From: Hal Finney (hal@finney.org)
Date: Tue May 07 2002 - 19:36:10 MDT
Eliezer writes:
> Now let's suppose that you use any of the quantum-cloning or
> quantum-teleportation techniques to produce two entangled photons. I
> transmit one photon to Boston and the other photon to London. Now I take
> the photon in Boston and send it through the second quantum experiment; I
> send it through a half-silvered mirror at A, let it possibly bounce off
> mirror B and mirror C, but now instead of mirror B and mirror C both
> sending the photon to recombine at half-silvered mirror D, mirror B and
> mirror C both send the photon bouncing back and forth along a long
> corridor of mirrors - long enough that there's a quarter second before the
> photon recombines at half-silvered mirror D.
>
> During this quarter second, the experimenters in London choose whether or
> not to perform a measurement on their photon - the photon that's entangled
> with the one in Boston. If they perform a measurement at this point, then
> as I understand quantum physics, both photons - the one in London and the
> one in Boston - should undergo a collapse of the wavefunction. If the
> photon in Boston undergoes a collapse of the wavefunction, it will have an
> equal chance of hitting target E or target F. If the photon in London is
> not measured, and the photon in Boston is thus not collapsed during that
> quarter second, the photon in Boston will hit target E with 100%
> probability.
Okay, I have done some reading about this and I think I understand
what happens. It is somewhat similar to the explanation of the other
experiment.
Basically I don't think you will see interference in the interferometer
when you give it the entangled photon. That is, whether you measure
the remote photon or not, the photon will always come out of the
interferometer with 50% chance to hit the targets E and F. This is
unlike the case with a properly chosen "local" photon, where you can
arrange for it to always hit E (or F, depending on the setup).
This can be explained in two ways, the easy one and the hard one.
The easy way is the same as before: to get quantum interference (which
manifests as suppressing one of the possible output paths so everything
goes to E), the paths have to be indistinguishable. The classic way of
making the paths distinguishable is to block one of them, and we know
in that case the interference vanishes and we get 50% output to E and F.
However sending in an entangled photon also allows for the paths to be
distinguished, in exactly the manner Eliezer proposes, by measuring the
other photon. Hence when we send in this kind of photon, there will be
no interference, just as when a path is blocked. We will always get 50%
output to E or F.
It might seem surprising that this special kind of photon has this
unusual property. Doesn't the interferometer work properly with all
kinds of photons? Well, no. The interferometer will only work properly
with certain kinds of photons.
Let us suppose that the interferometer splits the photons on the basis
of polarization, with vertically polarized photons taking the upper
path, and horizontally polarized photons taking the lower path. Now,
if we send in a diagonally polarized photon, it "takes both paths" and
interferes with itself and we can get all such photons to come out to E.
But if we send in a pure vertically or horizontally polarized photon, it
takes only one path, we know which one it takes, and it can't interfere
with itself. Such photons will get 50% output to E or F.
In fact I believe that for a given configuration of the experiment,
only carefully chosen polarizations of the incoming photons will assure
that everything goes to E. Slightly different polarizations will most
likely go to E but have a small chance of going to F. Very different
polarizations like pure H or V will have the 50% chance of where they
end up.
Given this behavior, which is glossed over in descriptions of the
interferometer, it should not be so surprising that sending in the kind
of photon needed for Eliezer's experiment fails to show interference.
Now for the hard explanation, I will describe some of this in
terms of equations. I am basing this really on just one paper,
http://xxx.arxiv.org/abs/quant-ph/0103081, by Len Vaidman, who first
showed some of the paradoxical features of measurements with these
interferometers.
Let us arrange the system so that horizontally polarized photons H take
the lower path and vertically polarized photons V take the upper path.
Let us suppose the initial photon is in the state (H + V), an equal
superposition of the two. (I will ignore constant factors like 1/sqrt(2)
because they are so cumbersome to write.) This would correspond to a
diagonally polarized photon.
The initial beam splitter A splits the photon with the H component
taking the lower path and the V component taking the upper. The
mirrors B and C will preserve the state. The final combiner D is
designed so that photons from the two paths go to the two output
detectors E and F with the formulas:
H -> E + F
V -> E - F
Both of these mean that there is a 50-50 chance of going to E or F with
photons from either path, but because of the sign difference there is
the possibility of interference. Now simple algebra shows that
(H + V) -> (E + F + E - F) = E
(again, ignoring constant factors). Therefore we see that an input photon
in the state H + V, which is a diagonally polarized photon, will with 100%
probability end up at detector E. As an exercise you can determine that
a photon in the initial state (H - V), which is the opposite diagonal
polarization, will end up at detector F with certainty. You can also see
that pure H and pure V photons will go to E and F with 50% probability.
So what happens with entangled photons? Now we have two photons,
numbers 1 and 2, and they are in an entangled state. Given the requirements
of the experiment, where we can measure remotely which path the local
photon will take, they must be in entangled pure states of horizontal
and vertical polarization, like this:
H1 H2 + V1 V2
This represents a 50% superposition of two pure states: one where both
photons are horizontally polarized, and one where both photons are
vertically polarized. Now let's send photon 1 through our apparatus.
H1 H2 + V1 V2 -> ((E + F)H2 + (E - F)V2)
Here we have just done H1 -> E + F and V1 -> E - F.
= E H2 + F H2 + E V2 - F V2
= E (H2 + V2) + F (H2 - V2)
Well, I'm not sure there is much point to doing the algebra in detail.
The point is that it does not reduce to E, as it did when we sent in a
pure H + V photon. The actual result is entangled with the other photon.
The bottom line is that we will measure E or F with 50% probability, and
that will give us information about the state of photon 2.
Or if we know the state of photon 2 because we measured it, either pure
H2 or pure V2, then that tells us the state of photon 1 as pure H or V,
and as we saw above, pure H or V photons come out with 50% probability
to the two detectors.
So whether we measure the remote photon or not, we will get 50-50
output probabilities. That corresponds to an absence of interference.
Whatever you do remotely will not produce a detectable difference in
the local result, and so the experiment does not succeed in transmitting
information.
Hal
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