enter image description here

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.

  • $\begingroup$ Judging from the setup I think the wavefunctions of each slit are not equal depending on which photon is detected because the position of each photon (e.g. which slit it is coming from) is "imprinted" (as stated in arXiv:2111.09347v2) onto the photon which is generating it (since they are spatially sourced from different points of the slit). It may seem counterintuitive from a classical perspective since the "source" S is the same, but it makes sense that from a quantum-mechanical perspective the photon source position plays a significant role. $\endgroup$
    – ondas
    Jul 3 at 8:21
  • $\begingroup$ If you really want to understand these experiments you need to look at the math. Hand waving explanations can only go so far. $\endgroup$ Jul 3 at 10:11

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.