Symmetrical Delayed Choice Quantum Eraser

Question

This is a modification to a Kim et al.-type quantum eraser experiment. According to the research paper which proposed this modified experiment,

This modification makes the similarity to Bell-type experiments obvious. We could now either measure the which-way information on both sides ($D_{1/2}$, $U_{1/2}$), and find that the results are perfectly correlated. Or we could erase the which-way information on both sides ($D_{3/4}$, $U_{3/4}$) and find the results are also perfectly correlated. Or we could measure the which-way information on one side and not on the other (($D_{1/2}$and $U_{3/4}$) or ($D_{3/4}$ and $U_{1/2}$)) and find the results are entirely uncorrelated.

Why $D_3$ is correlated with only $U_3$ and $D_4$ is correlated only with $U_4$, but not a 50:50 correlation mix up of $D_3$ with $U_3$:$U_4$ and $D_4$ with $U_3$:$U_4$ despite the equal chances of a photon going to either $D_3$:$D_4$ or $U_3$:$U_4$?

(I do not see the difference in the optical paths that should cause this behaviour.)

Note that the notation of $D_{1-4}$ in this experiment is a little bit different than Kim et al. version of the experiment in which $D_{3/4}$ is used as which-way detectors in contrast to this experiment.

Answer 1

First of all, if $D_3$ was correlated with both $U_3$ and $U_4$ (and same for $D_4$) then there would in fact be no correlation.

As for finding the correlation, it is a straightforward calculation. The authors define the input state as:

$$ |\Psi\rangle = \frac{|U_1\rangle|D_1\rangle + |U_2\rangle|D_2\rangle}{\sqrt{2}} $$

Here $|U_k\rangle$ ($|D_k\rangle$) denotes that the photon would generate a click in detector $U_k$ ($D_k$) if its which-path information were to be measured.

The path-erasing detectors correspond to projecting the photons on the states:

$$ |D_3\rangle = \frac{|D_1\rangle + |D_2\rangle}{\sqrt{2}}, \qquad |D_4\rangle = \frac{|D_1\rangle - |D_2\rangle}{\sqrt{2}}\\ |U_3\rangle = \frac{|U_1\rangle + |U_2\rangle}{\sqrt{2}}, \qquad |U_4\rangle = \frac{|U_1\rangle - |U_2\rangle}{\sqrt{2}} $$

We can invert these expressions to find:

$$ |D_1\rangle = \frac{|D_3\rangle + |D_4\rangle}{\sqrt{2}}, \qquad |D_2\rangle = \frac{|D_3\rangle - |D_4\rangle}{\sqrt{2}}\\ |U_1\rangle = \frac{|U_3\rangle + |U_4\rangle}{\sqrt{2}}, \qquad |U_2\rangle = \frac{|U_3\rangle - |U_4\rangle}{\sqrt{2}} $$

Inserting these expressions into the input state $|\Psi\rangle$ gives:

$$ \begin{align} |\Psi\rangle &= \frac{1}{2\sqrt{2}}\bigl(|D_3\rangle + |D_4\rangle)(|U_3\rangle + |U_4\rangle) + (|D_3\rangle - |D_4\rangle)(|U_3\rangle - |U_4\rangle\bigr)\\ &= \frac{1}{2\sqrt{2}}\bigl(|D_3\rangle |U_3\rangle + |D_3\rangle |U_4\rangle + |D_4\rangle |U_3\rangle + |D_4\rangle |U_4\rangle\\ &+\hphantom{\frac{1}{2\sqrt{2}}\bigl(} |D_3\rangle |U_3\rangle - |D_3\rangle |U_4\rangle - |D_4\rangle |U_3\rangle + |D_4\rangle |U_4\rangle\bigr)\\ &= \frac{|U_3\rangle|D_3\rangle + |U_4\rangle|D_4\rangle}{\sqrt{2}} \end{align} $$

Here you see that a click in $D_3$ is always paired with a click in $U_3$ and so on.

This is a very long-winded way of confirming that the Bell state $|\Phi^+\rangle$ is correlated in the $\sigma_x$ basis. If you changed the relative phase between the two paths in one arm of the experiment you would change the basis, and thereby also the correlations.