# New version of double-slit experiment

I am trying to find a definite answer whether the following version of the double-slit experiment has ever been performed.

Calculate/observe what interference pattern should appear by emitting photons individually one-by-one through the double-slit barrier and onto a detector screen behind it, with all equipment being in a vacuum to make sure each photon does not interact with anything other than the barrier and the screen.

Then perform the same experiment but keep shifting the double-slit barrier slightly after each photon has been emitted but before it is supposed to go through the slits.

Will the interference pattern appear to look similar and in the same location as in the original static version of the experiment or will it be different?

Could it be that such experiment cannot be performed at all because it is not possible to tell whether and when each photon is actually emitted to be able to shift the barrier after the emission?

• Single photon experiments have been done youtube.com/watch?v=GzbKb59my3U . A more expensive experiment ( vacuum etc) . I think you will just blur the interference if you move the slits, due to the geometry angular changes. Jun 22, 2014 at 13:58
• @Carl Wigtoft Jun what did u mean by photons can see in future ? Jun 27, 2014 at 15:18
• @Carl Witthoft is my idea in the question possible ? Interference pattern will be different ? Jun 27, 2014 at 15:29
• @CarlWitthoft you can know the photon has been emitted if you use generator of pairs of entangled photons. Then to delay the photon to have some time to move the slit you can pass it through multiple reflectors, and the experiment then seems possible. Jun 28, 2014 at 7:21
• @MurtuzaVadharia in fact, it's not necessary to do this with photons — similar results will be with massive particles, but, as a bonus, you don't have complications with delaying the particle arrival at the slit to move the slit first (just slow it down). Jun 28, 2014 at 7:30

It is not possible to be completely sure there is only one photon in the apparatus and in the quantum sense this is even an ambivalent question to ask. But to the question "Is there a source of photons for which we almost always detect one outcoming photon in a given time interval?" the answer is yes, single photon sources. As far as I know, it is however always a continuous source of photons, that is, it keeps spitting "antibunched" photons out continuously and just a single photon cannot be produced.

As far as I know, experiments with single photon sources have however not found any deviation from quantum mechanical predictions and the "single photons" behave completely as quantum particles. Historical electron double slit experiments were usually done in vacuum.

I don't know whether a "moving slit" experiment has been done, but there are some issues with it such as collecting a large number of dots forming the interference pattern. Obviously, you would like to get the full pattern of a photon which came out at $t_0$ and the double slit was moved from $x(t_0)$ to $x(t_0)+v\Delta t$. However, in a naive setting you will get a superposition of patterns from double slits at different position as the source keeps spitting out photons at different times when the slit is at different initial positions.

The time $t_0$ of the "photon spit" is generally also not settable accurately due to quantum principles. So in any case, I believe the resulting image would basically be blurred and inconclusive which is the reason the experiment has probably never been done.

As to a formal treatment, we could idealise the situation by taking the free particle with boundary conditions of the classical double-slit experiment and making the boundary conditions of the double slit time dependent and solve the time-dependent wave equation (a massless Klein-Gordon equation would be okay). However, as typical corrections of the perturbed solution would be $v/c$, where $v$ is the velocity of the double slit, a quasi-static treatment would most probably be sufficient. The probability density of finding a photon on the display would then just be the interference pattern of the non-moving double slit experiment translated in time according to the translation of the double slit.

Detecting the time of the incidence of the display could in principle resolve the issue of blurriness and a repeated super-sensitive experiment might reconstruct the time-frozen interference pattern and investigate any deviations from the predictions of a less idealized time-dependent wave equation (accounting e.g. for the interaction with the slit wall). Nevertheless, keep in mind that we do not know any kind of classical trajectory of a photon and thus do not know the time it took to go through the apparatus, and we also do not know at which exact time it left the source, so any formulation of alternative predictions would be very difficult.

EDIT: There is indeed a theory defining sharp states of quantum particles, the De Broglie-Bohm theory and in the non-static case, the formulation using the quantum potential would perhaps be useful.

However, these so called "hidden variable theories" also just speak in terms of probability distributions since any attempt to measure the particle state seriously disturbs the "piloting wave".

Just imagine how to actually measure the photon initial time and position, you cannot "see" it, you have to "bump" into it. But with what we can "bump" into it? Say with another photon, whose initial position and fly-out time we once again do not know! This "measurement" process can only converge to a limited certainty expressed by the Heisenberg uncertainty principle. This follows from the fact that there is a lower limit on the size of the particle/wavepacket you can use for measurement.

So in our case, if we tried to measure the initial momentum and position of the photon at a given time, our "probing" particle just could not retrieve everything without disturbing the certainty in other variables. This is why in the end De-Broglie Bohm theory reproduces all the results of quantum mechanics almost to the dot.

The term "frozen interference patterns" refers only to the fact that our wave-function will be time-dependent:

$\psi(x,t)$

and so the probability density of finding a particle on a given position on the display $x_d$:

$\rho(x_d,t) = |\psi(x_d,t)|^2$.

(In the model I have proposed, it will just be an interference patter moving along the display). However, in one run of the experiment, we will only get one dot for every $t_1,t_2,t_3,...$ on the display. We would however like to recover $\rho(x_d,t_1),\rho(x_d,t_2)...$ (the "time-frozen" interference patterns) which is possible only with very carefully rerunning the experiment again and again and getting more and more dots at various times. We can then get e.g. the "time-frozen" $\rho(x_d,t_1)$ by plotting only detections which happened at $t_1$.

• Can we SUPPOSE to modify quantum principles to define photon spit for this case ? Just can we detect the photons during the photon spit ? Just by detecting them we may obtain the path through which we nay get the source ? But problem is there if we detect the photon however but it would be difficult to find the path to get to the source . For this I have thought that if we first assume about the indirect source if it is not the source just cancel it ans we have to use try and error method but it would take much time , if I am correct please help me to make it better . Jun 30, 2014 at 6:30
• What is time frozen interference pattern I have heard it but I am not able to recall it ? Jun 30, 2014 at 6:32
• I did not understand completely all the questions from your first comment but I tried to edit my answer and elaborate on some of them. Could you maybe describe what you mean by "indirect source"?
– Void
Jun 30, 2014 at 9:22

Wherever you place a screen, even in the position where later you will place the double-slot barrier, you will receive photons (in the meaning of particles). What is not observable but theoretically explained is the wave character of the photon during the transition through the double-slit barrier.

You pointed out that "each photon does not interact with anything other than the barrier and the screen". How about perception that the interaction between the photon and the electromagnetic field of the double-slit barrier produces the fringes on the observation screen. This explanation is not better or worse when the perception that is an interference of one photon with itself during transition. Then more for the single electron experiments.

If you following up until this point may you find it right to say that the photon "feel" the barrier only in a very short distance and on the way to the barrier it's not important where the barrier is located or is the barrier available at all. Therefore the fringes on the screen will be blurred if you move the screen and this doesn't depending from the moment you will do it.

• I liked your second para but still not getting it. Jun 26, 2014 at 15:42
• Sorry what you mean with "liked but still not getting it"? Jun 26, 2014 at 17:09
• I am not getting about the interaction between photons and electromagnetic field producing fringes on screen and can you explain last two lines of second paragraph ? Jun 27, 2014 at 4:28
• When an photon hits the surface of a barrier it comes to an interaction between them. At the end photons of a certain bandwidth (frequencies) will be reflected. Consequently there is a small zone between direct interaction with the barrier and no interaction. In this zone the photons will be diffracted. Quantum field theory tells us that fields are quantized. That is the reason why photons during the transition of the slit drift into some areas which we see on the observation screen as fringes. And be sure that fringes you get near every edge, not only behind slits or double-slits. Jun 27, 2014 at 19:32
• After explanations above a short answer for your question. Will you move the screen before the photon is emitted or will you move the screen in the time between the emission and the distance the photon interact with the barrier, you get the same result. Jun 27, 2014 at 19:44