# Deriving probability in Bell's inequality

Measuring the spin with an arbitrary angle $$\theta$$ with respect to the plane xz we obtain the spin operator (multiplying Pauli matrices per the projection $$(\cos\theta,\sin\theta)$$)

$$\hat{\mathrm{S}}_{\hat{n}}=\sin(\theta)\hat{\mathrm{S}}_x+\cos(\theta)\hat{\mathrm{S}}_y=\frac{\hbar}{2}\left[\begin{array}{cc} \cos (\theta)& \sin(\theta) \\ \sin (\theta)& -\cos(\theta) \end{array}\right]$$

and when diagonalized, thr eigenvectors and eigenvalues found are

$$\left[\begin{array}{l} \cos(\theta/2)\\ \sin(\theta/2) \end{array}\right],\quad\left[\begin{array}{c} -\sin(\theta/2)\\ \cos(\theta/2) \end{array}\right]$$

you can write the up component from the z axis basis to the arbitrary direction basis:

$$\left[\begin{array}{l} 1 \\ 0 \end{array}\right]=\cos(\theta/2)\left[\begin{array}{l} \cos (\theta/2) \\ \sin (\theta/2) \end{array}\right]-\sin(\theta/2)\left[\begin{array}{c} -\sin (\theta/2) \\ \cos (\theta/2) \end{array}\right]$$

and the down:

$$\left[\begin{array}{l} 0 \\ 1 \end{array}\right]=\sin(\theta/2)\left[\begin{array}{l} \cos (\theta/2) \\ \sin (\theta/2) \end{array}\right]+\cos(\theta/2)\left[\begin{array}{c} -\sin (\theta/2) \\ \cos (\theta/2) \end{array}\right]$$

In Bell type experiments you measure first with an angle $$\theta$$ the first particle and $$\theta'$$ the second. Substituting $$\theta$$ for $$\theta'$$ everything is the same.

The state of the 2 particles needs to be up-down or down-up for conservation of angular momentum:

$$|\psi\rangle=\frac{1}{\sqrt{2}}(|ud\rangle_z-|du\rangle_z)$$

So the probability of for instance up-up (up in each particle in each direction set by each angle) is:

$$P=|_{\theta}\langle u|_{\theta'}\langle u|\psi\rangle|^2=\ldots=\frac{1}{2}|\sin(\theta/2)\cos(\theta'/2)-\sin(\theta'/2)\cos(\theta/2)|^2=\frac{1}{2}\sin^2(\Delta/2)$$

with $$\Delta=\theta'-\theta$$.

But my question is: if the initial supersposition had a change in the pase and we had a plus sign, i.e.,

$$|\psi\rangle=\frac{1}{\sqrt{2}}(|ud\rangle_z+|du\rangle_z)$$

we would obtain a sum of angles, not a difference. And the result the books show is a difference of angles, so you need the initial state with a minus. How is this possible? How is it that the phase is so important? Am I missing something?

The phase is important! If you change the phase in the initial state, you no longer have the property that the state takes the form $$|ud\rangle+|du\rangle$$ in any basis. It is only the singlet state (i.e., the one with the minus sign) that has this property.

It is still possible to show that any state with any relative phase $$|ud\rangle+\exp(i\phi)|du\rangle$$ is maximally entangled, it's just that the correlations you expect from the pairs of measurements will be different.

• Thanks, but I still can't understand why the singlet state is the only taking the form $|ud\rangle+|du\rangle$ in any basis. Sure I need more background. Could you please provide a reference? Commented Jul 14, 2023 at 7:07
• @David that is a calculation you can do! Write your up and down eigenstates in your $\theta$ basis, as you have done, construct $|ud\rangle_\theta-|du\rangle_\theta$, and show that it is equal to the construction in your original basis $|ud\rangle_z-|du\rangle_z$. Much better than a reference Commented Jul 14, 2023 at 13:39
• @David similar answer physics.stackexchange.com/a/175940/291677. Then further reading physics.stackexchange.com/a/421274/291677 Commented Jul 14, 2023 at 13:41