rangle \pm \mathrm {i}\left\langle D \rho ^{\frac{1}{2}},C \rho ^{\frac{1}{2}}\right\rangle \\= & {} \left\| C\rho ^{\frac{1}{2}}\right\| _2^2+\left\| \mathrm {i}D\rho ^{\frac{1}{2}}\right\| _2^2\mp \mathrm {i}\mathrm {Tr}(\rho [C,D])\\= & {} V_{\rho }(A)+V_{\rho }(B)\mp \mathrm {i}\mathrm {Tr}(\rho [A,B]). \end{aligned}$$

On the other hand, since \(\rho \rho _{\bot }=0,\)

$$\begin{aligned} \left\| \rho ^\frac{1}{2}(A\pm \mathrm {i}B)\rho _{\bot }^\frac{1}{2}\right\| _2^2= & {} \left\| \rho ^\frac{1}{2}(C\pm \mathrm {i}D)\rho _{\bot }^\frac{1}{2}\right\| _2^2\\\le & {} \left\| \rho ^\frac{1}{2}(C\pm \mathrm {i}D)\right\| _2^2\cdot \left\| \rho _{\bot }^\frac{1}{2}\right\| ^2\\\le & {} \left\| \rho ^\frac{1}{2}(C\pm \mathrm {i}D)\right\| _2^2=\left\| (C\mp \mathrm {i} D)\rho ^{\frac{1}{2}}\right\| _2^2. \end{aligned}$$

Therefore, we obtain the conclusion. \(\square \)

It is easy to compute that \(V_\rho (A)+V_\rho (B)\mp \mathrm {i}\mathrm {Tr}(\rho [A,B])=\mathrm {Cov}_{\rho }(A\pm \mathrm {i}B, A\mp \mathrm {i}B).\) Thus, by the use of Theorem 2, we can get the following corollary.

Corollary 1

Let \(\rho \in O(H), A,B\in S(H),\) and \(\exists \ \rho _{\bot }\in D(H), \mathrm {s.t.}\ \rho \rho _{\bot }=0.\) Then,

$$\begin{aligned} \mathrm {Cov}_{\rho }(A\pm \mathrm {i}B, A\mp \mathrm {i}B)\ge \left\| \rho ^\frac{1}{2}(A\pm \mathrm {i}B)\rho _{\bot }^\frac{1}{2}\right\| _2^2. \end{aligned}$$

Remark 1

If we take \(\rho =|\psi \rangle \langle \psi |,\rho _\bot =\left| \psi ^\bot \right\rangle \left\langle \psi ^\bot \right| \) , then the inequality in Theorem 2 can degenerate to (3).

Theorem 3

Let \(\rho \in O(H), A,B\in S(H),\) and \(\exists \ \rho _{\bot }\in D(H), \mathrm {s.t.}\ \rho \rho _{\bot }=0.\) Then,

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)\rho _{\bot }^\frac{1}{2}\right\| _2^2. \end{aligned}$$

Proof

Denote \(C=A-\mathrm {Tr}(\rho A), D=B-\mathrm {Tr}(\rho B).\) So \(\left\| C\rho ^{\frac{1}{2}}\right\| _2^2=V_{\rho }(A), \) and \(\left\| D\rho ^{\frac{1}{2}}\right\| _2^2=V_{\rho }(B).\) Using the parallelogram law in Hilbert space, we have

$$\begin{aligned} \left\| C\rho ^{\frac{1}{2}}\right\| _2^2+\left\| D\rho ^{\frac{1}{2}}\right\| _2^2=\frac{1}{2}\left[ \left\| (C+D)\rho ^{\frac{1}{2}}\right\| _2^2 +\left\| (C-D)\rho ^{\frac{1}{2}}\right\| _2^2\right] \ge \frac{1}{2}\left\| (C+D)\rho ^{\frac{1}{2}}\right\| _2^2. \end{aligned}$$

On the other hand, since \(\rho \rho _{\bot }=0,\)

$$\begin{aligned} \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)\rho _{\bot }^\frac{1}{2}\right\| _2^2= & {} \frac{1}{2}\left\| \rho ^\frac{1}{2}(C+D)\rho _{\bot }^\frac{1}{2}\right\| _2^2\\\le & {} \frac{1}{2}\left\| \rho ^\frac{1}{2}(C+D)\right\| _2^2\cdot \left\| \rho _{\bot }^\frac{1}{2}\right\| ^2\\\le & {} \frac{1}{2}\left\| \rho ^\frac{1}{2}(C+D)\right\| _2^2. \end{aligned}$$

Therefore, we obtain the conclusion. \(\square \)

Corollary 2

Let \(|\psi \rangle ,\left| \psi ^\bot \right\rangle \) be orthogonal unit vectors on H, \(A,B\in S(H)\). Then,

$$\begin{aligned} \varDelta A^2+\varDelta B^2\ge \frac{1}{2}\left| \langle \psi ^{\bot }|(A+B)|\psi \rangle \right| ^2. \end{aligned}$$
(9)

Proof

Take \(\rho =|\psi \rangle \langle \psi |,\rho _\bot =|\psi ^\bot \rangle \langle \psi ^\bot |\) in Theorem 3, then \(V_\rho (A)=\varDelta A^2,\) \(V_\rho (B)=\varDelta B^2,\) and

$$\begin{aligned} \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)\rho _{\bot }^\frac{1}{2}\right\| _2^2= \frac{1}{2}\left| \langle \psi ^{\bot }|(A+B)|\psi \rangle \right| ^2. \end{aligned}$$

Therefore, using Theorem 3, we get the conclusion. \(\square \)

In Corollary 2, if we choose \(|\psi ^{\bot }\rangle =\left| \psi _{A+B}^{\bot }\right\rangle \), then the inequality (9) degenerates to (4).

Corollary 3

Let \(\rho \in D(H),\) and \(\rho =\sum _i\lambda _i|\psi _i\rangle \langle \psi _i|.\) If there exists \(|\psi ^\bot \rangle \) such that \(\langle \psi ^\bot |\psi _i\rangle =0,\forall i\), then

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)|\psi ^{\bot }\rangle \right\| _2^2. \end{aligned}$$

Proof

Put \(\rho _{\bot }=|\psi ^\bot \rangle \langle \psi ^\bot |\), then \( \rho \rho _{\bot }=0.\) By the use of Theorem 3, we have

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)\rho _{\bot }^\frac{1}{2}\right\| _2^2 =\frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)|\psi ^{\bot }\rangle \right\| _2^2. \end{aligned}$$

This completes the proof. \(\square \)

Proposition 1

Let \(\rho \in D(H),A,B\in S(H).\) Then,

$$\begin{aligned} \sqrt{V_{\rho }(A+B)}\le \sqrt{V_{\rho }(A)}+\sqrt{V_{\rho }(B)}. \end{aligned}$$

Proof

Denote \(C=A-\mathrm {Tr}(\rho A), D=B-\mathrm {Tr}(\rho B).\) So \(\left\| C\rho ^{\frac{1}{2}}\right\| _2^2=V_{\rho }(A), \) and \(\left\| D\rho ^{\frac{1}{2}}\right\| _2^2=V_{\rho }(B).\) Thus,

$$\begin{aligned} V_\rho (A+B)=\left\| (C+D)\rho ^{\frac{1}{2}}\right\| _2^2= & {} \left\| C\rho ^{\frac{1}{2}}\right\| _2^2+\left\| D\rho ^{\frac{1}{2}}\right\| _2^2+2\mathrm {Re}{\left\langle C\rho ^{\frac{1}{2}},D\rho ^{\frac{1}{2}}\right\rangle }\\\le & {} \left\| C\rho ^{\frac{1}{2}}\right\| _2^2+\left\| D\rho ^{\frac{1}{2}}\right\| _2^2+2\left| \left\langle C\rho ^{\frac{1}{2}},D\rho ^{\frac{1}{2}}\right\rangle \right| . \end{aligned}$$

Further using Schwarz inequality, we have

$$\begin{aligned} V_\rho (A+B)\le & {} V_\rho (A)+V_\rho (B)+2\left\| C\rho ^{\frac{1}{2}}\right\| _2^{\frac{1}{2}}\cdot \left\| D\rho ^{\frac{1}{2}}\right\| _2^{\frac{1}{2}}\\= & {} V_\rho (A)+V_\rho (B)+2\sqrt{V_\rho (A)}\sqrt{V_\rho (B)}\\= & {} \left( \sqrt{V_\rho (A)}+\sqrt{V_\rho (B)}\right) ^2, \end{aligned}$$

which yields the conclusion. \(\square \)

In general, for N observables \(A_1,A_2,\cdots ,A_N\) and mixed state \(\rho \), we have

$$\begin{aligned} \sqrt{V_\rho \left( \sum _{i=1}^NA_i\right) }\le \sum _{i=1}^N\sqrt{V_{\rho }(A_i)}. \end{aligned}$$

Holevo derived the following useful relation [24]:

$$\begin{aligned} \sqrt{V_{\rho }(A)}+\sqrt{V_{\rho '}(A)}\ge \left| \left( E_{\rho }(A)-E_{\rho '}(A)\right) \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\rho '^{\frac{1}{2}}\right) \right| \Big /\sqrt{2\left( 1-\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\rho '^{\frac{1}{2}}\right) \right) }, \end{aligned}$$
(10)

where \(E_{\rho }(A)=\mathrm {Tr}(\rho A), E_{\rho '}(A)=\mathrm {Tr}(\rho ' A)\) are the expectation values of A on the states \(\rho \) and \(\rho '\), respectively. By using the square-modulus inequality and following a procedure analogous to the one employed by Holevo to derive the inequality (10), we can get the following relation.

Lemma 2

Let \(\rho \in D(H), \varphi \in (-\frac{\pi }{2},\frac{\pi }{2}), \) and \(\lambda \) be any real number. If \(\sigma \in L^2(H)\) satisfies \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0,\) \(\Vert \sigma \Vert _2=1\), and put

$$\begin{aligned} A_{\varphi }=\cos \varphi \cdot \rho ^{\frac{1}{2}}+e^{\mathrm {i}\lambda }\sin \varphi \cdot \sigma , \,\rho '=A_{\varphi }A_{\varphi }^*, \end{aligned}$$
(11)

then

  1. (i)

    \(\rho '\) is a state.

  2. (ii)

    \(\mathrm {Tr}\left( A_{\varphi }^*A_{\varphi }\right) =1.\)

  3. (iii)

    \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}A_{\varphi }\right) =\cos \varphi .\)

  4. (iv)

    In the sense of \(\Vert \cdot \Vert _2,\) we have \(\lim \limits _{\varphi \rightarrow 0}A_{\varphi }=\rho ^{\frac{1}{2}},\) \(\lim \limits _{\varphi \rightarrow 0}A_{\varphi }^{*}=\rho ^{\frac{1}{2}},\) and \(\lim \limits _{\varphi \rightarrow 0}\rho '=\rho .\)

Theorem 4

Let \(\rho \in D(H), A,B\in S(H).\) If \(\sigma \in L^2(H)\) satisfies \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\), then

$$\begin{aligned} \sqrt{V_{\rho }(A)V_{\rho }(B)}\ge \pm \frac{\mathrm {i}}{2}\mathrm {Tr}(\rho [A,B])\Bigg /\left[ 1-\frac{1}{2} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2\right] . \end{aligned}$$
(12)

Proof

Let Eq. (11) be valid. Start from the following inequality

$$\begin{aligned} \left\| c_A(A-a)\rho ^{\frac{1}{2}}\pm \mathrm {i} c_B(B-b')A_{\varphi }+c\left( \rho ^{\frac{1}{2}}-A_{\varphi }\right) \right\| _2^2\ge 0, \end{aligned}$$
(13)

where \(b'=\mathrm {Tr}(\rho 'B), a=\mathrm {Tr}(\rho A),\) \(c_A, c_B,c\) real constants. Calculating the square modulus, we find

$$\begin{aligned} c_A^2 V_{\rho }(A)+c_B^2 V_{\rho '}(B)\ge -c^2r-c_Ac_Bc\delta \pm \mathrm {i}c_Ac_Bk, \end{aligned}$$
(14)

with \(V_{\rho }(A)\) and \(V_{\rho '}(B)\) the variance of A and B on \(\rho \) and \(\rho '\), respectively, and where

$$\begin{aligned}&r=2\left( 1-\cos \varphi \right) , \,\,\, k=2\mathrm {i}\mathrm {Im}\left\{ \mathrm {Tr}\left( A_{\varphi }^{*}(B-b')(A-a)\rho ^{\frac{1}{2}}\right) \right\} ,\\&\delta =2\mathrm {Re}\left\{ {\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{a-A}{c_B}\pm \mathrm {i}\frac{B-b'}{c_A}\right) A_{\varphi }\right) }\right\} . \end{aligned}$$

Now choose the value of c that maximizes the right-hand side of (14), namely \(c=-\frac{c_Ac_B\delta }{2r}\), then (14) becomes

$$\begin{aligned} c_A^2 V_{\rho }(A)+c_B^2 V_{\rho '}(B)\ge \frac{(c_Ac_B\delta )^2}{4r}\pm \mathrm {i}c_Ac_Bk. \end{aligned}$$
(15)

Put \(c_A=\sqrt{ V_{\rho '}(B)},c_B=-\sqrt{ V_{\rho }(A)},\) then

$$\begin{aligned} \sqrt{ V_{\rho }(A)V_{\rho '}(B)}\ge \frac{\sqrt{ V_{\rho }(A)V_{\rho '}(B)}\delta ^2}{8r}\mp \frac{\mathrm {i}}{2}k. \end{aligned}$$
(16)

Denote \(b=\mathrm {Tr}(\rho B)\), and by the use of Lemma 2, we have

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}k= & {} \lim \limits _{\varphi \rightarrow 0}2\mathrm {i}\mathrm {Im}\left\{ \mathrm {Tr}\left( A_{\varphi }^{*}(B-b')(A-a)\rho ^{\frac{1}{2}}\right) \right\} \\= & {} 2\mathrm {i}\mathrm {Im}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}(B-b)(A-a)\rho ^{\frac{1}{2}}\right) \right\} \\= & {} 2\mathrm {i}\mathrm {Im}\left\{ \mathrm {Tr}\left( \rho (B-b)(A-a)\right) \right\} \\= & {} 2\mathrm {i}\mathrm {Im}\{\mathrm {Cov}_{\rho }(B,A)\}\\= & {} \mathrm {Cov}_{\rho }(B,A)-\mathrm {Cov}_{\rho }(A,B)\\= & {} -\mathrm {Tr}(\rho [A,B]), \end{aligned}$$

and \(\lim \limits _{\varphi \rightarrow 0}V_{\rho '}(B)=V_{\rho }(B)\). In the following, we compute the limit of \(\frac{\delta ^2}{8r}.\)

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}\frac{\delta ^2}{8r}= & {} \lim \limits _{\varphi \rightarrow 0}\frac{\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{ A-a}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B-b'}{\sqrt{V_{\rho '}(B)}}\right) A_{\varphi }\right) \right\} \right) ^2}{4\left( 1-\cos \varphi \right) }\\= & {} \lim \limits _{\varphi \rightarrow 0}\frac{\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A-a}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B-b}{\sqrt{V_{\rho }(B)}}\right) A_{\varphi }\right) \right\} \right) ^2}{4\left( 1-\cos \varphi \right) }. \end{aligned}$$

Denote \(D=\frac{A-a}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B-b}{\sqrt{V_{\rho }(B)}},\) then

$$\begin{aligned} \frac{\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A-a}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B-b}{\sqrt{V_{\rho }(B)}}\right) A_{\varphi }\right) \right\} \right) ^2}{4\left( 1-\cos \varphi \right) } =\frac{\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}DA_{\varphi }\right) \right\} \right) ^2}{4\left( 1-\cos \varphi \right) }. \end{aligned}$$

Note that

$$\begin{aligned} \mathrm {Tr}\left( \rho ^{\frac{1}{2}}DA_{\varphi }\right) =\mathrm {Tr}\left( \rho ^{\frac{1}{2}}D(\cos \varphi \cdot \rho ^{\frac{1}{2}} +e^{\mathrm {i}\lambda }\sin \varphi \cdot \sigma )\right) =e^{\mathrm {i}\lambda }\sin \varphi \mathrm {Tr}\left( \rho ^{\frac{1}{2}}D\sigma \right) . \end{aligned}$$

We can choose appropriate \(\lambda \) so that the term \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}DA_{\varphi }\right) \) is real. Therefore, \(\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}DA_{\varphi }\right) \right\} \right) ^2= \sin ^2\varphi \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}D\sigma \right) \right| ^2,\) which yields

$$\begin{aligned} \frac{\left( \mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}DA_{\varphi }\right) \right\} \right) ^2}{4\left( 1-\cos \varphi \right) }= \frac{\sin ^2\varphi \cdot \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}D\sigma \right) \right| ^2}{4(1-\cos \varphi )}, \end{aligned}$$

and so

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}\frac{\delta ^2}{8r}= & {} \frac{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}D\sigma \right) \right| ^2}{2} =\frac{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A-a}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B-b}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}{2}\\= & {} \frac{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}{2}. \end{aligned}$$

The last equality holds because of \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0.\) Hence, the inequality (16) becomes

$$\begin{aligned} \sqrt{ V_{\rho }(A)V_{\rho }(B)}\ge \sqrt{ V_{\rho }(A)V_{\rho }(B)}\cdot \frac{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}{2}\pm \frac{\mathrm {i}}{2}\mathrm {Tr}(\rho [A,B]), \end{aligned}$$

which is equivalent to the conclusion. \(\square \)

Remark 2

If we take \(\rho =|\psi \rangle \langle \psi |, \sigma =|\psi ^\bot \rangle \langle \psi |\), where \(\langle \psi |\psi ^\bot \rangle =0,\) then \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\). It is easy to compute \(V_\rho (A)=\varDelta A^2\), \(V_\rho (B)=\varDelta B^2\), and \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) =\left\langle \psi |\left( \frac{A}{\varDelta A}\pm \mathrm {i} \frac{B}{\varDelta B}\right) |\psi ^\bot \right\rangle ,\) which recover to the inequality (5).

Remark 3

Let

$$\begin{aligned} \rho = \left( \begin{array}{cc} \frac{1}{4} &{} 0\\ 0&{} \frac{3}{4} \\ \end{array}\right) , A= \left( \begin{array}{cc} 0 &{} \mathrm {i} \\ -\mathrm {i} &{}0 \\ \end{array}\right) , B= \left( \begin{array}{cc} 0 &{} 1 \\ 1 &{} 0 \\ \end{array}\right) , \sigma = \left( \begin{array}{cc} \frac{\sqrt{3}}{2} &{} 0 \\ 0 &{} -\frac{1}{2}\\ \end{array}\right) . \end{aligned}$$

Then, \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\). It is easy to compute that \(\sqrt{V_{\rho }(A)V_{\rho }(B)}=1,\) and the right of (12)

$$\begin{aligned} \pm \frac{\mathrm {i}}{2}\mathrm {Tr}(\rho [A,B])\Bigg /\left[ 1-\frac{1}{2} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2\right] =\pm \frac{16}{29}. \end{aligned}$$

Thus, the inequality (12) holds.

In the following, we will give the other improved Schrödinger uncertainty relation, by choosing an arbitrary phase factor \(e^{\mathrm {i}\tau }\) in place of the imaginary constant \(\mathrm {i}\) in (13).

Theorem 5

Let \(\rho \in D(H), A,B\in S(H).\) If \(\sigma \in L^2(H)\) satisfies \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\), then

$$\begin{aligned} V_{\rho }(A)V_{\rho }(B)\ge \left| \mathrm {Cov}_{\rho }(A,B)\right| ^2\Bigg /\left[ 1-\frac{1}{2} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}+e^{\mathrm {i}\alpha }\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2\right] ^2, \end{aligned}$$
(17)

where \(\alpha \) satisfies \(e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B)=|\mathrm {Cov}_{\rho }(A,B)|.\)

Proof

Let Eq. (11) be valid. Start from the following inequality

$$\begin{aligned} \left\| c_A(A-a)\rho ^{\frac{1}{2}}-c_B e^{\mathrm {i}\tau }(B-b')A_{\varphi }+c\left( \rho ^{\frac{1}{2}}-A_{\varphi }\right) \right\| _2^2\ge 0, \end{aligned}$$
(18)

where \(b'=\mathrm {Tr}(\rho 'B), a=\mathrm {Tr}(\rho A)\) and \(c_A, c_B,c\) real constants. Calculating the square modulus, we find

$$\begin{aligned} c_A^2 V_{\rho }(A)+c_B^2 V_{\rho '}(B)\ge -c^2r+c c_Ac_B\delta +c_Ac_Bk, \end{aligned}$$
(19)

with \(V_{\rho }(A)\) and \(V_{\rho '}(B)\) the variance of A and B on \(\rho \) and \(\rho '\), respectively, and where

$$\begin{aligned} r= & {} 2\left( 1-\cos \varphi \right) ,\\ k= & {} 2\mathrm {Re}\left\{ e^{\mathrm {i}\tau }\mathrm {Tr}\left( \rho ^{\frac{1}{2}}(A-a)(B-b')A_{\varphi }\right) \right\} ,\\ \delta= & {} 2\mathrm {Re}\left\{ {\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A-a}{c_B}+e^{\mathrm {i}\tau }\frac{B-b'}{c_A}\right) A_{\varphi }\right) }\right\} . \end{aligned}$$

Now choose the value of c that maximizes the right-hand side of (19), namely \(c=\frac{c_Ac_B\delta }{2r}\), then (19) becomes

$$\begin{aligned} c_A^2 V_{\rho }(A)+c_B^2 V_{\rho '}(B)\ge \frac{(c_Ac_B\delta )^2}{4r}+c_Ac_Bk. \end{aligned}$$
(20)

Put \(c_A=\sqrt{ V_{\rho '}(B)},c_B=\sqrt{ V_{\rho }(A)},\) then

$$\begin{aligned} \sqrt{V_{\rho }(A)V_{\rho '}(B)}\ge \frac{\sqrt{ V_{\rho }(A)V_{\rho '}(B)}\delta ^2}{8r}+\frac{k}{2}. \end{aligned}$$
(21)

Denote \(b=\mathrm {Tr}(\rho B)\), and by the use of Lemma 2, we get

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}\frac{k}{2}= & {} \lim \limits _{\varphi \rightarrow 0}\mathrm {Re}\left\{ e^{\mathrm {i}\tau }\mathrm {Tr}\left( \rho ^{\frac{1}{2}}(A-a)(B-b')A_{\varphi }\right) \right\} \\= & {} \mathrm {Re}\left\{ e^{\mathrm {i}\tau }\mathrm {Cov}_{\rho }(A,B)\right\} , \end{aligned}$$

and \(\lim \limits _{\varphi \rightarrow 0}V_{\rho '}(B)=V_{\rho }(B)\). In a similar way to the proof of Theorem 4, it is easy to obtain that

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}\frac{\delta ^2}{8r} =\frac{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}+e^{\mathrm {i}\tau }\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}{2}. \end{aligned}$$

We can choose appropriate \(\tau =\alpha \), so that \(e^{\mathrm {i}\tau }\mathrm {Cov}_{\rho }(A,B)\) is real and can be written as \(|\mathrm {Cov}_{\rho }(A,B)|.\) Thus, (21) can be rewritten as

$$\begin{aligned} \sqrt{V_{\rho }(A)V_{\rho }(B)}\ge \frac{|\mathrm {Cov}_{\rho }(A,B)|}{1-\frac{1}{2}\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}} +e^{\mathrm {i}\alpha }\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}. \end{aligned}$$

This completes the proof. \(\square \)

Note that

$$\begin{aligned} \mathrm {Cov}_{\rho }(A,B)= & {} \mathrm {Tr}(\rho AB)-\mathrm {Tr}(\rho A)\mathrm {Tr}(\rho B)\\= & {} \frac{1}{2}\mathrm {Tr}(\rho [A,B])+\frac{1}{2}\mathrm {Tr}(\rho \{A,B\})-\mathrm {Tr}(\rho A)\mathrm {Tr}(\rho B). \end{aligned}$$

Therefore, the inequality (17) can be represented as

$$\begin{aligned} \sqrt{V_{\rho }(A)V_{\rho }(B)}\ge \frac{\left| \frac{1}{2}\mathrm {Tr}(\rho [A,B])+\frac{1}{2}\mathrm {Tr}(\rho \{A,B\})-\mathrm {Tr}(\rho A)\mathrm {Tr}(\rho B)\right| }{1-\frac{1}{2}\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}} +e^{\mathrm {i}\alpha }\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}. \end{aligned}$$

Take \(c_A=c_B=1\) in (20); using the same procedure described above, we can obtain

$$\begin{aligned} V_{\rho }(A)+V_{\rho }(B)\ge & {} 2\left| \mathrm {Cov}_{\rho }(A,B)\right| +\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A+e^{\mathrm {i}\alpha }B\right) \sigma \right) \right| ^2\\= & {} \left| \mathrm {Tr}\left( \rho [A,B]\right) +\mathrm {Tr}\left( \rho \{A,B\}\right) -2\mathrm {Tr}(\rho A)\mathrm {Tr}(\rho B)\right| \\&+\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}( A+e^{\mathrm {i}\alpha }B)\sigma \right) \right| ^2. \end{aligned}$$

Removing the last term in the above inequality, we find the inequality

$$\begin{aligned} V_{\rho }(A)+V_{\rho }(B)\ge \left| \mathrm {Tr}\left( \rho [A,B]\right) +\mathrm {Tr}\left( \rho \{A,B\}\right) -2\mathrm {Tr}(\rho A)\mathrm {Tr}(\rho B)\right| . \end{aligned}$$
(22)

3 Uncertainty relations for three observables

One may generalize the Schrödinger uncertainty relation to three observables trivially, since

$$\begin{aligned}&V_{\rho }(A)+V_{\rho }(B)\ge 2\left| \mathrm {Cov_{\rho }(A,B)}\right| ,\\&V_{\rho }(B)+V_{\rho }(C)\ge 2\left| \mathrm {Cov_{\rho }(B,C)}\right| ,\\&V_{\rho }(A)+V_{\rho }(C)\ge 2\left| \mathrm {Cov_{\rho }(A,C)}\right| , \end{aligned}$$

we have

$$\begin{aligned} V_{\rho }(A)+V_{\rho }(B)+V_{\rho }(C)\ge \left| \mathrm {Cov_{\rho }(A,B)}\right| +\left| \mathrm {Cov_{\rho }(B,C)}\right| +\left| \mathrm {Cov_{\rho }(A,C)}\right| . \end{aligned}$$
(23)

The lower bound of (23) can be null. So we will prove the following more stringent inequality exists.

Theorem 6

Let \(\rho \in D(H), A,B\in S(H)\) and \(\alpha , \beta \) be any real numbers. If \(\sigma \in L^2(H)\) satisfies \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\), then

$$\begin{aligned}&V_{\rho }(A)+V_{\rho }(B)+V_{\rho }(C) \ge \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A+e^{-\mathrm {i}\alpha }B+e^{-\mathrm {i}\beta }C\right) \sigma \right) \right| ^2\nonumber \\&\quad -2\mathrm {Re}\left\{ e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B)+e^{\mathrm {i}\beta }\mathrm {Cov}_{\rho }(A,C)+e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\right\} . \end{aligned}$$
(24)

Proof

Let Eq. (11) be valid. To the inequality (24), we introduce a general inequality

$$\begin{aligned} \left\| c_1(A-a)\rho ^{\frac{1}{2}}+c_2 e^{\mathrm {i}\alpha }(B-b)\rho ^{\frac{1}{2}}+c_3e^{\mathrm {i}\beta }(C-c)\rho ^{\frac{1}{2}}+t\left( \rho ^{\frac{1}{2}}-A_{\varphi }\right) \right\| _2^2\ge 0, \end{aligned}$$

where \(a=\mathrm {Tr}(\rho A), b=\mathrm {Tr}(\rho B),c=\mathrm {Tr}(\rho C)\) and \(c_1, c_2,c_3, t\) real constants. Calculating the square modulus, we find

$$\begin{aligned} c_1^2 V_{\rho }(A)+c_2^2 V_{\rho }(B)+c_3^2 V_{\rho }(C)\ge -t^2r+c_1c_2c_3\delta t+c_1c_2c_3k, \end{aligned}$$
(25)

where

$$\begin{aligned}&r=2\left( 1-\cos \varphi \right) ,\\&k=-2\mathrm {Re}\left\{ \frac{1}{c_3}e^{\mathrm {i}\alpha } \mathrm {Cov}_{\rho }(A,B)+\frac{1}{c_2}e^{\mathrm {i}\beta }\mathrm {Cov}_{\rho }(A,C) +\frac{1}{c_1}e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\right\} ,\\&\delta =2\mathrm {Re}\left\{ {\mathrm {Tr}\left( \rho ^{\frac{1}{2}} \left( \frac{A-a}{c_2c_3}+e^{-\mathrm {i}\alpha }\frac{B-b}{c_1c_3} +e^{-\mathrm {i}\beta }\frac{C-c}{c_1c_2}\right) A_{\varphi }\right) }\right\} . \end{aligned}$$

Now choose the value of t that maximizes the right-hand side of (25), namely \(t=\frac{c_1c_2c_3\delta }{2r}\), then (25) becomes

$$\begin{aligned} c_1^2 V_{\rho }(A)+c_2^2 V_{\rho }(B)+c_3^2 V_{\rho }(C)\ge \frac{(c_1c_2c_3\delta )^2}{4r}+c_1c_2c_3k. \end{aligned}$$
(26)

Put \(c_1=c_2=c_3=1,\) then (26) can be represented as

$$\begin{aligned} V_{\rho }(A)+V_{\rho }(B)+V_{\rho }(C)\ge \frac{\delta ^2}{4r}+k, \end{aligned}$$

with

$$\begin{aligned}&r=2\left( 1-\cos \varphi \right) ,\\&k=-2\mathrm {Re}\left\{ e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B)+e^{\mathrm {i}\beta }\mathrm {Cov}_{\rho }(A,C)+e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\right\} ,\\&\delta =2\mathrm {Re}\left\{ \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A-a+e^{-\mathrm {i}\alpha }(B-b)+e^{-\mathrm {i}\beta }(C-c)\right) A_{\varphi }\right) \right\} . \end{aligned}$$

Similar to the proof of Theorem 4, if we take \(D_0=A-a+e^{-\mathrm {i}\alpha }(B-b)+e^{-\mathrm {i}\beta }(C-c),\) and then

$$\begin{aligned} \lim \limits _{\varphi \rightarrow 0}\frac{\delta ^2}{4r}= & {} \lim \limits _{\varphi \rightarrow 0}\frac{\left( \mathrm {Re}\left\{ \mathrm {Tr} \left( \rho ^{\frac{1}{2}}D_0A_{\varphi }\right) \right\} \right) ^2}{2\left( 1-\cos \varphi \right) }\\= & {} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}D_0\sigma \right) \right| ^2\\= & {} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A-a+e^{-\mathrm {i}\alpha }(B-b) +e^{-\mathrm {i}\beta }(C-c)\right) \sigma \right) \right| ^2\\= & {} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A+e^{-\mathrm {i}\alpha }B +e^{-\mathrm {i}\beta }C\right) \sigma \right) \right| ^2. \end{aligned}$$

Therefore, we can obtain the inequality (24). \(\square \)

Remark 4

The inequality (24) can reduce to the inequality (6) when we choose an appropriate state \(\rho \), a bound operator \(\sigma \), and real constants \(\alpha , \beta \).

Indeed, if we put \(\rho =|\psi \rangle \langle \psi |, \sigma =|\psi ^{\perp }\rangle \langle \psi |, \alpha =\pm \frac{2\pi }{3},\beta =\pm \frac{4\pi }{3},\) then \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\), so the inequality (24) holds. Clearly,

$$\begin{aligned}&V_{\rho }(A)=\varDelta A^2,\,\, V_{\rho }(B)=\varDelta B^2, \,\, V_{\rho }(C)=\varDelta C^2,\,\, \mathrm {Cov}_{\rho }(A,B)=\langle AB\rangle -\langle A\rangle \langle B\rangle ,\\&\mathrm {Cov}_{\rho }(A,C)=\langle AC\rangle -\langle A\rangle \langle C\rangle , \,\, \mathrm {Cov}_{\rho }(B,C)=\langle BC\rangle -\langle B\rangle \langle C\rangle . \end{aligned}$$

It is easy to see that

$$\begin{aligned} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( A+e^{-\mathrm {i}\alpha }B+e^{-\mathrm {i}\beta }C\right) \sigma \right) \right| ^2=\left| \left\langle \psi \left| \left( A+ e^{\mp \mathrm {i}\frac{2\pi }{3}}B+e^{\mp \mathrm {i}\frac{4\pi }{3}}C\right) \right| \psi ^{\bot }\right\rangle \right| ^2. \end{aligned}$$

Moreover,

$$\begin{aligned} k= & {} -2\mathrm {Re}\left\{ e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B)+e^{\mathrm {i}\beta } \mathrm {Cov}_{\rho }(A,C)+e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\right\} \\= & {} -2\mathrm {Re}\left\{ e^{\pm \mathrm {i}\frac{2\pi }{3}}(\langle AB\rangle -\langle A\rangle \langle B\rangle )+e^{\pm \mathrm {i}\frac{4\pi }{3}} (\langle AC\rangle -\langle A\rangle \langle C\rangle )+e^{\pm \mathrm {i}\frac{2\pi }{3}}(\langle BC\rangle -\langle B\rangle \langle C\rangle )\right\} \\= & {} \frac{1}{2}\langle \{A,B,C\}\rangle \mp \mathrm {i}\frac{\sqrt{3}}{2}\langle [A,B,C]\rangle -\langle A\rangle \langle B\rangle -\langle A\rangle \langle C\rangle -\langle B\rangle \langle C\rangle , \end{aligned}$$

where \(\langle \{A,B,C\}\rangle =\langle \{A,B\}\rangle +\langle \{A,C\}\rangle +\langle \{B,C\}\rangle \). Note that

$$\begin{aligned} \varDelta (A+B+C)^2=\varDelta A^2+\varDelta B^2+\varDelta C^2+\langle \{A,B,C\}\rangle -2(\langle A\rangle \langle B\rangle +\langle A\rangle \langle C\rangle +\langle B\rangle \langle C\rangle ). \end{aligned}$$

Thus, \(k=\frac{1}{2}\left( \varDelta (A+B+C)^2-\left( \varDelta A^2+\varDelta B^2+\varDelta C^2\right) \right) \mp \mathrm {i}\frac{\sqrt{3}}{2}\langle [A,B,C]\rangle .\) Therefore, we can obtain the following inequality:

$$\begin{aligned}&\varDelta A^2+\varDelta B^2+\varDelta C^2\ge \frac{1}{3}\varDelta (A+B+C)^2\mp \mathrm {i}\frac{\sqrt{3}}{3}\langle [A,B,C]\rangle +\frac{2}{3}\nonumber \\&\left| \left\langle \psi \left| \left( A+ e^{\mp \mathrm {i} \frac{2\pi }{3}}B+e^{\mp \mathrm {i}\frac{4\pi }{3}}C\right) \right| \psi ^{\bot }\right\rangle \right| ^2, \end{aligned}$$

which is accordant to (6).

Remark 5

If we put \(c_1=\sqrt{V_{\rho }(B)V_{\rho }(C)}, c_2=\sqrt{V_{\rho }(A)V_{\rho }(C)},c_3=\sqrt{V_{\rho }(A)V_{\rho }(B)}\) in the inequality (25), then

$$\begin{aligned} V_{\rho }(A)V_{\rho }(B)V_{\rho }(C)\le \frac{4r(3-k)}{\delta ^2}; \end{aligned}$$

using the same procedure described above, we can obtain the following uncertainty relation:

$$\begin{aligned}&V_{\rho }(A)V_{\rho }(B)V_{\rho }(C)\\\le & {} \frac{3\sqrt{V_{\rho }(A)V_{\rho }(B)V_{\rho }(C)}+2\mathrm {Re} \{\sqrt{V_{\rho }(C)}e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B) +\sqrt{V_{\rho }(B)}e^{\mathrm {i}\beta }\mathrm {Cov}_{\rho }(A,C) +\sqrt{V_{\rho }(A)}e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\}}{\left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}} +e^{-\mathrm {i}\alpha }\frac{B}{\sqrt{V_{\rho }(B)}}+e^{-\mathrm {i}\beta } \frac{C}{\sqrt{V_{\rho }(C)}}\right) \sigma \right) \right| ^2}. \end{aligned}$$

Remark 6

If we put \(c_1=\sqrt{V_{\rho }(A)}, c_2=\sqrt{V_{\rho }(B)},c_3=\sqrt{V_{\rho }(C)}\) in the inequality (25), then using the same procedure described above, the following uncertainty relation

$$\begin{aligned}&V_{\rho }^2(A)+V_{\rho }^2(B)+V_{\rho }^2(C)\\\ge & {} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \sqrt{V_{\rho }(A)}A +e^{-\mathrm {i}\alpha }\sqrt{V_{\rho }(B)}B+e^{-\mathrm {i}\beta }\sqrt{V_{\rho }(C)}C\right) \sigma \right) \right| ^2\\&-2\mathrm {Re}\left\{ \sqrt{V_{\rho }(A)V_{\rho }(B)}e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B) +\sqrt{V_{\rho }(A)V_{\rho }(C)}e^{\mathrm {i}\beta }\mathrm {Cov}_{\rho }(A,C)+\sqrt{V_{\rho }(B)V_{\rho }(C)}e^{\mathrm {i}(\beta -\alpha )}\mathrm {Cov}_{\rho }(B,C)\right\} \end{aligned}$$

is valid.

4 Conclusions

In this paper, we gave two stronger uncertainty relations, relating to the sum of variances with respect to density matrix:

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \pm \mathrm {i}\mathrm {Tr}(\rho [A,B])+\left\| \rho ^\frac{1}{2}(A\pm \mathrm {i}B)\rho _{\bot }^\frac{1}{2}\right\| _2^2, \end{aligned}$$

and

$$\begin{aligned} V_\rho (A)+V_\rho (B)\ge \frac{1}{2}\left\| \rho ^\frac{1}{2}(A+B)\rho _{\bot }^\frac{1}{2}\right\| _2^2, \end{aligned}$$

where \(\rho _{\bot }\in D(H),\mathrm {s.t.}\ \rho \rho _{\bot }=0.\) These lower bounds are guaranteed to be nontrivial whenever the two observables are incompatible on the state of the system. If we choose appropriate \(\rho \) and \(\rho _\bot \), the above results can reduce to (3) and (4), respectively, which were given by Maccone and Pati in [22]. In addition, a new bound for the sum of two variances of pure states is derived by theorem 3, that is corollary 3 including Maccone–Pati’s inequality (4) as a special case if we choose \(|\psi ^{\bot }\rangle =\left| \psi _{A+B}^{\bot }\right\rangle \).

Moreover, two stronger uncertainty relations in terms of the product of the variances of two observables were established:

$$\begin{aligned} \sqrt{V_{\rho }(A)V_{\rho }(B)}\ge \frac{\pm \frac{\mathrm {i}}{2}\mathrm {Tr}(\rho [A,B])}{1-\frac{1}{2} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}\pm \mathrm {i}\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2}, \end{aligned}$$

which can recover to the inequality (5) when we choose appropriate \(\rho \) and \(\sigma \), and

$$\begin{aligned} V_{\rho }(A)V_{\rho }(B)\ge \frac{\left| \mathrm {Cov}_{\rho }(A,B)\right| ^2}{\left[ 1-\frac{1}{2} \left| \mathrm {Tr}\left( \rho ^{\frac{1}{2}}\left( \frac{A}{\sqrt{V_{\rho }(A)}}+e^{\mathrm {i}\alpha }\frac{B}{\sqrt{V_{\rho }(B)}}\right) \sigma \right) \right| ^2\right] ^2}, \end{aligned}$$

where \(\sigma \) satisfies \(\mathrm {Tr}\left( \rho ^{\frac{1}{2}}\sigma \right) =0\) and \(\Vert \sigma \Vert _2=1\), \(\alpha \) satisfies \(e^{\mathrm {i}\alpha }\mathrm {Cov}_{\rho }(A,B)=|\mathrm {Cov}_{\rho }(A,B)|.\) We use the square-modulus inequality and follow a procedure analogous to the one employed by [22]. However, the proof of theorem 4 is more complex, and lemma 2 is the key point to the proof. The limit given here is taken in the sense of \(\Vert \cdot \Vert _2.\)

Several stronger uncertainty relations for three observables were established, relating to the sum and product of variances with respect to density matrix, respectively. If we choose an appropriate state \(\rho \), a bound operator \(\sigma \), and real constants \(\alpha , \beta \), the inequality (24) can reduce to the inequality (6) which was given by Song and Qiao in [25].