# Singlet state

### Compound spin projections

We have previously shown that spin operators associated with a pair of electrons are given by \begin{align*}\Pi_{\boldsymbol{\sigma}_m \equiv +1} &= \frac{\mathbf{I} + \sigma_m}{2}\\\Pi_{\boldsymbol{\tau}_n \equiv +1} &= \frac{\mathbf{I} + \tau_n}{2}\end{align*} where $\hat{m}, \hat{n}$ are given directions in normal space and so the probability that a pair of electrons in the singlet state $$\ket{S} = \tfrac{1}{\sqrt{2}}\left(\ket{ud} - \ket{du}\right)$$ will be measured to be in the two directions is $$\bra{S}\frac{\mathbf{I} + \sigma_m}{2} \frac{\mathbf{I} + \tau_n}{2}\ket{S}$$

### Varying the angle

Figure 5.4 - Propositions about the singlet state

Recall, from figure 5.4, in the test of Bell's theorem, we measured the state in directions with an angle of either $45^{\circ}$ or $90^{\circ}$.

For example, the set $A \setminus B$ corresponded to the $1$st electron being measured to be up ($+1$) in the $\hat{z}$ direction and the $2$nd electron being measured to be up in the $\hat{w} = \tfrac{1}{\sqrt{2}}\left(\hat{z} + \hat{x}\right)$ direction.

Figure 6.1 - Varying the angle

We can generalise this result to arbitrary angles between the two directions. Let $$\hat{\theta} = \hat{z}\cos\theta + \hat{x}\sin\theta$$

The corresponding projection (for the $2$nd electron) is then \begin{align*}\Pi_{\boldsymbol{\tau}_\theta \equiv +1} &= \frac{\mathbf{I} + \boldsymbol{\tau}_\theta}{2}\\&= \frac{\mathbf{I} + \boldsymbol{\tau}_z \cos\theta + \boldsymbol{\tau}_x \sin\theta}{2}\end{align*}

The probability that the two electrons of the singlet state will be measured to be up in the directions $\hat{z}, \hat{\theta}$ is then \begin{align*}\bra{S}\Pi_{\boldsymbol{\sigma}_z \equiv +1} \Pi_{\boldsymbol{\tau}_\theta \equiv +1} \ket{S} &= \bra{S} \frac{\mathbf{I} + \boldsymbol{\sigma}_z}{2} \frac{\mathbf{I} + \boldsymbol{\tau}_z \cos\theta + \boldsymbol{\tau}_x \sin\theta}{2} \ket{S}\\&= \tfrac{1}{4}\begin{bmatrix}0&1&-1&0\end{bmatrix}\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix}1 + \cos\theta & \sin\theta & 0 & 0\\ \sin\theta & 1 - \cos\theta & 0 & 0\\ 0 & 0 & 1 + \cos\theta & \sin\theta\\ 0 & 0 & \sin\theta & 1 - \cos\theta \end{bmatrix}\begin{bmatrix}0\\1\\-1\\0\end{bmatrix}\\&= \tfrac{1}{4}\begin{bmatrix}0&1&-1&0\end{bmatrix}\begin{bmatrix}1 + \cos\theta & \sin\theta & 0 & 0\\ \sin\theta & 1 - \cos\theta & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0\\1\\-1\\0\end{bmatrix}\\&= \tfrac{1}{4}\begin{bmatrix}0&1&-1&0\end{bmatrix}\begin{bmatrix}\sin\theta\\1 - \cos\theta\\0\\0\end{bmatrix}\\&= \frac{1 - \cos\theta}{4}\end{align*}

Figure 6.2 - Probability as a function of the angle

Thus the probability is only a function of the angle, $\theta$, between the two directions, as plotted in figure 6.2 (courtesy of Wolfram|Alpha).

There is zero probability that the electrons will both be measured up in the same direction.

The maximum probability of 50% occurs when $\theta = \pi$, or $\hat{\theta} = -\hat{z}$, which means measuring up in opposite directions. The other possibility, given $\theta = \pi$, is that they are both measured down in opposite directions, and that also has a 50% chance of occurring.

Finally, we can check the results we got from Bell's theorem, \begin{align*}2\tfrac{1}{4}\left(1 - \cos 45^\circ\right) &= \tfrac{1}{2}\left(1 - \tfrac{1}{\sqrt{2}}\right) \approx 0.146\\\tfrac{1}{4}\left(1 - \cos 90^\circ\right) &= \tfrac{1}{4}\left(1 - 0\right) = 0.25\end{align*}