I thought of an experiment we could do to demonstrate the weakness of
the DI. That's better than just arguments per se against DI.
Briefly, it shows we can recover some information about the
relationship between superpositions even after they've been fully
converted into a "mixture," a mixture that DI treats as equivalent to
a classical mixture. The outcome is already predictable by known
quantum optics, so I assembled a case from existing knowledge. (It
still should be verified.) I hold therefore that the "quantum
measurement paradox" remains unresolved, perhaps the deepest mystery
about the nature of our world.
No awkward text diagrams but my proposal is easy to describe. I use
"*" for multiplication from use of some spelled varibles. (The math is
less simple but not hard to work through.) Synopsis: despite
scrambling the phase relations, we can recover the unequal amplitudes
originally input into a recombining beamsplitter. That could not
happen if the input became a true mixture, rather than just an
apparent mixture. Consider a Mach-Zehnder interferometer (as
typically diagrammed) with a first beamsplitter BS1 that need not
divide intensity equally. For example, intensity along bottom leg L1
= 64% and top leg L2 = 36%. Hence, relative amplitudes are a = 0.8
and b = 0.6. We can represent what happens to a single photon as:
state a|1> goes along the bottom and state bu|2>, along the top. U is
the phase difference between |2> and |1>, as complex exponential. In
a properly balanced MZ with a *clean* path, u = 0. Then we recombine
the beams at BS2, a 50/50 splitter/recombiner. It combines relative
phases in the usual way for such work, with output from one outer face
called channel A2; and from the other outer face: channel B2. (This
keeps numbering consistent and allows easy reference to original
output from BS1.) I will just use "s" for [sqrt(2)]/2 ~ 0.7. A half-
silver mirror reduces intensity to 1/2 and thus multiplies amplitude
by s. Hence,
(1) CA2 superposition = s[(a|1> + bu|2>], and CB2 superposition = s[(a|
1> + bu''|2>].
Here u'' is the angle u + 180°, because input from L2 is reflected out
CB with 180° relative phase difference. (It must be, to show bright/
dark interference equivalent fringes.)
We find intensity (and photon statistics) by inserting into
(2) I = A^2 + B^2 + 2AB cos theta,
showing involved amplitudes A and B. If a = b and p = 0 degrees, then
the intensity out of CA2 = 1/4 + 1/4 + 1/2 = 1, and out of CB2 = 1/4 +
1/4 - 1/2 = 0. That is equivalent to bright and dark fringes. If a =
0.8 and b = 0.6, we get CA2 = 0.32 + 0.18 + 0.48 = 0.98 and CB2 = 0.32
+ 0.18 - 0.48 = 0.02. If there is a specific phase difference
introduced between a and b, then the relative intensities change but
must add to one. If we introduce photons one by one into such a
device, the statistics of detection are the same. To make the
argument and experiment about the wave function of a single photon,
that is what we'd do. (Despite some states of unclear photon number,
an effective "one photon at a time" *can* be introduced into such a
device. It means basically, one net "click" from arrays of ideal
photon detectors covering all avenues of escape.)
However, let's introduce complete decoherence into the picture, in a
manner like C. Orzel uses to supposedly model decoherence in e.g the
post at http://scienceblogs.com/principles/2008/11/manyworlds_and_decoherence.php.
(We had quite an argument there and elsewhere. I admit being testy
sometimes but think I'm in the right.) Now we have to integrate over
a range of p and q, and divide by the range to get the mean value.
This is easy for a uniform distribution of phase differences, since
(3) integral (Eq. 2) d theta = (A^2 + B^2)theta + 2AB sin theta + C.
We pick a range that completely scrambles the phases ("complete
decoherence") such as between +/- pi, substitute into the integral,
and divide by the range 2pi. Then (since sine of each limit = 0) we
find the result out of either channel is simply A^2 + B^2 = (a^2 +
b^2)/2. This destroyed the statistical interference pattern of photon
hits, even in the case a <> b. The output acts like a "mixture" of
photons entering BS2 from either L1 or L2 but not "both at the same
time." So if e.g. BS1 had sent a photon along L1 64% of the time and
along L2 36% of the time, each would have 50/50 chance of reflecting
or passing through BS2. That output would be 50/50 from CA2 and CB2.
A DI follower would say (following an ensemble interpretation): what
would have been a coherent superposition is now a mere "mixture"
despite being comprised from interacting waves. They would use the
positivist yet post-modern sounding line that "we couldn't tell the
difference, so the output should be regarded as being the same as a
'real' mixture." Hence, somehow we don't have to worry about why a hit
occurred at the CA counter instead of the CB counter, when under old-
fashioned (!) QM there are wave amplitudes (usually) at both counters
and a mysterious collapse was needed to sweep the whole big mess into
one little atom that absorbed it all. Their argument sounds circular
(what causes any "statistics" in the first place instead of
distributed amplitudes, to allow comparing one set of stats to another
etc.), and I'm fortified by seeing similar misgivings from e.g. Roger
Penrose. But DI is popular because it lets the perplexed brush off
their worries about paradoxical features of reality.
So, is the decodant view really apt - even in their own terms? I think
not. Let's recombine outputs CA2 and CB2 into BS3, which is just like
BS2. The wave reflected into CB3 will again turn the phasor angle by
pi, which reverts u'' back into u. Since intensities are again cut in
half, and phasors at opposite angles cancel out (destructive
interference), the new output that combines the beams is like this:
CA3 = s[CA2 + CB2] = s[s[(a|1> + bu|2>] + s[(a|1> + bu''|2>]] = a|1>
CB3 = s[CA2 + (angle pi rotation)CB2] = s[s[(a)|1> + bu|2>] + s[a(pi)|
1> + bu|2>]] = b|2>
Therefore, from BS3 we recover the original amplitudes (and hence,
same statistics) coming out of BS1! That information was hidden in the
relationship between the wave outputs from BS2, not showing in the raw
statistics of hits from the channels. This could not happen from a
mixture, since photons that came from either CA2 or CB2 but not "both
at the same time" would just scatter as individuals from BS3. Their
statistics from BS3 would be 50/50 output instead of a^2, b^2. This
demonstrates the continued wave nature of the output from BS2, despite
total mixing of phases which allegedly destroys the superposed
character of the photon wave function. So now we know, in each run of
this experiment there must simultaneously be two wavefunctions
existing together in our little ol' universe - not one photon going
one way in ours and another photon going the other path in "another
universe." The recovery of amplitudes from BS3 shows that. One of
the deep mysteries of reality remains a challenge.
IMHO, Bye bye decodance! Put it in the doghouse ...
--
Regards, and Happy New Year/Decade (so they say)!
Neil
--
Bob May
rmay at nethere.com
http: slash /nav.to slash bobmay
http: slash /bobmay dot astronomy.net
Why is everything rushing away from everything else?
No Center
http://www.astro.ucla.edu/~wright/nocenter.html
http://www.astro.ucla.edu/~wright/infpoint.html
Also see Ned Wright's Cosmology Tutorial
http://www.astro.ucla.edu/~wright/cosmolog.htm
http://www.astro.ucla.edu/~wright/cosmology_faq.html
http://www.astro.ucla.edu/~wright/CosmoCalc.html
WMAP: Foundations of the Big Bang theory
http://map.gsfc.nasa.gov/m_uni.html
WMAP: Tests of Big Bang Cosmology
http://map.gsfc.nasa.gov/m_uni/uni_101bbtest.html
Probably trying to move to a better neighborhood.