# Question on Double Slit experiment

1. Aug 5, 2007

### 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 travelling 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.

2. Aug 6, 2007

### country boy

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.

3. Aug 7, 2007

### prunthaban

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?

Last edited: Aug 7, 2007
4. Aug 7, 2007

### 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.

5. Aug 8, 2007

### country boy

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.

6. Aug 11, 2007

### 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.

7. Aug 11, 2007

### lightarrow

What do you mean?

8. Aug 15, 2007

### premagg

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

9. Aug 16, 2007

### lightarrow

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!)

10. Aug 16, 2007

### marlon

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

marlon

11. Aug 16, 2007

### nrqed

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?

12. Aug 16, 2007

### lightarrow

What does it mean that the photon is not divisible?

13. Aug 16, 2007

### cesiumfrog

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

14. Aug 17, 2007

### marlon

I refered 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

15. Aug 17, 2007

### lightarrow

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.

16. Aug 17, 2007

### marlon

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

17. Aug 17, 2007

### lightarrow

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?

18. Aug 17, 2007

### marlon

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

marlon

19. Aug 17, 2007

### lightarrow

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

20. Aug 17, 2007

### marlon

What does this have to do with your previous question ?

21. Aug 17, 2007

### lightarrow

If you don't want, no problem.

22. Aug 18, 2007

### marlon

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 wanna know where you are going with this, that's all.

marlon

23. Aug 18, 2007

### lightarrow

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.

24. Aug 18, 2007

### 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.

25. Aug 19, 2007

### marlon

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.

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.

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 ?

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.

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