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- From: aephraim@physics.Berkeley.EDU (Aephraim M. Steinberg)
- Newsgroups: sci.physics
- Subject: Re: cos(reflection phase beam splitter) = 0 ? NO FTL?
- Message-ID: <16h5taINNhi9@agate.berkeley.edu>
- Date: 14 Aug 92 20:43:22 GMT
- References: <BsyoHE.37H@well.sf.ca.us>
- Organization: University of California, Berkeley
- Lines: 144
- 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
-