Categories: Physics, Quantum information, Quantum mechanics.

CHSH inequality

The Clauser-Horne-Shimony-Holt (CHSH) inequality is an alternative proof of Bell’s theorem, which takes a slightly different approach and is more useful in practice.

Suppose there is a local hidden variable (LHV) λ\lambda with an unknown probability density ρ\rho:

∫ρ(λ)dλ=1ρ(λ)≥0\begin{aligned} \int \rho(\lambda) \dd{\lambda} = 1 \qquad \quad \rho(\lambda) \ge 0 \end{aligned}

Given two spin-1/2 particles AA and BB, measuring their spins along arbitrary directions a⃗\vec{a} and b⃗\vec{b} would give each an eigenvalue ±1\pm 1. We write this as:

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

If AA and BB start in an entangled Bell state, e.g. ∣Ψ−⟩\ket{\Psi^{-}}, then we expect a correlation between their measurements results. The product of the outcomes of AA and BB is:

⟨AaBb⟩≡∫ρ(λ) A(a⃗,λ) B(b⃗,λ)dλ\begin{aligned} \Expval{A_a B_b} \equiv \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

So far, we have taken the same path as for proving Bell’s inequality, but for the CHSH inequality we must now diverge.

Deriving the inequality

Consider four spin directions, two for AA called a⃗1\vec{a}_1 and a⃗2\vec{a}_2, and two for BB called b⃗1\vec{b}_1 and b⃗2\vec{b}_2. Let us introduce the following abbreviations:

A1≡A(a⃗1,λ)A2≡A(a⃗2,λ)B1≡B(b⃗1,λ)B2≡B(b⃗2,λ)\begin{aligned} A_1 \equiv A(\vec{a}_1, \lambda) \qquad \quad A_2 \equiv A(\vec{a}_2, \lambda) \qquad \quad B_1 \equiv B(\vec{b}_1, \lambda) \qquad \quad B_2 \equiv B(\vec{b}_2, \lambda) \end{aligned}

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

⟨A1B1⟩−⟨A1B2⟩=∫ρ(λ)(A1B1−A1B2)dλ\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:

⟨A1B1⟩−⟨A1B2⟩=∫ρ(λ)(A1B1−A1B2±A1B1A2B2∓A1B1A2B2)dλ=∫ρ(λ)A1B1(1±A2B2)dλ− ⁣∫ρ(λ)A1B2(1±A2B1)dλ\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:

∣⟨A1B1⟩−⟨A1B2⟩∣=∣ ⁣∫ρ(λ)A1B1(1±A2B2)dλ− ⁣∫ρ(λ)A1B2(1±A2B1)dλ ⁣∣≤∣ ⁣∫ρ(λ)A1B1(1±A2B2)dλ ⁣∣+∣ ⁣∫ρ(λ)A1B2(1±A2B1)dλ ⁣∣\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 the spin eigenvalues of AA and BB is always either −1-1 or +1+1 for all directions, we can reduce this to:

∣⟨A1B1⟩−⟨A1B2⟩∣≤∫ρ(λ)∣A1B1∣(1±A2B2)dλ+ ⁣∫ρ(λ)∣A1B2∣(1±A2B1)dλ≤∫ρ(λ)(1±A2B2)dλ+ ⁣∫ρ(λ)(1±A2B1)dλ\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:

∣⟨A1B1⟩−⟨A1B2⟩∣≤2±⟨A2B2⟩±⟨A2B1⟩\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:

∣⟨A1B1⟩−⟨A1B2⟩∣≤2±(⟨A2B2⟩+⟨A2B1⟩)≤2−∣⟨A2B2⟩+⟨A2B1⟩∣\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:

2≥∣⟨A1B1⟩−⟨A1B2⟩∣+∣⟨A2B2⟩+⟨A2B1⟩∣≥∣⟨A1B1⟩−⟨A1B2⟩+⟨A2B2⟩+⟨A2B1⟩∣\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 SS, and measures the correlation between the spins of AA and BB:

S≡⟨A2B1⟩+⟨A2B2⟩+⟨A1B1⟩−⟨A1B2⟩\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 SS for LHV-based theories:

∣S∣≤2\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=⟨S^⟩S = \expval{\hat{S}}:

S^=A^2⊗B^1+A^2⊗B^2+A^1⊗B^1−A^1⊗B^2\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. A^1\hat{A}_1 is the Pauli matrix for the a⃗1\vec{a}_1-direction. The square of this operator is then given by:

S^2=A^22⊗B^12+A^22⊗B^1B^2+A^2A^1⊗B^12−A^2A^1⊗B^1B^2+A^22⊗B^2B^1+A^22⊗B^22+A^2A^1⊗B^2B^1−A^2A^1⊗B^22+A^1A^2⊗B^12+A^1A^2⊗B^1B^2+A^12⊗B^12−A^12⊗B^1B^2−A^1A^2⊗B^2B^1−A^1A^2⊗B^22−A^12⊗B^2B^1+A^12⊗B^22=A^22⊗B^12+A^22⊗B^22+A^12⊗B^12+A^12⊗B^22+A^22⊗{B^1,B^2}−A^12⊗{B^1,B^2}+{A^1,A^2}⊗B^12−{A^1,A^2}⊗B^22+A^1A^2⊗[B^1,B^2]−A^2A^1⊗[B^1,B^2]\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. A^12=I^\hat{A}_1^2 = \hat{I}. Therefore S^2\hat{S}^2 reduces to:

S^2=4 (I^⊗I^)+[A^1,A^2]⊗[B^1,B^2]\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 ∥S^2∥\norm{\hat{S}^2} of this operator is the largest possible expectation value ⟨S^2⟩\expval{\hat{S}^2}, which is the same as its largest eigenvalue. It is given by:

∥S^2∥=4+∥[A^1,A^2]⊗[B^1,B^2]∥≤4+∥[A^1,A^2]∥∥[B^1,B^2]∥\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:

∥[A^1,A^2]∥=∥A^1A^2−A^2A^1∥≤∥A^1A^2∥+∥A^2A^1∥≤2∥A^1A^2∥≤2\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 ∥[B^1,B^2]∥≤2\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2 for the same reason. The norm is the largest eigenvalue, therefore:

∥S^2∥≤4+2⋅2=8  ⟹  ∥S^∥≤8=22\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 2\sqrt{2}:

∣S∣≤22\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:

A^1=σ^zA^2=σ^xB^1=σ^z+σ^x2B^2=σ^z−σ^x2\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}

Fundamental quantum mechanics says that ⟨AaBb⟩=−a⃗⋅b⃗\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}, so S=22S = 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.