# Why is glass QM transparent ?

1. May 19, 2007

Why is glass "QM transparent"?

Sorry if this is a basic question.

Why is a wave state preserved through it's interaction with glass, mirrors, air, etc.? Why doesn't reflection (absorbtion then re-emission) fix the position and collapse the function? It's isn't really the same photon that is re-emitted, is it? Well, apparently it has to be a copy at the very least, but what preserves state info in the copy process?

2. May 19, 2007

### ZapperZ

Staff Emeritus
Please read one of the entries in our FAQ in the General Physics forum.

Zz.

3. May 19, 2007

### da_willem

I believe that both reflection and transmission, does make the wavefunction collapse. As you might knwo from optics, the transmission of e.g. light through glass, is the continuous absorption and emssion of photons by atoms (this is what makes that the velocity of light differs somewhat from the photon velocity c). So in this sense the state info is lost at each absorption, also when we speak of reflection.

What then happens can be visualised from a number of different perspectives:

-classically you could say a shperical EM wave is emitted from the atoms which absorb light. There is only constructive interference in the forward direction (draw a picture of the wavefronts of some emitting molecules to see this).
-you could say a photon is emitted in a random direction and only those in the formward direction interfere constructively
-from a larger persepctive you could use the path integral method: a photon takes every possible path at the same time, but only stationary paths, where it 'interferes constructively with other paths' are the ones we see it take.

But the botom line is that the state information is lost in the absorption process but because there is this preferential direction of the initial propagation direction, constructive interefrence also takes place in a prefered direction.

4. May 19, 2007

Huh? This isn't a general physics question, but even so, I don't find an answer there. Did you have something in mind?

5. May 19, 2007

I thought "spin", etc. was preserved somehow, i.e. entanglement experiments.
Ah, that's one piece of the puzzle.

I guess what I'm missing is not how a photon reaches the destination at all, but how "every possible path" can include re-emission. It's like a pony express rider, the pony keeps getting replaced at the check points, but the message gets through. What's the "rider and his message" for bouncing photons?

It's like each re-emission emits a photon fully entangled with the absorbed photon, but better.

What's the difference in the way a mirror reflects a photon, collapsing its wave function, so that it doesn't count as a 'measurement', whereas getting re-emitted off a backdrop (so that we see a dot) is a 'measurement' that sets 'which-path' and presumably sets/changes state info?

I'm thinking of the experiment where a photon is sent through a partial mirror, then each branch has a down-convertor, and each branch from the down-converter is mirrored back, and after all this, the path integral still refers to the path started by a photon which no longer exists.

6. May 19, 2007

### da_willem

It is related to the question, whether photons move slower in a solid medium, but I agree that this read doesn't really answer your question.

7. May 19, 2007

### da_willem

Photons that don't get absorbed just follow a straight line. For the ones that get absorbed, I guess you have for the amplitude the product of two path integrals: one ending at a certain molecule and one starting there. Remember you can alsways cut a path integral open by inserting a complete set of states, i.e.

$$<q_f, T|q_i, 0>=\int dq <q_f, T | q, t><q,t|q_i,0>$$
$$\int [dq]_{q_i, 0} ^{q_f, T} exp(i \int _0 ^T L dt) = \int dq \int [dq]_{q, t} ^{q_f, T} exp(i \int _t ^T L dt) \int [dq]_{q_i, 0} ^{q, t} exp(i \int _0 ^t L dt)$$

So , I think that for paths that end on a certain molecule (the q,t being q', t', the coordinates of this event) the re-emitted paths (1st integral rhs) interfere with paths of nonabsorbed photons (going straight on) only constructive in the forward direction. Or in the direction indicated by Braggs law in case of reflection.

8. May 19, 2007

How can you interfer constructively in the direction of Braggs law, when no nonaborbed photon goes in that direction? I can't quite tell, but Braggs law might imply that you get a wave crest in the interference with the incoming wave, at an angle (of reflection)? Braggs law applys to a crystal lattice, and interference between waves from different depths in the crystal, but maybe the same thing is happening with mirrors.

Also, does the constructive interference somehow drag along state information with it (i.e. entangled spin)?

9. May 19, 2007

### ZapperZ

Staff Emeritus
Note: ALL our physics forums FAQ are in the General Physics section. It doesn't mean that it only covers general physics. If you have read it, you would have noticed that it covers a lot of things.

You're asking about a particular quantum mechanical aspect of transmission/reflection through a solid medium. This then requires the consideration of the photon picture interacting with the solid medium, i.e. the medium now must also be considered. Thus, the entry on photon transmission through solid would be something relevant.

Zz.

10. May 19, 2007

### da_willem

Mirrors work differently indeed, I was refereing to the reflection of X-rays. In mirrors the electronic structure starts vibrating and emits radiation like a dipole. A dipole radiates in preferred directions that turn out to be such that the angle of incidence is equal to the angle of reflection.

But I presume you can also describe this in the same way, using the path integral method.

Of course all conservation laws apply, so if the excited atoms return back to their original state, the quantum numbers of the outgoing photon are the same as that of the incoming photon.

11. May 19, 2007

Ok, now I understand how the phonons ZapperZ was trying to point me at are relevant (in addition to forcing timely re-emission?).

And thus you get your direction for the constructive interference with the emitted wave?
It's weird to think of "spin" as conserved in the same way as energy and momentum. I can see where the energy is stored in the excitation of the atom, and how it's ok if the direction of the momentum changes, but spin and everything else makes me wonder how its filing system works.

Hmm, you're implying that entanglement is preserved by conservation laws?

12. May 19, 2007

### da_willem

Me too, one of the weirdest things of qm. That there is this thing spin that look so much like something that rotates (it even has a magnetic moment associated with it) but nothing does. For a differnt view however read the article of Hans C. Ohanian "What is spin" in the American Journal of Physics!

Not sure what you mean.

13. May 19, 2007

Photons [must?] preserve/maintain their entanglements through absorbtion and re-emission, i.e. down a fiber optic cable. Since entanglement isn't a quantum number, or even something that seems like a conservable thing, it makes me think there is still something missing from the picture.

14. May 19, 2007

### da_willem

My intuition tells me that absorption destroys an entangled state, by being a nonunitary process, not sure...

15. May 19, 2007

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16. May 19, 2007

Ya, mine too, but how else would you get an entangled photon to a detector, especially if you want it delayed (I think they used fiber cables for the delay, and surely there must be mirrors).

17. May 20, 2007

I've still got this bee in my bonnet:

Hmm. Maybe the backdrop doesn't count as a measurement either, but instead the final resting place of the photon (a retina or CCD, or some brick somewhere that doesn't re-emit it)? So that would mean that the photon is still in an indeterminate state after it has left the backdrop on which is has supposedly created an interference pattern. So, when we're looking at the backdrop and seeing the pattern, what's really happening is the wave is collapsing in our retina, the position of which is determined by the probability crest, which is *influenced* by the interference probability crests of the reflection points on the backdrop?

The interference pattern we think we see on backdrop is then to be considered misleading, since the path integral is still in full play after the backdrop? Am I making any sense?

I'm trying to keep intact my understanding of why there is nothing special about measuring a photon.

18. May 22, 2007

19. May 23, 2007

### da_willem

The collapse of the wavefunction is what happens at a 'measurement', it maybe even defines what in qm we mean by a measurement. I explained that the reflecton of a photon off a mirror also involves re-emission and although it doesn't leave a (lasting) spot.

The final resting place of a photon sounds to me as a collapse of the photon wavefunction, and so a measurement.

Ok, this is a nice philosophical thought, but is absolutely not what happens. The photon is absorbed and a new photon is then re-emitted. As time passes on the wavefunction spreads out and the possible measurement position becomes uncertain again, it might just be on your retina.

Not really, there's actually two photons and two path integrals (or one factored in two) involved. Good luck figuring this out!

20. May 23, 2007

### vanesch

Staff Emeritus
All this collapsing leads to confusion, and one touches here the heart of the measurement problem. When does a wavefunction physically collapse ?
Answer: there is not the slightest hint of a physical collapse of the wavefunction in specific *processes*. So one cannot say that "upon reflection, the wavefunction collapses", because that would for instance mean, that we cannot describe reflection quantum-mechanically. You cannot say that upon absorption/re-emission, collapse occurs, or a non-unitary process occurs, because that would mean that we have no quantum-mechanical means to describe absorption and re-emission!

So when can you safely collapse your wavefunction ? WHEN INFORMATION IS IRREVERSIBLY RECORDED ABOUT THE EVENT SOMEWHERE.
This "somewhere" can be the environment, or a tape, or whatever ; and the important point is that you have to consider this as *irreversible* (so that, with sufficient effort, you might track it long after the experiment is over).

That said, you might also sometimes obtain results with collapsed wavefunctions even though you "shouldn't". In this case, this simply means that for a good or a bad reason, the way you look at the setup is too coarse-grained to observe any potential interference effect (which might be there!). As such, the "collapsed wavefunction" is nothing but a calculational approximation, which might nevertheless be very good.

As a reflection of a photon on a heavy mirror *doesn't leave any trace*, you cannot say that it "collapses the wavefunction". In fact, a mirror typically keeps coherence intact, so it is a very bad idea to collapse the wavefunction upon (coherent) reflection.
On the other hand, absorption and re-emission usually (but not always!) can be considered "incoherent". Most of the time, the phase relation gets too involved to be measurable and you can just as well collapse your wavefunction, as most of the time, you will not be able to distill the interference effects.
But a famous counter-example to this is the laser! In a laser, there is *coherent* emission, after there was absorption. So you see that the trick of "absorption and re-emission collapse the wavefunction" doesn't always work.

Again, you can only safely collapse your wavefunction upon the *irreversible* registration of a certain event in one way or another. In *that* case, you will not make any error. In all other cases, you're making an approximation (neglecting potential interference terms), which might, or might not be justified.

EDIT: why is this so ?
Well, given that the information is irreversibly recorded in the environment, or in the apparatus, or on a tape, you can in fact consider that whatever you are going to do/measure next, you will measure it in CORRELATION with this (wanted or unwanted) outcome. You will not be able to UNDO this result, and consider interferences between its result and its opposite. If the result is irreversibly recorded, whether you look at it or not, you cannot do any interference experiment anymore concerning this result. As such, you can replace the superposition in the wavefunction by a statistical mixture.

This is nothing else but the mechanism of decoherence.

EDIT2: Example of when you can collapse a photon wavefunction upon reflection on a mirror: consider a *very light* mirror which is freely floating. Upon photon reflection, the mirror is set into motion, a motion which you can detect later on (by letting it float, say, for an hour and measure the displacement). If part of the photon wavefunction was reflecting upon this mirror, then you can safely project this, and you will not find any interference pattern with the rest of its wavefunction. So in this case, "reflection on a mirror collapses the wavefunction", simply because it left some trace in the environment (motion of the mirror).
However, if you now fix this mirror to a table, and you do this again, you cannot collapse the wavefunction anymore. You can make the light reflecting on this fixed mirror interfere with the rest of its wavefunction. So here, "reflexion on a fixed mirror doesn't collapse the wavefunction".

Last edited: May 23, 2007