Hal gives the correct explanation of what's going on. In general,  all you
have to do to analyze the problem is to consider all contributions to a
particular state and add up the amplitudes. The absolute value squared of
the amplitude gives the probability, which may or may not contain an
interference term.

A simple minded formal description can be given as follows:

If you pass a photon through two slits then close to the screen its state
would be of the form:

Integral over z of [1 + Exp(i delta(z))] |s,z>

Here z denotes the postion on the screen, s is just a label for the photon
and Exp(i delta(z)) is the phase shift between the two paths which gives
rise to the interference term. Delta(z) will be zero exactly inbetween the
two slits and will be nonzero elsewhere. The probability of having the
photon at z is obtained (up to normalization) by taking the absolute value
squared of the prefactor of |s,z>, which is 2 + 2 cos(delta(z)).

Let's do the same for the two entagled photons. The entangled photon pair
can be denoted as:

 |p_x,s_y> + |p_y,s_x>

here x and y denote the polarization states.

If you pass s through the two slits then the state becomes:

 [1 + Exp(i delta(z))] |p_x,s_y,z> +  [1 + Exp(i delta(z))] ||p_y,s_x,z>

This has to be integrated over z, but let's focus only at the contribution
at some fixed position z. The prefactors of both state vectors |p_x,s_y,z>
and |p_y,s_x,z> are of the same form as in the single photon case and thus
you get an interference term Cos(delta(z) as above. If you put the quarter
wave plate in then instead of

 [1 + Exp(i delta(z))] |p_x,s_y,z> you get:

 |p_x,s_r,z> +  Exp(i delta(z))  |p_x,s_l,z>

And the complete state vector becomes:

|p_x,s_r,z> +  Exp(i delta(z))  |p_x,s_l,z>+

|p_y,s_l,z> +  Exp(i delta(z))  |p_y,s_r,z>

All the four state vectors are orthogonal. The probability that you detect
the photon s at z is just the sum of the absolute value squared of the four
terms, which is constant and doesn't contain an interference term.

Now let's pass photon p through the polarizer (45 degrees w.r.t. x). This
amounts to measuring the polarization state of photon p in the basis |p_x> +
p_y> and |p_x> - p_y>. If you don't discard one of these two states and keep
them both then, as Hal rightly points out, nothing changes. If you

|p_x> = |a> + |b>;

|p_y> = |a> - |b>

in the state vector above you get:

|a,s_r,z> +  Exp(i delta(z))  |a,s_l,z>+

|a,s_l,z> +  Exp(i delta(z))  |a,s_r,z>+

|b,s_r,z> +  Exp(i delta(z))  |b,s_l,z>+

- |b,s_l,z> -  Exp(i delta(z))  |b,s_r,z> = (collecting the prefactors of
like state vectors)


[1 + Exp(i delta(z))]|a,s_r,z> + [1 + Exp(i delta(z))]|a,s_l,z> +

[1 - Exp(i delta(z))]|b,s_r,z> - [1 -  Exp(i delta(z))] |b,s_l,z>

The absolute value squared of the terms with the p photons in the |a> state
contains the term 2 Cos (delta(z)), but for the b terms this is - 2
Cos(delta(z). If you don't observe the polarization of the p photon, the
interference terms would thus cancel. This is obvious since all we have done
is to write down the same state in a different basis. But if you do observe
the polarization of the p photon, then can measure the probability of
detecting s at position z and p in polarization state |a> then you have to
add up the absolute value squared of |a,s_r,z> and |a,s_l,z>. Then you do
get the Cos(delta(z) interference term.

----- Original Message ----- 
From: ""Hal Finney"" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <everything-list@eskimo.com>
Sent: Wednesday, October 12, 2005 08:16 PM
Subject: Re: Quantum theory of measurement

> Ben Goertzel writes about:
> > http://grad.physics.sunysb.edu/~amarch/
> >
> > The questions I have regard the replacement of the Coincidence Counter
> > here on: CC) in the  above experiment with a more complicated apparatus.
> >
> > What if we replace the CC with one of the following:
> >
> > 1) a carefully sealed, exquisitely well insulated box with a printer
> > it.  The printer is  hooked up so that it prints, on paper, an exact
> > of everything that comes into the CC.  Then,  to "erase" the printed
> > the whole box is melted, or annihilated using nuclear explosives, or
> > whatever.
> The CC is not what is "erased".  Rather, the so-called erasure happens
> to the photons while they are flying through the apparatus.  Nothing in
> the experiment proposes erasing data in the CC.  So I don't really see
> what you are getting at.
> > What will the outcome be in these experiments?
> It won't make any difference, because the CC is not used in the way you
> imagine.  It doesn't have to produce a record and it doesn't have to erase
> any records.
> Let me tell you what really happens in the experiment above.  It is
> actually not so mystical as people try to make it sound.
> We start off with the s photon going through a 2 slit experiment and
> getting interference.  That is standard.
> Now we put two different polarization rotations in front of the two slits
> and interference goes away.  The web page author professes amazement,
> but it is not really that surprising.  After all, interference between two
> photons would typically be affected by messing with their polarizations.
> It is not all that surprising that putting polarizers into the paths
> could mess up the interference.
> But now comes the impressive part.  He puts a polarizer in front of the
> other photon, the p photon, and suddenly the interference comes back!
> Surely something amazing and non-local has happened now, right?
> Not really.  This new polarizer will eliminate some of the p photons.
> They won't get through.  The result is that we will throw out some
> of the measurements of s photons, because if the p photon got eaten
> by its polarizer, the CC doesn't trigger as there is no coincidence.
> (This is the real reason for the CC in this experiment.)
> So now we are discarding some of the s photon measurements, and keeping
> some.  It turns out that the ones we keep do show an interference pattern.
> If we had added back in the ones we discarded, it would blur out the
> interference fringes and there would be no pattern.
> The point is, there is no change to the s photon when we put the polarizer
> over by p.  Its results do not visibly change from non-interference
> to interference, as the web page might imply.  (If that did happen,
> we'd have the basis for a faster than light communicator.)  No, all
> that is happening is that we are choosing to throw out half the data,
> and the half we keep does show interference.
> The only point of the CC, then, is to tell us which half of the data
> to throw out of the s photon measurements.  Destroying the CC and all
> of the other crazy things you suggest have nothing to do with the
> experiment.  The CC is not what is "erased" and it does not create a
> permanent record.  It is only there to tell us whether a p photon got
> through its polarizer or not, so that we know whether to throw away
> the s photon measurement.
> Hal Finney

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