Introduction to Bell’s Inequality in Quantum Mechanics ^†

Abstract

A pedagogical introduction to Bell’s inequality in Quantum Mechanics is presented. Several examples, ranging from spin $1/2$ to coherent and squeezed states are worked out. The generalization to Mermin’s inequalities and to GHZ states is also outlined.

1. Introduction

The Bell inequality [1,2,3] is a milestone of Quantum Mechanics. It highlights in a surprising way the existence of strong correlations between the components of a composite system. These correlations cannot be reproduced by any local hidden variable theory, surviving over very long distances. They are the signature of one of the most fascinating phenomena of the quantum world: the entanglement. The experimental confirmation of the violation of the Bell inequality by the pioneering works of A. Aspect, J. Clauser and A. Zeilinger [4,5,6,7,8,9,10,11,12,13] gives to Quantum Mechanics the status of the unique theory able to consistently confront the intricacies and the complexity of the entanglement. The whole story goes back to the famous paper by Einstein–Podolsky–Rosen [14]. This work is considered the birth of the study of the entanglement. The phenomenon was so challenging that it led Einstein–Podolsky–Rosen to argue that Quantum Mechanics was an incomplete theory. This statement marked the beginning of a long and difficult debate on the foundations of Quantum Theory. It was only in 1964, the year of the publication of Bell’s work [1], that the debate was elucidated. Bell was able to derive a certain inequality whose violation would imply that Quantum Mechanics could not be reconciled with any other local hidden variable formulation. Bell’s inequality is at the origin of the great achievements that have been obtained so far in many areas: Quantum Computation, Quantum Teleportation, Quantum Cryptography, etc. The purpose of these notes is to present, in an introductory way, the general features of a version of the Bell inequality called the Bell–CHSH inequality [4]. Many examples will be worked out, ranging from spin $1/2$ states to coherent and squeezed states. The generalization of the Bell–CHSH inequality to multipartite systems, known as the Mermin inequalities, will be also outlined. The material is organized in a self-contained way. Students with a basic knowledge of Quantum Mechanics will be able to reproduce all calculations and solve the proposed exercises.

2. Entangled and Product States: Spin 1/2 States

The concept of entangled states can be illustrated in a simple way by making use of spin $1/2$ states. Let us begin by considering a bipartite system, $AB$, made up of two spins $1/2$. The Hilbert space of such a system is $\mathcal{H}={\mathcal{H}}_{\mathcal{A}}\otimes {\mathcal{H}}_{\mathcal{B}}$, where ${\mathcal{H}}_A$ and ${\mathcal{H}}_{\mathcal{B}}$ are the Hilbert spaces of A and B, respectively. It is customary to refer to the capital letters $(A,B)$ as Alice and Bob (two agents measuring the system in their respective Labs). An orthonormal basis of $\mathcal{H}={\mathcal{H}}_{\mathcal{A}}\otimes {\mathcal{H}}_{\mathcal{B}}$ contains four elements: ${\left(\right|+\rangle }_A{\otimes |+\rangle }_B{,|+\rangle }_A{\otimes |-\rangle }_B{,|-\rangle }_A{\otimes |+\rangle }_B{,|-\rangle }_A{\otimes |-\rangle }_B)$1 Any state $|\psi \rangle$ of $\mathcal{H}$ can be written as a linear combination of the elements of the basis

$${\left|\right\{\psi ⟩=a_1|+⟩}_A\otimes |+{⟩}_B+a_2(|+{⟩}_A\otimes |-{⟩}_B+a_3|-{⟩}_A\otimes |+{⟩}_B+a_4|-{⟩}_A\otimes |-{⟩}_B,$$

(3)

where, from the normalization condition, $\langle \psi |\psi \rangle =1$, the coefficients $(a_1,a_2,a_3,a_4)$ fulfill the constraint

$$|a_1{|}^2+|a_2{|}^2+|a_3{|}^2+{\left|a_4\right|}^2=1.$$

(4)

The state $|\psi \rangle$ is said to be a product state if it can be written as the tensor product of states belonging to ${\mathcal{H}}_{\mathcal{A}}$ and ${\mathcal{H}}_B$, i.e.,

$$|\psi \rangle ={|\psi \rangle }_A\otimes {|\psi \rangle }_B,$$

(5)

where ${\left(\right|\psi \rangle }_A{,|\psi \rangle }_B)$ stand for vectors of ${\mathcal{H}}_{\mathcal{A}}$, ${\mathcal{H}}_{\mathcal{B}}$:

$${|\psi \rangle }_A={\left(c_1\right|+\rangle }_A+c_2{|-\rangle }_A{),|\psi \rangle }_B={\left(b_1\right|+\rangle }_B+b_2{|-\rangle }_B).$$

(6)

One sees thus that $|\psi \rangle$ in Equation (3) is a product state if $(a_1,a_2,a_3,a_4)$ can be written as

$$a_1=c_1{b}_1,a_2=c_1{b}_2,a_3=c_2{b}_1,a_4=c_2{b}_2.$$

(7)

The above condition can be linked to a $2\times 2$ determinant made up of $(a_1,a_2,a_3,a_4)$. Let us encode the coefficients of the expansion (3) in the matrix $\mathcal{A}$

$$\mathcal{A}=\left[\begin{array}{cc}{a}_1& a_2\\ a_3& a_4\end{array}\right]$$

(8)

Conditions (7) imply that the determinant of $\mathcal{A}$ vanishes:

$$\det \mathcal{A}=a_1{a}_4-a_2{a}_3=c_1{b}_1{c}_2{b}_2-c_1{b}_2{c}_2{b}_1=0.$$

(9)

This equation gives a quick way of checking if a given state can be written as a product state or not. Suppose that the determinant of $\mathcal{A}$ does not vanish. As a consequence, Equation (7) cannot be fulfilled, so that the state cannot be written as a product state. In such a case, i.e., when $|\psi \rangle$ cannot be written as in Equation (7), one says that it is entangled:

$$|\psi \rangle \mathrm{entangled}\Rightarrow |\psi \rangle \ne {|\psi \rangle }_A\otimes {|\psi \rangle }_B$$

(10)

Exercise Show that the state

$$\begin{array}{ccc}\hfill {|\psi \rangle }_N& =& {\displaystyle \frac{1}{\sqrt{N}}}\left({|+\rangle }_A{\otimes |+\rangle }_B{+|+\rangle }_A{\otimes |-\rangle }_B{+|-\rangle }_A\otimes {|+\rangle }_B\right)+{\displaystyle \frac{\sqrt{N-3}}{\sqrt{N}}}{|-\rangle }_A\otimes {|-\rangle }_B,\hfill \\ & N& \mathrm{integer},N\ge 3,\hfill \end{array}$$

(11)

is entangled for $N\ge 5$. What happens for $N=3,4$?

The Bell States

Out of the four elements ${\left(\right|+\rangle }_A{\otimes |+\rangle }_B,{\left(\right|+\rangle }_A{\otimes |-\rangle }_B{,|-\rangle }_A{\otimes |+\rangle }_B{,|-\rangle }_A{\otimes |-\rangle }_B)$ one can construct the following four entangled states $|{\varphi }_{\alpha }\rangle ,\alpha =0,1,2,3$, called the Bell states, namely

$$\begin{array}{ccc}\hfill |{\varphi }_0\rangle & =& {\displaystyle \frac{1}{\sqrt{2}}}\left({|+\rangle }_A{\otimes |+\rangle }_B{+|-\rangle }_A\otimes {|-\rangle }_B\right),\hfill \\ \hfill |{\varphi }_1\rangle & =& {\displaystyle \frac{1}{\sqrt{2}}}\left({|+\rangle }_A{\otimes |+\rangle }_B{-|-\rangle }_A\otimes {|-\rangle }_B\right),\hfill \\ \hfill |{\varphi }_2\rangle & =& {\displaystyle \frac{1}{\sqrt{2}}}\left({|+\rangle }_A{\otimes |-\rangle }_B{-|-\rangle }_A\otimes {|+\rangle }_B\right),\hfill \\ \hfill |{\varphi }_3\rangle & =& {\displaystyle \frac{1}{\sqrt{2}}}\left({|+\rangle }_A{\otimes |-\rangle }_B{+|-\rangle }_A\otimes {|+\rangle }_B\right).\hfill \end{array}$$

(12)

A quick computation shows that

$$\langle {\varphi }_{\alpha }|{\varphi }_{\beta }\rangle ={\delta }_{\alpha \beta },$$

(13)

so that the Bell states form an orthonormal basis of entangled states. The state ${\varphi }_2$ is called singlet, due to the fact that it has zero total spin:

$$\overrightarrow{S}|{\varphi }_2\rangle =({\overrightarrow{S}}_A+{\overrightarrow{S}}_B)|{\varphi }_2\rangle =0.$$

(14)

Exercise Prove Equation (14).

Exercise Express the four elements ${\left(\right|+\rangle }_A{\otimes |+\rangle }_B,{\left(\right|+\rangle }_A{\otimes |-\rangle }_B{,|-\rangle }_A{\otimes |+\rangle }_B{,|-\rangle }_A{\otimes |-\rangle }_B)$ in terms of the Bell entangled basis.

3. The Bell–CHSH Inequality

The meaning of the Bell–CHSH inequality can be captured by the following simplified argument.2 Alice and Bob are located in their respective laboratories, which are supposed to be far enough apart so that Alices’s measurements do not interfere with Bob’s measurements and vice-versa. In particular, as emphasized in [1], the settings of Alice’s device do not depend on those of Bob’s device and vice-versa. The measurements carried out by both Alice and Bob have only two possible outcomes: $\pm 1$. One says that they are dichotomic measurements, i.e., measurements, for instance, of the type yes/no, up/down, … The measurements of $(A,A^{\prime })$ and $(B,B^{\prime })$ are repeated n times. Let $(a_j,a_j^{\prime },b_j,b_j^{\prime })$ stand for the outcomes of the j-th measurement. Let us consider the combination

$${\langle (A+A^{\prime })B+(A-A^{\prime })B^{\prime }\rangle }_{av},$$

(15)

where the brackets ${\langle ·\rangle }_{av}$ mean

$${\langle (A+A^{\prime })B+(A-A^{\prime })B^{\prime }\rangle }_{av}={\displaystyle \frac{1}{n}}\sum _{j=1}^n\left((a_j+a_j^{\prime })b_j+(a_j-a_j^{\prime })b_j^{\prime }\right),$$

(16)

with

$$a_j^2={a^{\prime }}_j^2=b_j^2={b^{\prime }}_j^2=1$$

(17)

It holds that

$$|(a_j+a_j^{\prime })b_j+(a_j-a_j^{\prime })b_j^{\prime }|\le 2\forall j.$$

(18)

In fact, making use of the triangle inequality, it follows

$$|(a_j+a_j^{\prime })b_j+(a_j-a_j^{\prime })b_j^{\prime }|\le \left|(a_j+a_j^{\prime })b_j\right|+\left|(a_j-a_j^{\prime })b_j^{\prime }\right|=\left|(a_j+a_j^{\prime })\right|+\left|(a_j-a_j^{\prime })\right|.$$

(19)

since $|b_j|=|b_j^{\prime }|=1$. Moreover, the maximum value of $|a_j+a_j^{\prime }|$ is equal to 2, corresponding to $a_j=a_j^{\prime }=\pm 1$. However, when $|a_j+a_j^{\prime }|=2$, one observes that $|a_j-a_j^{\prime }|$ vanishes, and vice-versa, showing thus the validity of Equation (18). Therefore,

$$\left|{\langle (A+A^{\prime })B+(A-A^{\prime })B^{\prime }\rangle }_{av}\right|\le 2.$$

(20)

Let us now turn to Quantum Mechanics. Here, we deal with Hermitian dichotomic operators, $(A,A^{\prime })$ and $(B,B^{\prime })$, called Bell’s operators. Alice’s operators $(A,A^{\prime })$ act on the Hilbert space ${\mathcal{H}}_{\mathcal{A}}$, while Bob’s operators $(B,B^{\prime })$ act on ${\mathcal{H}}_B$. These operators are required to fulfill the following conditions:

$$\begin{array}{ccc}\hfill A& =& A^{†},A^{\prime }={A^{\prime }}^{†},B=B^{†},B^{\prime }={B^{\prime }}^{†},\hfill \\ \hfill A^2& =& {A^{\prime }}^2=B^2={B^{\prime }}^2=1,\hfill \end{array}$$

(21)

and

$$\begin{array}{ccc}\hfill \left[A,B\right]& =& 0,\left[A^{\prime },B\right]=0,\left[A,B^{\prime }\right]=0,\left[A^{\prime },B^{\prime }\right]=0.\hfill \\ \hfill \left[A,A^{\prime }\right]& \ne & 0,\left[B,B^{\prime }\right]\ne 0.\hfill \end{array}$$

(22)

From Equation (22), one sees that Alice’s operators commute with Bob’s operators. More precisely, Alice and Bob are implicitly meant to be space-like separated. As such, measurements performed by Alice do not interfere with Bob’s measurements. On the other hand, $(A,A^{\prime })$ have a non-vanishing commutator. The same holds for $(B,B^{\prime })$. This last condition is easy to understand. For instance, in the case of spin $1/2$, it means that Alice and Bob cannot perform two simultaneous measurements of their own spin in two different directions. In Quantum Mechanics, the analogue of the combination (15) is provided by the correlator

$$\langle \psi \left|\mathcal{C}\right|\psi \rangle ,$$

(23)

where $|\psi \rangle$ stands for the normalized state of the system, $\langle \psi |\psi \rangle =1$, and $\mathcal{C}$ is given by the Hermitian Bell–CHSH operator

$$\mathcal{C}=(A+A^{\prime })\otimes B+(A-A^{\prime })\otimes B^{\prime },$$

(24)

The question that is addressed is whether the bound (20) holds at the quantum level; that is, if

$$|\langle \psi |\mathcal{C}|\psi \rangle |\le 2?.$$

(25)

The answer given by Quantum Mechanics is astonishing. If the state $|\psi \rangle$ is a product state, the bound (18) still holds. However, if the state $|\psi \rangle$ is entangled, then the correlator (23) turns out to be strictly greater than 2, namely

$${|\langle \psi |\mathcal{C}|\psi \rangle |}_{\mathit{entang}}>2.$$

(26)

In this case, one speaks of a violation of the Bell–CHSH inequality.

Example of Violation of the Bell–CHSH Inequality

As a first example of violation of the Bell–CHSH inequality, we consider the case in which Alice and Bob share the first Bell entangled state $|{\varphi }_0\rangle$, Equation (12). The first task is that of identifying the four Bell dichotomic operators $(A,A^{\prime },B,B^{\prime })$. To that end, we shall introduce the following operators

$$\begin{array}{ccc}\hfill {A|+\rangle }_A& =& e^{i\alpha }{|-\rangle }_A{,A|-\rangle }_A=e^{-i\alpha }{|+\rangle }_A,\hfill \\ \hfill A^{\prime }{|+\rangle }_A& =& e^{i{\alpha }^{\prime }}{|-\rangle }_A,A^{\prime }{|-\rangle }_A=e^{-i{\alpha }^{\prime }}{|+\rangle }_A,\hfill \\ \hfill {B|+\rangle }_B& =& e^{i\beta }{|-\rangle }_B{,B|-\rangle }_B=e^{-i\beta }{|+\rangle }_B,\hfill \\ \hfill B^{\prime }{|+\rangle }_B& =& e^{i{\beta }^{\prime }}{|-\rangle }_B,B^{\prime }{|-\rangle }_B=e^{-i{\beta }^{\prime }}{|+\rangle }_B,\hfill \end{array}$$

(27)

where $(\alpha ,{\alpha }^{\prime },\beta ,{\beta }^{\prime })$ are arbitrary parameters. It is easy to show that the operators in Equation (27) obey all conditions of Equations (21) and (22). For example,

$$A^2{|+\rangle }_A=e^{i\alpha }{A|-\rangle }_A=e^{i\alpha }{e}^{-i\alpha }{|+\rangle }_A={|+\rangle }_A.$$

(28)

The next step is that of evaluating $A\otimes B|{\varphi }_0\rangle$. From Equation (27), we get

$$A\otimes B|{\varphi }_0\rangle ={\displaystyle \frac{1}{\sqrt{2}}}\left(e^{i(\alpha +\beta )}{|-\rangle }_A{\otimes |-\rangle }_B+e^{-i(\alpha +\beta )}{|+\rangle }_A\otimes {|+\rangle }_B\right).$$

(29)

Thus,

$$\langle {\varphi }_0|A\otimes B|{\varphi }_0\rangle =cos(\alpha +\beta ).$$

(30)

For the Bell–CHSH correlator, one obtains

$$\begin{array}{ccc}\hfill \langle {\varphi }_0\left|\mathcal{C}\right|{\varphi }_0\rangle & =& \langle {\varphi }_0|(A+A^{\prime })\otimes B+(A-A^{\prime })\otimes B^{\prime }|{\varphi }_0\rangle =cos(\alpha +\beta )+cos({\alpha }^{\prime }+\beta )\hfill \\ & +& cos(\alpha +{\beta }^{\prime })-cos({\alpha }^{\prime }+{\beta }^{\prime }).\hfill \end{array}$$

(31)

Expression (31) depends on the four arbitrary parameters $(\alpha ,{\alpha }^{\prime },\beta ,{\beta }^{\prime })$. It remains to investigate if a choice exists of $(\alpha ,{\alpha }^{\prime },\beta ,{\beta }^{\prime })$, which gives rise to a violation of the Bell–CHSH inequality, i.e., $|\langle {\varphi }_0|\mathcal{C}|{\varphi }_0\rangle |>2$. The standard choice that one finds in textbooks [15,16,17,18] is

$$\alpha =0,{\alpha }^{\prime }={\displaystyle \frac{\pi }{2}},\beta =-{\displaystyle \frac{\pi }{4}},{\beta }^{\prime }={\displaystyle \frac{\pi }{4}},$$

(32)

yielding

$$|\langle {\varphi }_0\left|\mathcal{C}\right|{\varphi }_0\rangle |=2\sqrt{2}=2.8284.$$

(33)

We see that the Bell state $|{\varphi }_0\rangle$ yields a violation of the Bell–CHSH inequality. As we shall see in the next Sections, the value $2\sqrt{2}$, known as Tsirelson’s bound [19], is the maximum value attainable by the Bell–CHSH correlator $|\langle \psi |\mathcal{C}|\psi \rangle |$. Replacing $|{\varphi }_0\rangle$ by any other Bell’s state yields the same value for the violation: $2\sqrt{2}$. All four Bell’s states violate maximally the Bell–CHSH inequality. Let us conclude this Section with a remark on the Bell operators of Equation (27). These operators are a particular case of a more general set of dichotomic operators built out with the three Pauli matrices

$${\sigma }_x=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right],{\sigma }_y=\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right],{\sigma }_z=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right].$$

(34)

From

$${\sigma }_i{\sigma }_j={\delta }_{ij}+i{\epsilon }_{ijk}{\sigma }_k,$$

(35)

it follows that the operator $(\widehat{n}·\overrightarrow{\sigma })$, where $\widehat{n}$ is a unit vector, is dichotomic and Hermitian

$${(\widehat{n}·\overrightarrow{\sigma })}^2=1,{(\widehat{n}·\overrightarrow{\sigma })}^{†}=(\widehat{n}·\overrightarrow{\sigma }).$$

(36)

Using polar coordinates, $\widehat{n}=(sin\theta cos\phi ,sin\theta sin\phi ,cos\theta )$, we have

$$\widehat{n}·\overrightarrow{\sigma }=\left[\begin{array}{cc}cos\theta & e^{-i\phi }sin\theta \\ e^{i\phi }sin\theta & -cos\theta \end{array}\right].$$

(37)

For the Bell operators, we get the more general form

$$\begin{array}{ccc}\hfill {A|+\rangle }_A& =& {(cos\left(\theta \right)|+\rangle }_A+e^{i\alpha }{sin\left(\theta \right)|-\rangle }_A),\hfill \\ \hfill {A|-\rangle }_A& =& {(-cos\left(\theta \right)|-\rangle }_A+e^{-i\alpha }{sin\left(\theta \right)|+\rangle }_A),\hfill \\ \hfill A^{\prime }{|+\rangle }_A& =& {(cos\left({\theta }^{\prime }\right)|+\rangle }_A+e^{i{\alpha }^{\prime }}sin\left({\theta }^{\prime }\right){|-\rangle }_A),\hfill \\ \hfill A^{\prime }{|-\rangle }_A& =& {(-cos\left({\theta }^{\prime }\right)|-\rangle }_A+e^{-i{\alpha }^{\prime }}sin\left({\theta }^{\prime }\right){|+\rangle }_A),\hfill \\ \hfill {B|+\rangle }_B& =& {(cos\left(\omega \right)|+\rangle }_B+e^{i\beta }{sin\left(\omega \right)|-\rangle }_B),\hfill \\ \hfill {B|-\rangle }_B& =& {(-cos\left(\omega \right)|-\rangle }_B+e^{-i\beta }{sin\left(\omega \right)|+\rangle }_B),\hfill \\ \hfill B^{\prime }{|+\rangle }_B& =& {(cos\left({\omega }^{\prime }\right)|+\rangle }_B+e^{i{\beta }^{\prime }}sin\left({\omega }^{\prime }\right){|-\rangle }_B),\hfill \\ \hfill B^{\prime }{|-\rangle }_B& =& {(-cos\left({\omega }^{\prime }\right)|-\rangle }_B+e^{-i{\beta }^{\prime }}sin\left({\omega }^{\prime }\right){|+\rangle }_B),\hfill \end{array}$$

(38)

where $(\alpha ,{\alpha }^{\prime },\theta ,{\theta }^{\prime })$ and $(\beta ,{\beta }^{\prime },\omega ,{\omega }^{\prime })$ are arbitrary parameters. We see that the operators of Equation (27) correspond to $\theta ={\theta }^{\prime }=\omega ={\omega }^{\prime }={\displaystyle \frac{\pi }{2}}$. Repeating now the previous calculation using the operators (38), one gets the slightly different expression

$$\begin{array}{ccc}\hfill \langle {\varphi }_0\left|\mathcal{C}\right|{\varphi }_0\rangle & =& cos\left(\theta \right)cos\left(\omega \right)+sin\left(\theta \right)sin\left(\omega \right)cos(\alpha +\beta )\hfill \\ & +& cos\left({\theta }^{\prime }\right)cos\left(\omega \right)+sin\left({\theta }^{\prime }\right)sin\left(\omega \right)cos({\alpha }^{\prime }+\beta )\hfill \\ & +& cos\left(\theta \right)cos\left({\omega }^{\prime }\right)+sin\left(\theta \right)sin\left({\omega }^{\prime }\right)cos(\alpha +{\beta }^{\prime })\hfill \\ & -& cos\left({\theta }^{\prime }\right)cos\left({\omega }^{\prime }\right)-sin\left({\theta }^{\prime }\right)sin\left({\omega }^{\prime }\right)cos({\alpha }^{\prime }+{\beta }^{\prime }).\hfill \end{array}$$

(39)

Though, it turns out that the maximum value attained by (39) is precisely $2\sqrt{2}$, namely

$$|\langle {\varphi }_0\left|\mathcal{C}\right|{\varphi }_0{\rangle |}_{max}=2\sqrt{2}.$$

(40)

This result might be considered as a nice check of Tsirelson’s bound. Comment In the case of spin $1/2$, for which the Hilbert space is of dimension two, the characterization of the Bell operators, Equations (27) and (38), is relatively simple. Though, as the dimension of the Hilbert space increases, the task becomes more complex. For the purposes of the present notes, we shall rely mostly on the use of Equation (27), suitably adapted to higher dimensional Hilbert spaces.

Exercise Consider the product state

$$|\psi \rangle ={|+\rangle }_A\otimes {|-\rangle }_B.$$

(41)

Show, by using Equation (38), that the Bell–CHSH correlator does not exhibit violation, i.e.,

$$|\langle \psi |\mathcal{C}|\psi \rangle |\le 2.$$

(42)

Exercise Consider the state $|{\varphi }_0\rangle$, Equation (12), and the following four angles

$$\alpha =0,{\alpha }^{\prime }={\displaystyle \frac{\pi }{2}},\beta =-\epsilon ,{\beta }^{\prime }=\epsilon ,$$

(43)

where $\epsilon >0$ is infinitesimal. Is there violation of the Bell–CHSH inequality?

4. Tsirelson’s Bound

As we have seen, the so-called Tsirelson bound [19] tells us that the maximum value attainable by the Bell–CHSH correlator is $2\sqrt{2}$. Due to its general validity, it is helpful to give a proof of this statement. It relies only on the algebraic features of the Bell operators, as expressed by Equations (21) and (22). Let us start by recalling a few properties of the operator norm [20]. Let $\mathcal{D}:\mathcal{H}\to \mathcal{H}$ be a bounded operator defined on a Hilbert space $\mathcal{H}$. The operator norm of $\mathcal{D}$ is defined as

$$\left|\right|\mathcal{D}\left|\right|=sup\left\{{\displaystyle \frac{\left|\right|\mathcal{D}x\left|\right|}{\left|\right|x\left|\right|}},x\ne 0,x\in \mathcal{H}\right\}.$$

(44)

where sup stands for the supremum3 and $\left|\right|x\left|\right|$ is the norm of the vector x: ${\left|\right|x\left|\right|}^2=\langle x|x\rangle$. The operator norm fulfills many properties, namely

• from the definition of the operator norm, it follows that

  $$\left|\right|\mathcal{D}x\left|\right|\le \left|\right|\mathcal{D}\left|\right|\left|\right|x\left|\right|.$$

  (45)
• the triangle inequality

  $$\left|\right|{\mathcal{D}}_1+{\mathcal{D}}_2\left|\right|\le \left|\right|{\mathcal{D}}_1\left|\right|+\left|\right|{\mathcal{D}}_2\left|\right|.$$

  (46)
• the so-called continuity of the norm

  $$\left|\right|{\mathcal{D}}_1{\mathcal{D}}_2\left|\right|\le \left|\right|{\mathcal{D}}_1\left|\right|\left|\right|{\mathcal{D}}_2\left|\right|.$$

  (47)
• if $\mathcal{D}=\mathcal{T}\otimes {\mathcal{T}}^{\prime }$, then

  $${\left|\right|\mathcal{D}\left|\right|}_{{\mathcal{H}}_T\otimes {\mathcal{H}}_{T^{\prime }}}={\left|\right|\mathcal{T}\left|\right|}_{{\mathcal{H}}_T}\left|\right|{\mathcal{T}}^{\prime }{\left|\right|}_{{\mathcal{H}}_{T^{\prime }}}.$$

  (48)
• if $\mathcal{D}$ is Hermitian, ${\mathcal{D}}^{†}=\mathcal{D}$, it holds that

  $$\left|\right|{\mathcal{D}}^2\left|\right|={\left|\right|\mathcal{D}\left|\right|}^2,$$

  (49)

In particular, the last equation follows by observing that, from the Cauchy–Schwarz inequality, one has

$${\left|\right|\mathcal{D}x\left|\right|}^2=\langle \mathcal{D}x|\mathcal{D}x\rangle =\langle x|{\mathcal{D}}^{2}x\rangle \le \left|\right|x\left|\right|\left|\right|{\mathcal{D}}^{2}x\left|\right|\le {\left|\right|x\left|\right|}^2\left|\right|{\mathcal{D}}^2\left|\right|.$$

(50)

Moreover, from the property

$${\left(sup\left\{{\displaystyle \frac{\left|\right|\mathcal{D}x\left|\right|}{\left|\right|x\left|\right|}},x\ne 0,x\in \mathcal{H}\right\}\right)}^2=sup\left\{{\left({\displaystyle \frac{\left|\right|\mathcal{D}x\left|\right|}{\left|\right|x\left|\right|}}\right)}^2,x\ne 0,x\in \mathcal{H}\right\},$$

(51)

it follows

$${\left|\right|\mathcal{D}\left|\right|}^2\le \left|\right|{\mathcal{D}}^2\left|\right|.$$

(52)

Also, from the continuity of the norm, we get

$$\left|\right|{\mathcal{D}}^2\left|\right|=\left|\right|\mathcal{D}\mathcal{D}\left|\right|\le \left|\right|\mathcal{D}\left|\right|\left|\right|\mathcal{D}\left|\right|={\left|\right|\mathcal{D}\left|\right|}^2,$$

(53)

thus showing Equation (49). We can proceed now to prove Tsirelson’s statement. Let us consider the Hermitian Bell–CHSH operator $\mathcal{C}$, Equation (24). From Equations (21) and (22), it follows

$${\mathcal{C}}^{†}\mathcal{C}={\mathcal{C}}^2=4I-\left[A,A^{\prime }\right]\otimes \left[B,B^{\prime }\right],$$

(54)

where $I=I_A\otimes I_B$ is the identity. Taking the norm operator and applying the triangle inequality, one gets

$$\left|\right|{\mathcal{C}}^2\left|\right|\le 4+\left|\right|\left[A,A^{\prime }\right]\left|\right|\left|\right|\left[B,B^{\prime }\right]\left|\right|.$$

(55)

Since

$$\left|\right|\left[A,A^{\prime }\right]\left|\right|\le 2\left|\right|A\left|\right|\left|\right|A^{\prime }\left|\right|=2,\left|\right|\left[B,B^{\prime }\right]\left|\right|\le 2\left|\right|B\left|\right|\left|\right|B^{\prime }\left|\right|=2,$$

(56)

one finds

$$\left|\right|{\mathcal{C}}^2\left|\right|\le 8.$$

(57)

Let $|\psi \rangle$ stand for any normalized state, $\left|\right|\psi \left|\right|=1$. From the Cauchy–Schwartz inequality, we get

$${|\langle \psi |\mathcal{C}|\psi \rangle |}^2\le {\left|\right|\mathcal{C}\psi \left|\right|}^2\le {\left|\right|\mathcal{C}\left|\right|}^2\le 8,$$

(58)

so that Tsirelson’s bound follows, namely

$$|\langle \psi |\mathcal{C}|\psi \rangle |\le 2\sqrt{2},$$

(59)

for any state $|\psi \rangle$.

Comment Equation (55) exhibits a nice property of the operator $\mathcal{C}$. In the classical case, the quantities $(A,A^{\prime })$ and $(B,B^{\prime })$ commute, i.e., $[A,A^{\prime }]=[B,B^{\prime }]=0$. The classical bound $\left|\right|\mathcal{C}\left|\right|\le 2$ is recovered. Nevertheless, at the quantum level, $(A,A^{\prime })$ and $(B,B^{\prime })$ become non-commuting quantities, a key feature in order to have violation.

Exercise Show that

$${\left(sup\left\{{\displaystyle \frac{\left|\right|\mathcal{D}x\left|\right|}{\left|\right|x\left|\right|}},x\ne 0,x\in \mathcal{H}\right\}\right)}^2=sup\left\{{\left({\displaystyle \frac{\left|\right|\mathcal{D}x\left|\right|}{\left|\right|x\left|\right|}}\right)}^2,x\ne 0,x\in \mathcal{H}\right\}.$$

(60)

5. Gisin’s Theorem

Gisin’s theorem [21] is an important result about the violation of the Bell–CHSH inequality. It states that any entangled pure state violates the Bell–CHSH inequality. The theorem means that, given an entangled pure state, it is always possible to find Bell’s operators $(A,A^{\prime })$ and $(B,B^{\prime })$ such that a violation occurs. Instead of reproducing Gisin’s original argument, we adopt the strategy of providing a check of the theorem. We shall choose the expression given in Equation (11) as the entangled state. As the value of the integer N becomes large, the last term is the dominant one. As a consequence, for large values of N, the state becomes close to a product state. Therefore, as N increases, we expect that the violation of the Bell–CHSH inequality decreases, approaching asymptotically the classical bound 2. In order to verify Gisin’s theorem it suffices to show that the Bell–CHSH inequality is still violated for large N, even if the size of the violation is small. After all, the state remains entangled, although in a rather weak way. Let us proceed with the evaluation of the Bell–CHSH correlator. In this case, we shall make use of the general expression for the Bell operators given in Equation (38). It turns out that

$$\begin{array}{ccc}\hfill \langle \psi |A\otimes {B|\psi \rangle }_N& =& {\displaystyle \frac{cos\left(\theta \right)cos\left(\omega \right)}{N}}\left[N-4\right]+{\displaystyle \frac{2cos\left(\theta \right)sin\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\beta \right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)cos\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\alpha \right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)sin\left(\omega \right)}{N}}\left[\sqrt{N-3}cos(\alpha +\beta )+cos(\alpha -\beta )\right].\hfill \end{array}$$

(61)

Therefore, for the Bell–CHSH correlator, one finds

$$\begin{array}{ccc}\hfill {\langle \psi \left|\mathcal{C}\right|\psi \rangle }_N& =& {\displaystyle \frac{cos\left(\theta \right)cos\left(\omega \right)}{N}}\left[N-4\right]+{\displaystyle \frac{2cos\left(\theta \right)sin\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\beta \right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)cos\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\alpha \right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)sin\left(\omega \right)}{N}}\left[\sqrt{N-3}cos(\alpha +\beta )+cos(\alpha -\beta )\right]\hfill \\ & +& {\displaystyle \frac{cos\left({\theta }^{\prime }\right)cos\left(\omega \right)}{N}}\left[N-4\right]\hfill \\ & +& {\displaystyle \frac{2cos\left({\theta }^{\prime }\right)sin\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\beta \right)\hfill \\ & +& {\displaystyle \frac{2sin\left({\theta }^{\prime }\right)cos\left(\omega \right)}{N}}\left[1-\sqrt{N-3}\right]cos\left({\alpha }^{\prime }\right)\hfill \\ & +& {\displaystyle \frac{2sin\left({\theta }^{\prime }\right)sin\left(\omega \right)}{N}}\left[\sqrt{N-3}cos({\alpha }^{\prime }+\beta )+cos({\alpha }^{\prime }-\beta )\right]\hfill \\ & +& {\displaystyle \frac{cos\left(\theta \right)cos\left({\omega }^{\prime }\right)}{N}}\left[N-4\right]\hfill \\ & +& {\displaystyle \frac{2cos\left(\theta \right)sin\left({\omega }^{\prime }\right)}{N}}\left[1-\sqrt{N-3}\right]cos\left({\beta }^{\prime }\right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)cos\left({\omega }^{\prime }\right)}{N}}\left[1-\sqrt{N-3}\right]cos\left(\alpha \right)\hfill \\ & +& {\displaystyle \frac{2sin\left(\theta \right)sin\left({\omega }^{\prime }\right)}{N}}\left[\sqrt{N-3}cos(\alpha +{\beta }^{\prime })+cos(\alpha -{\beta }^{\prime })\right]\hfill \\ & -& {\displaystyle \frac{cos\left({\theta }^{\prime }\right)cos\left({\omega }^{\prime }\right)}{N}}\left[N-4\right]\hfill \\ & -& {\displaystyle \frac{2cos\left({\theta }^{\prime }\right)sin\left({\omega }^{\prime }\right)}{N}}\left[1-\sqrt{N-3}\right]cos\left({\beta }^{\prime }\right)\hfill \\ & -& {\displaystyle \frac{2sin\left({\theta }^{\prime }\right)cos\left({\omega }^{\prime }\right)}{N}}\left[1-\sqrt{N-3}\right]cos\left({\alpha }^{\prime }\right)\hfill \\ & -& {\displaystyle \frac{2sin\left({\theta }^{\prime }\right)sin\left({\omega }^{\prime }\right)}{N}}\left[\sqrt{N-3}cos({\alpha }^{\prime }+{\beta }^{\prime })+cos({\alpha }^{\prime }-{\beta }^{\prime })\right].\hfill \end{array}$$

(62)

In

Table 1. The value of the Bell–CHSH correlator $\langle \mathcal{C}\rangle$ for different choices of N.

N	Bell–CHSH Correlator $\langle \mathcal{C}\rangle$
3	2.403
4	2
10	2.10555
${10}^2$	2.03108
${10}^3$	2.00374
${10}^4$	2.00039
${10}^5$	2.00004

, we report the value of the Bell–CHSH correlator $\langle \mathcal{C}\rangle$ for different choices of N, starting from $N=3$, corresponding to the biggest violation. As expected, the violation of the Bell–CHSH inequality decreases as N becomes large. Though, it is still visible for $N={10}^5$.

One sees that, for $N=4$, there is no violation. This is easily understood by observing that the state (11) becomes a product state for $N=4$, i.e.,

$${|\psi \rangle }_4={\displaystyle \frac{1}{2}}\left({|+\rangle }_A+{|-\rangle }_A\right)\otimes \left({|+\rangle }_B+{|-\rangle }_B\right).$$

(63)

Comment Gisin’s theorem holds only for pure entangled states. For mixed states, the task is more difficult. Here, it suffices to mention that there are examples of mixed entangled states that do not exhibit violation of the Bell–CHSH inequality. These states are known as the Werner states [15,16,17,18].

Exercise Consider the entangled state

$$|\psi \rangle ={\displaystyle \frac{1}{\sqrt{1+r^2}}}\left({|+\rangle }_A{\otimes |-\rangle }_B{+r|-\rangle }_A\otimes {|+\rangle }_B\right),r\ne \pm 1.$$

(64)

Check out Gisin’s theorem.

6. The Bell–CHSH Inequality for Spin 1 and Spin 3/2

After having discussed the Bell–CHSH inequality for spin $1/2$, it is instructive to move to spin 1 and spin $3/2$. This will give to the reader a concrete idea of what will be the violation for an arbitrary spin j. For spin 1, the Hilbert space is of dimension three. The spin matrices are given by

$$J_x={\displaystyle \frac{1}{\sqrt{2}}}\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right],J_y={\displaystyle \frac{1}{\sqrt{2}}}\left[\begin{array}{ccc}0& -i& 0\\ i& 0& 0\\ 0& i& 0\end{array}\right],J_z=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right],$$

(65)

with

$$\left[J_i,J_j\right]=i{\epsilon }_{ijk}{J}_k.$$

(66)

An orthonormal basis is given by the three eigenstates of $J_z$, $\left(\right|1\rangle ,|0\rangle ,|-1\rangle )$, corresponding to the eigenvalues $(1,0,-1)$. To investigate the Bell–CHSH correlator, we shall consider a bipartite system $AB$ built out with two spins 1 and select the singlet state

$$|{\psi }_1\rangle ={\displaystyle \frac{1}{\sqrt{3}}}\left({|1\rangle }_A{\otimes |-1\rangle }_B-{|0\rangle }_A\otimes {|0\rangle }_B{+|-1\rangle }_A\otimes {|1\rangle }_B\right).$$

(67)

Before going any further, we need to define the Bell operators. We shall rely on Equation (27), suitably adapted to the case of spin 1, [22,23] namely

$$\begin{array}{ccc}\hfill {A|1\rangle }_A& =& e^{i\alpha }{|-1\rangle }_A{,A|-1\rangle }_A=e^{-i\alpha }{|1\rangle }_A{,A|0\rangle }_A={|0\rangle }_A\hfill \\ \hfill A^{\prime }{|1\rangle }_A& =& e^{i{\alpha }^{\prime }}{|-1\rangle }_A,A^{\prime }{|-1\rangle }_A=e^{-i{\alpha }^{\prime }}{|1\rangle }_A,A^{\prime }{|0\rangle }_A={|0\rangle }_A\hfill \\ \hfill {B|1\rangle }_B& =& e^{i\beta }{|-1\rangle }_B{,B|-1\rangle }_B=e^{-i\beta }{|1\rangle }_B{,B|0\rangle }_B={|0\rangle }_B\hfill \\ \hfill B^{\prime }{|1\rangle }_B& =& e^{i{\beta }^{\prime }}{|-1\rangle }_B,B^{\prime }{|-1\rangle }_B=e^{-i{\beta }^{\prime }}{|1\rangle }_B,B^{\prime }{|0\rangle }_B={|0\rangle }_B.\hfill \end{array}$$

(68)

Proceeding as before, one finds

$$\langle {\psi }_1\left|AB\right|{\psi }_1\rangle ={\displaystyle \frac{1}{3}}\left(1+2cos(\alpha -\beta )\right),$$

(69)

so that, for the Bell–CHSH inequality, one has

$$\langle {\psi }_1\left|\mathcal{C}\right|{\psi }_1\rangle ={\displaystyle \frac{2}{3}}\left(1+cos(\alpha -\beta )+cos({\alpha }^{\prime }-\beta )+cos(\alpha -{\beta }^{\prime })-cos({\alpha }^{\prime }-{\beta }^{\prime })\right),$$

(70)

whose maximum value is

$$\langle {\psi }_1\left|\mathcal{C}\right|{\psi }_1{\rangle }_{max}={\displaystyle \frac{2}{3}}(1+2\sqrt{2})=2.55228,$$

(71)

Unlike the case of the spin $1/2$, for spin 1 Tsirelson’s bound is not attained. As shown in [22,23], this is a general result, valid for any integer spin j. It can be understood in terms of the pairing mechanism outlined in [22]. When the spin is half-integer, i.e., ${\displaystyle \frac{2s+1}{2}}$, $s=0,1$, etc., the dimension of the Hilbert space is even, i.e., $(2s+2)$, so that the basis states can all be combined in pairs. Let us give an explicit example of the pairing mechanism for spin $3/2$. We have four states: $(|{\displaystyle \frac{3}{2}}\rangle ,|{\displaystyle \frac{1}{2}}\rangle ,|-{\displaystyle \frac{3}{2}}\rangle ,|-{\displaystyle \frac{1}{2}}\rangle )$. Out of these states, we can form two pairs; for example, $(|{\displaystyle \frac{3}{2}}\rangle ,|-{\displaystyle \frac{3}{2}}\rangle )$ and $(|{\displaystyle \frac{1}{2}}\rangle ,|-{\displaystyle \frac{1}{2}}\rangle )$. For the Bell operators, we have

$$\begin{array}{ccc}\hfill A|{\displaystyle \frac{3}{2}}{\rangle }_A& =& e^{i{\alpha }_1}|-{\displaystyle \frac{3}{2}}{\rangle }_A,A|-{\displaystyle \frac{3}{2}}{\rangle }_A=e^{-i{\alpha }_1}|{\displaystyle \frac{3}{2}}{\rangle }_A\hfill \\ \hfill A|{\displaystyle \frac{1}{2}}{\rangle }_A& =& e^{i{\alpha }_2}|-{\displaystyle \frac{1}{2}}{\rangle }_A,A|-{\displaystyle \frac{1}{2}}{\rangle }_A=e^{-i{\alpha }_2}|{\displaystyle \frac{1}{2}}{\rangle }_A,\hfill \end{array}$$

(72)

with similar expressions for $(A^{\prime },B,B^{\prime })$. To examine the violation of the Bell–CHSH inequality in the case of spin $3/2$, we start from the expression of the singlet state for an arbitrary spin j [22,23]

$$|{\psi }_j\rangle ={\displaystyle \frac{1}{{(2j+1)}^{\frac{1}{2}}}}\sum _{m=-j}^{m=j}{(-1)}^{j-m}{|m\rangle }_A\otimes {|-m\rangle }_B.$$

(73)

Thus,

$$|{\psi }_{{\displaystyle \frac{3}{2}}}\rangle ={\displaystyle \frac{1}{2}}\left(|{\displaystyle \frac{3}{2}}{\rangle }_A\otimes |-{\displaystyle \frac{3}{2}}{\rangle }_B-|{\displaystyle \frac{1}{2}}{\rangle }_A\otimes |-{\displaystyle \frac{1}{2}}{\rangle }_B+|-{\displaystyle \frac{1}{2}}{\rangle }_A\otimes |{\displaystyle \frac{1}{2}}{\rangle }_B-|-{\displaystyle \frac{3}{2}}{\rangle }_A\otimes |{\displaystyle \frac{3}{2}}{\rangle }_B\right).$$

(74)

An easy computation shows that

$$\langle {\psi }_{{\displaystyle \frac{3}{2}}}\left|AB\right|{\psi }_{{\displaystyle \frac{3}{2}}}\rangle =-{\displaystyle \frac{1}{2}}\left(cos({\alpha }_1-{\beta }_1)+cos({\alpha }_2-{\beta }_2)\right)$$

(75)

yielding the following expression for the spin $3/2$ Bell–CHSH correlator

$$\begin{array}{ccc}\hfill \langle {\psi }_{{\displaystyle \frac{3}{2}}}|\mathcal{C}|{\psi }_{{\displaystyle \frac{3}{2}}}\rangle & =& -{\displaystyle \frac{1}{2}}\left(cos({\alpha }_1-{\beta }_1)+cos({\alpha }_2-{\beta }_2)\right)-{\displaystyle \frac{1}{2}}\left(cos({{\alpha }_1}^{\prime }-{\beta }_1)+cos({{\alpha }_2}^{\prime }-{\beta }_2)\right)\hfill \\ & -& {\displaystyle \frac{1}{2}}\left(cos({\alpha }_1-{{\beta }_1}^{\prime })+cos({\alpha }_2-{{\beta }_2}^{\prime })\right)+{\displaystyle \frac{1}{2}}\left(cos({{\alpha }_1}^{\prime }-{{\beta }_1}^{\prime })+cos({\alpha }_2-{{\beta }_2}^{\prime })\right)\hfill \end{array}$$

(76)

We see, therefore, that each pair gives a maximum contribution equal to $2\sqrt{2}$:

$$|\langle {\psi }_{{\displaystyle \frac{3}{2}}}\left|\mathcal{C}\right|{\psi }_{{\displaystyle \frac{3}{2}}}{\rangle |}_{max}={\displaystyle \frac{1}{2}}4\sqrt{2}=2\sqrt{2},$$

(77)

so that Tsirelson’s bound is attained for half-integer spins. Instead, in the case of integer spins, the dimension of the Hilbert space is odd, i.e., $(2j+1),j=1,2,3$, etc. The pairing mechanism is not perfect. A state, say the state $|0\rangle$, see Equation (68), will always be left unpaired, decreasing a little bit the maximum value of the violation. As a consequence, Tsirelson’s bound will not be attained for integer spins. Exercise Consider the singlet state $|{\psi }_j\rangle$ given in Equation (73). Show that

• for any integer spin $j\ge 1$

  $$|\langle {\psi }_j\left|\mathcal{C}\right|{\psi }_j{\rangle |}_{max}={\displaystyle \frac{2}{(2j+1)}}\left(1+2j\sqrt{2})\right)<2\sqrt{2}.$$

  (78)
• for any half-integer spin

  $$|\langle {\psi }_j\left|\mathcal{C}\right|{\psi }_j{\rangle |}_{max}=2\sqrt{2}.$$

  (79)

7. Coherent States

So far, we have studied the violation of the Bell–CHSH inequality for finite dimensional Hilbert spaces. We face now the more complex case of infinite dimensional Hilbert spaces, considering the examples of the coherent and squeezed states [24,25]. The so called coherent states are an important tool of Quantum Mechanics, with a variety of applications in many areas as diverse as Quantum Optics, Condensed Matter Physics, Nuclear Physics and Elementary Particle Physics. In order to introduce coherent states, let us recall a few basic features of the infinite dimensional Fock space for the Harmonic oscillator

$$\begin{array}{ccc}\hfill H& =& \omega (a^{†}a+{\displaystyle \frac{1}{2}})\hfill \\ \hfill \left[a,a^{†}\right]& =& 1,[a,a]=[a^{†},a^{†}]=0.\hfill \end{array}$$

(80)

The vacuum Fock state $|0\rangle$ is defined by the condition

$$a|0\rangle =0,\langle 0|0\rangle =1.$$

(81)

For the Fock basis, we have

$$|n\rangle ={\displaystyle \frac{{\left(a^{†}\right)}^n}{\sqrt{n!}}}|0\rangle ,\langle n^{\prime }|n\rangle ={\delta }_{n^{\prime }n},$$

(82)

with

$$a|n\rangle =\sqrt{n}|n-1\rangle ,a^{†}|n\rangle =\sqrt{n+1}|n+1\rangle ,H|n\rangle =\omega (n+{\displaystyle \frac{1}{2}})|n\rangle ,$$

(83)

and

$$\sum _{n=0}^{\infty }|n\rangle \langle n|=⊮.$$

(84)

In order to introduce the coherent states, it is useful to make use of the displacement operator

$$\mathcal{D}\left(z\right)=e^{z{a}^{†}-z^{*}a},$$

(85)

where $z=x+iy$ is a complex number. From

$$e^X{e}^Y=e^{(X+Y+{\displaystyle \frac{1}{2}}[X,Y])},\mathrm{valid}\mathrm{when}[X,[X,Y]]=[Y,[X,Y]]=0$$

(86)

it follows that $\mathcal{D}\left(z\right)$ is an unitary operator

$$\mathcal{D}\left(z\right){\mathcal{D}}^{†}\left(z\right)={\mathcal{D}}^{†}\left(z\right)\mathcal{D}\left(z\right)=I,$$

(87)

and that it can be cast in the equivalent form

$$\mathcal{D}\left(z\right)=e^{-{\displaystyle \frac{{\left|z\right|}^2}{2}}}{e}^{z{a}^{†}}{e}^{-z^{*}a}.$$

(88)

Moreover, from

$$e^{X}Y{e}^{-X}=Y+\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k!}}\underset{\mathrm{k} \mathrm{times}}{\underbrace{\left[X,\left[\cdots \cdots \left[X,Y\right]\cdots \cdots \right],\right]}},$$

(89)

it follows that

$${\mathcal{D}}^{†}\left(z\right)a\mathcal{D}\left(z\right)=a+z.$$

(90)

We have now all ingredients to introduce the coherent states $\left\{\right|z\rangle \}$. These states are obtained by acting with the displacement operator $\mathcal{D}\left(z\right)$ on the Fock vacuum $|0\rangle$:

$$|z\rangle =\mathcal{D}\left(z\right)|0\rangle =e^{-{\displaystyle \frac{{\left|z\right|}^2}{2}}}\sum _{n=0}^{\infty }{\displaystyle \frac{z^n}{\sqrt{n!}}}|n\rangle .$$

(91)

Since $\mathcal{D}\left(z\right)$ is unitary, we have

$$\langle z|z\rangle =\langle 0\left|{\mathcal{D}}^{†}\left(z\right)\mathcal{D}\left(z\right)\right|0\rangle =1.$$

(92)

Also,

$$a|z\rangle =a\mathcal{D}\left(z\right)|0\rangle =\mathcal{D}\left(z\right)\left({\mathcal{D}}^{†}\left(z\right)a\mathcal{D}\left(z\right)\right)|0\rangle =\mathcal{D}\left(z\right)(a+z)|0\rangle =z\mathcal{D}\left(z\right)|0\rangle =z|z\rangle .$$

(93)

We see that the coherent state $|z\rangle$ is eigenstate of the annihilation operator a with eigenvalue given by the complex number z. Equation (93) expresses a fundamental property of the coherent states. Exercise Show the resolution of the unity

$$\int {\displaystyle \frac{d^{2}z}{\pi }}|z\rangle \langle z|=⊮.$$

(94)

7.1. The Pseudospin Operators

In view of the study of entangled coherent states and of the ensuing violation of the Bell–CHSH inequality, it is worth providing a discussion on the so-called pseudospin operators [26,27,28]. To that end, we divide the basis states of the Fock space in even and odd states, namely

$$|2n\rangle ={\displaystyle \frac{{\left(a^{†}\right)}^{2n}}{\sqrt{\left(2n\right)!}}}|0\rangle ,|2n+1\rangle ={\displaystyle \frac{{\left(a^{†}\right)}^{2n+1}}{\sqrt{(2n+1)!}}}|0\rangle ,n=0,1,2,\dots .$$

(95)

We now introduce the three operators, defined as

$$\begin{array}{ccc}\hfill s_x^{\left(n\right)}& =& |2n+1\rangle \langle 2n|+|2n\rangle \langle 2n+1|,\hfill \\ \hfill s_y^{\left(n\right)}& =& i\left(|2n\rangle \langle 2n+1|-|2n+1\rangle \langle 2n|\right),\hfill \\ \hfill s_z^{\left(n\right)}& =& |2n+1\rangle \langle 2n+1|-|2n\rangle \langle 2n|.\hfill \end{array}$$

(96)

An easy calculation shows that

$$\begin{array}{cc}\hfill \left[s_x^{\left(n\right)},s_y^{\left(n\right)}\right]& =2i{s}_z^{\left(n\right)},\hfill \\ \hfill \left[s_y^{\left(n\right)},s_z^{\left(n\right)}\right]& =2i{s}_x^{\left(n\right)},\hfill \\ \hfill \left[s_z^{\left(n\right)},s_x^{\left(n\right)}\right]& =2i{s}_y^{\left(n\right)}.\hfill \end{array}$$

(97)

It follows thus that

$$s_x=\sum _{n=0}^{\infty }{s}_x^{\left(n\right)},s_y=\sum _{n=0}^{\infty }{s}_y^{\left(n\right)},s_z=\sum _{n=0}^{\infty }{s}_z^{\left(n\right)},$$

(98)

obey the same algebraic relations of the spin $1/2$ Pauli matrices

$$\begin{array}{c}\hfill \left[s_x,s_y\right]=2i{s}_z,\left[s_y,s_z\right]=2i{s}_x,\left[s_z,s_x\right]=2i{s}_y,\end{array}$$

(99)

from which the name pseudospin follows. In particular, from expressions (96), one observes that the introduction of the pseudospin operators can be related, once again, to a pairing mechanism in Hilbert space, a pair being given by two states, namely $\left(\right|2n\rangle ,|2n+1\rangle )$, with $n=0,1,2,\cdots$ Each pair gives rise to a set of operators, $(s_x^{\left(n\right)},s_y^{\left(n\right)},s_z^{\left(n\right)})$, which obey the algebra of the spin $1/2$ Pauli matrices. As we shall see in the sequel, these operators will be very helpful in the discussion of the entanglement properties of coherent as well as of squeezed states.

7.2. Entangled Coherent States

We are now ready to investigate an example of an entangled coherent state; see [29,30] and refs. therein. We start by considering a bipartite system $\mathcal{H}={\mathcal{H}}_A\otimes {\mathcal{H}}_B$, where ${\mathcal{H}}_A$ and ${\mathcal{H}}_B$ denote the Hilbert spaces of two Harmonic oscillators. An entangled coherent state can be introduced by means of the antisymmetric combination

$$\begin{array}{c}\hfill |{\psi }_a\rangle =\mathcal{N}\left[{|\eta \rangle }_A\otimes {|\sigma \rangle }_B+e^{i\varphi }{|-\eta \rangle }_A\otimes {|-\sigma \rangle }_B\right],\end{array}$$

(100)

where $\varphi$ is an arbitrary phase factor, $(\eta ,\sigma )$ are real parameters and ${\left(\right|\eta \rangle }_A{,|\sigma \rangle }_B)$ are coherent states defined, respectively, in ${\mathcal{H}}_A$ and ${\mathcal{H}}_B$:

$$\begin{array}{ccc}\hfill {|\eta \rangle }_A& =& {\mathcal{D}}_A{\left(\eta \right)|0\rangle }_A=e^{-{\displaystyle \frac{{\eta }^2}{2}}}{e}^{\eta a^{†}}{e}^{-\eta a}{|0\rangle }_A\hfill \\ \hfill {|\sigma \rangle }_B& =& {\mathcal{D}}_B{\left(\sigma \right)|0\rangle }_B=e^{-{\displaystyle \frac{{\sigma }^2}{2}}}{e}^{\sigma b^{†}}{e}^{-\sigma b}{|0\rangle }_B,\hfill \end{array}$$

(101)

with $(a,a^{†})$, $(b,b^{†})$ the creation and annihilation operators of ${\mathcal{H}}_A$ and ${\mathcal{H}}_B$. For the normalization factor $\mathcal{N}$, one finds

$$\begin{array}{c}\hfill \mathcal{N}={\displaystyle \frac{1}{\sqrt{2}}}{\left(1+cos\varphi exp\left[-2\left({\eta }^2+{\sigma }^2\right)\right]\right)}^{-1/2}.\end{array}$$

(102)

We move then to the construction of Bell’s operators $(A,A^{\prime },B,B^{\prime })$. To that aim, we rely on the definition of the pseudospin operators given in the previous section and define

$$\begin{array}{cc}\hfill {A|2n\rangle }_A& =e^{i\alpha }{|2n+1\rangle }_A{,A|2n+1,\rangle }_A=e^{-i\alpha }{|2n\rangle }_A\hfill \\ \hfill A^{\prime }{|2n\rangle }_A& =e^{i{\alpha }^{\prime }}{|2n+1\rangle }_A,A^{\prime }{|2n+1,\rangle }_A=e^{-i{\alpha }^{\prime }}{|2n\rangle }_A\hfill \\ \hfill {B|2m\rangle }_B& =e^{i\beta }{|2m+1\rangle }_B{,B|2m+1,\rangle }_B=e^{-i\beta }{|2m\rangle }_B\hfill \\ \hfill B^{\prime }{|2m\rangle }_B& =e^{i{\beta }^{\prime }}{|2m+1\rangle }_B,B^{\prime }{|2m+1,\rangle }_B=e^{-i{\beta }^{\prime }}{|2m\rangle }_B.\hfill \end{array}$$

(103)

These operators fulfill all required conditions, Equations (21) and (22). In terms of pesusdospin operators, we have

$$A=\overrightarrow{u}·{\overrightarrow{s}}_A\otimes I_B,$$

(104)

where $\overrightarrow{u}$ is the unit vector

$$\overrightarrow{u}=\left(cos\left(a\right),-sin\left(a\right),0\right),\overrightarrow{u}·\overrightarrow{u}=1,$$

(105)

Similar expressions hold for the operators $(A^{\prime },B,B^{\prime })$.

Comment Although a more general form for the operators $(A,A^{\prime },B,B^{\prime })$ can be introduced, see Equation (38), expressions (103) will enable us to already capture the violation of the Bell–CHSH inequality for both coherent and squeezed states. In order to evaluate the Bell–CHSH correlator when the state $|{\psi }_a\rangle$ is employed, i.e.,

$$\langle {\psi }_a\left|\mathcal{C}\right|{\psi }_a\rangle =\langle {\psi }_a|(A+A^{\prime })\otimes B+(A-A^{\prime })\otimes B^{\prime }|{\psi }_a\rangle .$$

(106)

we look at the term $\langle {\psi }_a|A\otimes B|{\psi }_a\rangle$. Making use of the properties of the Fock basis, it turns out that

$$\langle {\psi }_a|A\otimes B{\psi }_a{|\rangle }_A=4{\Omega }_a\left(cos\left(\alpha \right)cos\left(\beta \right)-cos\left(\varphi \right)sin\left(\alpha \right)sin\left(\beta \right)\right)\Delta (\eta ,\sigma ),$$

(107)

with

$$\Delta (\eta ,\sigma )=\sum _{n,m=0}^{\infty }\left[{\displaystyle \frac{{\eta }^{4n+1}{\sigma }^{4m+1}}{\sqrt{\left(2n\right)!(2n+1)!\left(2m\right)!(2m+1)!}}}\right],$$

(108)

end

$$\begin{array}{c}\hfill {\Omega }_a={\displaystyle \frac{exp\left[-\left({\eta }^2+{\sigma }^2\right)\right]}{1+cos\varphi exp\left[-2\left({\eta }^2+{\sigma }^2\right)\right]}}.\end{array}$$

(109)

Therefore, for the Bell–CHSH correlator, we obtain

$$\begin{array}{ccc}\hfill \langle {\psi }_a\left|\mathcal{C}\right|{\psi }_a\rangle & =& 4{\Omega }_a\Delta (\eta ,\sigma )\left(cos\left(\alpha \right)cos\left(\beta \right)-cos\left(\varphi \right)sin\left(\alpha \right)sin\left(\beta \right)\right)\hfill \\ & +& 4{\Omega }_a\Delta (\eta ,\sigma )\left(cos\left({\alpha }^{\prime }\right)cos\left(\beta \right)-cos\left(\varphi \right)sin\left({\alpha }^{\prime }\right)sin\left(\beta \right)\right)\hfill \\ & +& 4{\Omega }_a\Delta (\eta ,\sigma )\left(cos\left(\alpha \right)cos\left({\beta }^{\prime }\right)-cos\left(\varphi \right)sin\left(\alpha \right)sin\left({\beta }^{\prime }\right)\right)\hfill \\ & -& 4{\Omega }_a\Delta (\eta ,\sigma )\left(cos\left({\alpha }^{\prime }\right)cos\left({\beta }^{\prime }\right)+cos\left(\varphi \right)sin\left({\alpha }^{\prime }\right)sin\left(\beta \right)\right)\hfill \end{array}$$

(110)

Although this expression looks pretty simple, the infinite sum cannot be written in a closed form. To analyze the violation, we choose to work with $\varphi =\pi$ and $(\alpha =0,$${\alpha }^{\prime }=\pi /2,\beta =+\pi /4,{\beta }^{\prime }=-\pi /4)$. Firstly, we consider the first contribution of the sum, taking the $n=m=0$ term. Moreover, setting $\sigma =\eta$, it can be written as

$$\begin{array}{c}\hfill \langle {\psi }_a|A\otimes B|{\psi }_a{\rangle |}_{n=m=0,\sigma =\eta }={\displaystyle \frac{4\sqrt{2}{\eta }^2}{sinh\left(2{\eta }^2\right)}}.\end{array}$$

(111)

For small values of $\eta$, expression (111) is already close to $2\sqrt{2}$. We face then the full expression, Equation (110). Taking $\eta =\sigma =0.1$ and performing a numerical analysis of the sum, we already find $\langle \mathcal{C}\rangle =2.8284\simeq 2\sqrt{2}$. We increase now the values of $(\eta ,\sigma )$. For $\eta =\sigma =1$, we still have $\langle \mathcal{C}\rangle =2.6678$. It is worth noticing that a violation of the Bell–CHSH inequality takes place even in the case without a relative phase, that is, when $\varphi =0$. In fact, taking $\varphi =0$, one can consider a slightly different set of parameters: $(\alpha =0,{\alpha }^{\prime }=\pi /2,\beta =-\pi /4,{\beta }^{\prime }=+\pi /4)$. Thus, for $\eta =\sigma =0.7$ one finds $\langle \mathcal{C}\rangle =2.0895$, see [30] for a more general discussion. In conclusion, one can state that the entangled coherent state (100) exhibits a violation of the Bell–CHSH correlator for a wide range of the parameters. Exercise Consider the symmetric entangled coherent state

$$\begin{array}{c}\hfill {|\psi \rangle }_s={\mathcal{N}}_s\left[{|\eta \rangle }_A\otimes {|\sigma \rangle }_B+e^{i\varphi }{|\sigma \rangle }_A\otimes {|\eta \rangle }_B\right].\end{array}$$

(112)

Evaluate the normalization constant ${\mathcal{N}}_s$ and discuss the violation of the Bell–CHSH inequality. Exercise Consider the Schrödinger cat states, i.e.,,

$$\begin{array}{c}\hfill {|\psi \rangle }_{\pm }=C_{\pm }\left[{|\eta \rangle }_{\pm }\otimes {|\sigma \rangle }_{\pm }+e^{i\varphi }{|\sigma \rangle }_{\pm }\otimes {|\eta \rangle }_{\pm }\right],\end{array}$$

(113)

where $C_{\pm }$ is a normalization factor and

$$\begin{array}{c}\hfill {|\eta \rangle }_{\pm }=N_{\pm }\left[|\eta \rangle \pm |-\eta \rangle \right]..\end{array}$$

(114)

Evaluate the normalization factors and analyze the violation of the Bell–CHSH inequality.

8. Squeezed States

Similarly to coherent states, the squeezed states [24,31] find applications in many areas. These states can be defined as

$$|\lambda \rangle =\sqrt{1-{\lambda }^2}\sum _{n=0}^{\infty }{\lambda }^n|n_a{n}_b\rangle ,0<\lambda <1,$$

(115)

where $|n_a{n}_b\rangle$ is a shorthand notation for

$$|n_a{n}_b\rangle ={\displaystyle \frac{{\left(a^{†}\right)}^n}{\sqrt{n!}}}\otimes {\displaystyle \frac{{\left(b^{†}\right)}^n}{\sqrt{n!}}}|0\rangle ,|0\rangle ={|0\rangle }_A\otimes {|0\rangle }_B,$$

(116)

where $(a^{†},b^{†})$ stand for the creation operators of ${\mathcal{H}}_A$ and ${\mathcal{H}}_B$. The parameter $\lambda$ is known as the squeezing parameter. Let us check that the state $|\lambda \rangle$ is correctly normalized. From

$$\langle n_a{n}_b|m_a{m}_b\rangle ={\delta }_{nm}$$

(117)

it follows that

$$\langle \lambda |\lambda \rangle =(1-{\lambda }^2)\sum _{n=0}^{\infty }{\lambda }^{2n}={\displaystyle \frac{1-{\lambda }^2}{1-{\lambda }^2}}=1.$$

(118)

To investigate the violation of the Bell–CHSH inequality, we rewrite the squeezed state in terms of even and odd Fock states, namely

$$|\lambda \rangle =\sqrt{1-{\lambda }^2}\left(\sum _{n=0}^{\infty }{\lambda }^{2n}{|\left(2n\right)}_a{\left(2n\right)}_b\rangle +\sum _{n=0}^{\infty }{\lambda }^{2n+1}|{(2n+1)}_a{(2n+1)}_b\rangle \right).$$

(119)

Making use of the Bell operators of Equation (103), it turns out that

$$A\otimes B|\lambda \rangle =\sqrt{1-{\lambda }^2}\left(e^{i(\alpha +\beta )}\sum _{n=0}^{\infty }{\lambda }^{2n}{|(2n+1)}_a{(2n+1)}_b\rangle +e^{-i(\alpha +\beta )}\sum _{n=0}^{\infty }{\lambda }^{2n+1}|{\left(2n\right)}_a{\left(2n\right)}_b\rangle \right).$$

(120)

Therefore,

$$\langle \lambda |A\otimes B|\lambda \rangle =(1-{\lambda }^2)2\lambda cos(\alpha +\beta )\sum _{n=0}^{\infty }{\lambda }^{4n}={\displaystyle \frac{2\lambda }{1+{\lambda }^2}}cos(\alpha +\beta ).$$

(121)

As a consequence, the Bell–CHSH correlator is found

$$\langle \lambda \left|\mathcal{C}\right|\lambda \rangle =\langle \lambda |(A+A^{\prime })\otimes B+(A-A^{\prime })\otimes B^{\prime }|\lambda \rangle ={\displaystyle \frac{2\lambda }{1+{\lambda }^2}}\left(cos(\alpha +\beta )+cos({\alpha }^{\prime }+\beta )+cos(\alpha +{\beta }^{\prime })-cos({\alpha }^{\prime }+{\beta }^{\prime })\right).$$

(122)

The angular contribution attains its maximum value for

$$\alpha =0,\beta =-{\displaystyle \frac{\pi }{4}},{\alpha }^{\prime }={\displaystyle \frac{\pi }{2}},{\beta }^{\prime }={\displaystyle \frac{\pi }{4}},$$

(123)

yielding

$$\langle \lambda \left|\mathcal{C}\right|\lambda \rangle ={\displaystyle \frac{4\sqrt{2}\lambda }{1+{\lambda }^2}}.$$

(124)

There is violation whenever

$$|\langle \lambda |\mathcal{C}|\lambda \rangle |>2,$$

(125)

which occurs for

$$\sqrt{2}-1<\lambda <1.$$

(126)

The maximum violation is attained when $\lambda$ is very close to 1:

$${|\langle \lambda |\mathcal{C}|\lambda \rangle |}_{\lambda \approx 1}\approx 2\sqrt{2}.$$

(127)

9. Mermin’s Inequalities and GHZ States

As we have seen in the previous examples, the formulation of the Bell–CHSH inequality is valid for a bipartite system $AB$. Nevertheless, there are generalizations of the Bell–CHSH inequality for multipartite systems [32], see also [31] and refs therein. These generalized inequalities are known as the Mermin inequalities. Let us discuss in detail the case of the Mermin inequality for a three partite system $ABC$. The Hilbert space of the system is $\mathcal{H}={\mathcal{H}}_A\otimes {\mathcal{H}}_B\otimes {\mathcal{H}}_C$. Proceeding as in Section 3, we consider three pairs of classical quantities $(a,a^{\prime }),(b,b^{\prime }),(c,c^{\prime })$, taking values $\pm 1$, i.e.,

$$a^2=a^{\prime 2}=b^2=b^{\prime 2}=c^2=c^{\prime 2}=1.$$

(128)

Let us prove that the absolute value of the combination

$$a^{\prime }bc+a{b}^{\prime }c+ab{c}^{\prime }-a^{\prime }{b}^{\prime }{c}^{\prime },$$

(129)

is bounded by 2. We observe that the pair $(c,c^{\prime })$ can take the four values $(1,1),(1,-1),$$(-1,1),(-1,-1)$. Though, since we are considering the absolute value, we can analyze only two cases: $(1,1)$ and $(1,-1)$. It is easy to check out that in both cases one goes back to the classical combination already encountered in the case of the Bell–CHSH inequality, Section 3. In fact, in the first case, we have

$$|a^{\prime }bc+a{b}^{\prime }c+ab{c}^{\prime }-a^{\prime }{b}^{\prime }{c}^{\prime }{|}_{(1,1)}=|a^{\prime }b+a{b}^{\prime }+ab-a^{\prime }{b}^{\prime }|=|(a+a^{\prime })b+(a-a^{\prime })b|\le 2.$$

(130)

Similarly, in the second case, one finds

$$\begin{array}{ccc}\hfill |a^{\prime }bc+a{b}^{\prime }c+ab{c}^{\prime }-a^{\prime }{b}^{\prime }{c}^{\prime }{|}_{(1,-1)}& =& |a^{\prime }b+a{b}^{\prime }-ab+a^{\prime }{b}^{\prime }|\hfill \\ & =& |(a^{\prime }-a)b+(a+a^{\prime })b^{\prime }|\le |a^{\prime }-a|+|a+a^{\prime }|\le 2\hfill \\ \hfill \end{array}$$

(131)

Thus, on classical grounds, we have

$$|a^{\prime }bc+a{b}^{\prime }c+ab{c}^{\prime }-a^{\prime }{b}^{\prime }{c}^{\prime }|\le 2.$$

(132)

At the quantum level, $(a,a^{\prime })$, $(b,b^{\prime })$, $(c,c^{\prime })$ become dichotomic Hermitian operators fulfilling the conditions

$$\begin{array}{ccc}\hfill A& =& A^{†},A^{\prime }={A^{\prime }}^{†},B=B^{†},B^{\prime }={B^{\prime }}^{†},C=C^{†},C^{\prime }={C^{\prime }}^{†}\hfill \\ \hfill A^2& =& {A^{\prime }}^2=B^2={B^{\prime }}^2=C^2={C^{\prime }}^2=1,\hfill \end{array}$$

(133)

and

$$\begin{array}{ccc}\hfill \left[A,B\right]& =& 0,\left[A^{\prime },B\right]=0,\left[A,B^{\prime }\right]=0,\left[A^{\prime },B^{\prime }\right]=0.\left[A,C\right]=0,\left[A^{\prime },C\right]=0\hfill \\ \hfill \left[B,C\right]& =& 0,\left[B^{\prime },C\right]=0,\left[B,C^{\prime }\right]=0,\left[B^{\prime },C^{\prime }\right]=0.\hfill \\ \hfill \left[A,A^{\prime }\right]& \ne & 0,\left[B,B^{\prime }\right]\ne 0,\left[C,C^{\prime }\right]\ne 0.\hfill \end{array}$$

(134)

Following [32], we introduce the Hermitian operator

$${\mathcal{M}}_3={\mathcal{M}}_3^{†}=A^{\prime }\otimes B\otimes C+A\otimes B^{\prime }\otimes C+A\otimes B\otimes C^{\prime }-A^{\prime }\otimes B^{\prime }\otimes C^{\prime }.$$

(135)

Repeating the same procedure employed in the derivation of Tsirelson’s bound, it follows

$${\mathcal{M}}_3^2=4I-\left[A,A^{\prime }\right]\otimes \left[B,B^{\prime }\right]-\left[A,A^{\prime }\right]\otimes \left[C,C^{\prime }\right]-\left[B,B^{\prime }\right]\otimes \left[C,C^{\prime }\right].$$

(136)

Using the properties of the operator norm and recalling that each commutator contributes with a factor 2, we get

$$\left|\right|{\mathcal{M}}_3^2\left|\right|=\left|\right|{\mathcal{M}}_3{\left|\right|}^2\le 16.$$

(137)

As a consequence, for any normalized state $|\psi \rangle$, $\langle \psi |\psi \rangle =1$, we have

$$|\langle \psi |{\mathcal{M}}_3|\psi \rangle |\le 4.$$

(138)

The operator ${\mathcal{M}}_3$ is called the Mermin operator of order three. One speaks of a violation of the Mermin inequality whenever

$$2<|\langle \psi |{\mathcal{M}}_3|\psi \rangle |\le 4.$$

(139)

Let us look now at the analogue of the Bell states in the case of the ${\mathcal{M}}_3$ inequality. These states are known as the Greenberger–Horne–Zeilinger states [33], simply called GHZ states, given by the expression

$$|{\psi }_{GHZ}\rangle ={\displaystyle \frac{{|+\rangle }_A{\otimes |+\rangle }_B{\otimes |+\rangle }_C{-|-\rangle }_A{\otimes |-\rangle }_B\otimes {|-\rangle }_C}{\sqrt{2}}}.$$

(140)

From

$$\begin{array}{ccc}\hfill {A|+\rangle }_A& =& e^{i\alpha }{|-\rangle }_A{,A|-\rangle }_A=e^{-i\alpha }{|+\rangle }_A,\hfill \\ \hfill A^{\prime }{|+\rangle }_A& =& e^{i{\alpha }^{\prime }}{|-\rangle }_A,A^{\prime }{|-\rangle }_A=e^{-i{\alpha }^{\prime }}{|+\rangle }_A,\hfill \\ \hfill {B|+\rangle }_B& =& e^{i\beta }{|-\rangle }_B{,B|-\rangle }_B=e^{-i\beta }{|+\rangle }_B,\hfill \\ \hfill B^{\prime }{|+\rangle }_B& =& e^{i{\beta }^{\prime }}{|-\rangle }_B,B^{\prime }{|-\rangle }_B=e^{-i{\beta }^{\prime }}{|+\rangle }_B,\hfill \\ \hfill {C|+\rangle }_C& =& e^{i\gamma }{|-\rangle }_C{,C|-\rangle }_C=e^{-i\gamma }{|+\rangle }_C,\hfill \\ \hfill C^{\prime }{|+\rangle }_C& =& e^{i{\gamma }^{\prime }}{|-\rangle }_C,C^{\prime }{|-\rangle }_C=e^{-i{\gamma }^{\prime }}{|+\rangle }_C,\hfill \end{array}$$

(141)

one obtains

$$ABC|{\psi }_{GHZ}\rangle ={\displaystyle \frac{-e^{-i(\alpha +\beta +\gamma )}{|+\rangle }_A{\otimes |+\rangle }_B{\otimes |+\rangle }_C+e^{i(\alpha +\beta +\gamma )}{|-\rangle }_A{\otimes |-\rangle }_B\otimes {|-\rangle }_C}{\sqrt{2}}},$$

(142)

so that

$$\langle {\psi }_{GHZ}\left|ABC\right|{\psi }_{GHZ}\rangle =cos(\alpha +\beta +\gamma ).$$

(143)

Thus, for the Mermin inequality, we get

$$\langle {\psi }_{GHZ}|{\mathcal{M}}_3|{\psi }_{GHZ}\rangle =cos({\alpha }^{\prime }+\beta +\gamma )+cos(\alpha +{\beta }^{\prime }+\gamma )+cos(\alpha +\beta +{\gamma }^{\prime })-cos({\alpha }^{\prime }+{\beta }^{\prime }+{\gamma }^{\prime }).$$

(144)

Setting

$$\alpha =0,{\alpha }^{\prime }={\displaystyle \frac{\pi }{2}},\beta =-{\displaystyle \frac{\pi }{4}},{\beta }^{\prime }={\displaystyle \frac{\pi }{4}},\gamma =-{\displaystyle \frac{\pi }{4}},{\gamma }^{\prime }={\displaystyle \frac{\pi }{4}},$$

(145)

one gets maximum violation, i.e.,

$$|\langle {\psi }_{GHZ}|{\mathcal{M}}_3|{\psi }_{GHZ}{\rangle |}_{max}=4.$$

(146)

We see thus that the GHZ state, Equation (149), maximizes the Mermin inequality of order three.

Exercise Consider the Mermin operator of order 4

$$\begin{array}{ccc}\hfill 2{\mathcal{M}}_4& =& -ABCD+(A^{\prime }BCD+A{B}^{\prime }CD+AB{C}^{\prime }D+ABC{D}^{\prime })\hfill \\ & +& (A^{\prime }{B}^{\prime }CD+A^{\prime }B{C}^{\prime }D+A^{\prime }BC{D}^{\prime })\hfill \\ & +& (A{B}^{\prime }CD+A{B}^{\prime }C{D}^{\prime }+AB{C}^{\prime }{D}^{\prime })\hfill \\ & -& (A^{\prime }{B}^{\prime }{C}^{\prime }D+A^{\prime }{B}^{\prime }C{D}^{\prime }+A^{\prime }B{C}^{\prime }{D}^{\prime }+A{B}^{\prime }{C}^{\prime }{D}^{\prime })-A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime },\hfill \end{array}$$

(147)

where $(A,A^{\prime },B,B^{\prime },C,C^{\prime }D,D^{\prime })$ are dichotomic Hermitian operators. Show that, for any normalized state $|\psi \rangle$, there is violation whenever

$$2<\left|\psi \right|{\mathcal{M}}_4|\psi \rangle |\le 4\sqrt{2}$$

(148)

Consider the GHZ state

$$|{\psi }_{GHZ}{\rangle }_4={\displaystyle \frac{{|+\rangle }_A{\otimes |+\rangle }_B{\otimes |+\rangle }_C{\otimes |+\rangle }_D{-|-\rangle }_A{\otimes |-\rangle }_B{\otimes |-\rangle }_C\otimes {|-\rangle }_D}{\sqrt{2}}}.$$

(149)

Show that $|{\psi }_{GHZ}{\rangle }_4$ gives maximum violation of the ${\mathcal{M}}_4$-inequality [34].

10. Conclusions

Since its discovery, Bell’s inequality has been the source of renewed investigations. In these notes, we have provided a concise and self-contained account on a version of the Bell inequality, known as the Bell–CHSH inequality.

Many examples have been worked out, covering both finite and infinite dimensional Hilbert spaces.

The extension of the Bell–CHSH inequality to the multipartite Mermin inequalities has also been outlined.

We hope that the material will be helpful to graduate as well as undergraduate students aiming at pursuing research projects on this fascinating topic.