Quantum vs Classical multiphoton interference on double slit

In summary: This quantity is zero if the two photons have the same phase (in which case ##|A'B'>+|B'A'>##), and it becomes very large if the two photons have different phases (in which case ##|A'B'>-|B'A'>##).What this tells us is that, for a wave function near the screen corresponding to a photon passing through slit 1, the probability of photon 1 hitting the screen at point 1 is:##(\phi_i A'\psi_i)+(-1)^{2}\left(\frac{1}{|\phi_i A'|}\right)
  • #1
MichPod
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TL;DR Summary
Can there be classical interference without quantum? How? Is there mistake in my calculations showing there is no interference?
For a setup analogous to double-slit experiment, do I understand it correctly that if we have two photons, one going only through slit A and another going only through slit B simultaneously (or vice versa as they have a symmetrical wave function), then there will be no interference pattern created when this experiment is repeated many times, even providing the photons have the same phase?

If that is correct, how would then two coherent laser beams interfere, while one passing through the slit A and another through the slit B?
I.e. if in the experiment with two photons they roughly could hit any points on the screen, how then dark fringes where photons cannot land would appear for two laser beams?
 
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  • #2
I believe it is not meaningful to talk of interference between two photons.
I find it makes most sense to think of photons as observations - experimental outcomes - rather than as inputs. The interference in a double slit pattern is not interference between designated photons but interference between two probability fields* - each field being associated with one of the two slits. The probability fields tell us the probability of observing a 'photon strike' at each point on the screen.

Accordingly, we can't meaningfully talk about a setup with two photons crossing the slits simultaneously, because that implies that they are observed at the slit, so we no longer have waves and probability fields, just a couple of photon observations. Furthermore, the act of observing them likely destroys them. Either way, there would be nothing left to interfere.

If we directed two laser beams at the slits, I imagine that, if the beams were narrower than the slits, they would simply pass through and make two tiny dots on the screen. No interference.

I expect what happens if the beams are wider than the slits would be more interesting. Somebody that understands more about lasers would need to answer that.

* They are actually 'pre-probability' fields. At each point on the screen we need to add the fields and then square, to get a probability. Further, the 'probability' is actually an intensity rather than a probability: the probability of observing a photon strike in a particular square nanometre in a nanosecond.
 
  • #3
MichPod said:
For a setup analogous to double-slit experiment, do I understand it correctly that if we have two photons, one going only through slit A and another going only through slit B simultaneously (or vice versa as they have a symmetrical wave function), then there will be no interference pattern created when this experiment is repeated many times, even providing the photons have the same phase?

No. If the phases of the two photons are the same (more precisely, the correlation between them is precisely controlled), and if the photons are emitted at the same time, then you can get interference.

The real issue is those two"if"s. Setting up two separate sources of photons and controlling them so that (a) they emit photons at the same time, and (b) the phases of the emitted photons have a precisely controlled correlation, is really, really, really, really hard.

andrewkirk said:
I believe it is not meaningful to talk of interference between two photons.

It is, under the appropriate conditions--but the conditions you have to set up to make it meaningful are extremely difficult. See above.
 
  • #4
PeterDonis said:
If the phases of the two photons are the same (more precisely, the correlation between them is precisely controlled), and if the photons are emitted at the same time, then you can get interference.

I probably got my calculations wrong, as I got that two photons cannot interfere on the two slits in my setup. It's been a long time since I did anything of this sort and I was never good at QM, so I might easily make a mistake. Could please somebody review? For what I could see:

If at the slits the wave function of two photons is ##|AB>+|BA>## where ##|A>## and ##|B>## are wave functions of a singular photon at 1st and 2nd slits respectively,

I then try to calculate a probability the first photon hits a particular point at the screen. When a photon passing through the slit 1 is near the screen, it's wave function is ##A'=\sum|\psi_i>## and for a photon passing through the slit 2 ##B'=\sum|\phi_i>## where ##|\psi_i>## and ##|\phi_i>## correspond to a component of the wave function corresponding to the detection of a photon at the point ##i## of the screen.

Therefore for a wave function near the screen ##|A'B'>+|B'A'>## the component corresponding to hitting the point ##i## by the first photon is:

##|\psi_i B'>+|\phi_i A'>##

The magnitude of this vector (which is the probability of photon 1 hitting the screen at the point 1) is:

##(<\psi_i B'|+<\phi_i A'|)(|\psi_i B'>+|\phi_i A'>) = 1 + 1 + 2 Re(<\phi_i A'|\psi_i B'>)##

For a point corresponding, for instance, to the destructive self-interference of a single photon ##|\psi_i>=-|\phi_i>##

and so ##Re(<\phi_i A'|\psi_i B'>)=Re(-<\psi_i A'|\psi_i B'>)=-Re(<A'|B'>)##

I then expect that ##<A'|B'>=<A|B>## as |A'> and |B'> are a result of unitary evolution of |A> and |B>, so their product remains invariant in time, and then I expect that ##<A|B>=0## as they are non-overlapping in the space, while being at separate slits.

Therefore ##2 Re(<\phi_i A'|\psi_i B'>)=0## and the probability I am looking at is constant (for other points we'll get exactly the same result, which is quite obvious). So I must conclude there is no variations of probability for the first photon hitting the screen, nor for the second one (by analogy).
 
  • #5
MichPod said:
If at the slits the wave function of two photons is ##|AB>+|BA>## where ##|A>## and ##|B>## are wave functions of a singular photon at 1st and 2nd slits respectively,

That's not correct. If both photons can go through either slit, then you have four possibilities: AA, AB, BA, and BB.

You also need to calculate the probability of having photons hit the screen for pairs of points, since there are two photons so there will be two photon hits on the screen for each run of the experiment. In other words, you need to calculate probabilities of the form "probability that a photon will hit at point ##i## and a photon will hit at point ##j##". Note that you cannot tell which photon hit is from which source, so it makes no sense to ask, for example, what the probability is of photon #1 hitting a particular point on the screen.
 
  • #6
PeterDonis said:
That's not correct. If both photons can go through either slit, then you have four possibilities: AA, AB, BA, and BB.

I actually meant that one photon goes through slit 1 and another through slit 2. And the same for lasers - one beam through slit 1, 2nd beam through the slit 2.

For 4 possibilities you are talking above I think we have

## |AA>+|AB>+|BA>+|BB> = (|A>+|B>)(|A>+|B>)##

which I think just makes two self-interfering photons or "two experiments" with single photons.

Anyway, I meant just photons coming from different slits.

PeterDonis said:
You also need to calculate the probability of having photons hit the screen for pairs of points, since there are two photons so there will be two photon hits on the screen for each run of the experiment. In other words, you need to calculate probabilities of the form "probability that a photon will hit at point ##i## and a photon will hit at point ##j##". Note that you cannot tell which photon hit is from which source, so it makes no sense to ask, for example, what the probability is of photon #1 hitting a particular point on the screen.

I first took a symmetrized wave function, then I thought I could speak about "photon 1" and "photon 2" without any risk. And the probability I calculated was the first photon hits the screen at point ##i## while there is no condition on which point the second photon hits the screen. Which allows us to see a distribution of one of the photons. And by "photon 1" I do not mean the photon passing through slit 1, I mean the photon staying at the first position in my |ket>.
 
  • #7
MichPod said:
I actually meant that one photon goes through slit 1 and another through slit 2.

If you are treating the two photons as exactly in phase, then you cannot tell which photon from which source goes through which slit. That means you can't say "one photon goes through slit 1 and one photon goes through slit 2". You have to treat both photons as going through both slits.

Note that all this is assuming both photon sources are such that a photon from either one could go through both slits. If by "one photon goes through slit 1 and one photon goes through slit 2", you mean that you have each source aimed so that it only goes through one slit, then of course you won't get any interference, because you've eliminated it by giving each photon the equivalent of "which-slit" information: aiming each photon source at just one slit is the same as putting a detector behind each slit to signal which slit the photon went through in the single photon case.
 
  • #8
MichPod said:
I meant just photons coming from different slits.

If both photon sources are such that a photon from them can go through either slit, then you cannot assume that the two photons that hit the screen come from different slits. They could both come from the same slit.
 
  • #9
MichPod said:
which I think just makes two self-interfering photons or "two experiments" with single photons.

The field equation for photons is linear, so for two photon sources set up so that a photon from either one can go through both slits, the setup is equivalent to a superposition of two single photon experiments, yes.
 
  • #10
PeterDonis said:
If you are treating the two photons as exactly in phase, then you cannot tell which photon from which source goes through which slit. That means you can't say "one photon goes through slit 1 and one photon goes through slit 2". You have to treat both photons as going through both slits.

I think this is addressed by considering symmetrical |AB>+|BA> wavefunction of two photons, where |A> corresponds to a photon passing via the 1st slit and |B> to the 2nd. No, I cannot tell which photon went through which slit. Yes, I can tell that 2 photons are coming through different slits.

PeterDonis said:
aiming each photon source at just one slit is the same as putting a detector behind each slit to signal which slit the photon went through in the single photon case.

I am not sure about this argument, but if you agree that for my setup with two photons there will be no interference (as also my calculations might show), then I would return to my previous question - how then would be able to see the interference if we fire not two photons, but coherent laser beams, one through slit A and another through slit B? I.e. if there is no interference on quantum level, how are we going to have classical interference of two light beams?

Disclaimer: I am not sure whether my calculations are correct and whether there will be no two photon interference in my setup.
 
  • #11
MichPod said:
I think this is addressed by considering symmetrical |AB>+|BA> wavefunction of two photons

No, it isn't, because it assumes that one photon goes through each slit. You are ignoring the possibility that both photons go through the same slit.

MichPod said:
Yes, I can tell that 2 photons are coming through different slits.

How?
 
  • #12
MichPod said:
if you agree that for my setup with two photons there will be no interference

I am not clear about exactly what you intend your setup to be. Is each photon source emitting photons that can go through both slits?

MichPod said:
if we fire not two photons, but coherent laser beams

What's the difference?

A laser can't even be analyzed classically anyway; it depends on quantum effects to work at all.

MichPod said:
if there is no interference on quantum level, how are we going to have classical interference of two light beams?

If there is no interference, there is no interference. I don't understand what you think the difference is between "classical interference" and "quantum interference".
 
  • #13
PeterDonis said:
No, it isn't, because it assumes that one photon goes through each slit. You are ignoring the possibility that both photons go through the same slit.

I am not "ignoring", it's just not the experimental setup I am considering. I might just use a spontaneous parametric down-conversion (SPDC) device, the same one used in the delayed quantum eraser experiment to produce two photons, then route them via a lens to try to see an interference pattern. As SPDC creates two photons orthogonally polarized, I'll of course need to rotate their polarization directions to align them in order to get interference, if such interference is possible at all.
PeterDonis said:
How?

Because the projection of ##|AB>+|BA> ## onto ##|AA>## state is zero, there is no chance my state describes two photons passing the first slit together.
 
  • #14
PeterDonis said:
I am not clear about exactly what you intend your setup to be. Is each photon source emitting photons that can go through both slits?

No. The first source emits a photon which can only go through the slit 1, second source emits a photon which can go only through the slit 2.

PeterDonis said:
What's the difference?
A laser can't even be analyzed classically anyway; it depends on quantum effects to work at all.
If there is no interference, there is no interference. I don't understand what you think the difference is between "classical interference" and "quantum interference".

Ok, whatever. Suppose we have two coherent light beams which can be described precisely enough as classical sinusoidal wave. I remember, there must be corresponding quantum distribution of photon states, just forgot the name. Will we see then the interference? If yes and if we cannot see it just for two photons - how is it possible?
 
  • #15
MichPod said:
I am not "ignoring", it's just not the experimental setup I am considering.

Then you need to clarify what experimental setup you are considering, because at this point I have no idea what it is.

MichPod said:
I might just use a spontaneous parametric down-conversion (SPDC) device, the same one used in the delayed quantum eraser experiment to produce two photons, then route them via a lens to try to see an interference pattern. As SPDC creates two photons orthogonally polarized, I'll of course need to rotate their polarization directions to align them in order to get interference, if such interference is possible at all.

None of this helps me understand what experimental setup you intend.

MichPod said:
Because the projection of ##|AB>+|BA> ## onto ##|AA>## state is zero, there is no chance my state describes two photons passing the first slit together.

That assumes that your state is the correct quantum state for the two-photon system in your experimental setup. I am not aware of any experimental setup that involves a double slit and two photons for which your state would be the correct quantum state for the two-photon system. So, again, at this point I have no idea what experimental setup you intend.
 
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  • #16
MichPod said:
Suppose we have two coherent light beams which can be described precisely enough as classical sinusoidal wave... Will we see then the interference?
Yes, if you really can maintain the coherence. This is just the linearity of ordinary classical electrodynamics - if two electromagnetic waves are impinging on the same point, the amplitude of the E and B fields at that point will just be the sum of the individual E and B fields.
If yes and if we cannot see it just for two photons - how is it possible?
A one-photon or a two-photon state is a different thing than a coherent light beam. If you have a fixed number of photons, whether zero, one, two, or two million, you don't have a coherent plane wave.
 
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  • #17
Nugatory said:
If you have a fixed number of photons, whether zero, one, two, or two million, you don't have a coherent plane wave.

Thanks. What do you think about my conclusion/calculations that two photons cannot interfere in my setup?
And if we have coherent light beams with some distribution of the number of photons, how does it help them to interfere if we consider this interference on the quantum level? I thought considering just two photons may be enough to make the conclusions, but I am not 100% sure.
 
  • #18
"Therefore for a wave function near the screen ##|A'B'>+|B'A'>##"

That was my mistake, it's wrong to denote the value of the wave function "near the screen" as a vector (ket-vector), as the values of the wave function near the screen are just a section of the whole wave function. If calculated for the values of the wave functions, then not ##<A'|B'>##, but the integral appears (the integration is done along the screen)
##\int A'^\star(x)B'(x)dx## which is non-zero.

The conclusion is - two photons can interfere.
 
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1. What is the difference between quantum and classical multiphoton interference on a double slit?

Quantum and classical multiphoton interference on a double slit refer to the phenomenon of light behaving as both a wave and a particle as it passes through two narrow slits. The main difference between the two is that in quantum interference, individual photons can interfere with each other, while in classical interference, the overall intensity of the light is affected by the interference pattern.

2. How is quantum multiphoton interference on a double slit observed?

Quantum multiphoton interference on a double slit is observed by sending individual photons through the double slit and measuring the resulting interference pattern on a screen. This pattern will show areas of high and low intensity, indicating the constructive and destructive interference of the photons.

3. Can classical multiphoton interference on a double slit be explained by classical wave theory?

No, classical multiphoton interference on a double slit cannot be fully explained by classical wave theory. While classical wave theory can predict the overall intensity of the interference pattern, it cannot explain the individual interference patterns of each photon. This requires the use of quantum mechanics.

4. How does the number of photons affect the interference pattern in quantum multiphoton interference on a double slit?

The number of photons affects the interference pattern in quantum multiphoton interference by increasing the visibility of the interference pattern. As more photons are sent through the double slit, the interference pattern becomes more defined and the areas of high and low intensity become more distinct.

5. What applications does understanding quantum vs classical multiphoton interference on a double slit have?

Understanding quantum vs classical multiphoton interference on a double slit has various applications in fields such as quantum computing, cryptography, and imaging. It also helps us better understand the nature of light and its behavior, leading to advancements in technology and scientific research.

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