# What would the CHSH inequality be equal to if the two qubits were separable?

I am writing an Excel spreadsheet to work through the matrix algebra for a simple Bell's game with the following parameters.

$$\newcommand{\bra}{\langle #1|}\newcommand{\ket}{|#1\rangle}A_\pm = \frac{1}{2}\times(I \pm Z)$$

$$A'_\pm = \frac{1}{2}\times(I \pm X)$$

$$B_\pm = \frac{1}{2}\times(I \pm \frac{1}{\sqrt{2}}\times(X+Z))$$

$$B'_\pm- = \frac{1}{2}\times(I \pm \frac{1}{\sqrt{2}}\times(X-Z))$$

For completion, I will define the matrices:

$$I = \begin{bmatrix}1&0\\0&1\end{bmatrix},\qquad X=\begin{bmatrix}0&1\\1&0\end{bmatrix},\qquad Z = \begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$

The expectation value is calculated using the following: Use matrix multiplication to calculate the probability of the outcome. If the outcomes are both positive or both negative, ie A+B+, then the outcome is +1. If the one is positive and the other negative (or vice versa) then the value is -1.

Sample calc: $$Pr(A_+,B_+) = \bra{Phi} A\otimes{B} \ket{Phi}$$

$$ = Pr(A_+,B_+)\times1 + Pr(A_-,B_-)\times1 + Pr(A_+,B_-)\times(-1) + Pr(A_-,B_+)\times(-1)$$

The CHSH Inequality states: $$-++ \leq 2$$

What is typically shown is that when Phi equals an entangled state, the outcome violates the CHSH Inequality.

For example, running through the above calcs with Phi equal to:

$$Phi = \frac{\ket{00}}{\sqrt{2}} + \frac{\ket{11}}{\sqrt{2}}$$

You can show the expectation value is $$2\sqrt{2}$$.

I then calculated the sum with Phi equal to:

$$Phi = \ket{00}$$

The expectation value I got is $$\sqrt{2}$$.

My question is two parts.

• First, is this value correct?

• Second, I want to know how to state in words the calculation being done when the states are separated. It seems to me, in the last situation, this is equivalent to what would happen if the bits were classical bits that both happened to be 0. I imagine a scenario where Alice and Bob are both given a classical bit and then asked to measure it. The sum value I am showing is the outcome of the game when the bits they receive are both 0 (played many times to generate a sample size sufficient to obtain the probabilities).

Yes, it is. If you expand out the calculation you're doing, this is the same as $$\sqrt{2}\langle\psi|X\otimes X+Z\otimes Z|\psi\rangle$$ for any two-qubit state $$|\psi\rangle$$. In the particular case $$|\psi\rangle=|00\rangle$$, it's easy to extract that this is $$\sqrt{2}$$. Indeed, for any separable state of the form $$|\psi\rangle=(\cos\theta_1|0\rangle+\sin\theta_1|1\rangle)\otimes(\cos\theta_2|0\rangle+\sin\theta_2|1\rangle),$$ you get the same answer.
Not really. If the bits are classical, the only measurement you can make on them is in the $$Z$$ basis. You don't have all these funky measurements that are performed, so it isn't really comparable in that way. The only way I see to describe it is simply as two independent single-qubit experiments.