Question on Double Slit experiment

[prunthaban]

I have a simple question on double slit experiment,
In double slit experiment, after the light rays pass through the slits, place two beam splitters (down converters) which split the light into two rays, one going towards the screen and other going orthogonal to it. Now my question is do we get interference on the screen?
(Remember that I do nothing to the orthogonally traveling light. I leave it undisturbed)

The simple answer seems to be 'Yes' because we are in no way trying to obtain 'which-path' information. We just split the beam. But I believe the answer is 'No'.
I would like to know the correct answer.
If I am correct, placing reflecting mirrors do not affect interference. Similarly placing beam-splitters also should not affect it (But this contradicts at times). But placing polarizers do affect interference because it gives an 'identity' to the photon.

 

[country boy]

I have a simple question on double slit experiment,
In double slit experiment, after the light rays pass through the slits, place two beam splitters (down converters) which split the light into two rays, one going towards the screen and other going orthogonal to it. Now my question is do we get interference on the screen?
(Remember that I do nothing to the orthogonally traveling light. I leave it undisturbed)

The simple answer seems to be 'Yes' because we are in no way trying to obtain 'which-path' information. We just split the beam. But I believe the answer is 'No'.
I would like to know the correct answer.
If I am correct, placing reflecting mirrors do not affect interference. Similarly placing beam-splitters also should not affect it (But this contradicts at times). But placing polarizers do affect interference because it gives an 'identity' to the photon.

Just to clarify, are you placing a different beamsplitter behind each of the two slits? If so, how are the beams recombined toward the screen? I'm just trying to get the layout right before attempting to answer your question.

 

[prunthaban]

Just to clarify, are you placing a different beamsplitter behind each of the two slits? If so, how are the beams recombined toward the screen? I'm just trying to get the layout right before attempting to answer your question.

Sorry, my intention is to place down-converters. Not beam-splitters. down-converters will make 2 photons out of one. So one of them always travel towards the screen. The other one will travel orthogonal to it. Like this I have placed down-converters in front of both slits. So two rays of down-converted photons still hit the screen (forming interference or not I am not sure). Only the idler photons travel orthogonal to the original beam.
We have something like,

...|...|
...|...|
A<--- []...[]---> B
...|...|
...|...|
..._______
...screen

Ignore the dots. They just represent empty space.
Here [] is the down converter. A and B are idler photons. I am not disturbing them in anyway. They just travel into empty space. Will I get interference on the screen?

 

[cesiumfrog]

Answer is no. It doesn't matter whether you actually measure the which path information from the idler beam, or whether you just "could have" measured it, or even (as far as raw aggregate data is concerned) whether you at some point manipulated the idler beam so that you "can't in principle" measure it (eg. quantum eraser), the signal photons still produce no pattern on the screen.

 

[country boy]

Sorry, my intention is to place down-converters. Not beam-splitters. ... Will I get interference on the screen?

I'm assuming that in your diagram the slits are at the same locations as the down converters. The only pertinent question is whether the two beams going to the screen are coherently related. Are the down-converted photons in a fixed frequency and phase with respect to the photons incident on the slits? If so, there will be an interference pattern at the screen. If not, there won't.

 

[premagg]

I think that the inerference will occur because it is a wave phenomenon.If two waves are reaching,the interfernce must occur.Although the intensity may be disturbed.
One more thing,are photons really divisible?I think that as they are fundamental so they are not.

 

[lightarrow]

I think that the inerference will occur because it is a wave phenomenon.If two waves are reaching,the interfernce must occur.Although the intensity may be disturbed.
One more thing,are photons really divisible?I think that as they are fundamental so they are not.

What do you mean?

 

[premagg]

[red]down-converters will make 2 photons out of one[/red]
i MEANT THIS

 

[lightarrow]

[red]down-converters will make 2 photons out of one[/red]
i MEANT THIS

Ok, but if you send, one photon at a time, light against a screen with two slits and put two detectors behind the slits, you can't detect the photon simultaneously in the two detectors. Why?
(it's just a question!)

 

[marlon]

Ok, but if you send, one photon at a time, light against a screen with two slits and put two detectors behind the slits, you can't detect the photon simultaneously in the two detectors. Why?
(it's just a question!)

But this has nothing to do withthe photon nature (or ANY other elementary particle for that matter) itself. This is a measurement property !

marlon

 

[nrqed]

Answer is no. It doesn't matter whether you actually measure the which path information from the idler beam, or whether you just "could have" measured it, or even (as far as raw aggregate data is concerned) whether you at some point manipulated the idler beam so that you "can't in principle" measure it (eg. quantum eraser), the signal photons still produce no pattern on the screen.

And the reason for this is what you explained in our other thread, that the down converters introduce a random phase in the photons so that the signal photons are no longer coherent. Right?

 

[lightarrow]

But this has nothing to do withthe photon nature (or ANY other elementary particle for that matter) itself. This is a measurement property !

marlon

What does it mean that the photon is not divisible?

 

[cesiumfrog]

the down converters introduce a random phase in the photons so that the signal photons are no longer coherent. Right?

Because the crystals work by a process of *spontaneous* (ie. random in time, like nuclear decay) emission.

 

[marlon]

What does it mean that the photon is not divisible?

I referred to the measurement property because your question involved the breakdown of superposition upon measurement. That is why you detect the photon at one slit only. This has nothing to do with "the photon being divisible". Actually, for the above reason, the divisibility is a concept that does not apply to a photon.


marlon

 

[lightarrow]

I referred to the measurement property because your question involved the breakdown of superposition upon measurement. That is why you detect the photon at one slit only. This has nothing to do with "the photon being divisible". Actually, for the above reason, the divisibility is a concept that does not apply to a photon.
marlon

Ok, I just would like to understand better what exactly mean "if you send one photon at a time, you can't detect the photon simultaneously at the two detectors" as written in some QM books.

 

[marlon]

Ok, I just would like to understand better what exactly mean "if you send one photon at a time, you can't detect the photon simultaneously at the two detectors" as written in some QM books.

Err, i could have sworn i just answered to that question : breakdown of superposition of the particle's wavefunction upon measurement. This is a basic ingredient of the QM formalism.

marlon

 

[lightarrow]

Err, i could have sworn i just answered to that question : breakdown of superposition of the particle's wavefunction upon measurement. This is a basic ingredient of the QM formalism.

marlon

Thanks. However my question was actually intended in this way: if the two detectors click simultaneously, it's said it's because there arrived two photons. How do I know that it was sent only one photon or two photons, apart from what the detectors registered?

 

[marlon]

Thanks. However my question was actually intended in this way: if the two detectors click simultaneously, it's said it's because there arrived two photons. How do I know that it was sent only one photon or two photons, apart from what the detectors registered?

Because you are performing TWO measurements here. Again, once you perform ONE (thus this also happens with TWO) measurement, the wavefunction collapse has occurred and you either detect the photon through slit 1 or slit 2.

marlon

 

[lightarrow]

Because you are performing TWO measurements here. Again, once you perform ONE (thus this also happens with TWO) measurement, the wavefunction collapse has occurred and you either detect the photon through slit 1 or slit 2.

marlon

So, how can we say that the photon is not the click of the detector?

 

[marlon]

So, how can we say that the photon is not the click of the detector?

What does this have to do with your previous question ?

 

[lightarrow]

What does this have to do with your previous question ?

It's your diplomatic way not to answer the question?
If you don't want, no problem.

 

[marlon]

It's your diplomatic way not to answer the question?
If you don't want, no problem.

Well, it's just that you have this tendency of jumping from one topic to another and back. I just answered your question on the two detectors/the photon non-divisibility and now you come up with another question. You don't take the effort of explaing why you are asking this and within what context of our discussion. I want to know where you are going with this, that's all.

marlon

 

[lightarrow]

Well, it's just that you have this tendency of jumping from one topic to another and back. I just answered your question on the two detectors/the photon non-divisibility and now you come up with another question. You don't take the effort of explaing why you are asking this and within what context of our discussion. I want to know where you are going with this, that's all.

marlon

Yes, it's true I sometimes jump from one concept to another (they are all linked in my mind but clearly it's not so obvious for the others).

Essentially, I'm trying to understand better what is a photon. So, every information about it is useful for me. The (non) divisibility of a photon is related with:
1. how the photon is detected.
2. what happens from when a photon is sent from the source to when it reach the detector (if this is what really happens).
3. how and why energy is conserved in this process.

The problem I intended to understand now, is this: if we can say, experimentally, that we send a single photon from the source A, *and then* we detect it some distance apart, at B, it's a thing; if, instead, we cannot say to have sent one single photon from A, before having detected it at B, it's another thing. In this last case it doesn't seem so obvious to me that only one of many detectors at B clicks "because only one photon was sent", and not simply because it wasn't sent any particle, and the detectors respond to an EM field, with a very low probability to click because the field intensity is very low.

 

[vinm300]

Lightarrow aked : "Ok, I just would like to understand better what exactly mean "if you send one photon at a time, you can't detect the photon simultaneously at the two detectors" as written in some QM books."

I think the QM book is saying that if you send one photon at a time but make no attempt at measurement then you get an interferance pattern becuase
'Under the Copenhagen interpretation of quantum theory, an individual photon is seen as passing through both slits at once, and interfering with itself, producing the interference pattern.' (wikipedia)

BUT you can't detect the photon at the two detectors !

Any attempt at measurent causes the probability wave function (ie the single photon going through both slits) to collapse.

 

[marlon]

1. how the photon is detected.

I already had an extensive discussion with you on this in the past. We talked about the photon interaction with the detector. Actually, this is not important to the fundamental meaning of the double slit experiment : ie the particle/wave duality.

2. what happens from when a photon is sent from the source to when it reach the detector (if this is what really happens).

We do not know unless we measure the photon in between source and detector. But, in that case, no interference pattern will be detected because of the wavefunction collapse due to the measurement.

3. how and why energy is conserved in this process.

Energy conservation is ALWAYS respected. The observed momentum values will respect the demands from this law. Besides, why do you think this is not the case ?

The problem I intended to understand now, is this: if we can say, experimentally, that we send a single photon from the source A, *and then* we detect it some distance apart, at B,

Well, if we observe it at the detector B. QM does not state that there is absolute certainty this will happen each time you send out a photon. Look at the probability distribution at the detector. That's the entire point of the double slit experiment.

sent any particle, and the detectors respond to an EM field, with a very low probability to click because the field intensity is very low.

Now you are overcomplicating things here. Keep in mind that, originally, the double slit exp was a thought exp. One of the assumptions was that there are no other particles being present other than photons. In more a more realistic situation, photons can be detected and recognised thanks to the conservation of energy law (momentum values for example). That's how exotic particles get recognized in (sub)nuclear physics.

But really, i don't see what this has to do with your trying to understand photon nature.

marlon

 

[lightarrow]

Energy conservation is ALWAYS respected. The observed momentum values will respect the demands from this law. Besides, why do you think this is not the case ?

Example: photoelectric effect. The classical description cannot explain (among the other things) the very low time delay between source switching on and photoelectrons detection, because of energy conservation principle. But we know that energy is conserved on average in time; are we totally sure it's also conserved for every instant of time? Or this is a non-written postulate?

Well, if we observe it at the detector B. QM does not state that there is absolute certainty this will happen each time you send out a photon. Look at the probability distribution at the detector. That's the entire point of the double slit experiment.

You mean: "because the sent photon can go somewhere else than in B?" If this were the meaning, I agree, but it was not this that I intended. What I intended is: do we say that a photon was sent just because it's detected at some distance from the source after have switched on it, or because we can be sure to have sent a photon, before having detected it at some distance from the source? What I "see" in my mind is just an electromagnetic field generated from the source, sent in the form of waves, and I "see" photons as quanta of the energy exchange, only in the interaction between the EM field and the detector. No flying particles (at least no point-like ones).