Musings

We are indeed studying an entangled state in a product of two-state systems. But we are measuring the operators $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\otimes\mathbb{1}$ and $\mathbb{1}\otimes\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ (in the basis where $|\text{up}\rangle=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ and $|\text{down}\rangle=\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$).

I am sure you can construct the corresponding projection operators and compute the probabilities for the various outcomes of those measurements.

It was swept up with a bunch of spam. Restored from backup now.

I suppose I should have started with an exposition of the difference between classical probabilities (there’s a 50% chance that Schrödinger’s cat is alive or dead) and quantum-mechanical superposition.

The entangled state is a particular kind of quantum-mechanical superposition.

Distinguishing between the two requires “non-commuting observables” — a measurement of a second observable that can change the result of measuring the first observable. In the case at hand, the act of measuring the polarization of the photon at a $45^\circ$ axis changes what the result for measuring whether it is horizontally/vertically polarized.

This criteria makes sense indeed, is it correct to interpret it as an “action at distance” ?

No more than the classical correlations that I talked about in my introduction constitute “action at a distance.”

So if I understood correctly, the second photon polarization probabilities can change when the first photon mirror’s orientation change ?

We know things about the first photon (and hence also about the second) that we didn’t before we made the measurement on the first photon. The funny thing about non-commuting observables is that they also wipe out things that we previously knew about the system.

Just to go back to the polaroid filters, remember that they work solely by absorbing photons (of one polarization, while letting the other polarization pass).

So imagine a setup with a pair of filters, one oriented horizontally and the other vertically. No photons pass through the pair: those which are not absorbed by the first are absorbed by the second.

But now put a third filter in between the two, oriented at $45^\circ$. Even though all it does is absorb photons, now suddenly light can pass through the system, whereas none could before.

Fair enough. I wasn’t trying to rule out hidden variable theories (which, if you are willing to entertain sufficiently baroque ones, can be an arduous task), but rather to simply state (in the terms of conventional quantum mechanics) what “entanglement” is and how it differs from mere classical correlation.

Once we know what quantum physicists mean by entanglement, then we can discuss whether there are “alternative” theories that yield the same result.

Sorry, but no.

The state of the photon pair (in an obvious basis) is $$\rho_c=\frac{1}{2}\begin{pmatrix} 0&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&0 \end{pmatrix}$$ for the correlated photons and $$\rho_e=\frac{1}{2}\begin{pmatrix} 0&0&0&0\\ 0&1&-1&0\\ 0&-1&1&0\\ 0&0&0&0 \end{pmatrix}$$ for the entangled state (a pure state).

In the original configuration, with their polaroid filters aligned vertically and horizontally, the projection matrices for Adam and Betty’s measurements are $$P_A=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0&\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix},\qquad P_B=\begin{pmatrix} 1&0&0&0\\ 0&0&0&0&\\ 0&0&1&0\\ 0&0&0&0 \end{pmatrix}$$ These satisfy $$\begin{split} Tr P_A \rho_c=Tr P_B\rho_c=Tr P_A P_B \rho_c&=1/2\\ Tr P_A \rho_e=Tr P_B\rho_e=Tr P_A P_B \rho_e&=1/2 \end{split}$$ as described. When we rotate Adam and Betty’s polaroid filters by $45^\circ$, their new projection matrices are $$\tilde{P}_A=\frac{1}{2}\begin{pmatrix} 1&0&-1&0\\ 0&1&0&-1\\ -1&0&1&0\\ 0&-1&0&1 \end{pmatrix},\qquad \tilde{P}_B=\frac{1}{2}\begin{pmatrix} 1&1&0&0\\ 1&1&0&0\\ 0&0&1&1\\ 0&0&1&1 \end{pmatrix}$$ and these new projection matrices satisfy $$\begin{split} Tr \tilde{P}_A\rho_c=\tilde{P}_B\rho_c&=1/2,\qquad Tr\tilde{P}_A\tilde{P}_B\rho_c=1/4\\ Tr \tilde{P}_A\rho_e=\tilde{P}_B\rho_e&=1/2,\qquad Tr\tilde{P}_A\tilde{P}_B\rho_e=1/2\\ \end{split}$$ exactly as described in my post.

I am not sure what the original source of your confusion was, but I hope this clears it up.

Just to be clear, there’s nothing special about $45^\circ$. You can write down the projection matrices $P_A(\theta),P_B(\theta)$ which interpolate between $P_A=P_A(0),\; P_B=P_B(0)$ and $\tilde{P}_A=P_A(\pi/4),\; \tilde{P}_B=P_B(\pi/4)$.

What you will find (exercise for the reader) is $$\begin{split} Tr(P_A(\theta)\rho_e)&=Tr(P_B(\theta)\rho_e)=Tr(P_A(\theta)P_B(\theta)\rho_e)=1/2\\ Tr(P_A(\theta)\rho_c)&=Tr(P_B(\theta)\rho_c)=1/2,\qquad Tr(P_A(\theta)P_B(\theta)\rho_c)=\tfrac{1}{8}\bigl(3+\cos(4\theta)\bigr)\\ \end{split}$$