one_stinky_bum said:
Didn't mean to bring the thread back from the dead, but I worked on a QE experiment last summer - maybe our paper would be helpful. We've fixed the problem in the explanation that the earlier version had.
http://marcus.whitman.edu/~beckmk/QM/qe/qe.pdf
Ashifi.
Nice paper :-)
However, what I find strange in these reasonings is that one insists on ERASING information. Of course, this somehow comes down to the same thing, but I find it more mysterious that "erased" information gives rise to interference. Usually one needs to EXTRACT INCOMPATIBLE information in order to obtain interference.
What I mean is the following. In the "classical" quantum erasure experiment where one can use the idler polarisation "to find out which path the signal photon took" or where one can "erase that information", what one actually needs when "erasing that information" is to extract information of an incompatible variable in order to subselect a sample which does show interference. Of course, when one extracts that information, and because of the incompatibility of that measurement with the "which path" measurement, one has in fact also lost every possibility to recover that which path information, so one can say, with some poetic liberty, that one has "erased" that which path information. But what one really needed was to EXTRACT other information !
A simple way to demonstrate this, is in the interference or not of the two beams obtained from the signal beam. It is not because the idler beam is dumped into a block of hot graphite (which also erases all possibilities of recovering the which path information from that beam), that suddenly an interference pattern appears at the signal side. It is because you need to have a coincident CLICK of the the idler under 45 degrees that you can select one of the two subsamples at the signal side. So I'd say that in this case, it is clear that you *need to extract* information, and that "erasure" is not sufficient (or has in fact anything to do with it). But, as I said, it is true that the extraction of the needed information makes you make an incompatible measurement with the "which path" measurement ; and as such you did erase this information, as a side effect.
The reason you need information extraction is that in ALL these potential "which path" experiments, you have lack of interference, or, what comes down to the same, a superposition of two shifted interference patterns (because sin^2 + cos^2 = 1). In order to separate out one of both subsamples (the cos^2 one, or the sin^2 one), you need extra information. It is the extraction of that extra information which is incompatible with the extraction of the "which path" information.
As you point out in the paper, around equation (2), you need to know whether you will be in the +45 or the -45 branch, because it are these two branches which give rise to the two interference patterns (the cos^2 and the sin^2).
I don't, however agree with the explanation on top of page 9: indeed, if you DUMP the idler beam in a hot graphite block, you erased the information also. That doesn't make the interference pattern appear at the signal side.
I don't agree to what is stated below equation (4) either: both density matrices are equal, and as such, the mixed states are identical. It is an error that is easily made (I've been guilty of that myself and been corrected for it a few times): different statistical compositions of pure states can give rise to identical mixtures. In quantum statistics, if the density matrices are identical, the mixtures are identical, even if you composed them by lumping together different pure states. You have the liberty to write the mixed state in the hh + vv way, or in the ++ + -- (45 degree) way. Both are diagonal density matrices with 1/2 on the diagonal, and as you know, a unitary transformation keeps such a scalar matrix a scalar diagonal matrix.
The reason why these mixtures are physically identical is that ALL expectation values of ALL possible measurements (which constitute all what is observable) are given by Tr(rho A). So if the rho's are identical, there is NO WAY to distinguish the two mixtures, hence they are physically identical.
You are cheating if you look at the "interference of fringes and anti-fringes" because to do that you have to synchronize with your 1 Hz generator. If you want the 1 Hz generator to make a "mixed state", then you shouldn't analyse results on the 1 Hz scale, but you should accumulate data over many cycles ; otherwise you're not working with the mixture, but with the individual |HH> and |VV> states ; now THERE is of course a difference between a |HH> state and a |45 45> state of course. There is no difference between the mixtures. But you didn't really make a mixture because the 1 Hz scale is too remote from the frequency of the light or the time constants of the detectors, and there are many simple techniques to recover the pure states from the data. The reason is that you are working with an analysis that supposes stationary random processes and that your 1 Hz modulated choice is not a stationary process (except if you look onto it on a timescale which is very long compared to 1 second).
cheers,
Patrick.