Exploring Wave Properties of Photons in Double-Slit Experiment

[ballzac]

Okay, I had a rather 'interesting' 'discussion' with a professor at my university yesterday. Nothing that he said made any sense to me at all, and I feel I understand the principles quite well, so I came away from the 'discussion' feeling that maybe I understand it better than HE does. I don't want to be arrogant, and it's hard to believe that someone in his position could misunderstand something basic, so I then think maybe I really do not understand. So I'm going to ask here what others believe.

I was performing an experiment. It was the double-slit experiment using one photon at a time. I said that there is no way to be absolutely sure that there is only one photon in the tube at a time, and I wanted to know if there was some kind of statistical analysis that could be applied to determine the likelyhood of two photons being in the tube at the same time. In the lab manual it said to set it up so that, ON AVERAGE, there is only one photon in the tube at a time. I figure that without quantifying what this actually means, the results of the experiment are meaningless. Anyway, he grilled me for ages about why there can only be one photon in there at a time, and I just couldn't see the logic. I could turn down the bulb so low that I only record two photons after a year, and I still don't know that they didn't both come out at the same time.

Anyway, the really weird part came next. He told me that the value of counts that I had calculated as a maximum value was too low. I had about 137 counts per second as my maximum value. I explained that I was trying to, as the manual suggested, have on average one photon IN THE TUBE at a time, not just at the end where the detector is. He told me that I should have about 10^9 counts as my maximum value . So I was like wait, we are using a slit in front of the detector. I move the slit using a micrometer 50 micron at a time. If the slit records one photon there could have been many more that 'interfered with each other' before they hit the end where the detector is. He said they don't count because if they don't enter the detector slit they die and they might as well have never existed.

I was getting a bit confused because this didn't make sense to me. I said that I am trying to determine that one photon can be aware of both slits. Let's assume on the contrary that it is the case that the interference pattern is caused by many photons going through the slits and then 'bouncing' off each other. In this case, they will cause an interference pattern, but only one might reach the detector slit. In which case this professor would say that only one photon had been in there. So I don't find that convincing.

Then what he said to me was, "They DON'T bounce off each other." I said, "Yeah, I know." And then tried to explain that I was using it as a way to explain that I think it DOES matter if there are photons in there that aren't detected, and then again explained my idea. Again, he told me that I don't understand the experiment and they don't just bounce off each other. I said that I never said they do, and he told me I just did. So again I tried to explain what I was getting at, using a different approach.

Maybe he is right and for some reason the photons that don't enter the detector slit don't matter, but he certainly never gave me a convincing argument. All he did was patronize me. I mean, I learned in high school that one photon can have wave properties. I don't think I would be able to get to third year without understanding that. Maybe there are other things that I don't understand about the experiment. If so I am all ears. Hopefully I've explained this all well enough. It's hard to explain without waving my hands, hehe.

 

[cragar]

im not 100% sure , but i don't see how anyone could know that only one photon
was hitting the detector.

 

[ballzac]

He did concede that due to the inefficiency (4%) of the detector, that I would have to have fewer counts observed than the number of photons that were required. However, he maintained that photons that miss the narrow detector slit are unimportant. That is what I mainly have issue with. There is an intire 10 cm^2 or so that the interference pattern covers. I can't remember how wide the detector slit is, maybe 500 micron. I believe that for the results of this experiment to prove the wave nature of a single photon, we need to make sure that only one photon is going through the two slits at a time (or as was suggested in the manual, making sure only one is in the tube at a time). I don't see how measuring only one photon at a time is any guarantee when only a tiny fraction of the photons inside the tube will go through the slit.

 

[cragar]

can u do this expirement with electrons

 

[ballzac]

can u do this expirement with electrons

I don't know if we have the apparatus, why?

 

[cragar]

maybe it would be easier to control , u prolly know more about this than me.

 

[chrisphd]

NO your wrong! Photon's don't bounce off each other!

 

[cragar]

i don't think he said that

 

[chrisphd]

Only kidding.

 

[cragar]

got ya

 

[PenKnight]

I like to try and help.
How do you know you are detecting more than one photon?

 

[ballzac]

I like to try and help.
How do you know you are detecting more than one photon?

I'm not sure I understand your question. The photomultiplier is connected to a pulse generator that is set to record pulses and it takes a reading over ten seconds and then returns the number of pulses it received per second. The photomultiplier has an efficiency of 4%, so if the reading on the pulse generator says 100, you can say that approximately 2500 photons hit the photomultiplier per second on average.

 

[chrisphd]

When you say "detector slit" what are you referring to?

 

[ballzac]

There is a photomultiplier that is fixed at the end of the tube. In front of it is a single slit that blocks 'all' photons except those that pass through the slit. The slit is mounted in a micrometer. You move the slit to a position using the micrometer, if you count a high number of photons at a specific point, it indicates a maximum. Does that explain it? The detector slit is there so you can place it at a point to measure the brightness at that position. The photomultiplier does not move just the slit. But the number that reach the photomultiplier indicates how many went through the slit.

 

[alxm]

I was performing an experiment. It was the double-slit experiment using one photon at a time. I said that there is no way to be absolutely sure that there is only one photon in the tube at a time, and I wanted to know if there was some kind of statistical analysis that could be applied to determine the likelyhood of two photons being in the tube at the same time.

Pretty basic statistics would be enough to cover this. You can likely find that out to an arbitrary degree of confidence. If you don't know the relevant statistics then you should probably accept the statements which were probably made by folks who do.

(...) we are using a slit in front of the detector. If the slit records one photon there could have been many more that 'interfered with each other' before they hit the end where the detector is.

And in what way do you propose they 'interfere' with each other at the slit that stops them from getting through? It's not like people moving through a doorway!

Let's assume on the contrary that it is the case that the interference pattern is caused by many photons going through the slits and then 'bouncing' off each other.

Why? When it's well known that 1) photons do _not_ 'bounce' off each other in any way. And 2) Even if they did, it would then cause a fairly random scattering pattern - not a well-ordered interference pattern.

Then what he said to me was, "They DON'T bounce off each other." I said, "Yeah, I know." And then tried to explain that I was using it as a way to explain that I think it DOES matter if there are photons in there that aren't detected, and then again explained my idea.

And what is that idea? You have not explained it here either. Photons do not bounce off each other literally. And if you are using that as an analogy for some other interaction - what interaction are you talking about? All you've given us is scattering-which-isn't-scattering that relies on some unspecified force which would, for unspecified reasons, explain the result you got.

Whereas the guy who's probably supervised this lab a million times, knows exactly the result you're supposed to be seeing and getting, simply thinks you've screwed up somewhere.

And you're telling him that he's wrong, you're right. You didn't screw up and it's all due to some physical effect that the guy with the degree doesn't understand? And you think he was patronizing towards you?

It's hard to explain without waving my hands, hehe.

Oh there was plenty of http://en.wikipedia.org/wiki/Handwaving" in your post.

 

[chrisphd]

alxm says: Why? When it's well known that 1) photons do _not_ 'bounce' off each other in any way.

Isn't this experiment meant to be demonstrating that point alxm. You cannot therefore assume that photons don't bounce off each other, as that is what you are trying to test. It does not make sense to have to assume photons don't bounce off each other in order to prove photons don't bounce off each other.

 

[ballzac]

alxm says: Why? When it's well known that 1) photons do _not_ 'bounce' off each other in any way.

Isn't this experiment meant to be demonstrating that point alxm. You cannot therefore assume that photons don't bounce off each other, as that is what you are trying to test. It does not make sense to have to assume photons don't bounce off each other in order to prove photons don't bounce off each other.

Thank you. That is my point. I can prove that the square root of two is irrational by first assuming that it isn't, and then proving that is not the case. If you say, let root 2 = p/q, and then just say that it cant, you haven't proven anything.

Pretty basic statistics would be enough to cover this. You can likely find that out to an arbitrary degree of confidence. If you don't know the relevant statistics then you should probably accept the statements which were probably made by folks who do.

Yes. I agree with this, and it is that confidence level that was interested in finding out, as I believe it is important. But the professor said that it is IMPOSSIBLE for two photons to be in there at the same time, and I don't understand what, if anything, prevents this. Unfortunately I have not studied any statistics, so yes, I would trust in an analysis by someone who has, but he did not offer me any response based on statistics. He just said that if it takes 10^-9 seconds for a photon to travel down the length of the tube, then as long as there are no more than 10^9 photons recorded in one second there is no way that two can be in there at the same time. I don't understand why. It seems to me that his reasoning assumes that the timing of the production of photons is equally spaced. Is this the case? It was only a light bulb, not a synchrotron or anything.

And in what way do you propose they 'interfere' with each other at the slit that stops them from getting through? It's not like people moving through a doorway!

I'm obviously not very good at explaining this, and I'm sorry for that. I never thought there was anything stopping them getting through. I'm not sure why you thought that. What I mean is that the detector slit is very narrow and many, many photons can hit other points at the detector end of the tube, but where the slit is not positioned, before one goes through the slit. There is nothing inherent in the slit to make all the photons go into it.

Why? When it's well known that 1) photons do _not_ 'bounce' off each other in any way. And 2) Even if they did, it would then cause a fairly random scattering pattern - not a well-ordered interference pattern.

I've already covered this above, but I will reiterate in case I have caused confusion with my poor explanations. The experiment is meant to prove that one photon can interfere with itself, or to be somehow 'aware' of both slits. If we cannot show that there was only one photon in there at a time, then the experiment does not prove this. Forget that I even mentioned this 'bouncing off' or interference BETWEEN different photons as I can also approach this from another angle. Say we have interference working the way that it actually does. So each individual photon has wave properties and can interfere with itself. If we have many photons in the tube at once, we will still build up an interference pattern. Let's say is takes 10^-9 seconds for a photon to traverse the length of the tube. We could have 10^12 photons reach the end of the tube in one second, meaning that there had to be more that one photon in the tube at some stage during that second. If only one in every 10^3 photons hits the detector slit, then we will count 10^9 photons, and if we assume that the ones that don't go in the detector slit don't count, we will assume that only one photon was in the tube at a time. But we started by saying that this isn't the case. Obviously my reasoning is wrong, but I don't see where I have gone wrong.

And what is that idea? You have not explained it here either. Photons do not bounce off each other literally. And if you are using that as an analogy for some other interaction - what interaction are you talking about? All you've given us is scattering-which-isn't-scattering that relies on some unspecified force which would, for unspecified reasons, explain the result you got.

I was not talking about any specific interaction. It is not an interaction that actually happens. But the only way to determine that it is NOT SOME kind of interaction BETWEEN photons, is to ensure that only one photon is in the tube at a time.

Whereas the guy who's probably supervised this lab a million times, knows exactly the result you're supposed to be seeing and getting, simply thinks you've screwed up somewhere.

There is nothing wrong with my results. I got exactly the result that one would expect, both with the bulb down as low as I thought it should be, and also when I turned it up after he told me to.

And you're telling him that he's wrong, you're right. You didn't screw up and it's all due to some physical effect that the guy with the degree doesn't understand? And you think he was patronizing towards you?

Oh there was plenty of http://en.wikipedia.org/wiki/Handwaving" in your post.

I'm really sorry if I caused any offence. That was not my intention. I have a lot of respect for anyone who has been able to get a degree of ANY kind, let alone to get as far as becoming a professor. Although I think people are fallable, and I also think being a good scientist involves being critical and skeptical. I personally did not understanding his logic, and I want to understand better. I did not say I am right. I said that part of me feels like I am right, the other part feels like I must be wrong. The problem I had with the lecturer was that he didn't seem to care as much about helping me understand, as he did about just telling me I was wrong. I don't think there was really any need for you to have a go at me. Physics is meant to be a collaboration, not a competition.

 

[cragar]

if you are only emitting one photon then there would be nothing to bounce off of .

 

[ballzac]

I don't know why I didn't think to do this before, instead of explaining the apparatus in words. Here is the manufacturer's website, with a diagram of the apparatus: http://www.teachspin.com/instruments/two_slit/experiments.shtml

Also, there is a bit down the page that supports the first part of what I was saying,

"The probability of there being two photons in flight simultaneously is thus negligible. All of the interference effects observed in the apparatus can thus be comfortable ascribed to the behavior of individual photons, coming through the single slit one at a time!"

I figured that that probability must be negligible. All I wanted to do is to quantify it. This guy was saying that there is absolutely no way that two photons can be in there at the same time. Negligible might mean it won't happen in a million years, but isn't the same as being physically impossible.

In the same section of that webpage, "INTERPRETING THE DATA TO VALIDATE SINGLE PHOTON PRODUCTION", it uses logic similar to what the professor was using, but it is ambiguous as to whether they are talking about having the detector slit in place or not. I would imagine this logic to work if the detector slit was taken out, to record 5% of ALL the photons that reach the end of the tube. In the experiment I performed, the slit was in place the whole time, and the professor said that photons that miss the slit don't matter.

My contention is that if we record a certain number of photons with the slit in place, we will record many more if the slit is taken out. I don't believe that taking the slit out will cause more photons to be produced upstream, so I conclude that those extra photons are in the tube even when they are not recorded because the slit is in place. If we only count the ones that go through the slit as being in the tube, then we will underestimate how many are in the tube at one time and thus turn the bulb up too high and invalidate the results.

Not saying I'm right, just trying to understand better.

if you are only emitting one photon then there would be nothing to bounce off of .

Not sure what you are getting at sorry.EDIT: Actually it does say elsewhere on the webpage that the detector slit should be removed along with the double slit and the slit blocker for determining single photon. The lab manual we use at uni does NOT tell us to do this, and the professor was well aware that the detector slit was in place.

 

[PenKnight]

I asked becuase you assume ( I assume that you assumed :S) that the all photons will be dectected as they hit the photomulitpler.

I used a photomulitpler too in one of my experiments and been told that there a dead time for this device. It's the time required for charging up to dectect the next photon. So that any other photons will be ignored within the dead time. This way you are only getting one photon at a time.

"The drawback, however, is that not every photon incident on the primary surface is counted either because of less-than-perfect efficiency of the photomultiplier, or because a second photon can arrive at the photomultiplier during the "dead time" associated with a first photon and never be noticed." ...wiki

http://en.wikipedia.org/wiki/Photomultiplier

Then
What's the effect of haing two photons if I can only dectect one? and
How would both photons hit the slits at the same time? <- only way to cause problems

 

[ballzac]

I asked becuase you assume ( I assume that you assumed :S) that the all photons will be dectected as they hit the photomulitpler.

I used a photomulitpler too in one of my experiments and been told that there a dead time for this device. It's the time required for charging up to dectect the next photon. So that any other photons will be ignored within the dead time. This way you are only getting one photon at a time.

"The drawback, however, is that not every photon incident on the primary surface is counted either because of less-than-perfect efficiency of the photomultiplier, or because a second photon can arrive at the photomultiplier during the "dead time" associated with a first photon and never be noticed." ...wiki

http://en.wikipedia.org/wiki/Photomultiplier

Then
What's the effect of haing two photons if I can only dectect one? and
How would both photons hit the slits at the same time? <- only way to cause problems

I assumed that any kind of effect like that would be factored into the inefficiency of 4%, or 5% as it says on the website. But if it isn't, then that seems to me to be even MORE reason to turn the bulb down rather than up, because there may be many more photons than what is detected.

As to your last remark, it is hard to speculate on what WOULD happen if photons DID interfere with each other, and I guess that is why people have had a problem with my 'proof by contradiction' kind of approach. But yes, it would be logical to think that they would have to enter the slits at the SAME time for there to be a problem. However, the manual suggested that there should be only one photon in the tube at a time, and I think this is a good way to ensure that there can be no interaction between photons in the experiment.

 

[PenKnight]

I made a mistake by saying that both photons will have to hit the slit at the same time.I completely disregarded the wave behavior of light.
How can you tell if two photons are arriving at the slit at the same time? It would be just one larger photon = superposition of the two, and it's this photon that will interact with itself in some sense.

If you turned up the intensity, would you be able to get more than 10^9 counts or close to it? Completely saturate the photomultiplier and see what happens. ^^. But check if there a limit that you shouldn't exceed.
Are you still performing this experiment?

Penknights's thought:
It's weird how the wave splits at the slits. I wonder if there's any reflections since the wave spread all over the wall with the slits. Would the be intensity lost due to this. Is the intensity half as it splits?

 

[DrChinese]

I made a mistake by saying that both photons will have to hit the slit at the same time.I completely disregarded the wave behavior of light.
How can you tell if two photons are arriving at the slit at the same time? It would be just one larger photon = superposition of the two, and it's this photon that will interact with itself in some sense.

Just to be clear: in a normal double slit setup, there is self-interference with a single photon. It does NOT take 2 to have interference. You could have 2, but that is not the cause of interference. The wave nature of a single photon will be enough. Photons come in discrete units, such as 1, 2, 3, etc. There is no such thing as half a photon.

 

[ballzac]

I approached my quantum mechanics lecturer today and explained all this to him, and he agrees with me. So now I'm happy to trust myself and him over this other guy. I find it quite upsetting that there are people with such a dogmatic and narrowminded way of thinking about physics, so I'm happy to put this behind me and hopefully I do not have to deal with this guy too much in the future. Thanks for all the replies.

I made a mistake by saying that both photons will have to hit the slit at the same time.I completely disregarded the wave behavior of light.
How can you tell if two photons are arriving at the slit at the same time? It would be just one larger photon = superposition of the two, and it's this photon that will interact with itself in some sense.

If you turned up the intensity, would you be able to get more than 10^9 counts or close to it? Completely saturate the photomultiplier and see what happens. ^^. But check if there a limit that you shouldn't exceed.
Are you still performing this experiment?

Penknights's thought:
It's weird how the wave splits at the slits. I wonder if there's any reflections since the wave spread all over the wall with the slits. Would the be intensity lost due to this. Is the intensity half as it splits?

I don't know if a photon can be 'larger'. You will just have two photons. Both will hit the detector end with positions determined by the probability amplitudes for the wave and the given slits.

I am actually finished using the apparatus now, and don't care to spend more time with it.

If you look at the experimental setup, there is a slit-blocker after the double-slit to prevent these reflections. I believe that, due to the size of the slit blocker, there is minimal diffraction from this.

If you mean is the intensity half what it was originally, it would be much less because the wave diffracts from each of the two slits.

 

[chrisphd]

What university are you from Ballzac?

 

[ballzac]

What university are you from Ballzac?

I'm not sure I feel comfortable disclosing that information after bagging a member of staff.:uhh: