Can one photon interfere with another photon?

[tim1608]

Hi Everyone

This may sound like a question with an obvious answer but with my current knowledge, it is not obvious to me. Can one photon interfere with another photon or can each photon only interfere with itself?

Most people on this forum will know that Young's Slits produces interference patterns even with "one-at-a-time" photons. This means that it is not necessary for one photon to interfere with another photon in order for Young's Slits to produce an interference pattern.

One way of looking at this question is as follows: Imagine that you have two seemingly identical beams of parallel light which are as powerful as each other and have exactly the same pure wavelength. Let's say the two beams are coming from two unmarked emitters so that you can't tell exactly how each emitter is producing its beam but you have been told that one of the emitters is a laser device which produces completely phase-aligned light and the other emitter contains just a normal but powerful LED which produces light which is not phase-aligned. Can Young's Slits tell the difference between the two beams? Can the normal but powerful LED produce visible interference patterns?

If it is not possible to tell the difference between the two beams using Young's Slits then I would say that this would mean that one photon cannot interfere with another photon. What do the experts on this forum think?

Thank you very much.

 

[bhobba]

This may sound like a question with an obvious answer but with my current knowledge, it is not obvious to me. Can one photon interfere with another photon or can each photon only interfere with itself?

No.

Its obvious why not when you see a correct analysis of the double slit experiment:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

But I also have to say since all photons are equivalent the question is in some sense ambiguous.

Thanks
Bill

 

[DrChinese]

I have always seen this particular question as a bit complex (and having nothing to do with a double slit setup). Some assert that independent photons cannot interfere, and others call it the other way. Here is one article:

http://arxiv.org/abs/1112.1943
Quantum interference between two single photons of different microwave frequencies

 

[tim1608]

Hi Bill

Thank you very much for your reply. I would be very grateful if you could clarify your answer by answering the following two questions:

1. Does a beam of light have to be phase-aligned in order to produce visible interference patterns?

2. Can two independent laser beams produce visible interference patterns when shone onto the same spot?

If I am interpreting your answer correctly, then I think your answer to both of the above two questions ought to be "no".

Thank you very much.

 

[stevendaryl]

One thing that I am uncertain about when it comes to interference is understanding the relationship between quantum interference and classical interference. The electromagnetic field, being a tensor-valued field, exhibits interference classically: You add up the fields due to two (or more) sources to get a resulting field. The intensity for the resulting field will in some places be less than either field separately (destructive interference) and will in some places be greater than either field separately (constructive interference). This classical notion of interference, I would think, still applies quantum-mechanically. An E&M wave could be understood as the effect of many, many photons, and it seems to me that it would interfere with a second E&M wave consisting of many other photons.

 

[tim1608]

Thank you DrChinese and Stevendaryl for your replies.

I am getting the impression that there does not seem to be a clear answer to my question. I wish I had access to an appropriately equipped lab so that I could attempt to find out the answer myself. How difficult would it be to build a device which uses a powerful pure-wavelength LED to produce a non-phase-aligned beam which looks exactly like a laser beam?

 

[Cthugha]

One way of looking at this question is as follows: Imagine that you have two seemingly identical beams of parallel light which are as powerful as each other and have exactly the same pure wavelength. Let's say the two beams are coming from two unmarked emitters so that you can't tell exactly how each emitter is producing its beam but you have been told that one of the emitters is a laser device which produces completely phase-aligned light and the other emitter contains just a normal but powerful LED which produces light which is not phase-aligned. Can Young's Slits tell the difference between the two beams? Can the normal but powerful LED produce visible interference patterns?

Young's double slit experiment is fully classical and works on the field level. In fact it is a measurement of spatial coherence, which is more or less the angular size of the light source as seen from the double slits. As a consequence you can take any light source in the world and will get exactly the same interference pattern if you

a) filter it spectrally such that you get the same spectral distribution
b) filter it in real space using a pinhole such that you get a point-like light source
c) adjust it to the same brightness

So the double slit cannot distinguish between different light sources. However, there are other kinds of measurements which can do that. The double slit just measures first order coherence (see http://en.wikipedia.org/wiki/Degree_of_coherence).

If you go to higher oder coherence, you will indeed find that multi-photon interference is possible. However, one needs conditional measurements to test that. Usually you split your light field, put it on two photodiodes and measure the conditional probability to detect a photon at a delay tau at detector 2, if you already detected a photon at time 0 at detector 1. The simplest and best known two-photon interference effect is the so-called Hong-Ou-Mandel dip (http://en.wikipedia.org/wiki/Hong–Ou–Mandel_effect) which is a standard experiment to check whether two single photons are indistinguishable or not.

However, one should keep in mind that this kind of interference is very different from what one typically has in mind when thinking about double slit experiments.

 

[bhobba]

Does a beam of light have to be phase-aligned in order to produce visible interference patterns?

Of course not. You can do it with ordinary light that is anything but.

Can two independent laser beams produce visible interference patterns when shone onto the same spot?

No - the thing I am pretty sure is if they are correlated - independent laser beams are by definition not correlated.

Thanks
Bill

 

[bhobba]

I am getting the impression that there does not seem to be a clear answer to my question.

Like I said since photons are indistinguishable the question is a bit ambiguous - the out being if they are entangled in some way.

I do not think uncorrelated photons can interfere.

Thanks
Bill

 

[tim1608]

Firstly, thank you Bill for your latest replies.

Thank you Cthugha for your reply. To be honest, the end of your penultimate paragraph went a bit beyond my knowledge of physics and I think that the Wikipedia article for the Hong-Ou-Mandel effect does not give a very good explanation for a layman. However the gist I get is that under certain circumstances, two separate photons can "extinguish each other" as Wikipedia puts it when they meet at a beam splitter. If this is correct, where does the energy go?

Edit: Am I correct that photon extinguishment would happen when one photon is reflected and the other passes right through?

Edit: Can you give a layman's explanation for the different orders of coherence?

Thank you very much.

 

[Cthugha]

To be honest, the end of your penultimate paragraph went a bit beyond my knowledge of physics and I think that the Wikipedia article for the Hong-Ou-Mandel effect does not give a very good explanation for a layman. However the gist I get is that under certain circumstances, two separate photons can "extinguish each other" as Wikipedia puts it. If this is correct, where does the energy go?

Right. It is not written too well. Let me sum it up.

You have a beam splitter with two input ports (I1 and I2), two output ports (O1 and O2) and two indistinguishable photons (P1 and P2) arrive at two different entry ports. Now there are 4 possibilities:

a) P1 gets reflected, P2 gets transmitted: Both photons leave via O2.
b) P1 gets transmitted, P2 gets reflected: Both photons leave via O1.
c) Both photons get transmitted: One photon leaves via O1, one photon leaves via O2.
d) Both photons get reflected: One photon leaves via O1, one photon leaves via O2.

c) and d) lead to the same final situation. So if the photons are indistinguishable, you need to add the probability amplitudes for these two paths and square afterwards to get the probability for this event. If the photons are distinguishable, you just take the square for each single path and just add the probabilities. Due to the phase shift occurring upon reflection at a beam splitter, the probability amplitudes for c) and d) will always interfere destructively and this event will not take place.

So, the photons do not extinguish each other, but they will form a bunch of two photons leaving via the same exit port. It will never happen that you find one photon emerging from each output port. This is an interference process involving two photons. However, it is usually considered as misleading terminology to consider this as the interference OF two photons. One should rather consider it as an interference of probability amplitudes for events containing more than one photon.

A classic paper discussing that issue is "Can Two-Photon Interference be Considered the Interference of Two Photons?" (Phys. Rev. Lett. 77, 1917 (1996)). Also available from NIST, if you do not have a subscription for PRL: http://physics.nist.gov/Divisions/Div844/publications/migdall/psm96_twophoton_interference.pdf

 

[snorkack]

Compare a pair of radio antennae.
They certainly can operate completely independently.
Yet you can easily make them operate at pretty exactly same frequency, and at a specific phase relationship.
So radio wave photons from two independent but synchronized antennae can certainly give interference pattern.

 

[tim1608]

Hi Cthugha and Bill

Firstly, Cthugha, thank you very much for your latest reply. I think it is somewhat clearer than the Wikipedia article.

since all photons are equivalent

Like I said since photons are indistinguishable

Bill, I just want to clear something up here. What precisely do you mean by "equivalent" and "indistinguishable"? Do you mean just simply identical or do you mean something more than this? I may be wrong but it sounds slightly as if you think that there is actually no such thing as a completely individual, discrete photon. Am I correct?

Do you think that on some level (a level which does not effect Young's Slits but which possibly does effect the Hong-Ou-Mandel effect), all photons in the Universe of exactly the same wavelength share the same vast communal wavefunction?

Thank you very much.

You have a beam splitter with two input ports (I1 and I2), two output ports (O1 and O2) and two indistinguishable photons (P1 and P2) arrive at two different entry ports. Now there are 4 possibilities:

a) P1 gets reflected, P2 gets transmitted: Both photons leave via O2.
b) P1 gets transmitted, P2 gets reflected: Both photons leave via O1.
c) Both photons get transmitted: One photon leaves via O1, one photon leaves via O2.
d) Both photons get reflected: One photon leaves via O1, one photon leaves via O2.

c) and d) lead to the same final situation. So if the photons are indistinguishable, you need to add the probability amplitudes for these two paths and square afterwards to get the probability for this event. If the photons are distinguishable, you just take the square for each single path and just add the probabilities. Due to the phase shift occurring upon reflection at a beam splitter, the probability amplitudes for c) and d) will always interfere destructively and this event will not take place.

So, the photons do not extinguish each other, but they will form a bunch of two photons leaving via the same exit port. It will never happen that you find one photon emerging from each output port. This is an interference process involving two photons. However, it is usually considered as misleading terminology to consider this as the interference OF two photons. One should rather consider it as an interference of probability amplitudes for events containing more than one photon.

Cthugha, I Have five questions I would like to ask about this:

1. In a typical Hong-Ou-Mandel experimental setup, are two separate emitters used to generate the two photons entering on either side of the beam splitter?

2. In order to "interfere" with each other, do the two photons have to enter the beam splitter at exactly the same place on either side of the beam splitter? If yes, how is this precision achieved?

3. If one of the beams is already phase-shifted by half a wavelength before entering the beam splitter, does this mean that instead of possibilities c) and d) not taking place, possibilities a) and b) don't take place instead?

4. When the two photons leave the beam splitter, are they side-by-side or is one behind the other or are they positioned relative to each other in some other way? How do we know that two photons are leaving the beam splitter?

5. Do you think it is possible that the true explanation of the Hong-Ou-Mandel effect may actually have nothing to do with the square root of probabilities? I am thinking that it might be the case that a more classical mechanism is operating whereby the beam splitter contains a vast number of tiny molecules which behave as randomly directed "one-way valves" (like in plumbing) which only allow photons to travel in one direction. It may be that these molecules act differently on photons of different wavelengths. Do you think that I might be on to something here?

Thank you very much.

 

[stevendaryl]

Like I said since photons are indistinguishable the question is a bit ambiguous - the out being if they are entangled in some way.

I do not think uncorrelated photons can interfere.

As I said in a previous post, you can certainly have classical interference between electromagnetic waves. If you have two sources of E&M waves, they will constructively or destructively interfere at a point depending on the difference of the two phases. What does that mean for photon interference? I'm not sure. I don't know what the description of a propagating macroscopic electromagnetic field would be in terms of photons.

 

[tim1608]

I don't know what the description of a propagating macroscopic electromagnetic field would be in terms of photons.

Thank you, Stevendaryl for your post. I may be wrong here but I think that the square of the field strength at each point in a "propagating macroscopic electromagnetic field" corresponding to an individual photon is proportional to the probability value of the same point in the photon's probability distribution.

 

[Cthugha]

1. In a typical Hong-Ou-Mandel experimental setup, are two separate emitters used to generate the two photons entering on either side of the beam splitter?

One could do that, but the easiest way is to take consecutive single photons emitted from the same source and a delay line.

2. In order to "interfere" with each other, do the two photons have to enter the beam splitter at exactly the same place on either side of the beam splitter? If yes, how is this precision achieved?

You need reasonable overlap of the modes. In reality each lab will have their own technique to achieve that. One way lies in coupling the single photons into fibers. This way you can first couple a bright laser into the fibers, adjust the optics such that the overlap at the fiber output is ideal and then you can remove the laser an start coupling the weak single photon signal into the fibers.

3. If one of the beams is already phase-shifted by half a wavelength before entering the beam splitter, does this mean that instead of possibilities c) and d) not taking place, possibilities a) and b) don't take place instead?

No. The phase shift occurs in the two-photon probability amplitude between transmission/transmission and reflection/reflection and does not really depend on the initial phase (which is ill defined for single photons anyway). The Mandel/Wolf has a good description of this topic.

4. When the two photons leave the beam splitter, are they side-by-side or is one behind the other or are they positioned relative to each other in some other way? How do we know that two photons are leaving the beam splitter?

You get a n=2 Fock state containing two absolutely indistinguishable photons.

5. Do you think it is possible that the true explanation of the Hong-Ou-Mandel effect may actually have nothing to do with the square root of probabilities? I am thinking that it might be the case that a more classical mechanism is operating whereby the beam splitter contains a vast number of tiny molecules which behave as randomly directed "one-way valves" (like in plumbing) which only allow photons to travel in one direction. It may be that these molecules act differently on photons of different wavelengths. Do you think that I might be on to something here?

No. You are absolutely not on to something. You will get both photons leaving via exit port 1 50% of the time and both photons leaving via the other exit port 50% of the time. This does not depend on the wavelength, the material used for the beam splitter or even the beam splitter design as long as the beam splitter is 50/50. This has been verfied experimentally thousands of times in very different settings and for very different sources. Please do not make the mistake of guessing and formulating personal theories before actually learning in detail about a topic.

 

[tim1608]

Hi Cthugha

Thank you very much for your reply.

One could do that, but the easiest way is to take consecutive single photons emitted from the same source and a delay line.

Am I correct in thinking that the delay line would be applied to one of the two paths to ensure that two probability sub-distributions of the same photon cannot arrive at the beam splitter at the same time?

You need reasonable overlap of the modes.

I am not entirely sure of what exactly you mean by the "modes". Do you mean the areas on the beam splitter where the probability distributions of the photons will land? Am I correct that by using fibers or some other method, these areas will be made as small as possible?

By-the-way, was my reply to Stevendaryl correct? (Post #15)

Thank you very much.

 

[carrz]

I have always seen this particular question as a bit complex (and having nothing to do with a double slit setup). Some assert that independent photons cannot interfere, and others call it the other way. Here is one article:

If photons cannot interfere with photons, then what in the world are they interfering with in a double slit experiment? There is nothing supposed to be there but the slit and the photons, and if it is not the photons, it must be the slit. Would edge-diffraction not produce the same pattern even without any interference?

http://arxiv.org/abs/1112.1943
Quantum interference between two single photons of different microwave frequencies

They start talking about interference and end up talking about entanglement. What does quantum entanglement have anything to do with wave interference?

 

[D H]

If photons cannot interfere with photons, then what in the world are they interfering with in a double slit experiment?

Themselves? Nothing? You choose. It's best not to look at photons as sometimes a particle, sometimes a wave. They are quantum mechanical objects. Trying to make a classical analogy just doesn't work sometimes.

Suppose you perform the double slit experiment with the incoming light reduced to such an extent that only one photon at a time passes through the slits. You'll still see an interference pattern arise over time as you accumulate where those individual photons are observed to hit the detector. With what do those single photons interfere? Certainly not other photons; there are no other photons in these single photon / double slit experiments. Yet the interference pattern still arises.

 

[DrChinese]

If photons cannot interfere with photons, then what in the world are they interfering with in a double slit experiment? There is nothing supposed to be there but the slit and the photons, and if it is not the photons, it must be the slit. Would edge-diffraction not produce the same pattern even without any interference?

In any double slit setup, it is SELF interference. That is not usually obvious because there are so many photons present. It is much more obvious when something with a rest mass is diffracted through the slits. Self interference is the quantum effect being observed. Please note that interference is also a classical wave property, which makes it somewhat confusing to distinguish one from the other.

Edge diffraction has nothing to do with an interference pattern either way, that only determines the shape of the bars. The only thing that determines the interference pattern is whether which-slit information is potentially available.

 

[tim1608]

If photons cannot interfere with photons, then what in the world are they interfering with in a double slit experiment?

Hi Carrz

When DrChinese was saying "independent photons cannot interfere", he was referring to the assertion of some scientists that two or more individual photons cannot interfere with each other. I might be wrong but if I understand it correctly, what instead may be happening is that part of the probability distribution of an individual photon is interfering with another part of the same probability distribution. Or putting it another way, two probability sub-distributions of the same photon are interfering with each other.

 

[carrz]

In any double slit setup, it is SELF interference. That is not usually obvious because there are so many photons present. It is much more obvious when something with a rest mass is diffracted through the slits. Self interference is the quantum effect being observed. Please note that interference is also a classical wave property, which makes it somewhat confusing to distinguish one from the other.

Ok, so be it. Classically though, waves interfere with their amplitudes, but photon amplitude is not a measure of a single photon, but rather a measure of the quantity of photons. It's like a single photon is interacting with itself to produce less or more of itself, rather than to shrink or expand in its frequency or wavelength. Right?

 

[tim1608]

photon amplitude is not a measure of a single photon, but rather a measure of the quantity of photons.

I might be wrong but I think that the electromagnetic amplitude at a particular point in space is a measure of the square root of how likely you are to find a photon there. It is about probability, not quantity.

 

[carrz]

I might be wrong but I think that the electromagnetic amplitude at a particular point in space is a measure of the square root of how likely you are to find a photon there. It is about probability, not quantity.

I think that probability still comes as a direct consequence of the number of photons. More photons - more likely to find one. Who knows?

 

[DrChinese]

... if I understand it correctly, what instead may be happening is that part of the probability distribution of an individual photon is interfering with another part of the same probability distribution. Or putting it another way, two probability sub-distributions of the same photon are interfering with each other.

Yup, that's pretty good...

 

[tim1608]

I think that probability still comes as a direct consequence of the number of photons. More photons - more likely to find one. Who knows?

In a rather complicated way, involving interference between two or more overlapping photonic probability distributions, you may be half right because if, for example the probability of finding Photon A in a particular location is 0.5 and the probability of finding Photon B in that same location is 0.5 then I think that the probability of finding <EDIT>[STRIKE]any[/STRIKE] either</EDIT> photon there is 0.75. But when only one photon is involved, the photon's position is described by a single probability distribution which I don't think has anything to do with quantity.

 

[DrChinese]

Ok, so be it. Classically though, waves interfere with their amplitudes, but photon amplitude is not a measure of a single photon, but rather a measure of the quantity of photons. It's like a single photon is interacting with itself to produce less or more of itself, rather than to shrink or expand in its frequency or wavelength. Right?

There is still just 1 photon. A series of these makes the pattern. Unlike with classical waves, there is no possibility of observing half a photon, nor one and a half photons.

 

[tim1608]

Yup, that's pretty good...

Thank you.

 

[carrz]

But when only one photon is involved, the photon's position is described by a single probability distribution which I don't think has anything to do with quantity.

So varying photon amplitude practically means changing its position/direction?

 

[tim1608]

So varying photon amplitude practically means changing its position/direction?

Hi Carrz

Not exactly. Varying a part of an individual photon's wavefunction and associated probability distribution changes the probability of finding the photon in the changed part of its probability distribution. What I think that maybe you don't understand (and I don't think anyone, including myself fully understands) is that before the photon is found (absorbed), not even the photon itself knows where it is within its probability distribution. It may sort of be smeared throughout its probability distribution in a way that I don't think anyone fully understands. But once it is found in a particular location, it is absorbed, along with its energy, into whatever found it and its wavefunction and associated probability distribution collapses (disappears). Between emission and absorption, the wavefunction and associated probability distribution continually evolve, taking into account where the photon could potentially have been absorbed but wasn't.

Edit: Please note the edit I have made to Post #26.

 

[carrz]

Between emission and absorption, the wavefunction and associated probability distribution continually evolve, taking into account where the photon could potentially have been absorbed but wasn't.

Look at it from the practical point of view. There is only one way we measure the interference. We measure it relative to photons position at the detector which depends on photons exit angle from the slits.

Since frequency and wavelength do not change it's the amplitude. And so we are left with this direct relation between the amplitude and the exit angle. Therefore, photon trajectory is a function of photon amplitude. It must be, there is nothing else.

 

[stevendaryl]

In any double slit setup, it is SELF interference. That is not usually obvious because there are so many photons present. It is much more obvious when something with a rest mass is diffracted through the slits. Self interference is the quantum effect being observed. Please note that interference is also a classical wave property, which makes it somewhat confusing to distinguish one from the other.

I'm confused about the relationship between classical and quantum interference.

If you do a classical analysis of the double slit experiment, Maxwell's equations predict a characteristic interference pattern, with light and dark regions. Classically, it doesn't matter whether the light sent through the slits is from the same source, or two different sources, as long as the two sources are in-phase. So imagine using two in-phase light sources to produce the double-slit diffraction pattern. You see characteristic light and dark regions. Now, dial down the intensity from each of the sources until you start seeing individual photons. I would expect that the "classical" diffraction pattern would gradually give way to a probabilistic treatment, where the light region would become an area with high probability of a photon appearing, and the dark region an area of low probability. But in this case, there are two different sources. So the interference pattern of the photons are not due to self-interference.

So would the interference pattern disappear in the case of separate sources, in the low-intensity limit?

 

[DrChinese]

I'm confused about the relationship between classical and quantum interference.

If you do a classical analysis of the double slit experiment, Maxwell's equations predict a characteristic interference pattern, with light and dark regions. Classically, it doesn't matter whether the light sent through the slits is from the same source, or two different sources, as long as the two sources are in-phase. So imagine using two in-phase light sources to produce the double-slit diffraction pattern. You see characteristic light and dark regions. Now, dial down the intensity from each of the sources until you start seeing individual photons. I would expect that the "classical" diffraction pattern would gradually give way to a probabilistic treatment, where the light region would become an area with high probability of a photon appearing, and the dark region an area of low probability. But in this case, there are two different sources. So the interference pattern of the photons are not due to self-interference.

So would the interference pattern disappear in the case of separate sources, in the low-intensity limit?

I wouldn't call it that way. The separate sources would still be building the pattern up. There gets to be a tricky spot as we drop the intensity, but when it is low enough, it should be clear that the interference is always from self-interference.

So to be specific: I can prepare a source of pairs of entangled photons called X, and another source of similar pairs called Y. I use the Alice of each pair to "herald" the arrival of the other (Bob) via a X detector and a Y detector, each with a precise timing mechanism. I route the Bob of the X and Y stream through fiber so that they are both precisely directed* on the same course to the double slit. Barring nothing else, there will be interference and it will be clear it is self interference because the Alice arrival times are clearly labeled on the X and Y detectors. The times should be far enough apart most of the time so the source is clear.

Any variation of the intensity should give the same result.

*And this would have a few special requirements, namely that the photon is no longer entangled as to position/momentum.

 

[carrz]

So would the interference pattern disappear in the case of separate sources, in the low-intensity limit?

I don't know of any experiment that produced interference pattern with two separate light sources. If photons can not interact with other photons that should actually be impossible. The real puzzle seems to be what the slits are doing to photons rather than what photons are doing themselves.

 

[DrChinese]

Look at it from the practical point of view. There is only one way we measure the interference. We measure it relative to photons position at the detector which depends on photons exit angle from the slits.

Since frequency and wavelength do not change it's the amplitude. And so we are left with this direct relation between the amplitude and the exit angle. Therefore, photon trajectory is a function of photon amplitude. It must be, there is nothing else.

You cannot say what the trajectory was. The momentum is smeared. A photon is indivisible in terms of detection (no half photons) and yet the probability at spots varies. But only one detection. So you cannot link trajectory (momentum) and amplitude in the manner you describe. Interference patterns are due to self-interference, and the path cannot be known for there to be such interference.

 

[DrChinese]

I don't know of any experiment that produced interference pattern with two separate light sources. If photons can not interact with other photons that should actually be impossible. The real puzzle seems to be what the slits are doing to photons rather than what photons are doing themselves.

See my post #33 above.

The slits do not have anything to do with interference other than to create the shape of the bar. Diamond shaped slits create (roughly) diamond shape bars, for example.

 

[tim1608]

The real puzzle seems to be what the slits are doing to photons rather than what photons are doing themselves.

Hi Carrz

As I see it, the slits split the probability distribution of a single photon into two probability sub-distributions which both belong to the same photon. These two sub-distributions then interfere with each other on the wavefunction level. (The probability value of any point in each sub-distribution is proportional to the square of the sub-wavefunction value at that point.) The two sub-wavefunctions are added together to create the interference pattern of constructive and destructive interference. The probability of finding a photon at any point in the resulting interference pattern is then proportional to the square of the resulting wavefunction value at that point.

 

[carrz]

The probability of finding a photon at any point in the resulting interference pattern is then proportional to the square of the resulting wavefunction value at that point.

Do you think that two identical photons emitted in exactly the same direction, will not end up detected at exactly the same position?

 

[carrz]

See my post #33 above.

The slits do not have anything to do with interference other than to create the shape of the bar. Diamond shaped slits create (roughly) diamond shape bars, for example.

Does polarization have any impact on the interference pattern? That is, do we get the same pattern regardless of whether the light source is not polarized or polarized in any arbitrary plane?

 

[DrChinese]

Do you think that two identical photons emitted in exactly the same direction, will not end up detected at exactly the same position?

Of course they don't in a double slit experiment. That is the point. These are quantum particles and they do not have classical-only attributes as you describe them.

Please keep in mind that the double slit experiment has been done with large molecules too. These are easily controlled individually. See for example:

http://arxiv.org/abs/quant-ph/0110012

It should be clear that quantum particles can exhibit the ability to self interfere. There are plenty of studies on this point.

 

[DrChinese]

Does polarization have any impact on the interference pattern? That is, do we get the same pattern regardless of whether the light source is not polarized or polarized in any arbitrary plane?

Generally, no differences due to polarization.

 

[tim1608]

Do you think that two identical photons emitted in exactly the same direction, will not end up detected at exactly the same position?

Hi Carrz

As I understand it, the emitter does not emit the photon as a particle that already knows its location. Instead, all that the emitter emits is a wavefunction which corresponds to an associated probability distribution which describes where the photon might be. The precise information about where the photon actually is and its direction, does not yet exist in any form at this point. The photon only obtains an exact location at the point of absorption.

 

[DrChinese]

As I understand it, the emitter does not emit the photon as a particle that already knows its location. Instead, all that the emitter emits is a wavefunction which corresponds to an associated probability distribution which describes where the photon might be. The precise information about where the photon actually is and its direction, does not yet exist in any form at this point. The photon only obtains an exact location at the point of absorption.

In a double slit setup, it could have arrived there by at least 2 paths (corresponding to the 2 slits). You could say that there are any number of paths >1 actually.

 

[Cthugha]

I don't know of any experiment that produced interference pattern with two separate light sources. If photons can not interact with other photons that should actually be impossible. The real puzzle seems to be what the slits are doing to photons rather than what photons are doing themselves.

These experiments are possible. See for example: Am. J. Phys. 68, 245 (2000) ("Interference fringes from stabilized diode lasers").

To see why it is simple to see interference using a double slit, but hard using two lasers can be found by looking at the math. Whether you will see constructive or destructive interference at some position depends on the relative phase of the two light fields in question. For lasers, the phase will drift away quite quickly. Even for extremely good lasers, that will happen on a scale of milliseconds. As a consequence there will always be an interference pattern, but it will change every few milliseconds. For common lasers, this will rather happen on a scale of nanoseconds. Our eyes are slow and will just average over the many patterns and their sum is no pattern at all. If you have a fast detector, you will be able to see it.
In a double slit, the phase difference is determined only by geometry. There will be a well defined path length difference between the two slits and any point on the detection screen. This path length difference corresponds to a phase difference which determines the shape of the interference pattern. Now, if the phase of the light beam changes, it changes the same way at both slits. As you take the difference, this phase drift just cancels out. Therefore, it is much easier to see an interference pattern using a double slit.

Am I correct in thinking that the delay line would be applied to one of the two paths to ensure that two probability sub-distributions of the same photon cannot arrive at the beam splitter at the same time?

You usually use pulsed excitation to make the system emit a single photon. The delay between two consecutive photons is typically something like 13 ns. Now you just use the delay line to compensate this temporal offset.

I am not entirely sure of what exactly you mean by the "modes". Do you mean the areas on the beam splitter where the probability distributions of the photons will land? Am I correct that by using fibers or some other method, these areas will be made as small as possible?

Just have the system emit many single photons and have a look at their averaged distribution in real space. You want this distribution to be the same for both beams. However, you do not want to make it too small. Adjustment becomes a nightmare if you do.

Not exactly. Varying a part of an individual photon's wavefunction and associated probability distribution changes the probability of finding the photon in the changed part of its probability distribution.

A bit of nitpicking: There are no wavefunctions for photons. Wavefunctions imply eigenstates which stay unchanged under the appropriate measurement. Measuring a photon will destroy it, so that concept does not work. You rather calculate probability amplitudes for different events.

Also, it is a good idea to rethink what interference really means. If you ask an authority in quantum optics, Nobel prize winner Roy Glauber, then he will tell you that "interference of photons" is a really bad terminology. He stated in "Quantum Optics and Heavy Ion Physics" (http://arxiv.org/abs/nucl-th/0604021): "First of all, the things that interfere are not the photons themselves, they are the probability amplitudes associated with different possible histories. You can obviously have
different histories that involve more than one photon at a time."

It is a good thing to follow this advice. In quantum optics one uses the same approach as elsewhere: Consider the initial state, consider the final state and all the indistinguishable ways to get from one to the other. Add them, get the square and you are done. One can imagine that simply for a single photons. However, one can also imagine that there may be situations where the physics is more complex and the single photon level is not sufficient. Consider a light source which will always emit two photons simultaneously, but in a completely random direction. A look at the single photon level will give you the mean intensity at each position, but there is no way to tell, whether the source emits pairs or not. A look at the two-photon level will show you that whenever a photon is detected at some point, the second photon will show up ath the same point. All these phenomena include coincidence counts involving more than one photon and include what is loosely called multi-photon interference. However, I strongly suggest to follow Glauber's advice and simply consider these effects inside a framework of interfering probability amplitudes for complicated final states.

By the way, the paper above and Glauber's Nobel lecture "100 years of light quanta" both contain some great insights. The one I cited is a bit more radical and provoking.

 

[atyy]

Also, it is a good idea to rethink what interference really means. If you ask an authority in quantum optics, Nobel prize winner Roy Glauber, then he will tell you that "interference of photons" is a really bad terminology. He stated in "Quantum Optics and Heavy Ion Physics" (http://arxiv.org/abs/nucl-th/0604021): "First of all, the things that interfere are not the photons themselves, they are the probability amplitudes associated with different possible histories. You can obviously have different histories that involve more than one photon at a time."

This seems to involve the path integral picture? If so, is it really correct, given that the path integral must usually be rotated into imaginary time?

 

[stevendaryl]

These experiments are possible. See for example: Am. J. Phys. 68, 245 (2000) ("Interference fringes from stabilized diode lasers").

To see why it is simple to see interference using a double slit, but hard using two lasers can be found by looking at the math. Whether you will see constructive or destructive interference at some position depends on the relative phase of the two light fields in question. For lasers, the phase will drift away quite quickly. Even for extremely good lasers, that will happen on a scale of milliseconds. As a consequence there will always be an interference pattern, but it will change every few milliseconds. For common lasers, this will rather happen on a scale of nanoseconds. Our eyes are slow and will just average over the many patterns and their sum is no pattern at all. If you have a fast detector, you will be able to see it.

That's what I suspected was true, but that seems to contradict claims that a photon "only interferes with itself".

 

[DrChinese]

... Also, it is a good idea to rethink what interference really means. If you ask an authority in quantum optics, Nobel prize winner Roy Glauber, then he will tell you that "interference of photons" is a really bad terminology. He stated in "Quantum Optics and Heavy Ion Physics" (http://arxiv.org/abs/nucl-th/0604021): "First of all, the things that interfere are not the photons themselves, they are the probability amplitudes associated with different possible histories. You can obviously have
different histories that involve more than one photon at a time."

It is a good thing to follow this advice. In quantum optics one uses the same approach as elsewhere: Consider the initial state, consider the final state and all the indistinguishable ways to get from one to the other. Add them, get the square and you are done. One can imagine that simply for a single photons. However, one can also imagine that there may be situations where the physics is more complex and the single photon level is not sufficient. Consider a light source which will always emit two photons simultaneously, but in a completely random direction. A look at the single photon level will give you the mean intensity at each position, but there is no way to tell, whether the source emits pairs or not. A look at the two-photon level will show you that whenever a photon is detected at some point, the second photon will show up ath the same point. All these phenomena include coincidence counts involving more than one photon and include what is loosely called multi-photon interference. However, I strongly suggest to follow Glauber's advice and simply consider these effects inside a framework of interfering probability amplitudes for complicated final states.

Great points! Can't go too wrong quoting Glauber.

 

[DrChinese]

That's what I suspected was true, but that seems to contradict claims that a photon "only interferes with itself".

In a double slit setup, I think the "only interferes with itself" view is fine. It is in other places that view is dubious.

But most of those setups are over my head.

 

[Cthugha]

This seems to involve the path integral picture? If so, is it really correct, given that the path integral must usually be rotated into imaginary time?

Well, you can take the approach via path integrals or via canonical quantization. "Really correct" is a complicated wording. If you ask an experimentalist, whatever gives predictions in accordance with the experiment (and does not make millions of assumptions) is correct.

That's what I suspected was true, but that seems to contradict claims that a photon "only interferes with itself".

Glauber has an opinion on that, too. The passage comes directly before the one I quoted earlier:
"When you read the first chapter of Dirac’s famous textbook in quantum mechanics [8], however, you are confronted with a very clear statement that rings in everyone’s memory. Dirac is talking about the intensity fringes in the Michelson interferometer, and he says,

Every photon then interferes only with itself. Interference between two different photons never occurs.

Now that simple statement, which has been treated as scripture, is absolute nonsense."

I guess you need a Nobel prize to be able to say it that way. As DrChinese already said correctly, in a common double slit self-interference is all you need.

 

[atyy]

Well, you can take the approach via path integrals or via canonical quantization. "Really correct" is a complicated wording. If you ask an experimentalist, whatever gives predictions in accordance with the experiment (and does not make millions of assumptions) is correct.

Looking at http://en.wikipedia.org/wiki/Hong–Ou–Mandel_effect, the final state is:

(c⁺ + d⁺)(c⁺ - d⁺)|0,0>_cd = (c⁺c⁺ + d⁺d⁺ - c⁺d⁺ + d⁺c⁺ )|0,0>_cd

And the interference is due to (- c⁺d⁺ + d⁺c⁺ )|0,0>_cd = 0 ?

So whereas this is commonly stated as intereference between two photons, Glauber's complaint is that it should be seen as intereference between two states, each with two photons? I do see probability amplitudes here, but not histories. So maybe histories has to refer to the path integral, but not the canonical formalism?