Can a single photon create an interference pattern in a double slit experiment?

[Logician]

Hi All,
I am trying too undestand the slit test which shows wave-particle duality. Please correct me if I am wrong on the basic info on the test. One photon is shot out and if there is one slit it will go through the slit and make a point on the detection medium behind the slit. However if both slits are open then there will be an interference pattern which suggests hat the photon has gone through both slits.

Here are my questions. How is it actually done to insure beyond any doubt that only one photon has been released. Also what happens to the photon after it strikes the detection medium. Third question is how do we KNOW that if an individual photon is out there that it will be detected. My understanding is this is done in a dark room but how do we know that thre aren't stray photons or other particles in the room which could interact with the detection medium in a way that makes the interferene pattern.

Thanks,
Logician

 

[mfb]

How is it actually done to insure beyond any doubt that only one photon has been released.

You don't have to make this sure, but you can and it has been done in various ways:

• Make the light source so dim that photons are extremely rare. The probability to get multiple photons at the same time becomes negligible
• Make the screen sensitive enough to detect individual photons - that gives a direct counting
• Use a single-photon source
• Use electrons, atoms or small molecules instead of light, where you can directly measure individual particles on their way to the double-slit due to their charge

Also what happens to the photon after it strikes the detection medium.

It gets absorbed.

Third question is how do we KNOW that if an individual photon is out there that it will be detected. My understanding is this is done in a dark room but how do we know that thre aren't stray photons or other particles in the room which could interact with the detection medium in a way that makes the interferene pattern.

There is always some background (usually not from light in the room, but from the detector itself), but this is measured with the light source off and subtracted.

There are no other particles that could make the interference pattern - how would those particles do that, or even know of the slits?

 

[afcsimoes]

In a real double-slit experiment, what type of device is used to "observe" a photon to pass? What is the physical principle underlying the operation of this device?

 

[vanhees71]

These are very valid questions, and unfortunately even many otherwise good textbooks get them wrong.

You don't have to make this sure, but you can and it has been done in various ways:

• Make the light source so dim that photons are extremely rare. The probability to get multiple photons at the same time becomes negligible

• 
  This is already incorrect. Simply dimming down a light source (be it a light bulb or a laser), doesn't give you single photons but still coherent states. On the average you may have even less than one photon, but then still most of the state is the vacuum, not a single-photon (Fock) state.
  
  

  [*]Make the screen sensitive enough to detect individual photons - that gives a direct counting

  Yes, but then still the source is not a stream of single photons. It's only very likely that the device "clicks" due to absorption of a single photon, but you cannot rule out the possibility that the "click" is due to two-photon absorption.
  
  

  [*]Use a single-photon source

  That's a very good explanation of the problem to create real single-photon Fock states.

 

[afcsimoes]

In a real double-slit experiment, what type of device is used to "observe" a photon to pass? What is the physical principle underlying the operation of this device?

I have put there questions because the know efect of the "observer" (human or device) on the kind of the end result.

 

[mfb]

This is already incorrect. Simply dimming down a light source (be it a light bulb or a laser), doesn't give you single photons but still coherent states. On the average you may have even less than one photon, but then still most of the state is the vacuum, not a single-photon (Fock) state.

Yes most of the time it is vacuum. So what? A vacuum won't give a hit at the screen. If your light source is very dim and you replace your slits by detectors, you will "never" detect two photons at the same time.

Yes, but then still the source is not a stream of single photons. It's only very likely that the device "clicks" due to absorption of a single photon, but you cannot rule out the possibility that the "click" is due to two-photon absorption.

True, but then you would need a theory that directs both photons to the exact same location every time. I'm not aware of such a theory.

That's a very good explanation of the problem to create real single-photon Fock states.

To use your argument: how can you rule out that a photon is not a tightly bound state of two particles without additional input?

 

[Cthugha]

Yes most of the time it is vacuum. So what? A vacuum won't give a hit at the screen. If your light source is very dim and you replace your slits by detectors, you will "never" detect two photons at the same time.

The problem is that for coherent states any detection event does not reduce the relative probability to detect a second photon. The concept of single photons is well defined and the key quantity here is the relative ratio of the photon number variance to the mean photon number and this does not change if you use a dim light source. One is classical, the other is as non-classical as it gets.

It may seem like nitpicking, but this is a central and fundamental concept of quantum optics and - loosely speaking - this insight is what got Glauber the Nobel prize. Many of my students had this misconception and they had a hard time "unlearning" it. I simply do not see the necessity to sacrifice correctness if a correct explanation is easy to give. Using dim coherent states gives you a classical Poissonian distribution with low, but finite probability for photon numbers larger than 1 with $g^{(2)}(0)=1$. Single photon sources give you an intrinsically non-classical photon number distribution with $g^{(2)}(0)<0.5$.

 

[mfb]

Using dim coherent states gives you a classical Poissonian distribution with low, but finite probability for photon numbers larger than 1 with $g^{(2)}(0)=1$. Single photon sources give you an intrinsically non-classical photon number distribution with $g^{(2)}(0)<0.5$.

Yes sure, that is a more mathematical way to say the same as I did in post #2. Where is the problem with the low, but finite probability?

 

[Cthugha]

Yes sure, that is a more mathematical way to say the same as I did in post #2. Where is the problem with the low, but finite probability?

The problem I see is:

How is it actually done to insure beyond any doubt that only one photon has been released.

the "beyond any doubt". Using low intensity coherent light is ok in a double slit because nothing in the experiment is at all sensitive to whether you use single photons or not. The physics is absolutely the same. So for the practical purpose of a double slit, having less than one photon on average is fine.

However, there are also experiments, where you need to really clarify beyond any doubt that at most one photon has been released. Think about quantum key distribution like in BB84. In that case dim coherent states are not sufficient as a the state remains unchanged after a photon detection event for coherent states. I have seen it more than once that people confuse these points because they did not understand that difference.

 

[mfb]

The problem I see is:

How is it actually done to insure beyond any doubt that only one photon has been released.

the "beyond any doubt". Using low intensity coherent light is ok in a double slit because nothing in the experiment is at all sensitive to whether you use single photons or not. The physics is absolutely the same. So for the practical purpose of a double slit, having less than one photon on average is fine.

That is easy to handle (way easier than various other experimental challenges). If you suspect that something happening in 1 out of 1000 measurements somehow modifies the whole distribution in a significant way, record the pattern with various intensities and extrapolate to "zero intensity" (where all events are single photons).
Experimental physics would be so easy if you just have things like that...

Note that single-photon sources are never perfect as well. They are in theory, but detection of the signal photon has background and inefficiencies. In the same way, the overall setup has a background and the detection is not 100% efficient. There are no perfect measurements.

However, there are also experiments, where you need to really clarify beyond any doubt that at most one photon has been released. Think about quantum key distribution like in BB84. In that case dim coherent states are not sufficient as a the state remains unchanged after a photon detection event for coherent states. I have seen it more than once that people confuse these points because they did not understand that difference.

I don't see the relation to this thread.

 

[Cthugha]

That is easy to handle (way easier than various other experimental challenges). If you suspect that something happening in 1 out of 1000 measurements somehow modifies the whole distribution in a significant way, record the pattern with various intensities and extrapolate to "zero intensity" (where all events are single photons).

If the op had asked how one could make sure that the effect of having more than one photon present is negligible in the double slit, I would agree. However, that was not his question. He asked how one can assure beyond doubt that only one photon has been released. This is simply impossible with coherent light.

Note that single-photon sources are never perfect as well. They are in theory, but detection of the signal photon has background and inefficiencies. In the same way, the overall setup has a background and the detection is not 100% efficient. There are no perfect measurements.

Of course. That is trivial. Also, detector efficiency does not really matter for coincidence measurements anyway. However, there is a fundamental difference between a Fock state and a dim coherent state. This difference is more or less the fundament of quantum optics and not a negligible small correction.

I don't see the relation to this thread.

See my first response. I tried to emphasize that you answered a different question. I am not the only person in this thread who pointed that out.

 

[mfb]

If the op had asked how one could make sure that the effect of having more than one photon present is negligible in the double slit, I would agree. However, that was not his question. He asked how one can assure beyond doubt that only one photon has been released. This is simply impossible with coherent light.

You cannot be 100% sure with any method. And a dim light source is easier to set up than a single-photon source.

Of course. That is trivial. Also, detector efficiency does not really matter for coincidence measurements anyway. However, there is a fundamental difference between a Fock state and a dim coherent state. This difference is more or less the fundament of quantum optics and not a negligible small correction.

I never questioned the difference - and this is difference is crucial if you want to distribute quantum keys. But we don't want to do this here. The relevant part of the difference (multi-photon events) is a small correction.

See my first response. I tried to emphasize that you answered a different question. I am not the only person in this thread who pointed that out.

I did not say "no experiment is perfect, but here are multiple possible approximations with various precision", right - I assumed that this is obvious for experiments. And you always have to find some way to handle those experimental issues in the data analysis.

 

[DrChinese]

If the op had asked how one could make sure that the effect of having more than one photon present is negligible in the double slit, I would agree. However, that was not his question. He asked how one can assure beyond doubt that only one photon has been released. This is simply impossible with coherent light.

I thought that it would be possible to get a single photon to a double slit which would be sufficiently coherent to create an interference pattern. You would use entangled photon pairs, which I realize is not coherent light to start with. And use the Alice stream to herald the Bob stream. But cannot you take the Bob stream and run it through some magic box with a pinhole exit (causing it to no longer be entangled - and of course I am kidding about a magic box) and then end up with a coherent source? You would not be obtaining any information about Alice at that point.

So I guess I am saying I don't suitably understand how a single photon is incoherent if you can diffract it through a point source and send to a double slit.

 

[Cthugha]

You cannot be 100% sure with any method. And a dim light source is easier to set up than a single-photon source.

Well, sure. No doubt about that.

I never questioned the difference - and this is difference is crucial if you want to distribute quantum keys. But we don't want to do this here. The relevant part of the difference (multi-photon events) is a small correction.

My point is that I am not sure what "we" want to do here. Of course you and I know what we are talking about. However, I do not have the slightest idea about the level of education of the op and his interests. It may well be that he is able to put things into context and understands that attenuated coherent states are a reasonable approximation for systems wchich are only sensible to mean photon numbers, but not their variance. It may well be that he did not get the fine details and (yes, I am exaggerating) jumps into the next discussion on single photon sources and claims that dim lasers are just as good. My point is just that pointing out this difference will not hurt.

I did not say "no experiment is perfect, but here are multiple possible approximations with various precision", right - I assumed that this is obvious for experiments. And you always have to find some way to handle those experimental issues in the data analysis.

Well, the problem is that the term "single photons" changed meaning drastically between the early days of quantum mechanics and today. Today it has become a well defined technical term. Weak coherent states get like 1 out of 10 defining properties of this state right - the small absolute multi-photon probability, but I understand that this approximation still seems appealing. As these forums are one of the rare exceptions which discuss quality physics, I have just tried to make it my standard to keep posts on a level that would make it through peer review in journals (where antibunching is the gold standard) and try to explicitly mention it if I use more simplified explanations - but that may be nitpicking on my side.

I thought that it would be possible to get a single photon to a double slit which would be sufficiently coherent to create an interference pattern. You would use entangled photon pairs, which I realize is not coherent light to start with. And use the Alice stream to herald the Bob stream. But cannot you take the Bob stream and run it through some magic box with a pinhole exit (causing it to no longer be entangled - and of course I am kidding about a magic box) and then end up with a coherent source? You would not be obtaining any information about Alice at that point.

So I guess I am saying I don't suitably understand how a single photon is incoherent if you can diffract it through a point source and send to a double slit.

This is tricky because of the reason I mentioned above. The term "coherence" has one meaning prior to the birth of quantum optics and two different meanings since then. It would be a bit off-topic here, but if you would like to discuss that point, I will be glad to discuss it in a separate topic or via PM.