# Violation of Bell’s theorem

### Bell's theorem

Bell's theorem is a result from set theory. In classical terms, it is unremarkable but we shall show that it does not hold for the singlet state - the simplest quantum system that exhibits entanglement.

It shouldn't be too surprising, since states in quantum theory are complex vectors, rather than elements of sets. The violation of Bell's theorem is a very simple way to see that there is no underlying classical interpretation of quantum mechanics. Historically, this was known before Bell, but his theorem is perhaps the most elegant demonstration.

Later, Alan Aspect proved the result for entangled photons rather than electrons. The measurements made on the photons were sufficiently simultaneous so that no light signal (information) could travel between them, hence completely eliminating any chance that the result was due to anything other than entanglement.

Figure 5.3 - Bell's theorem

Let $A,B,C \subset U$ be three finite subsets of some universal set $U$. We use the notation $$A^c = \left\{x \in U: x \notin A\right\}$$ for the complement of $A$, and then \begin{align*}A \setminus B &= A \cap B^c\\ &= \left\{x \in U: x \in A \text{ and } x \notin B\right\}\end{align*}

We now define a function, $N$, say that just counts the number of elements in a set. Then, Bell's theorem states that $$N\left(A \setminus B\right) + N\left(B \setminus C\right) \geq N\left(A \setminus C\right)$$

To prove it, note that, from figure 5.3, \begin{align*}N\left(A \setminus B\right) &= N(1) + N(4)\\N\left(B \setminus C\right) &= N(2) + N(3)\\N\left(A \setminus C\right) &= N(1) + N(2)\end{align*}

So, we get \begin{align*}N\left(A \setminus B\right) + N\left(B \setminus C\right) &= N(1) + N(4) + N(2) + N(3)\\&\geq N(1) + N(2)\\&= N\left(A \setminus C\right)\end{align*}

We can easily convert this into statements about probabilities of propositions. If $N(U)$ is the total number of elements in a universal set of propositions, then the probability that a particular proposition, $X$, is true is the function $$P(X) = \frac{N(X)}{N(U)}$$

Bell's theorem then becomes $$P\left(A \setminus B\right) + P\left(B \setminus C\right) \geq P\left(A \setminus C\right)$$

### Propositions about the singlet state

We shall consider the spin of the singlet state along the directions $$\hat{z}, \hat{x}, \hat{w} = \frac{\hat{z} + \hat{x}}{2}$$

Consider the following propositions,

• $A$: the spin of the first electron is measured to be $+1$ in the $\hat{z}$ direction.
• $B$: the spin of the first electron is measured to be $+1$ in the $\hat{w}$ direction.
• $C$: the spin of the first electron is measured to be $+1$ in the $\hat{x}$ direction.

We also need the complement propositions $B^c,C^c$ for the theorem. Due to the nature of the singlet state, if the spin of the first electron is not measured to be $+1$ in the $\hat{n}$ direction, say, then that must mean that the spin of the second electron is measured to be $+1$ in the $\hat{n}$ direction. Thus,

• $B^c$: the spin of the second electron is measured to be $+1$ in the $\hat{w}$ direction.
• $C^c$: the spin of the second electron is measured to be $+1$ in the $\hat{x}$ direction.

For ease of notation, write $$[\hat{m}, \hat{n}] \rightarrow \left\{\begin{matrix}1\text{st electron measured } +1 \text{ in } \hat{m} \text{ direction }\\2\text{nd electron measured } +1 \text{ in } \hat{n} \text{ direction }\end{matrix}\right.$$

Figure 5.4 - Propositions about the singlet state

We can now combine these propositions,

• $A \setminus B \rightarrow [\hat{z}, \hat{w}]$
• $B \setminus C \rightarrow [\hat{w}, \hat{x}]$
• $A \setminus C \rightarrow [\hat{z}, \hat{x}]$

We shall show that, contrary to Bell's theorem, the probabilities that these propositions are true satisfy $$P[\hat{z}, \hat{w}] + P[\hat{w}, \hat{x}] \not \geq P[\hat{z}, \hat{x}]$$

### Calculating the probabilities

We first note that rotational symmetry implies that $P[\hat{z}, \hat{w}] = P[\hat{w}, \hat{x}]$, since only the angle between the pair of electrons determines the probability and this angle is $45^{\circ}$ for both $P[\hat{z}, \hat{w}]$ and $P[\hat{w}, \hat{x}]$. So, in order to contradict Bell's theorem, we need to show that $$2P[\hat{z}, \hat{w}] \not\geq P[\hat{z}, \hat{x}]$$

The projection operator associated with the combined proposition $[\hat{m}, \hat{n}]$ is $$\frac{\mathbf{I} + \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_n}{2}$$ and the corresponding probability that the spin of each electron of the singlet state, $\ket{S}$ is measured as up in the given directions is $$P[\hat{m}, \hat{n}] = \bra{S}\frac{\mathbf{I} + \boldsymbol{\sigma}_m}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_n}{2}\ket{S}$$

Now, in matrix form, the general projection operator for the first electron is $$\frac{\mathbf{I} + \boldsymbol{\sigma}_m}{2} = \tfrac{1}{2} \begin{bmatrix} 1 + m_z & 0 & m_- & 0 \\ 0 & 1 + m_z & 0 & m_-\\ m_+ & 0 &1 - m_z & 0 \\ 0 & m_+ & 0 & 1 - m_z \end{bmatrix}$$

We only need to consider the $\hat{z}$ operator, $$\frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} = \tfrac{1}{2} \begin{bmatrix} 1 + 1 & 0 & 0 & 0 \\ 0 & 1 + 1 & 0 & 0\\ 0 & 0 &1 - 1 & 0 \\ 0 & 0 & 0 & 1 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

The general projection operator for the second electron is $$\frac{\mathbf{I} + \boldsymbol{\tau}_n}{2} = \tfrac{1}{2} \begin{bmatrix} 1 + n_z & n_- & 0 & 0 \\ n_+ & 1 - n_z & 0 & 0 \\ 0 & 0 & 1 + n_z & n_- \\ 0 & 0 & n_+ & 1 - n_z \end{bmatrix}$$

We need two second-electron operators, for the $\hat{w}, \hat{x}$ directions. They are $$\frac{\mathbf{I} + \boldsymbol{\tau}_w}{2} = \tfrac{1}{2} \begin{bmatrix} 1 + \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} & 0 & 0 \\ \tfrac{1}{\sqrt{2}} & 1 - \tfrac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 + \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} \\ 0 & 0 & \tfrac{1}{\sqrt{2}} & 1 - \tfrac{1}{\sqrt{2}} \end{bmatrix} = \frac{1}{2\sqrt{2}} \begin{bmatrix} \sqrt{2} + 1 & 1 & 0 & 0 \\ 1 & \sqrt{2} - 1 & 0 & 0 \\ 0 & 0 & \sqrt{2} + 1 & 1 \\ 0 & 0 & 1 & \sqrt{2} -1 \end{bmatrix}$$ and $$\frac{\mathbf{I} + \boldsymbol{\tau}_x}{2} = \tfrac{1}{2} \begin{bmatrix} 1 + 0 & 1 & 0 & 0 \\ 1 & 1 - 0 & 0 & 0 \\ 0 & 0 & 1 + 0 & 1 \\ 0 & 0 & 1 & 1 - 0 \end{bmatrix} = \tfrac{1}{2} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}$$

Thus, the two relevant compound projections in matrix form are \begin{align*}[\hat{z}, \hat{w}] &= \frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_w}{2}\\&= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \frac{1}{2\sqrt{2}} \begin{bmatrix} \sqrt{2} + 1 & 1 & 0 & 0 \\ 1 & \sqrt{2} - 1 & 0 & 0 \\ 0 & 0 & \sqrt{2} + 1 & 1 \\ 0 & 0 & 1 & \sqrt{2} -1 \end{bmatrix}\\&= \frac{1}{2\sqrt{2}} \begin{bmatrix} \sqrt{2} + 1 & 1 & 0 & 0 \\ 1 & \sqrt{2} - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\end{align*} and \begin{align*}[\hat{z}, \hat{x}] &= \frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_x}{2}\\&= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \tfrac{1}{2} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix}\\&= \tfrac{1}{2} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\end{align*}

The singlet state can be written as $$\ket{S} = \tfrac{1}{\sqrt{2}}\begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \end{bmatrix}$$ and so now we are in a position to calculate the probabilities \begin{align*}2P[\hat{z}, \hat{w}] &= 2\bra{S}\frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_w}{2}\ket{S}\\&= 2\left(\tfrac{1}{\sqrt{2}} \begin{bmatrix}0 & 1 & -1 & 0 \end{bmatrix}\right) \left(\frac{1}{2\sqrt{2}} \begin{bmatrix} \sqrt{2} + 1 & 1 & 0 & 0 \\ 1 & \sqrt{2} - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\right) \left(\tfrac{1}{\sqrt{2}}\begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\right)\\&= \frac{1}{2\sqrt{2}} \begin{bmatrix}0 & 1 & -1 & 0 \end{bmatrix} \begin{bmatrix}1 \\ \sqrt{2} - 1 \\ 0 \\ 0 \end{bmatrix}\\&= \frac{\sqrt{2} - 1}{2\sqrt{2}} \approx 0.146\end{align*}

Also, \begin{align*}P[\hat{z}, \hat{x}] &= \bra{S}\frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_x}{2}\ket{S}\\&= \left(\tfrac{1}{\sqrt{2}} \begin{bmatrix}0 & 1 & -1 & 0 \end{bmatrix}\right) \left(\tfrac{1}{2} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\right) \left(\tfrac{1}{\sqrt{2}}\begin{bmatrix}0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\right)\\&= \tfrac{1}{4} \begin{bmatrix}0 & 1 & -1 & 0 \end{bmatrix} \begin{bmatrix}1 \\ 1 \\ 0 \\ 0 \end{bmatrix}\\&= \tfrac{1}{4} = 0.25\end{align*}

Clearly, $$2P[\hat{z}, \hat{w}] \approx 0.146 \not\geq 0.25 = P[\hat{z}, \hat{x}]$$ hence we have shown that the singlet state violates Bell's theorem.