Mermin's EPR paradox explanation [closed]

Question

I am reading N. D. Mermin, Bringing home the atomic world: Quantum mysteries for anybody . The main part of the article is pretty clear to me. But I am not sure how quantum mechanics described in Appendix generates Cases (a) and (b) on p.941. Can someone explicate in mathematical details?

Answer 1

I would really help if you just wrote the cases in question out, so we don't have to do all the work.

I did not know the Red/Green box construction was do to David Mermin. His stuff is gold.

Anyway, "C" produces singlet state (up to normalization)

$$ \psi = \psi_{1,2} + \psi_{2,1} = |\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle $$ where the first (second) arrow labels the particles going to detector A (B).

We can write that in matrix form as: $$ \psi = \left[\begin{array}{c} 1 \\ 0 \end{array}\right]_1 \left[\begin{array}{c} 0\\1\end{array}\right]_2 - \left[\begin{array}{c} 0\\1\end{array}\right]_1 \left[\begin{array}{c} 1\\0\end{array}\right]_2 $$

(Note that the subscripts on the brackets indicate particle index, so these matrices are from different spaces, and don't matrix multiple with each other).

Now $C$ can randomize the angle of the spin quantization, so that state becomes

$$ \psi = \left[\begin{array}{c} \cos(\theta/2)\\\sin(\theta/2)\end{array}\right]_1 \left[\begin{array}{c}\cos((\theta+\pi)/2)\\\sin((\theta+\pi)/2)\end{array}\right]_2 - \left[\begin{array}{c}\cos((\theta+\pi)/2)\\\sin((\theta+\pi)/2)\end{array}\right]_1 \left[\begin{array}{c} \cos(\theta/2)\\\sin(\theta/2)\end{array}\right]_2 $$

where $\theta$ is the angle of quantization wrt to the arbitrary z axis.

Now we set up the switches, which I am not calling $(1,2,3)$ because obv. reasons. They're now $(i, j, k)$, and their alignment directions are (picking a starting angle without loss of generality):

$$ \phi_i = 0 $$ $$ \phi_j = 120^{\circ} $$ $$ \phi_k = 240^{\circ} $$

So if the switches are the same: $A_i=0, B_i=0$ (picking a switch position without loss of generality), and a random state is created. It is measured by detector $A$ (without loss of generality), and choses aligned ($+$) or anti aligned ($-$) with $\phi_i = 0$

Particle $A$ has probabilites

$$p_+ = \cos^2{(\theta-\phi_1)/2} = \cos^2(\theta/2)$$ $$p_- = \sin^2{(\theta-\phi_1)/2} = \sin^2(\theta/2)$$

Note that + (-) means red (green) in A and green (red) in B.

Suppose it is $+$, without loss of generality:

Then $A$ shows red and the state "collapses" (not my favorite term) to:

$$ \psi = \left[\begin{array}{c} 1 \\ 0 \end{array}\right] \left[\begin{array}{c} 0\\1\end{array}\right]$$

Now when $B$ measures, it see (-) with 100% probability and also shows red.

for different switches, since $\cos^2(\frac{120^{\circ}} 2) = \frac 1 4 $, I'll leave it as an exercise (in MathJax) to show the other case.