Categories: Physics, Quantum information, Quantum mechanics.

Bell’s theorem

Bell’s theorem states that the laws of quantum mechanics cannot be explained by theories built on so-called local hidden variables (LHVs).

Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$, in an entangled Bell state:

\begin{aligned} \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big) \end{aligned}

Since they are entangled, if we measure the $$z$$-spin of particle $$A$$, and find e.g. $$\ket{\uparrow}$$, then particle $$B$$ immediately takes the opposite state $$\ket{\downarrow}$$. The point is that this collapse is instant, regardless of the distance between $$A$$ and $$B$$.

Einstein called this effect “action-at-a-distance”, and used it as evidence that quantum mechanics is an incomplete theory. He said that there must be some hidden variable $$\lambda$$ that determines the outcome of measurements of $$A$$ and $$B$$ from the moment the entangled pair is created. However, according to Bell’s theorem, he was wrong.

To prove this, let us assume that Einstein was right, and some $$\lambda$$, which we cannot understand, let alone calculate or measure, controls the results. We want to know the spins of the entangled pair along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$, so the outcomes for particles $$A$$ and $$B$$ are:

\begin{aligned} A(\vec{a}, \lambda) = \pm 1 \qquad \quad B(\vec{b}, \lambda) = \pm 1 \end{aligned}

Where $$\pm 1$$ are the eigenvalues of the Pauli matrices in the chosen directions $$\vec{a}$$ and $$\vec{b}$$:

\begin{aligned} \hat{\sigma}_a &= \vec{a} \cdot \vec{\sigma} = a_x \hat{\sigma}_x + a_y \hat{\sigma}_y + a_z \hat{\sigma}_z \\ \hat{\sigma}_b &= \vec{b} \cdot \vec{\sigma} = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z \end{aligned}

Whether $$\lambda$$ is a scalar or a vector does not matter; we simply demand that it follows an unknown probability distribution $$\rho(\lambda)$$:

\begin{aligned} \int \rho(\lambda) \dd{\lambda} = 1 \qquad \quad \rho(\lambda) \ge 0 \end{aligned}

The product of the outcomes of $$A$$ and $$B$$ then has the following expectation value. Note that we only multiply $$A$$ and $$B$$ for shared $$\lambda$$-values: this is what makes it a local hidden variable:

\begin{aligned} \expval{A_a B_b} = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

From this, two inequalities can be derived, which both prove Bell’s theorem.

Bell inequality

If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins:

\begin{aligned} A(\vec{a}, \lambda) = A(\vec{b}, \lambda) = - B(\vec{b}, \lambda) \end{aligned}

The expectation value of the product can therefore be rewritten as follows:

\begin{aligned} \expval{A_a B_b} = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

Next, we introduce an arbitrary third direction $$\vec{c}$$, and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$:

\begin{aligned} \expval{A_a B_b} - \expval{A_a B_c} &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda} \\ &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

Inside the integral, the only factors that can be negative are the last two, and their product is $$\pm 1$$. Taking the absolute value of the whole left, and of the integrand on the right, we thus get:

\begin{aligned} \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big| &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda} \\ &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda} \end{aligned}

Since $$\rho(\lambda)$$ is a normalized probability density function, we arrive at the Bell inequality:

\begin{aligned} \boxed{ \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big| \le 1 + \expval{A_b B_c} } \end{aligned}

Any theory involving an LHV $$\lambda$$ must obey this inequality. The problem, however, is that quantum mechanics dictates the expectation values for the state $$\ket{\Psi^{-}}$$:

\begin{aligned} \expval{A_a B_b} = - \vec{a} \cdot \vec{b} \end{aligned}

Finding directions which violate the Bell inequality is easy: for example, if $$\vec{a}$$ and $$\vec{b}$$ are orthogonal, and $$\vec{c}$$ is at a $$\pi/4$$ angle to both of them, then the left becomes $$0.707$$ and the right $$0.293$$, which clearly disagrees with the inequality, meaning that LHVs are impossible.

CHSH inequality

The Clauser-Horne-Shimony-Holt or simply CHSH inequality takes a slightly different approach, and is more useful in practice.

Consider four spin directions, two for $$A$$ called $$\vec{a}_1$$ and $$\vec{a}_2$$, and two for $$B$$ called $$\vec{b}_1$$ and $$\vec{b}_2$$. Let us introduce the following abbreviations:

\begin{aligned} A_1 &= A(\vec{a}_1, \lambda) \qquad \quad A_2 = A(\vec{a}_2, \lambda) \\ B_1 &= B(\vec{b}_1, \lambda) \qquad \quad B_2 = B(\vec{b}_2, \lambda) \end{aligned}

From the definition of the expectation value, we know that the difference is given by:

\begin{aligned} \expval{A_1 B_1} - \expval{A_1 B_2} = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda} \end{aligned}

We introduce some new terms and rearrange the resulting expression:

\begin{aligned} \expval{A_1 B_1} - \expval{A_1 B_2} &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda} \\ &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \end{aligned}

Taking the absolute value of both sides and invoking the triangle inequality then yields:

\begin{aligned} \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| \\ &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg| + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg| \end{aligned}

Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$, we can reduce this to:

\begin{aligned} \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \\ &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \end{aligned}

Evaluating these integrals gives us the following inequality, which holds for both choices of $$\pm$$:

\begin{aligned} \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| &\le 2 \pm \expval{A_2 B_2} \pm \expval{A_2 B_1} \end{aligned}

We should choose the signs such that the right-hand side is as small as possible, that is:

\begin{aligned} \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| &\le 2 \pm \Big( \expval{A_2 B_2} + \expval{A_2 B_1} \Big) \\ &\le 2 - \Big| \expval{A_2 B_2} + \expval{A_2 B_1} \Big| \end{aligned}

Rearranging this and once again using the triangle inequality, we get the CHSH inequality:

\begin{aligned} 2 &\ge \Big| \expval{A_1 B_1} - \expval{A_1 B_2} \Big| + \Big| \expval{A_2 B_2} + \expval{A_2 B_1} \Big| \\ &\ge \Big| \expval{A_1 B_1} - \expval{A_1 B_2} + \expval{A_2 B_2} + \expval{A_2 B_1} \Big| \end{aligned}

The quantity on the right-hand side is sometimes called the CHSH quantity $$S$$, and measures the correlation between the spins of $$A$$ and $$B$$:

\begin{aligned} \boxed{ S \equiv \expval{A_2 B_1} + \expval{A_2 B_2} + \expval{A_1 B_1} - \expval{A_1 B_2} } \end{aligned}

The CHSH inequality places an upper bound on the magnitude of $$S$$ for LHV-based theories:

\begin{aligned} \boxed{ |S| \le 2 } \end{aligned}

Tsirelson’s bound

Quantum physics can violate the CHSH inequality, but by how much? Consider the following two-particle operator, whose expectation value is the CHSH quantity, i.e. $$S = \expval*{\hat{S}}$$:

\begin{aligned} \hat{S} = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2 \end{aligned}

Where $$\otimes$$ is the tensor product, and e.g. $$\hat{A}_1$$ is the Pauli matrix for the $$\vec{a}_1$$-direction. The square of this operator is then given by:

\begin{aligned} \hat{S}^2 = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_1 \hat{B}_2 + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1^2 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1 \hat{B}_2 \\ + &\hat{A}_2^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2^2 \\ + &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_1^2 + \hat{A}_1 \hat{A}_2 \otimes \hat{B}_1 \hat{B}_2 + \hat{A}_1^2 \otimes \hat{B}_1^2 - \hat{A}_1^2 \otimes \hat{B}_1 \hat{B}_2 \\ - &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_1 \hat{A}_2 \otimes \hat{B}_2^2 - \hat{A}_1^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_1^2 \otimes \hat{B}_2^2 \\ = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_1^2 \otimes \hat{B}_1^2 + \hat{A}_1^2 \otimes \hat{B}_2^2 \\ + &\hat{A}_2^2 \otimes \acomm*{\hat{B}_1}{\hat{B}_2} - \hat{A}_1^2 \otimes \acomm*{\hat{B}_1}{\hat{B}_2} + \acomm*{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_1^2 - \acomm*{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_2^2 \\ + &\hat{A}_1 \hat{A}_2 \otimes \comm*{\hat{B}_1}{\hat{B}_2} - \hat{A}_2 \hat{A}_1 \otimes \comm*{\hat{B}_1}{\hat{B}_2} \end{aligned}

Spin operators are unitary, so their square is the identity, e.g. $$\hat{A}_1^2 = \hat{I}$$. Therefore $$\hat{S}^2$$ reduces to:

\begin{aligned} \hat{S}^2 &= 4 \: (\hat{I} \otimes \hat{I}) + \comm*{\hat{A}_1}{\hat{A}_2} \otimes \comm*{\hat{B}_1}{\hat{B}_2} \end{aligned}

The norm $$\norm*{\hat{S}^2}$$ of this operator is the largest possible expectation value $$\expval*{\hat{S}^2}$$, which is the same as its largest eigenvalue. It is given by:

\begin{aligned} \norm{\hat{S}^2} &= 4 + \norm{\comm*{\hat{A}_1}{\hat{A}_2} \otimes \comm*{\hat{B}_1}{\hat{B}_2}} \\ &\le 4 + \norm{\comm*{\hat{A}_1}{\hat{A}_2}} \norm{\comm*{\hat{B}_1}{\hat{B}_2}} \end{aligned}

We find a bound for the norm of the commutators by using the triangle inequality, such that:

\begin{aligned} \norm{\comm*{\hat{A}_1}{\hat{A}_2}} = \norm{\hat{A}_1 \hat{A}_2 - \hat{A}_2 \hat{A}_1} \le \norm{\hat{A}_1 \hat{A}_2} + \norm{\hat{A}_2 \hat{A}_1} \le 2 \norm{\hat{A}_1 \hat{A}_2} \le 2 \end{aligned}

And $$\norm*{\comm*{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason. The norm is the largest eigenvalue, therefore:

\begin{aligned} \norm{\hat{S}^2} \le 4 + 2 \cdot 2 = 8 \quad \implies \quad \norm{\hat{S}} \le \sqrt{8} = 2 \sqrt{2} \end{aligned}

We thus arrive at Tsirelson’s bound, which states that quantum mechanics can violate the CHSH inequality by a factor of $$\sqrt{2}$$:

\begin{aligned} \boxed{ |S| \le 2 \sqrt{2} } \end{aligned}

Importantly, this is a tight bound, meaning that there exist certain spin measurement directions for which Tsirelson’s bound becomes an equality, for example:

\begin{aligned} \hat{A}_1 = \hat{\sigma}_z \qquad \hat{A}_2 = \hat{\sigma}_x \qquad \hat{B}_1 = \frac{\hat{\sigma}_z + \hat{\sigma}_x}{\sqrt{2}} \qquad \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}} \end{aligned}

Using the fact that $$\expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$, it can then be shown that $$S = 2 \sqrt{2}$$ in this case.

References

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
2. J.B. Brask, Quantum information: lecture notes, 2021, unpublished.