# B Photon and half-silvered mirror

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1. Apr 6, 2016

### entropy1

Suppose we have a photon fired at a half-silvered mirror. There is a 50% chance that the photon reflects, and a 50% chance it passes through, right?

Suppose there are two detectors, one that only detects reflected photons and one that detects only passed photons.

So my question is: does the photon take both paths? And if so, is a detection a collapse? OR: does the photon decide which path it will take?

I think I am beginning to understand what the problem with QM is.

2. Apr 6, 2016

### vanhees71

You'll find each photon only in one detector. Whether or not there is a collapse, it's up to you to decide, i.e., which interpretation of quantum theory you follow. I don't follow any collapse interpretation, because the collapse leads only to problems with causality ("instantaneous actions at a distance"), and it's unnecessary to use quantum theory to describe nature.

3. Apr 6, 2016

### entropy1

Indeed. But if we don't endorse collapse, we have to resort to interpretations like MWI, right? (The photon splits the universe in two?)

I don't know the details of it, but if I'm correct, an inferometer can detect interference between the two paths single photons travel by. If one path of the inferometer is blocked, the inteference disappears, right? Does that mean that a single photon takes both paths?

4. Apr 6, 2016

### f95toli

No, a single photon never takes "both" paths.
However, light rarely comes in the form of a stream of single photon. In fact, what you are describing is know as a Hanbury Brown-Twiss interferometer and is used to determine IF a source if light is emitting single photons or not. For a true single photon source you will find that the measurements (actually the correlations) sort of implies that the photon can only go "one way".

However, in the more general case of coherent (e.g. laser light) and thermal light the results are very different. Have a look at the wiki for the H B&T experiment. .

5. Apr 6, 2016

### entropy1

I believe I mean the Mach–Zehnder inferometer, except that, in my case, the mirrors have a negligable influence. The photon interferes with itself at the second hs-mirror, but not when one of the paths is blocked.

Last edited: Apr 6, 2016
6. Apr 8, 2016

### jfizzix

If there is any information in the universe that will tell you which path the photon went though, the interference will disappear.

Such information could only arise due to interactions with the photon in question (however indirectly), and interactions trade single-party coherence for two-party entanglement.

7. Apr 8, 2016

### StevieTNZ

I would say it takes both paths as "potentialities" until one of the detectors goes off. QM states at the half-silver mirror, it is in a superposition of being transmitted and reflected.

8. Apr 11, 2016

### f95toli

The M-Z interferometer is much more complicated than the H-B&T interferomter; and in this case QM is not only silent about what "really" happens to the photons between the source and detector, it is also difficult to get some intuition for what kind of result you would expect.

You could say that the photon interfers with "itself" but this does not really -in my personal view- help you understand what is going on. I am not sure there is any substitute for sitting down and just doing the maths.

The reason the H-B&T interferomter is a bit easier to understand is that it is quite easy to get some intuition for what kind of result you would expect as long as you are dealing with single photons; you can then think of the photons as taking one of the two possible paths (which does not of course mean that it does, but it works as a model).

Last edited: Apr 11, 2016
9. Apr 12, 2016

### Khashishi

Ultimately, quantum mechanics tells you what you will measure. It has limited applicability in telling you what happens where you can't make a measurement.

10. Apr 12, 2016

### Paul Colby

On the question of two detectors and a half silvered mirror. One might assume that if the photon was reflected then it wasn't transmitted and if it was transmitted, then it wasn't reflected. If this were a true statement then the counts in the two detectors would have an anti-correlation,

$\sigma_{12} = \langle n_1 n_2 \rangle - \langle n_1\rangle \langle n_2 \rangle = -\langle N \rangle / 4$

where $\langle N\rangle$ is the average number of photons per observation. However, for the vast majority of light sources, thermal, led, laser etc the counts in the detectors are uncorrelated, $\sigma_{12}=0$. This is very much a property of the source, not the photons.

11. Apr 12, 2016

### Staff: Mentor

No.

The formalism without any interpretation explains it.

That the photons takes all paths is Feynmans sum over histories interpretation. Its valid, and so are many others, but its not needed.

Although usually not mentioned the sum over histories approach, while following directly from the formalism, is actually a hidden variable interpretation - but of a rather novel type. The path is the hidden variable.

Thanks
Bill

12. Apr 14, 2016

### Staff: Mentor

A small point, and one that doesn't weaken the point you're making, but..... The formalism doesn't explain, it describes.

13. Apr 15, 2016

### entropy1

I still don´t understand the problem with interpretations precisely. I take it then that if there are interpretations of the formalism, you can pick one you like? Does that mean they are not relevant?

14. Apr 15, 2016

### Feeble Wonk

They are not relevant for the purposes of running the numbers and making measurement predictions. Each of the interpretations make the same predictions, and are therefore equally valid in terms of experimental verification. However, the interpretations are certainly relevant if one is concerned about "what" the mathematical formalism is describing. But those concerns are more esoteric than technical, so the professionals frequently disregard them.

15. Apr 15, 2016

### entropy1

I guess it would also result in ¨denominations¨ of particular interpretation-endorsers. What use are the various interpretations if they describe the same thing but differ at the same time?

However, I would like an interpretation that is not in some way absurd like all current ones seem to suffer from.

16. Apr 15, 2016

### A. Neumaier

Then you should not interpret things in terms of particles but in terms of fields. This minimizes the absurdities.

The particle picture is not really appropriate at subatomic sizes and classical intuitions about them strongly conflict with observations.

Whereas the field picture remains semiclassical for all easy-to-perform experiments. It requires experimental ingenuity to prepare the nonclassical states of the fields that show a strong deviation from the semiclassical field picture and behave in a counterintuitive way.

Last edited: Apr 16, 2016
17. Apr 15, 2016

### Staff: Mentor

Yes - or none at all - its your choice.

No.

Thanks
Bill

Last edited: Apr 16, 2016
18. Apr 16, 2016

### vanhees71

You only need the minimal interpretation for doing physics. You just take the Born rule as it is and consider the description of nature probabilistic. States thus describe ensembles (including coarse-grained macroscopic quantities of single many-body systems, i.e., averages over space-time grids). Of course, then the idea of a "state of the entire universe" and other esoterica are self-contradictory, but I'd take this as a feature rather than a bug of this point of view.

Every other interpretation is either at best philosophy of science or science fiction ;-).