Are photons really anti-correlated off their basis vectors?

[zincshow]

The wiki page https://en.wikipedia.org/wiki/Bell's theorem states the following which I agree with:

Suppose the two particles are perfectly anti-correlated—in the sense that whenever both measured in the same direction, one gets identically opposite outcomes, when both measured in opposite directions they always give the same outcome.

My question is whether photons really act in this manner. If you measure a pair of correlated vertical photons along their vertical axis in one location and horizontal axis in another location you will get perfect anti-correlation for each photon pair. But, if you measure vertical photons off the vertical axis (say 45 degrees one way and 45 degrees off the other way), you only get a statistical correlation for groups of photon pairs. Experiments are unable to show the above quote is true for individual photon pairs when measured off their basis vectors. Or do they? Why is this assumption made?

 

[Nugatory]

But, if you measure vertical photons off the vertical axis (say 45 degrees one way and 45 degrees off the other way), you only get a statistical correlation for groups of photon pairs. Experiments are unable to show the above quote is true for individual photon pairs when measured off their basis vectors. Or do they?

They do not. On the contrary, experiments show that no matter what angles you choose, as long as they are 90 degrees apart each individual pair will always be ideally correlated (or anticorrelated, depending on how the pair was prepared).

One implication is that there is no such thing as a "pair of correlated vertical photons". It's a pair of correlated photons, with no preexisting orientation.

 

[Strilanc]

They do not. On the contrary, experiments show that no matter what angles you choose, as long as they are 90 degrees apart each individual pair will always be ideally correlated (or anticorrelated, depending on how the pair was prepared).

One implication is that there is no such thing as a "pair of correlated vertical photons". It's a pair of correlated photons, with no preexisting orientation.

I thought that photons could only be all-axes-anti-correlated (the singlet state), not all-axes-correlated? The other entangled states making up the bell basis, the triplet states, all correspond to two-axies-correlated-one-axis-anti-correlated.

Let me explain. Suppose we represent the state of two entangled photons between Alice and Bob as a 2x2 unitary matrix $S$ where an Alice operation $U$ left-multiplies $S$ by $U$ while a Bob operation $U$ right-multiplies $S$ by $U^T$.

The singlet state $\left| 01 \right\rangle - \left| 10 \right\rangle$ corresponds to $S = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. If Alice and Bob each apply the same operation $U = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ to $S$ then the resulting state is:

$U \cdot S \cdot U^T$

$= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} a & c \\ b & d \end{bmatrix}$

$= \begin{bmatrix} b & -a \\ d & -c \end{bmatrix} \cdot \begin{bmatrix} a & c \\ b & d \end{bmatrix}$

$= \begin{bmatrix} b a - a b & b c - a d \\ d a - c b & d c - c d \end{bmatrix}$

$= \begin{bmatrix} 0 & -det(U) \\ det(U) & 0 \end{bmatrix}$

$= det(U) \cdot S$

$= e^{i \theta} \cdot S$ for some $\theta$ since $U$ must be unitary.

That is to say: when Alice and Bob apply the same operation, the singlet state is unchanged (except for an unobservable global phase factor). Measuring along a different axis is equivalent to applying an operation, so we conclude that no matter what axis they measure along they observe the same thing they would along the computational basis axis (Z): an anti-correlated result.

Contrast this to what happens in the state $\left| 00 \right\rangle + \left| 11 \right\rangle$, which would be the obvious candidate for an "always agree" state. This state corresponds to the matrix representation $S_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$, and applying the same operation to both sides gives:

$U \cdot S_2 \cdot U^T$

$= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} a & c \\ b & d \end{bmatrix}$

$= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix} a & c \\ b & d \end{bmatrix}$

$= \begin{bmatrix} a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \end{bmatrix}$

$\neq e^{i \theta} \cdot S_2$ because $a^2 + b^2$ may destructively interfere (e.g. try $U \propto \sqrt{X}$, with $a=d=\frac{i}{\sqrt{2}}$ and $b=c=\frac{1}{\sqrt{2}}$).

In this case, Alice and Bob applying the same operation can significantly affect the state. Y-axis rotations cancel each other out, but X-axis and Z-axis rotations get doubled up. So this is not an all-axes-agree state. The same is true of the remaining two bell basis states.

(That being said, suppose that Alice and Bob are not facing in the same direction. Instead, one of them turned around the Y axis to face the other. They perform rotations relative to the axes of their personal coordinate system, so turning around inverts X and Z rotations in the global coordinate system but not Y rotations. That can fix the doubling-up problem and cause applying the same operation to have no observable effect again... but the turning around also effectively puts us back into the singlet state!)