There are many questions on this site about the quantum eraser, but I think mine is not quite answered by any of the other answers on the topic.

Here's the setup:

enter image description here

My understanding of this experiment is the following. If a stream of photons is passed through a double-slit one at a time, but then passed through the BBO crystal, the photon is marked with which-way information due to the nonlinear interaction (basically the act of destroying a photon and making two new ones represents a measurement which collapses the wavefunction). After this the entangled pairs are split; one member of the pair is sent to a detection screen and the other member of the pair is sent into this setup with mirrors and beamsplitters and finally onto some click detectors $D_1, D_2, D_3, D_4$.

Based on this description, it seems obvious that if we look at the photons at the detector screen $D_0$ whose entangled pair photons hit $D_3$ or $D_4$, we'll see no interference pattern: in these cases we have which-way information and there should be no reason for interference.

What confuses me is what $D_0$ registers when you look at the photons whose pair photon hits $D_1$ or $D_2$. I understand that, due to the setup, the two paths have been recombined in the lower portion of the experiment, so that a click in $D_1$ or $D_2$ does not tell you which slit the photon went through anymore. But the upper photons are still marked with this information, so why should we recover an interference pattern? In other words, based on my understanding, the BBO crystal should prevent us from ever seeing an interference pattern.

  • $\begingroup$ Interference is the absence of self-interaction, so we only "get interference" because the field is linear. $\endgroup$ Commented Jun 8, 2023 at 14:26

3 Answers 3


$$ \newcommand{\ket}[1]{\left| #1 \right>} $$ Let S be a quantum system, and let M be another quantum system which behaves in some respects like a measuring device. Let $\ket{0}_S$ and $\ket{1}_S$ be a pair of orthogonal states of S. For example, they could correspond to a particle going through one slit or another. Let $\ket{{\bf 0}_n}_M$ and $\ket{{\bf 1}_n}_M$ be two orthogonal states of M, where the notation is to suggest that M might be a larger system than S.

Let $$ \ket{\psi(\alpha)} = \frac{1}{\sqrt{2}} \left( \ket{0}_S \ket{{\bf 0}_n}_M + e^{i\alpha} \ket{1}_S \ket{{\bf 1}_n}_M \right). $$ This state can be 'read' as the situation that S has first been placed in a superposition (e.g. left and right slits) with interference phase $\alpha$, and then it has interacted with M with the result that the two are now entangled. This interaction is somewhat like a measurement, in that if one wished to determine which slit S went through, it would suffice to examine M, and M could be a large system which is easy to examine.

The quantum eraser concept, in whatever form, has as its basic underlying ingredient the following identity: $$ \ket{\psi(\alpha)} \equiv \frac{1}{2}\left[\left(\ket{0}_S + e^{i\alpha} \ket{1}_S \right) \ket{{\bf +}_n}_M + \left(\ket{0}_S - e^{i\alpha} \ket{1}_S \right) \ket{{\bf -}_n}_M \right] \tag{1} $$ where $$ \ket{{\bf +}_n}_M \equiv \frac{1}{\sqrt{2}} \left( \ket{{\bf 0}_n}_M + \ket{{\bf 1}_n}_M \right), \qquad \ket{{\bf -}_n}_M \equiv \frac{1}{\sqrt{2}} \left( \ket{{\bf 0}_n}_M - \ket{{\bf 1}_n}_M \right). $$ We can 'read' the right hand side of (1) as telling us that if we observe M in the basis $\ket{+_n},\;\ket{-_n}$ and we find, for example, that the $(+)$ was obtained, then subsequent observations on S will be consistent with a superposition of S with interference phase $\alpha$. If in the other hand $(-)$ is obtained in the observation of M, then observations on S will be consistent with a superposition with interference phase $\alpha+\pi$.

In the present example, the crystal and the rest of the apparatus gives an example of M. The observation of a photon arriving in D1 or D2 corresponds to a measurement of M in the basis $\ket{+_n},\;\ket{-_n}$.

Note that it is best not to adopt the terminology of 'collapse of the wavefunction' in considering experiments of this kind. In particular, during the interaction with the BBO crystal there is no collapse, but there is an entanglement.


At the heart of your question is the supposition that:

"basically the act of destroying a photon and making two new ones represents a measurement which collapses the wavefunction"

This is not necessarily the case, if we consider the photon process per Feynman/Dirac, we have excited electron, photon creation and path determination (or vice versa), un-excited original electron and excited receiving electron. With an intermediary BBO crystal we would add another step with an additional excitation.

The wave function for the above processes may be one or two successive wave functions .... But the fact that a pattern is seen would be evidence for a single wave function at least for the photons on D1/D2.


In your 2nd paragraph the photon is NOT marked by its interaction with the BBO crystal, the photons are subject to the EM field which influences the path to take and the path taken is what effectively marks them. In this experiment not only have the photons taken slit A or B but have made a further path selection (due to EM field) as to proceed on a marked path or unmarked path (its 50/50 based on the beam splitter).

An excited atom in a laser is already disturbing the EM field (virtual photons), the creation and transit/path taken by the eventual real photon could be the result of these interactions .... it is well said by Feynman and Dirac that every photon determines its own path within the confines of the EM field. The EM field is constantly changing due to moving charges in atoms (ex. excited electrons and even somewhat unexcited electrons) and the photon creation is subject to all these forces. Richard Feynman proposed the path integral explanation, in summary a photon considers all paths (this could be virtual/forces). Most probable paths in his model is based on amplitude summing, considering phase and path length. One result is that paths that are "resonant" i.e. are integer multiples of the wavelength have higher probabilities.

The upper photons are not marked, we can NOT tell if they came from A or B until afterwards (when D3/D4 click) and this only true for the the ones that have no interference. What we are seeing is that some photons created in the laser were being affected by seeing (or having a choice) either A or B which results in the pattern ...... OR sometimes they just take a direct path to D3/D4, ie not being affected by the "double" slit at all .... just seeing one slit. ITS all the fault of the EM field and the quantum (means somewhat random with a pattern) outcomes.

What's interesting about this DS experiment is that adding the crystal and the beam splitters really had a dramatic affect of possible patterns observed. The photons were able to be affected by the 2 slits (pattern) or just only affected by slit (no pattern)!

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    $\begingroup$ Photons aren't living beings able to make choices, and with the amount you use that language I can't really understand your explanation. Could you revise your answer to be a little more precise? $\endgroup$ Commented Nov 22, 2021 at 17:52
  • $\begingroup$ @flevinBombastus I'll change some wording. But photons "make choices" means that only certain paths each with its own probability is allowed. It's all governed by EM forces (virtual photons) and energy exchange (real photons or just photons). We can never say exactly what one photon will do, the quantum nature adds some randomness, but significant patterns are observed. $\endgroup$ Commented Nov 24, 2021 at 0:59
  • $\begingroup$ It now sounds like you're saying the photons always go through one path or the other, which isn't a consistent interpretation of these experiments. Remember that the detectors can't be the path markers either, because we have to appeal to retrocausality to make that logically consistent. $\endgroup$ Commented Nov 24, 2021 at 13:26
  • $\begingroup$ That's exactly the point, photons do take one path or another and its exactly consistent with the results. Even in the simple DSE we can even place detectors in the dark areas of the pattern and conclude NO photons have taken this path. $\endgroup$ Commented Nov 24, 2021 at 14:20
  • $\begingroup$ Yes detectors can be path markers, here D3/D4 clearly indicate path. D1/D2 we do not know the path from slit A or B. For D0 we only infer the path if the count is coincident with D3/D4. $\endgroup$ Commented Nov 24, 2021 at 14:26

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