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Wave-particle duality in quantum optics
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Joined: 09 May 2006
Posts: 1

PostPosted: Tue May 09, 2006 7:31 am    Post subject: Re: Wave-particle duality in quantum optics Reply with quote

A |
B |

I did not go through your math, I would just like to know whether you
agree with the following statements
If A and B are from the same source (electrons , for instance), there
will be no interference pattern
on the screen if you put a detector to find out through which slit the
electron go through
However, if A and B are from two different sources (exactly identical),
then the interference will still be there even you put a detector (in
this case you done really need a detector to tell).
Another comment, when A and B are from the same source and you put a
detector to make the interference go away; however, the particle still
have wave property. It is not a classical particle.
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Stephen Parrott
science forum beginner

Joined: 12 May 2005
Posts: 9

PostPosted: Fri May 05, 2006 4:26 am    Post subject: Wave-particle duality in quantum optics Reply with quote

SUMMARY: This post discusses wave-particle duality in quantum optics.
I have questions about some papers presenting fundamental results
in this area and hope that someone may be able to answer them.

The wave-particle duality to be discussed
looks superficially similar to the Heisenberg uncertainty principle,
but is actually a completely different phenomenon.
However, it is described in the same language used by most
introductory quantum mechanics texts to introduce
the familiar uncertainty principle, so the latter is a convenient
place to start.

Most quantum mechanics texts introduce "wave-particle duality"
by describing a "two-slit" diffraction experiment like the following.
A stream of particles (photons will be the case of interest below) is
directed toward a screen with two slits.
Beyond this screen is a solid screen which registers when a particle hits.

| |
| |
| |
***---> | |
| |
| |
| |

If one determines in some way
through which slit each particle passes,
the image on the screen has two peaks---like a superposition of
two Gaussian distributions:

exp(-x^2) + exp(-(x-c)^2),

where x is a spatial coordinate and c a constant.

But if there is no means of determining through which
slit the particle passes, the pattern on the screen is
qualitatively different---like a diffaction pattern with many light and
dark "fringes". (This is hard to render in ASCII, but you have all seen
the pictures in textbooks.) Thus if one "looks" to see which slit
a particle goes through, the particles act like classical particles
(e.g., billiard balls), but otherwise the behavior is more like
that of light waves.

Let M denote the maximum intensity of the pattern
and m the minimum. Then the "visibility" V of the pattern
is defined by

V := (M-m)/(M+m) .

The visibility, which is between 0 and 1,
is a quantitative measure of how "wave-like"
is the particles' behavior. The maximum visibility of 1
indicates the most wave-like behavior.

The same language of "fringes" and "visibility" is used
to describe a different wave-particle duality principle in quantum optics.
Consider an interferometer set up as follows.

* U _ _ Phase Shifter _ _ _ _ _ _
\ / \ Polarization Detector 1
\ / \ /
\ / \ /
__\/__ Polarizing beam splitter_ /
BS \ \
\ / \
\ / \
L \_ _ _ Polarization Rotator _ _/ Polarization Detector 2

with the following notation:

The stars at the upper left of the diagram
represent a beam of incoming photons with
identical polarization (say "vertical"). This beam
is divided into two beams, U (upper) and L (lower),
by a 50-50 beam splitter BS (e.g., a half silvered mirror).
The dashed lines show the paths of the beams.
(The changes in direction are caused by reflection from mirrors
which are not shown.)

The "Polarization Rotator" can be set to rotate the polarization
of a photon in the lower beam by any desired angle.

The "Phase Shifter" changes the phase of the upper beam.
In quantum-mechanical notation it is generally taken to
multiply the wave-function by a complex exponential exp(i phi).
Physically, it can be accomplished by lengthening the upper
arm of the interferometer.

The polarizing beam splitter sends photons
with one polarization to Detector 1 and
those with the orthogonal polarization to Detector 2.
The detected polarizations don't have
to be "horizontal" and "vertical";
the polarizing beam splitter can be set to split the beam into
any desired orthogonal polarizations such
as +- 45 degrees to the horizontal.

In this setup, "interference fringes" are observed by varying the
phase shift in the upper arm. The "visibility" V := (M-m)/(M+m)
of the "fringes" is defined as before, where M and m represent the
maximum and minimum number of detections at Detector 1 per unit time.

Now comes the interesting part.
Assume that the polarizing beam splitter is set to output
"vertically" polarized photons to Detector 1 and "horizontally"
polarized photons to detector 2.
Consider various settings for the Polarization Rotator (PR)
in the lower arm. (Recall that the incoming photons are all
polarized "vertical".)

Case 1: PR rotates polarization by zero degrees
(i.e., does nothing).
So, the photons in the lower arm are still vertically
polarized, like those in the upper arm.

In this case, interference fringes with visibility 1
are observed, just as one would expect with a classical
light beam. This is because classically,
complete cancellation of the beams
could be obtained by shifting the phase by 180 degrees,
so that the minimum intensity m would be zero.

Case 2: PR rotates polarization by 90 degrees, changing the
initial vertical polarization in the lower arm
into horizontal.

In this case, one can determine the path of a detected
photon by its polarization. If detector 1 fires,
the photon took the upper path, and otherwise the lower.

i.e. M=m and V=0. The phase shifter in the upper arm
has NO effect on the detection rates.

Case 3: PR is set to rotate the lower arm polarization
to some intermediate value between 0 and 90 degrees.
Then one does see interference fringes, but their
visibility depends on the rotation. For a rotation
by D degrees, the observed visibility is V = cos(D).

Qualitatively, the fringe visibility seems to be inversely
related to the extent to which one can determine the photon's path
(i.e., through the upper or lower arm). If we can determine the path
surely as in Case 2, we see no fringes, i.e., V=0.
If either path is equally likely as in Case 1, we get maximum
visibility V=1. In an intermediate case like polarization rotation
through 85 degrees, when "horizonal" polarization is detected,
we can be almost sure that the photon took the lower arm,
and fringe visibility is small.

These are simplifications of experimental results reported in

P.D.D. Schwindt, P.G. Kwiat, and B.-G. Englert,
"Quantitative wave-particle duality and non-erasing quantum erasure",
Phys. Rev. A. (Sorry, I don't have the rest of the reference),
I'll call this paper "SKE".

The "quantitative" in the title refers to a definition of
"path knowledge" K which makes precise the qualitative expectation
that the better we can guess the path, the lower the fringe visibility.

This definition of "path knowledge" K is a little too involved
to give here in detail.
I'll just say that one determines (by calculation
or observation) the conditional probability of each path given
that detector i fires (i=1,2), and one "bets" on the more likely path.
The probability that one wins the bet (which can be computed or observed)
is called the "likelihood L", which is always between 1/2 (as in Case 1)
and 1 (as in Case 2). The "path knowledge" K is defined by

K := 2L-1, so 0 <= K <= 1.

An earlier paper of Englert,

B.-G. Englert, "Fringe visibility and which-way information",
Phys. Rev. Lett. 77 (1996), 2154-2157,

claims to establish the quantitative relation

(A) V^2 + K^2 <= 1,

with equality if and only if the input state to the interferometer
is a "pure" state.

The paper SKE just mentioned verifies this
experimentally, giving "*absolute* agreement at the percent level".
In particular, when the interferometer's input is a light beam so
dilute that on average only a single photon is in the interferometer
at any time, then theoretically and experimentally one has

(B) V^2 + K^2 = 1.

The relation (B) looks like an extremely fundamental relation,
something anyone interested in quantum optics ought to understand.
However, I cannot follow Englert's mathematics in detail.
The paper is vaguely written, and
the reader is left to guess at the meaning of key symbols.
None of the guesses which I have tried have produced a proof of (B).
All have led to contradictions.

I would very much like to hear from anyone who is familiar
with this paper, believes it correct, and can translate its
definitions into standard mathematical language.
The only mathematics in it is a little operator theory.
That was my field of mathematics before I became interested in physics,
so I should be able to follow this paper easily.

Following are a few closing remarks.

(1) The theoretical predictions for SKE's experiment can be obtained
in elementary ways without relying on Englert's paper
(though such analysis is not included in SKE).
The result is that for this experiment with a polarization
rotation of D degrees, V = cos(D) and K = sin (D),
so (B) is indeed predicted independently of Englert.

[This assumes SKE's definitions of K and V
in the context of their experiment. These
seem to me physically questionable,
but that's another issue.

My reservation about these definitions is that
in their determination of "knowledge" K, they
allow the detectors to detect polarization
in an *arbitrary* basis, but their calculation
of "visibility" V allows the detectors
to detect only "horizontal" and "vertical" polarizations.
This seems arbitary and inconsistent.
Englert's paper makes no such assumption,
so far as I can tell.]

(2) I'd also like to hear from anyone familiar with the SKE paper
and interested in discussing it.

I suspect that their Figure 4(a) for the polarization rotation
angle of 20 degrees may be badly incorrect. In particular,
it seems physically unreasonable
(and contrary to my calculations)
to suppose that for a polarization detection angle of zero degrees
(i.e. the detectors detect "horizontal" and "vertical" photons),
the fringe visibility vanishes, as the Figure indicates.
The only case in which fringes should vanish is when
one can predict the photon's path with certainty,
which only occurs with a 90 degree polarization rotation
(Case 2 above).

Also, Fig. 4(a) appears to contradict Figure 2,
which presents the same data point
(polarization rotation of 20 degrees, polarization analysis
angle of 0) in different format. Figure 2 shows a visiblity
of approximately V = 0.95 ~ cos(20) for this situation.

Perhaps the questionable curve in the Figure was accidentally
displaced. Has anyone seen an errata presenting the correct curve?

(3) Another relevant paper is

G. Bjork and Anders Karlsson, "Complementarity and
quantum erasure in welcher weg experiments",
Phys. Rev. A 58 (1998), 3477-3483,

which claims to generalize Englert's results.
Unfortunately, this is also vaguely written,
and it relies essentially on a part of Englert's
analysis which I've been unable to verify.

(4) The inverse relation (B) between fringe visibility and path
knowledge may look superficially like the uncertainty principle
of quantum mechanics, but Englert correctly points out that
they are unrelated.
If correct, (B) is a new physical principle.
That's why it's so important to understand its proof.

Stephen Parrott
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