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The double-slit "thought experiment" described by Feynman in Lectures on Physics Volume 3 section I-6 Watching the electrons consists in firing electrons through a double-slit to observe the interference of electron waves, and watching them after passing the slits with a light source placed behind the double-slit, at equal distance of each slit. As electric charges scatter light, one can "detect" which slit the electron went through if the photon wavelength is small enough.

Question: has this "thought experiment" been simulated by solving numerically the underlying Schrödinger equation? I am aware about numerical experiments of the double-slit, but did not find any including the interaction between the electrons and the photons just after the double-slit.

The numerical simulation can address other types of particles (replacing for example the photon by a slow electron to avoid relativistic equations), the crucial point being the simulation of the observation (here the photons being scattered by the electrons) and its effect on the wavefunction. Its interest could be in particular to better understand in which precise way the observation progressively becomes inoperative when the photon wavelength increases.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Sep 23, 2016 at 14:03
  • $\begingroup$ Something similar has actually been done experimentally with fullerenes, see doi: 10.1038/44348 Scattered photons would end up entangled with your electrons. Measuring those photons could destroy your interference pattern, if the measurement can tell you which slit the electron went through. $\endgroup$
    – Arturs C.
    Mar 22, 2017 at 11:07

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I'll pull my remarks from the (now migrated) comment thread in, since this has yet to be properly addressed. To answer your main question,

has this "thought experiment" been simulated by solving numerically the underlying Schrödinger equation?

I would say:

  • the precise simulation you propose (or close variations of it) probably has been performed at some point (probably multiple points, by multiple people);
  • it probably hasn't been published;
  • if it did get published, it won't make for a particularly interesting paper; and
  • if it hasn't been published, it's a mountain of a task to show credibly that there is no such paper in the literature.

Now, the reason for the above is that the simulation you propose just isn't very interesting. You say that

Its interest could be in particular to better understand in which precise way the observation progressively becomes inoperative when the photon wavelength increases.

but that is not the case: anything that such a simulation could show us, we already understand.

It is well understood, since the days of von Neumann, that within formal, unitary quantum mechanics, the effect of measurements is to cause entanglement: if the system is in a superposition of $|A=1\rangle$ and $|A=2\rangle$, say, and you measure $\hat A$, with some detector that goes to $\left|\uparrow\right>$ on $A=1$ and to $\left|\downarrow\right>$ on $A=2$, what you're really generating is the superposition $$ |\Psi\rangle = \alpha \left|A=1\right>\left|\uparrow\right> + \beta \left|A=2\right>\left|\downarrow\right>, $$ i.e. an entangled state between the system and the detector. This state can no longer show interference, because if you take the inner product between the two components, i.e. if you do $$ \bigg< \left|A=1\right>\left|\uparrow\right> ,\, \left|A=2\right>\left|\downarrow\right> \bigg> = \left<A=1|A=2\right> \left<\uparrow\middle|\downarrow\right> = 0 $$ you get zero, because the detector states are orthogonal. This completely kills any chance of interference, but the wavefunction hasn't "collapsed" (yet); if you do a projective measurement on the detector (forcing its wavefunction to "collapse", whatever that means), that extends to the system as well, but that is external to the simulation.

Let me reiterate the point: anything you could simulate, while keeping to unitary quantum mechanics, would completely fit within the scheme above, and it would not add to our understanding of the thought experiment.

If you do want your simulation to include some form of wavefunction "collapse" (or decoherence or whatever you want to call it), i.e. if you want your simulation to actually say anything useful about the measurement problem, then you're going to need to decide how you handle the information encoded in your which-way detector, and this is where your scheme goes south: in order to simulate anything at all, you essentially need to pre-bake some resolution of the measurement problem into your simulation. Whatever results you get back will just be a restatement of the premises you feed in, and will be subject to all the flaws of the premises.

Given this, all that such a simulation would provide is a very expensive visualization aid for processes that can already be understood analytically, and for which the main difficulty is conceptual. The creation of visualization aids is not without merit, but this one would offer very little towards the resolution of the true conceptual problems of the measurement problem, which have much more to do with what projective measurements mean than with photons and double slits.

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  • $\begingroup$ +1 great answer, particularly the last paragraph I would sign my name under. However I disagree with the second last paragraph. You could (practical realisability aside ;) ) put your detector and all the necessary environments in your your simultation. I don't see how that requires a pre-baked resolution of the measurement problem. The dynamics of such a system should surely be all there is to it? Of course this all hinges on realisability and if we could simulate such a system we would probably know the solution of the measurement problem, but that is really the reverse of your paragraph. $\endgroup$ Mar 23, 2017 at 10:10
  • $\begingroup$ @Mrphlng The thing is, you can try and get as fancy as you want with your simulated detector, but it still won't get you past the fact that (unless you're manually breaking unitarity) the simulation's outcome will be entanglement between the particle and the detectors, and the interpretation of that entangled state is precisely where the measurement problem lies. You end up essentially dancing around the Wigner's Friend paradox instead of tackling it head on on a conceptual level, like you ought to. $\endgroup$ Mar 23, 2017 at 10:24
  • $\begingroup$ Surely if the system you have is a good detector the collective probability distribution for your detector observable after decoherence has kicked in will just be very peaked and very stable with time? (i.e. classical without using that word). I don't see the conceptual issue. Or am I getting something wrong here? $\endgroup$ Mar 23, 2017 at 10:36
  • $\begingroup$ Example: if I shoot my photon on a photographic film they entangle for a bit, decoherence comes, photon runs away/gets absorbed somewhere, at the end the state of the photographic film has changed and there is a black spot somewhere. Also would be happy to continue this in chat if you're not too bored of the topic, since it seems to develop into a discussion that might be useful (at least for me ;) ) but not very relevant to the question. $\endgroup$ Mar 23, 2017 at 10:37
  • $\begingroup$ @Mrphlng I have nothing to add beyond my comment above and what's in the answer. What you propose presupposes a resolution to the measurement problem, so the simulation would not add anything. $\endgroup$ Mar 23, 2017 at 10:52
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The loss of interference is not due to collapse because collapse is a myth.

The loss of interference can be simulated by quantum mechanical models that include the interaction that transfers information about the measured system, which cause decoherence. These papers are not specifically about decoherence in electrons but they are about decoherence in general double slit type experiments

https://arxiv.org/abs/1606.09442

https://arxiv.org/abs/quant-ph/0310095

There are papers about dehoherence of electrons specifically in electronic Mach-Zehnder interferometers. These are a couple of samples of a very large literature on the topic:

https://arxiv.org/abs/0801.2338

https://arxiv.org/abs/1105.2587

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  • $\begingroup$ Thank you very much for the links. Additional links to the full papers would be very welcome! $\endgroup$
    – user130529
    Mar 23, 2017 at 7:53
  • $\begingroup$ You can get the full papers by clicking on the PDF link in the links above. the PDF link is in a box on the right. $\endgroup$
    – alanf
    Mar 23, 2017 at 8:25
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The effect Feynman describes is not calculated using the Schrödinger equation. Instead, it is a result of "wave function collapse" which is a result of the measurement of the electron's position. The collapse is a separate postulate that cannot be derived from the Schrödinger equation. This is the standard Copenhagen interpretation anyway.

Everett, in his "Relative State" paper, showed that in fact it can be derived from the Schrödinger equation, although you have to include the observer in your wave equation. This leads to the Many Worlds interpretation of quantum mechanics.

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  • $\begingroup$ Your second assert is exactly what I had in mind: using the Schrödinger equation including the observer in the numerical simulation (the observer here being simply the incident photon). As a side remark, I am not sure that including the observer necessarily leads to the multiverse interpretation. The validity of applying the Schrödinger equation to the whole system is independent of the interpretation (the multiverse interpretation doesn't seem to rely on facts). $\endgroup$
    – user130529
    Jan 23, 2017 at 11:23
  • $\begingroup$ @claudechuber right, it's just one way of seeing all that huge configuration space of the setup+observer: there are subspaces where the wavefunction of the setup collapsed in one way, other subspaces with another way of collapse etc.. $\endgroup$
    – Ruslan
    Jan 23, 2017 at 12:03
  • $\begingroup$ @claudechuber I didn't mean that including the observer leads to Many Worlds. I meant that including the observer was necessary in order to explain Feynman's effect without collapse. $\endgroup$
    – Bruce
    Jan 25, 2017 at 2:05
  • $\begingroup$ Thank you Bruce, I fully agree with you. It would be so exciting if one day someone performs such a numerical simulation! $\endgroup$
    – user130529
    Jan 25, 2017 at 7:45
  • $\begingroup$ There is no such thing as wave function collapse. $\endgroup$
    – my2cts
    Oct 13, 2021 at 17:36
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What Feynman here is talking about is the collapse of the wave function that occurs due to measurement.

Suppose, you fire a stream of electrons on a plate with two slits but don't observe which slit each electron passes through. In this situation, each electron is said to pass through both the upper slit and the lower slit resulting in the interference pattern on the screen.

Now, let's say you observe the path of each electron without actually disturbing its state (which is actually impossible, but bear with me here). In this case, you can precisely say which electron passed through which of the slits and can observe two piles of electrons on the screen.

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  • $\begingroup$ The wavefunction squared xx* is a probability distribution, not a prediction of a single measurement, but an accumulation of measurements. A single electron has a probability of going through one or the other. No one has seen an electron spread over space, as it is a point particle, even experimentally to a high accuracy. $\endgroup$
    – anna v
    Mar 22, 2017 at 11:45
  • $\begingroup$ It was a gross analogy, the cat used as an elaborate detector instead of a geiger counter, to plot the probability of decay, which would need many boxes with cats. In any case quantum mechanics has been clarified since S. times, and he is not a pope. Superposition is for the wavefunction, not for the measured particle which needs the square. $\endgroup$
    – anna v
    Mar 22, 2017 at 16:30
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So, I am about 6 years late to the punch, but here is a computational experiment setup. I'd like to code this up some day when I have enough time, it's quite doable.

Theory

You have two Hilbert spaces, one whose elements are the usual wavefunctions $\psi(x,y)\in\mathcal{H}_x$, and one which is your detector which says whether a particle passed through a left or right slit. Say your detector starts in state $(|L\rangle+|R\rangle)/\sqrt{2}$: a 50% superposition of the left and right slits. I'll denote this Hilbert space $\mathcal{H}_d=\operatorname{span}(|L\rangle,|R\rangle)$

The tensor product space of the particle and detector will be described by a function $\psi(x,y,d)$ where $d$ is one of $L$ or $R$. So for example if we had our initial wavepacket of our electron as $\psi_0(x,y)$ and the 50/50 detector state above, our initial full wavefunction would be $\psi(x,y,L)=\frac{1}{\sqrt{2}}\psi_0(x,y)$ and $\psi(x,y,R)=\frac{1}{\sqrt{2}}\psi_0(x,y)$.

For the Hamiltonian, the potential barrier acts as a potential on $\mathcal{H}_x$ and as identity on $\mathcal{H}_d$. So we have a $V(x,y)$ which is infinite except for two blocks cut out: the slits. I'll write the identity on $\mathcal{H}_d$ as $\mathbf{1}_d$.

To model the detection of particles, you can have a test function $h(x,y)$ which is $1$ inside the left slit, $-1$ inside the right slit, and $0$ elsewhere. If a particle is detected, we want to move the amplitude for detection to the correct state $|L\rangle$ or $|R\rangle$, so we should add a term something like the Pauli $\sigma_y$ matrix*. The full Hamiltonian will look like:

$$H=(p_x^2+p_y^2+V(x,y))\otimes \mathbf{1}_d + I h(x,y)\otimes \sigma_y$$

Where $I$ is the interaction strength.

Simulation

For numerical simulation, this turns into the following PDE ($m=\hbar=1$ WLOG): \begin{align*} i \partial_t \psi(x,y,L,t)&=(-\frac{1}{2}\partial_x^2-\frac{1}{2}\partial_y^2+V(x,y))\psi(x,y,L,t) -i I h(x,y)\psi(x,y,R,t)\\ i \partial_t \psi(x,y,R,t)&=(-\frac{1}{2}\partial_x^2-\frac{1}{2}\partial_y^2+V(x,y))\psi(x,y,R,t) +i I h(x,y)\psi(x,y,L,t) \end{align*}

You are basically simulating two separate wavefunctions, one for $L$ and one for $R$, but the term $I$ can mix the two wavefunctions and will do so in a way that disturbs interference. If $\psi(x,y,L)$ is plotted in blue and $\psi(x,y,R)$ is plotted in red, then for $I=0$ we see a purple pattern showing interference. For $I$ tuned to a nice value, we'll see the interference pattern disappearing, and a blue-purple-red bump with no interference pattern showing.

Comment on Interactions

The usual objection (also mentioned in Feynman) is that "well we're turning on an interaction $I$, it's no surprise that messes with the wavefunction." The point is that there's no way whatsoever to get position information out of the Hilbert space of the particle $\mathcal{H}_x$ and into the Hilbert space of the detector $\mathcal{H}_d$ without destroying the interference pattern. None, nada, zilch.


*I choose $\sigma_y$ because I know this will cause states to evolve like $e^{-i H t}$, and $-i \sigma_y t$ is $t\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$, which is a rotation matrix and is easier for me to visualize.

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When creating a "thought experiment" a main problem is to set up conditions that describe a realistic setup. The so called "Capacitor Paradox" Paradox involving two capacitors only comes up because the connection is set to have no inductance.

So what is the setup of the double slit experiment with electrons: The incoming electron's kinetic energy is about 50 keV. The slit's width about 0.5 µm, the distance 2 µm. The original paper: https://www.leifiphysik.de/quantenphysik/quantenobjekt-elektron/geschichte/originalarbeit-von-joensson shows relations and difficulties.

As it is extremely hard to get the pattern it is extremely simple to destroy it. But that doesnt explain, why the pattern exists.

As Jönsson wrote: there has to be a slot (in optic a grating on a transparent will work) as no material object is not covered with electrons, and indeed: the incoming electrons only see a sea of electrons and two areas with no electrons. In the classical world every electron will just pass through a single slot (if not reflected) and we will not see interference.

In the quantum world interaction between two electrons is quantized. And now we can see two cases:

The incoming electron interacts with a single electron in the sea and passes the slot. It will appear at a distinct place. The position of the sea electron is influenced by all the surrounding electrons so the geometry of the slits will be reflected in the average target points on the detector screen. So what we see is the interference between two electrons following the rules of quantum mechanics. Then many of the electrons will in average show the interference pattern.

The second case is: we see the sea of electrons as a condensed unit that results in one single, complex wave function and the interference of the incoming electron with this wave function creates the pattern.

However: there is a fundamental difference between a macroscopic interference (where a wave is just a analog wave) and microscopic interference (where a wave is constructed from elementary wave particles). And we should not forget that the double slit experiment only is surprising when not related to quantized interaction between particles. Only the integrating property of the screen makes a pattern from events. It's not at all a great idea to do statistics from single events.

Now, as I was thinking that folding wave functions involves the Fourier transform I googled "double slit Fourier transform" and got this hit: https://www.thefouriertransform.com/applications/diffraction3.php

That makes the double slit fully transparent ;-) . All electrons that create a spacial distribution apparently a double slit form a wave function which in turn folded with the delta function "electron" gives exactly the ft of the double slit alone. At least as far, as I can follow.

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  • $\begingroup$ if 1000 scientists each did a DSE experiment which only one electron ..... if they got together at a big meeting and compared/shared results ... voila the pattern would emerge! $\endgroup$ Aug 10 at 3:25
  • $\begingroup$ @PhysicsDave: can I interpret what you said as supporting? Be an experiment_two: 1000 scientists do an experiment_one . Then statistics about the outcome of experiment_two can be done. When it turns out that the results of statistics are independent of the number of experiment_one. then we are allowed to see experiment_one as normalized and introduce a propability for the outcome of the next experiment_one. Is that what you wanted to say? $\endgroup$
    – Erna
    Aug 10 at 7:00
  • $\begingroup$ Each experiment is independent . If 1000 scientists each flipped a coin and then met to discuss the results ...... voila they would find about 50/50 results distribution. Firing the electron thru the DSE is like flipping a coin ..... it has a probability of appearing at only certain positions. $\endgroup$ Aug 10 at 12:18
  • $\begingroup$ @physicsDave: can you agree that the target coordinate is a result of the probability of the interaction between the electron and at least one specific electron of the sea of electrons that actually form the potential that "is" the double slit? $\endgroup$
    – Erna
    Aug 12 at 10:10
  • $\begingroup$ Yes that sounds reasonable. Per Richard Feynman (path integral) if we consider all or even many paths, the paths with path lengths that are multiples of the wavelength are most probable (i.e. resonant). The EM field is very dynamic, even before the electron has left the starting electrode it is excited and already in theory interacting with every electron in the apparatus .... eventually one of the preferred "resonant" paths is taken/choosen which does correspond to a final electron. $\endgroup$ Aug 12 at 13:48
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The double-slit "thought experiment" ... consists in firing electrons through a double-slit ... and watching them after passing the slits with a light source placed behind the double-slit...

To generalize this it is said that any interaction with the electron changes its straight path of movement.

As electric charges scatter light, one can "detect" which slit the electron went through...

What else can deflect electrons from their path? Moving electrons could be influenced by EM radiation as well as by electric and magnetic fields. For magnetic fields this is described by the Lorentz force. The mentioned by you "Numerical Simulation of the Double Slit Interference with Ultracold Atoms" consists a lot of calculations Before the slits and Behind the slits. But what is about calculations of the electrons interaction With the slit or With the edges of the slit?

It is a strange thing that there has to be or better there is an interaction between the moving through the slit particles and the slits edges but this interaction is not an object of consideration.

Just my two cents, it has to be calculated the interaction between the moving particles and the magnetic and electric fields of the surface electrons of the (sharp) edges. Then it could be explained easily that an intensity distribution happens even behind single (sharp) edges and even over time with one by one moving particles.

Numerical simulation of the double-slit experiment including watching the electrons ...

... in their interaction with the slit seems to be the right way to bring the endless discussions of double slit experiments to a good end.

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