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.