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- Path: sparky!uunet!usc!rpi!uwm.edu!caen!destroyer!cs.ubc.ca!unixg.ubc.ca!ramsay
- From: ramsay@unixg.ubc.ca (Keith Ramsay)
- Newsgroups: sci.physics
- Subject: Re: Photon pair Mach-Zehnder Interferometer IV
- Date: 20 Nov 1992 04:46:51 GMT
- Organization: University of British Columbia, Vancouver, B.C., Canada
- Lines: 96
- Message-ID: <1ehqjrINN1eb@iskut.ucs.ubc.ca>
- References: <BxyF6I.76v@well.sf.ca.us>
- NNTP-Posting-Host: unixg.ubc.ca
-
- Sarfatti discusses a "Mach-Zehnder Interferometer":
- |* Note that phi is definite. There is no reason to assume that phi is
- |random as Ramsay requires.
-
- I do not so "require" for *all* experiments. I only require that if
- there are phase differences, that they be taken accurately into
- account.
-
- |According to Ramsay:
-
- I'd be careful attributing to me claims about an experiment which we
- haven't discussed yet!
-
- Let me suggest that in the future, we try to limit our use of
- "argument by analogy" between different experiments since it seems not
- to be helpful, in either direction.
-
- |Note that the relative phase shift of the beam splitter is not frozen in
- |stone like the Ten Commandments but adjusts itself globally to the boundary
- |conditions or "total experimental arrangement" - in this case the setting
- |on the variable phase plate.
-
- Who's the supplier of your optical equipment? :-) You're implying that
- the sort of interaction that the beam splitter has with photons
- passing through it varies somehow, without giving a reasonable
- physical basis for it. Beam splitters have optical properties (index
- of refraction, reflectivity, etc.), which determine the relative phase
- shift. If you claim that it changes, you'll have to explain why its
- optical properties have changed.
-
- |#4 Now insert a half-waveplate in the o-space path. The unitary operator
- |U(o:1/2) only acts locally and unitarily in the o-path because that is
- ^^^^
- |physically where the plate is. There is no half-wave plate in the e-path.
- |Ramsay seems to get confused on this, falsely applying the unitary operator
- |to the entire state - which is wrong standard quantum mechanics (SQM).
-
- No, Sarfatti. Remember that you yourself described the experiment
- having the operators act on the state of the photon (the "whole
- thing"), until it turned out to contradict your "indistinguishability"
- claim about photons in the two paths.
-
- Standard quantum mechanics talks about transformations of the state,
- and asserts that the operators producing them are unitary (until
- measurement, of course). Where does it say, "but in some cases you
- have to break the state space into pieces; the operator acts
- unitarily, but only on each piece"? I think this is just your invention.
-
- |Preamble:Ramsay's remarks about summing over phi are clearly wrong as he
- |has not conceptually distinguished between a 2-slit experiment and a Mach-
- |Zehnder experiment.
-
- This is inaccurate; I am aware of the difference.
-
- [diagrams and description]
- |The initial "entangled" (i.e. quantum-connected) photon-pair state is:
- |
- ||a,b> = {|a,e>|a,+>|b,+> + |a,o>|a,->|b,->}/sqrt2
- |
- |local unitary action of both phi-meter in e-path and 1/2 wave plate in o
- |path of photon a is "polarization disentanglement":
- |
- ||a,b> -> {e^iphi|a,e>|a,+>|b,+> + |a,o>|a,+>|b,->}/sqrt2
- |
- | = |a,+>{e^iphi|a,e>|b,+> + |a,o>|b,->}/sqrt2 = |a,b>'
-
- I trust you will agree that, more generally, the action of this
- optical apparatus upon a pair state
-
- |a,b> = z1 |a,e,+>|b,+> + z2 |a,o,->|b,+>
- +z3 |a,e,+>|b,-> + z4 |a,o,->|b,->
-
- with |z1|^2+|z2|^2+|z3|^2+|z4|^2=1, takes it to
-
- |a,b'> = e^{i*phi} z1 |a,e,+>|b,+> + z2 |a,o,+>|b,+>
- +e^{i*phi} z3 |a,e,+>|b,-> + z4 |a,o,+>|b,->.
-
- |The recombination at counter 1 induces
- |
- ||a,b>' -> |a,+>|1>{e^iphi|b,+> + |b,->}/sqrt2 = |a,b>'''
-
- I still see no reason to believe this, and moreover, I see no natural
- way to extend such a transformation to the entire state space (which
- makes sense, that is). Unless it is defined on the whole state space,
- it is not a well-defined operator.
-
- |<a,b|a,b> = '<a,b|a,b>' = '''<a,b|a,b>''' = 1
- |
- |so these transformations are unitary.
-
- This also has to hold true for the action upon all the other vectors
- of the state-space, or it isn't really unitary.
-
- Keith Ramsay "But I really think that frequent posters such as
- ramsay@unixg.ubc.ca myself, Dale, Scott, McIrvin and others are not
- crackpots; we are simply loudmouths."
-