From: aephraim@physics.Berkeley.EDU (Aephraim M. Steinberg)
Newsgroups: sci.physics
Subject: Re: cos(reflection phase beam splitter) = 0 ? NO FTL?
Date: 14 Aug 92 20:43:22 GMT
Organization: University of California, Berkeley
NNTP-Posting-Host: physics.berkeley.edu


I hadn't really planned to take up more bandwidth on this, but since
my email response to sarfatti got posted here, I feel like adding my
response to his response to my response...


In article <BsyoHE.37H@well.sf.ca.us> sarfatti@well.sf.ca.us (Jack Sarfatti) writes:
%>
%>
%>Message  1:
%>>From aephraim@physics.Berkeley.EDU Thu Aug 13 19:00:06 1992
%>>
%>>Dear Dr. Sarfatti,
%>>
%>>You write :
%>>>So that the locally observable probabilities to measure the
%>>receiver
%>>>photon polarization eigenvalues are conserved and we get
%>>the connection
%>>>communication equation
%>>>
%>>>p(e) - p(o) = sin2x cos(phi+m)cosb  (6)
%>>>
%>>>
%>>
%>>where you have defined b as the phase shift upon reflection
%>>at your 50/50 beam-splitter.(With respect to the
%>>transmission phase shift.)
%>>
%>>It can be shown from unitarity or from time-reversal
%>>symmetry that this phase is always plus or minus pi/2.(For
%>>any lossless beam-splitter, not just 50/50.) Thus cos(b) is
%>>always 0, so p(e)=p(o), and there is no communication.
%>>
%>> Aephraim.
%>>
%>
%>
%>Sarfatti response: At last a refutation that makes sense. If
%>true it would definitely mean that my device would not work.
%>On the other hand is it really true? Or is it a circular
%>argument? That is, is the "proof" that b =pi/2 really from
%>causality rather than unitarity or time reversal? From my
%>equations I do not see why it must be true. So I would like
%>to see the formal proof before I withdraw my claim. Then, if
%>it is true, we must see what happens if there is absorption.
%>
%>However, I cannot find any information that b = pi/2 in any
%>book on optics. To the contrary for external reflection from
%>a single surface the reflection phase shift is pi not pi/2
%>in discussions of Fresnel equations. For example Fig.6.2-2
%>p.206 of PHOTONICS by Saleh & Teich (Wiley 1991) says for TE
%>wave "External reflection. The reflection coefficient is
%>always real and negative, corresponding to a phase shift of
%>pi" The only case, and it is unusual, where the reflection
%>phase shift is pi/2 is in internal reflection of a TM wave
%>at a unique angle of incidence corresponding to total
%>internal reflection. Why should a beam splitter be so
%>special as not to obey Fresnel equations?
%>
%>
%
%

And I responded....



%From aephraim Fri Aug 14 12:53:23 1992
%To: sarfatti@well.sf.ca.us
%Subject: Re:  cosb = 0 contradicts Fresnel eqs?
%
%Born & Wolf treats it, Cohen-Tannoudji does it for general Schrodinger
%barriers, and Ou&Mandel had a paper on it a few years back.
%
%As for the Fresnel equations, think of the simplest beam-splitter,
%a thin sheet of glass which reflects 4% at each surface (we'll half-
%justifiably ignore multiple reflections for now).
%
%The amplitude for transmission is just exp(ikd) to lowest order, where
%k is the wavevector and d is the thickness.
%
%The amplitude for reflection off the front surface is +0.2, no pi phase shift.
%The amplitude for reflection off the back surface is -0.2exp(2ikd), where
%there is the pi phase shift you noted, but there are two
%path lengths of d.  If you add the two phasors, you can see that the phase
%of the resultant is the average of the two inital phases (2kd and pi),
%that is, kd+pi/2, that is, phi(r)=pi/2 + phi(t) as claimed.
%
%If you do it carefully for the infinite sum, you also get the right answer,
%of course, that's just a plausibility argument in response to your question
%about the Fresnel relations.
%
%The general argument is easiest by time-reversal symmetry.
%
%Suppose our input ports are I1 and I2, going into outputs O1 and O2.
%If the amplitude incident on I1 is 1 and on I2 is 0, t comes out in O1 and
%r comes out in O2.
%Let us now time-reverse this, assuming a symmetric beam-splitter (so r and
%t are the same from either side).
%Now t* comes in O1 and r* comes in O2, so t*t+r*r comes out I1 and
%t*r + r*t comes out I2.
%But we know nothing originally went into I2, so t*r + r*t = 2Re (t*r) = 0.
%That is, if we write t*r as |t||r|exp(iphi(r)-iphi(t)),
%cos(phi(r)-phi(t))=0, so the phases differ by plus or minus 90 degrees.
%There are other ways of showing this.
%If it's lossless but not symmetric, then the AVERAGE of the two different
%reflection phase shifts still differs from the transmission phase shift
%by 90 degrees.
%If it's lossy, you can put an upper bound on that cosine, you can derive
%it yourself or I can look for my old notes if you're curious.
%I don't think it's reasonable to think that the losses would HELP you in
%your scheme, though.
%(I admit I don't immediately understand how to demonstrate this latter
%guess though.)
%
%Hope this is clear,
%aephraim.
%
%

And in response to Brent Buckner's question about a prism in place of
a beam-splitter, note that the proof is completely general and the
MECHANISM of "recombination" is irrelevant.  All that's important is
the unitarity of the transfer matrix.

And if sarfatti's quotation of Saleh&Teich on all phase shifts being
pi in Fresnel reflection seems to go against my claim that one is pi
and the other is zero, I refer the reader to Feynman's QED.

The amplitude for reflection is
(n2-n1)/(n2+n1),
which is positive (zero phase shift) for n2>n1 and negative (pi phase
shift) for n2<n1.  (At normal incidence.)

Or maybe I got it backwards, but in this case, that's irrelevant.  :-)

aephraim

-- 
Aephraim M. Steinberg                    | "WHY must I treat the measuring
UCB Physics                              | device classically??  What will
aephraim@physics.berkeley.edu            | happen to me if I don't??"
                                         |       -- Eugene Wigner