# Polarization, then detection: collapse when?

1. Mar 16, 2008

### birulami

In the book 'Quantum Physics', Alastair Rae uses polarized photons and photon detectors to describe many effects observed in quantum physics. In his typical setup, photons are passed through a polarizer and exit it through two different channels depending on the polarization $\phi$ or $\phi+90^\circ$ they then have. Behind each channel, there is a photon detector to tell through which of the two channels "the" photon exited.

Now my question: Where does the wave function representing the photon collapse?
1. Does it happen in the polarizer, when the photon's polarization is decided to be either $\phi$ or $\phi+90^\circ$, or
2. does the wave function collapse only later in the detector?
I guess one answer could be that we don't know, but I would hope that there is a canonical, mostly agreed upon answer whether it is (1) or (2).

2. Mar 16, 2008

### DrChinese

I would say (1) when the photon passes through the polarizer. You could put additional polarizers between the first and the detector, and the photon would always act AS IF the first polarizer is the point at which it took on a specific polarization.

That said, no one really know much about the details of collapse. Is it even something that is physical? No one truly knows, and this element of QM is something that sometimes triggers criticism.

3. Mar 16, 2008

### jostpuur

How do these polarizers work? Is it reasonable to assume, that the photon interacts with the polarizer enough, that we can think the photon gets entangled with the polarizer?

4. Mar 16, 2008

Hi,
I do believe, that the real collapse of the function happens in the detector. If you assume, you'd split the light in a polarizer, and then you would join the two channels again, i'm quite sure you'd see an interference pattern (u'd have to rotate the two polarisations to the same direction, to make them interfere, but it's a technical detail). It means, that the photon spreads through both ways and the polariser has a purely passive function.

As it arrives to the detectors, something happens and it somehow collapses so we get an event only from one of the detectors. This process is not entirely understood yet (i believe), but there is a theory, called quantum decoherence theory. It says, that as soon as a system (photon) which is in quantum pure state (it's in both channels, or let's say, it's in coherent superposition of the two halfways), comes to interaction with a system in thermodynamical equilibrium (detector) - or let's say reservoir- , the purity of the state is somehow absorbed by the reservoir and the system comes to a mixed state - which means that in a particular measurement, we will see the system in a single state - it's not likely to detect the photon in both channels.

5. Mar 16, 2008

### DrChinese

Please keep in mind that you can run a polarization test with a polarizer or a (polarizing) beam-splitter. The beam splitter shows you that the photon goes one way or the other. So you don't have to imagine too hard that the decision to go one way must occur as it goes through the splitter - ergo, the same must be true for a polarizer. Most folks say that if you COULD have detected the difference, then collapse occurs. So the detector is optional.

6. Mar 17, 2008

### birulami

How does the beam splitter show you that? Don't you have to have two detectors behind the polarizing beam splitter. And only these show you that --- well, what exactly? One registers something and we only deduct that it must have been the photon having come out only one side of the detector.

I must admit that I have some sympathy for the idea as tomasko789 seems to describe. What if two (nearly) identical "parts" of the photon (or wave) leave the polarizer. The first one hitting a detector interacts with it and collapses the system into one part. It may be the first part then immediately releasing its energy in the detector, or it may be the second part, releasing its energy soon after when hitting its detector. Do you know whether this model contradicts any experiments?

7. Mar 17, 2008

### Ken G

My understanding is that of tomasko789's. A "collapse" has to be irreversible, at least that's why I would mean by a "collapse" (or by "which-polarization information"). It might not have to be a "detector" per se, but it has to couple the system to something complex enough that we would always be forced to average over what we don't know. When you do that averaging, you "collapse" the wavefunction (and I do mean you-- when you chose to describe an entangled subsystem as if it had its own wavefunction). If the polarizer or beam-splitter forced you to do that, it would be a source of "collapse", but if it did not and you could reverse its action, then it would not be. But a detection is irreversible-- you can't say "oops, that photon I detected is now undetected and back in play".

8. Mar 17, 2008

### DrChinese

I think they are arguing about this on another thread. There they have determined that it is not even the detector at which the collapse occurs... it is at the level of human consciousness.

So I think the point is: you can make it be wherever you like as long as you apply the "rules" correctly. Again, this is a controversial issue within quantum theory. As far as I know, there are no strong proofs that collapse has specific physical components that can be explained in a satisfying way.

9. Mar 18, 2008

### Ken G

I think in this regard it is quite important to provide a definition of what "collapse" means. On that other thread, it was being used to mean "actualization of some outcome", which is essentially a personal experience with philosophical overtones and is not clearly even a requirement to be able to do science (though I presume we all perceive it as being true). I prefer to use "collapse" to mean "irreversible destruction of all coherences between outcomes of an experiment", and it occurs as a possible stage of how we choose to treat a prediction on a subsystem whenever we do not wish to include all possible influences explicitly (i.e., unitarily). Note it is not necessary for one outcome to be actualized, a probability distribution meets the requirement, and note that it is not a uniquely real effect, it has to do with our chosen treatment of the situation. From where I'm standing, this approach removes all ambiguities as to "when it happens".

Last edited: Mar 18, 2008