5

A modified version of the quantum eraser experiment, consisting of just a photon source, double slit, photon-splitter crystal and a screen, capturing one of the entangled twins. No detectors, no mirrors, nothing. The other twin is lost.

Would this modified setup produce an interference pattern or not? (The sources I find are not clear on this.)

Here is the implication: If this modified experiment produces an interference pattern and the full quantum eraser experiment does not (when placed in the mode which unambiguously determines which slit the photon passed through, so the "eraser" part is missing), it means we can send information to the past. Imagine this: I send photons in a wave, split them, using the crystal and look for patterns on the screen, produced by the screen-side twin photon. The other twin I bounce between mirrors until I want to send the information to the past. Let's say the format of the information is binary - "$0$" encoded by NOT producing an interference pattern and "$1$" by producing an interference pattern. So if I want to send "$0$" I will add the detectors to the setup (resulting in the classic setup, with the eraser missing), and if I want to send "$1$" I will remove them (resulting in my modified setup).

This all hinges on the answer to the title: "Does simply putting a photon-splitting crystal after a double slit break the interference pattern?"

EDIT: As requested here is what I imagine and ask if the photons, hitting the screen (D0) will produce an interference pattern.The proposed setup

Duke William
  • 613
K. Kirilov
  • 629
  • A drawing of your proposed experiment would be helpful-- – S. McGrew Jan 04 '19 at 04:38
  • It's so easy to say "The other twin I bounce between mirrors", but reflection off a mirror is a non-negligible interaction, imparting momentum to the mirror. So your experiment already has a detector in it. – Ben Voigt Jan 04 '19 at 04:53
  • @BenVoigt Let's assume we can do the perfectly. If not - we will just extend the distances and make this a FTL communication problem, instead. – K. Kirilov Jan 04 '19 at 05:57
  • 1
    One important aspect to remember about the DCQE experiment, is that the pattern that the photons make on the main side only depends on what they encounter on the main side. That pattern is totally independent of whatever you do on the secondary side. You can put the original DCQE mirrors and detectors, remove everything, send the secondary photons to the moon and back, it does not matter. The pattern on the main side will not change. The interference pattern is only visible when looking at a specific subsets of the main photons, not when you look at all main photons. – fishinear Apr 14 '20 at 10:57

2 Answers2

4

Your proposed configuration does not show any interference.

In the Quantum Eraser experiment, what's getting "erased" is the which-way information, not the interference. The SPDC stage (which you term "photon splitting", though that's a confusing and misleading term that should not be used) creates which-way information, and the presence of that which-way information completely destroys the interference pattern.

The Quantum Eraser experiment restores the interference pattern by carefully erasing the which-way information in a coherent way (i.e. by taking coincidence counts against measurements on $D_1$ and $D_2$, in Wikipedia's notation).

Since your scheme does not do that, there will be no interference pattern.

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
  • doesn't what destroy a visible interference at $D_0$ is incoherence because the double-slit is placed before BBO hence photon has probability to hit two spatially separated locations on the crystal? Hence resulting path probabilistic photon incoherent as a result no visible interference at $D_0$. – Duke William May 25 '23 at 04:18
  • @DukeWilliam In short: no. (At least your first sentence. The second sentence is word salad.) If you want more details, ask separately. – Emilio Pisanty May 25 '23 at 21:25
  • SPDF can spit out an entangled pair from many locations on SPDF crystal. So it is a spectrum of coherent and incoherent probabilities. (these are probably not the right words, but you know what I mean). Doesn't that what cause $D_0$ plot to become random. Sum of all probabilities random – Duke William May 28 '23 at 10:37
  • @DukeWilliam No, I do not know what you mean. You need to ask separately. – Emilio Pisanty May 29 '23 at 11:53
  • I mean that at start, photon from the argon laser have probability to hit through the two slits into many locations of the BBO crystal around slits simultaneously. Because of wave nature of photon we can consider entangled pairs have probability to generate from many location of the BBO crystal. Now entangled pairs generated from nearby locations have more chance to be coherent than pairs generated from far away regions of the crystal. So it is the total sum of probabilities that cause the clump at $D_0$. Visible clump formed has all these probabilities of interference and no interference. – Duke William May 29 '23 at 14:12
  • @DukeWilliam I don't know how else to say this. Ask separately. Start a new thread, and ask your question there. I will not reply further in this thread. – Emilio Pisanty May 29 '23 at 16:27
  • Do you mean a new post (new question)? – Duke William May 29 '23 at 16:39
  • @DukeWilliam Yes, I mean a new question. – Emilio Pisanty May 29 '23 at 18:34
1

Here is a drawing of what happens before and after the crystal that can give spontaneous parametric down conversion, SPDC:

onetotwophotons

An SPDC scheme with the Type I output

The conversion efficiency of SPDC is typically very low, with the highest efficiency obtained is on the order of 4 pairs per $10^6$ incoming photons for PPLN in waveguides. However, if one half of the pair (the "signal") is detected at any time then its partner (the "idler") is known to be present.

This is a "crystal+photon" interaction, that allows for the two produced photons to be entangled. Note the term interaction.

The incoming photon in the double slit experiment comes with a specific wave function, which is the solution of the quantum mechanical problem "photon hitting two slits of specific distance apart with specific dimensions", and thus one can get the interference pattern one sees in single photon at a time experiments, as this in this one here..

Any interaction after passing the slits will change this wavefunction, and due to the probabilistic nature of quantum mechanics the original phases in the wavefunction of the incoming photons, which carry the memory of the double slits, will be lost whether interacting with an atom in a detector or with a whole crystal splitting into two.

This is seen clearly in this which way experiment with electrons:

“When the electron suffers inelastic scattering, it is localized; this means that its wavefunction collapses and after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons,” Frabboni said. “The experimental results show electrons through two slits (so two bright lines in the image when elastic and inelastic scattered electrons are collected) with negligible interference effects in the one-slit Fraunhofer diffraction pattern formed with elastic electrons.”

In the image in this answer, the photon scatters inelastically on the total crystal.

It is enough to interact, for the double slit interference pattern to be lost.

I cannot comment on the implications you wanted, but can state that the original incoming interference pattern (phases) are lost.

Community
  • 1
anna v
  • 233,453
  • Thank you. I just edited my question with a drawing, as suggested in the comments. I just wanted to let you know about the edit and make sure we are on the same page. So photons, hitting the screen will definitely not produce an interference pattern. – K. Kirilov Jan 04 '19 at 05:23
  • I do not know the dynamics of these crystals, there might be an interference pattern because it could be the crystal itself is a type of "double slit" , i.e that the phases of the generated photons are correlated , but it will not be the original double slit interference pattern. That will be lost. One could do an experiment just with the crystal, you do not need the slits. – anna v Jan 04 '19 at 05:30
  • The interference pattern is defined by the energy of the photon, the frequency. If different frequency photons come up with each "photon+crystal" there cannot be an interference pattern. And maybe this is the simplest explanation of no interference even starting with a double slit. – anna v Jan 04 '19 at 05:33
  • I'm pretty sure they use photons with the same frequency, so loss of the pattern due to varying inputs shouldn't be a concern. – K. Kirilov Jan 04 '19 at 05:39
  • It is not the input, it is the output photons, do they always split into two fixed frequencies? – anna v Jan 04 '19 at 05:40
  • Yes, they are, "This is done by a nonlinear optical crystal BBO (beta barium borate) that converts the photon (from either slit) into two identical, orthogonally polarized entangled photons with 1/2 the frequency of the original photon." They use identical inputs and they say they get identical outputs aswel. I mean, not just identical in the pair, but between pairs too. – K. Kirilov Jan 04 '19 at 05:47
  • then it is the phases that will define whether an interference pattern can be seen.As I said my background is not in crystal behavior. – anna v Jan 04 '19 at 06:55