Single photon double slit experiment

[Davephaelon]

TL;DR Summary

Energy loss between source and detector in single photon double slit experiment?

My question is does the photon that is absorbed by an atom on the detection screen have exactly the same energy as the photon that left the 'gun' source? Hence, does the wave packet representing a photon lose some of its energy when it impinges on the double slit barrier, so that when the diffracted photon is absorbed by an atom on the detection screen it is lower in energy than the one that emerged from the gun?

 

[PeroK]

Summary:: Energy loss between source and detector in single photon double slit experiment?

My question is does the photon that is absorbed by an atom on the detection screen have exactly the same energy as the photon that left the 'gun' source? Hence, does the wave packet representing a photon lose some of its energy when it impinges on the double slit barrier, so that when the diffracted photon is absorbed by an atom on the detection screen it is lower in energy than the one that emerged from the gun?

The frequency of light (hence the energy of a photon) does not change as a result of diffraction.

 

[DrChinese]

Hence, does the wave packet representing a photon lose some of its energy when it impinges on the double slit barrier, so that when the diffracted photon is absorbed by an atom on the detection screen it is lower in energy than the one that emerged from the gun?

Just to add to PeroK's correct answer: The diffraction of the light as it goes through a slit (or slits) is not due to "bouncing off" the edge of a slit. It is due to a form of photon self-interaction. Accordingly, no energy is transferred at the slit.

 

[PeterDonis]

The diffraction of the light as it goes through a slit (or slits) is not due to "bouncing off" the edge of a slit. It is due to a form of photon self-interaction.

It can't be just self-interaction, can it? If the slit weren't there, the light wouldn't diffract, so the diffraction is due to some form of interaction between the slit and the light. It's just not an interaction that transfers any energy.

 

[DrChinese]

It can't be just self-interaction, can it? If the slit weren't there, the light wouldn't diffract, so the diffraction is due to some form of interaction between the slit and the light. It's just not an interaction that transfers any energy.

It is true that the shape of the slit is essentially seen as the various bars when the pattern builds up. A diamond shaped slit yields diamond shaped bars on a screen, for example. But... there really is no contribution worth mentioning related to the material, thickness, etc of the slit material itself that I am aware of. Although I haven't seen any specific experiments where that was varied. Usually, the parameter is simply the width of the slit relative to the wavelength of the light.

As far as I am aware: the theoretical treatment simply sums the effect of the many paths the light could take from the source to the slit to the screen. And there are of course related reinforcement or cancellation effects. I would say this is no different than examining how a photon moves at any time (such as when it goes from one transparent material to another).

If there is a contribution from "bouncing" off the slit itself, it can't be much and is ignored in treatments I've seen.

 

[PeterDonis]

the theoretical treatment simply sums the effect of the many paths the light could take from the source to the slit to the screen.

Yes, but if you model it this way, those paths are not straight lines all the way. They change direction at the slit--or at least some of them do. (This is somewhat oversimplified--see further comments below.) They are not the same paths you would get if you just had light traveling in free space from source to detector with no slits in between.

Actually, in most treatments I've seen, the only paths that are considered are those originating at each slit, considered as a diffraction source--i.e., paths that go in all directions from each slit, with the path length from that slit to the point where the path intersects the detector determining the phase of the path at detection.

If there is a contribution from "bouncing" off the slit itself, it can't be much

To even check for such a contribution, I think one would have to construct a more detailed model of what actually happens at the slits--i.e., what converts the free space path from source to slit, which is idealized as a plane wave with wave fronts orthogonal to the screen the slits are in, to the paths used in the analysis I described above, which are paths from each slit treating that slit as a diffraction source. I am not familiar enough with the literature to say whether or not such a model has been developed.

What I was really wondering about was what you meant by "photon self-interaction", since that doesn't seem to me to correspond to anything in the models I'm familiar with. If all you meant was "sum different photon paths to get an overall amplitude", then I have no further questions.

 

[vanhees71]

Just to add to PeroK's correct answer: The diffraction of the light as it goes through a slit (or slits) is not due to "bouncing off" the edge of a slit. It is due to a form of photon self-interaction. Accordingly, no energy is transferred at the slit.

Photons do not self-interact (except for a very weak interaction due to quantum effects). Diffraction is indeed the result of the interaction of the electromagnetic field with the matter making up the slit. Indeed many photons get absorbed. Of course, it's way better to think about the slit and photon in terms of electromagnetic waves than to think as if photons were localized point-particle like objects (which they never are, because they don't even have a position observable in the usual sense).

 

[DrChinese]

If all you meant was "sum different photon paths to get an overall amplitude", then I have no further questions.

Yes exactly, nothing more, nothing less.

Regardless of whether we are talking about a single slit or a double slit setup: the light pattern hitting the screen is normally described without reference to any kind of physical interaction with the slit edge itself. It's just the possible paths the light could take to get to a particular spot, net of constructive or destructive effects, that determines relative intensity. And those won't normally involve interaction with the slit edge itself in a fashion in which momentum is transferred (although I guess there might be some paths in which that happens).

This video is the for OP, a basic discussion, although Dave may already be aware of this now that I reread his question:

 

[vanhees71]

Of course we describe the reaction of the em. wave with the material making up the slits. It's just simplified by the approximation to make the material total absorptive, i.e., we substitute it by integrating over the openings only in Kirchhoff's diffraction theory. Under Fraunhofer conditions this finally yields the result that the diffraction pattern is the Fourier transform of the openings.

 

[PeterDonis]

the light pattern hitting the screen is normally described without reference to any kind of physical interaction with the slit edge itself. It's just the possible paths the light could take to get to a particular spot

But the presence of the screen with the slits drastically restricts the possible paths from the source to the detector. That, all by itself, is a physical interaction with the slits (more precisely, with the screen and the slits). So I don't think it's correct to say that there is no reference at all to any kind of physical interaction with the slits. I think it would be better to say that there is no detail given about that interaction; it's just treated as a simple "light can only go through the slits, not through the screen, and each slit becomes the source of a diffraction waveform".

 

[DrChinese]

Of course we describe the reaction of the em. wave with the material making up the slits. It's just simplified by the approximation to make the material total absorptive, i.e., we substitute it by integrating over the openings only in Kirchhoff's diffraction theory. Under Fraunhofer conditions this finally yields the result that the diffraction pattern is the Fourier transform of the openings.

No one is disputing that the *open* portion of the slit is relevant, and I said so in #5. The issue at hand is whether the composition of the edge material, thickness of the edge material, coarseness of the edge material, etc. are the cause of most of the diffraction. I say they aren't, although there could be some minor contribution. I have never seen a treatment discussing those elements, although perhaps you have a reference. All discussions I have ever seen discuss only the relative width of the slit as it relates to the wavelength of the light.

 

[DrChinese]

But the presence of the screen with the slits drastically restricts the possible paths from the source to the detector. That, all by itself, is a physical interaction with the slits...

Of course the slits completely prevent some/most light from reaching the screen. That obviously has no impact on the light that does reach the screen, which is what we are discussing.

I am saying that there are essentially NO photons taking a path to the screen, in which that photon's path is affected by an interaction with the edge of the slit. Suppose the slit is a traditional bar. The diffraction pattern will be bars. Suppose the slit is instead the shape of a diamond. The bars will now be diamond shaped - each of them. Obviously that is not due to interaction with the edge itself.

Here is a well accepted discussion of the Y component after passing through a slit. No where is the slit edge itself a factor.

https://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf
https://arxiv.org/pdf/1009.2408.pdf

 

[PeterDonis]

I am saying that there are essentially NO photons taking a path to the screen, in which that photon's path is affected by an interaction with the edge of the slit.

Stated this way, what you are saying still seems ambiguous. However:

The issue at hand is whether the composition of the edge material, thickness of the edge material, coarseness of the edge material, etc. are the cause of most of the diffraction.

This makes it much clearer what you mean. I agree that these factors are mostly irrelevant.

 

[vanhees71]

No one is disputing that the *open* portion of the slit is relevant, and I said so in #5. The issue at hand is whether the composition of the edge material, thickness of the edge material, coarseness of the edge material, etc. are the cause of most of the diffraction. I say they aren't, although there could be some minor contribution. I have never seen a treatment discussing those elements, although perhaps you have a reference. All discussions I have ever seen discuss only the relative width of the slit as it relates to the wavelength of the light.

Well, it's a very much simplified treatment of the interaction of the electromagnetic field with the material making up the slits. That's all I'm saying. Exact diffraction theory is among the most complicated problems in classical electrodynamics already. It was solved for the most simple case of a half-space by Sommerfeld in his habilitation thesis.

Of course you are right in saying that the details about composition of the "edge material" etc. are of very limited importance for the understanding of diffraction.

 

[msumm21]

Is there no momentum exchange between the slit and photon? E.g. in the cases the photon is found further up on the screen (picture in post#8), was the slit pushed down slightly?

 

[DrChinese]

Is there no momentum exchange between the slit and photon? E.g. in the cases the photon is found further up on the screen (picture in post#8), was the slit pushed down slightly?

You are thinking in terms of conservation. If you read the references provided in post #12, you will see that momentum is not handled that way. For example, what is any quantum particle's y momentum when its y position is sharply defined? From Marcella:

"Because position and momentum are non-commuting observables, a particle passing through slits always has an uncertainty in its y-component of momentum. It can be scattered with anyone of the continuum of momentum eigenvalues py = p sinθ , where −π /2 ≤ θ ≤ π /2."

 

[msumm21]

If you read the references provided in post #12, you will see that momentum is not handled that way. For example, what is any quantum particle's y momentum when its y position is sharply defined?

Thanks. Let me try to explain my (flawed?) understanding. There's Y-momentum uncertainty that must grow when the photon's position is limited to the slit width. Before entering the slit, say the Y momentum distribution had mean=0 with a smaller variance and similarly the slit's momentum distribution had mean 0. Upon observing the photon up high in the picture in #8 (Y>0) can't we say the Y-momentum distribution of the slit now has mean < 0? And yes I guess I'm struggling to conserve Y-momentum!

 

[PeterDonis]

Before entering the slit, say the Y momentum distribution had mean=0 with a smaller variance

Yes, this would be a consequence of the preparation procedure at the source.

the slit's momentum distribution had mean 0

Yes, if we assume that the laboratory setup is to hold the screen with the slits in a fixed position and not allow it to move.

Upon observing the photon up high in the picture in #8 (Y>0) can't we say the Y-momentum distribution of the slit now has mean < 0?

No, because you're not measuring it. (And in any case the slit itself would transfer any Y-momentum to the material of the screen, which in turn would transfer it to the Earth since the screen is being held in place by something fixed to the Earth. So you're basically out of luck trying to measure it anyway with this setup.)

I guess I'm struggling to conserve Y-momentum!

If you don't measure it, you can't say it's conserved. You can't even say it has a definite value. This is one of the key things about QM that is counterintuitive and makes it very different from classical mechanics.

 

[vanhees71]

That's also not true. An observable is conserved if it commutes with the Hamiltonian. There has been a famous work by Kramers and Bohr in the very early days of modern quantum theory, just before the full theory has been developed by Heisenberg, Born, Jordan and Schrödinger and Dirac, where they assumed that the conservation laws only hold on the average. This has been refuted by an even more famous experiment by Bothe, applying his coincidence measurement method to the Compton effect, clearly demonstrating that energy and momentum conservation holds in each single event.

For a photon interacting with a grating you can of course neglect momentum exchange since the macroscopic slit is so heavy.

 

[PeterDonis]

An observable is conserved if it commutes with the Hamiltonian.

First, Y-momentum does not commute with the Hamiltonian in the case under consideration, since passing through the slits causes a change in the wave function that affects it.

Second, the rule you state only applies, strictly speaking, to a complete isolated system. An observable on a complete isolated system that commutes with the Hamiltonian of that complete isolatede system will be conserved, yes. But the photon alone in this experiment is not a complete isolated system, so no observable on the photon alone is a candidate for a conserved quantity. (A more precise statement of the above would involve degrees of freedom--the Y momentum of the photon is not an isolated degree of freedom.)

 

[vanhees71]

I have not claimed that the photon momentum is conserved in the double-slit experiment. I have only made the remark that a conservation law holds for an observable if it commutes with the Hamiltonian and that the probabilistic nature of quantum theory does not invalidate the corresponding conservation law, which holds for any single event though the quantity may fluctuate for the given prepared state of the object. This has been resolved with Bothe's coincidence experiment refuting the (in)famous Bohr-Kramers theory.

 

[PeterDonis]

I have not claimed that the photon momentum is conserved in the double-slit experiment.

Ok, good.

I have only made the remark that a conservation law holds for an observable if it commutes with the Hamiltonian

And I have given a clarification to that statement that makes it clear why it does not apply to the Y-momentum of the photon in the case under discussion here.

 

[msumm21]

No, because you're not measuring it. (And in any case the slit itself would transfer any Y-momentum to the material of the screen, which in turn would transfer it to the Earth since the screen is being held in place by something fixed to the Earth. So you're basically out of luck trying to measure it anyway with this setup.

So the mean Y-momentum of the slit remains 0, independent of the where the photon landed on the screen?

Also, couldn't we measure it theoretically, e.g. by measuring the location of a slit before and after? If doing this on Earth is complicating things with ground interaction, just assume this is all floating in space.

 

[PeterDonis]

So the mean Y-momentum of the slit remains 0, independent of the where the photon landed on the screen?

First you need to ask yourself how you know that the mean Y-momentum of the slit is 0 before the experiment. As I said, you know that because the screen with the slits in it is fixed to the laboratory and hence to the Earth. Since that remains true all through the experiment, the mean Y-momentum of the slit would remain the same as well.

couldn't we measure it theoretically, e.g. by measuring the location of a slit before and after?

You could in principle change the experiment to include measurements made on the screen with the slits in it to measure any momentum change. But doing that would also eliminate the interference.