Re: Question On Down Conversion.

["John S. Denker" <jsd@MONMOUTH.COM>]

Robert B Zannelli wrote:
> ... parametric down conversion of photons as
> they pass through an optically nonlinear crystal such as lithium iodate or
> barium borate. Here a single photon of frequency f_0 down converts to two
> photons in an entangled state whose frequencies f_1+f_2=f_0.
...
> how are the spin states for these photons correlated?

Oooh, what a spiffy question.

> I suspect that their polarization states should be orthogonal.

I see no physical basis for suspecting that.

1) In some sense the physics suggests that the symmetry of
what comes out should match the symmetry of what goes in,
but even then you have to be careful because there can be
broken  symmetries, as exemplified by a coin-toss:  the
probability distribution of outcomes is symmetric, but any
particular toss results in a non-symmetric outcome.

2) As usual, it pays to think about the physical principle
that produces the effect in question.  The lithium iodate
(or niobate) crystal changes its index, locally, in response
to the local E-field-squared.  So if you have an incident
wave of wavelength lambda !!and!! an applied static E-field,
you can create a grating with spacing lambda.  (Without the
static E-field, the grating spacing would be lambda/2.)

Roughly, the grating can be considered a _standing wave_ of
phonons, since a phonon is just a wavelike distortion of the
lattice, and you can imagine a grating of less-dense crystal
alternating with more-dense crystal.  (Actually it's not quite
plain old density, so phonon isn't exactly the right word, but
it evokes roughly the right idea.)

So, think about the process as involving _four_ waves:  the
incident photon, an incident phonon (from the grating), and
the two emergent photons.  AFAIK this is the only way to
make sense of the phenomenon.  For one thing, this predicts
(correctly) that momentum is transferred to the crystal. Also,
it predicts (correctly) that the outcome depends on the state
of motion of the crystal.  In contrast, if you were to think
of it as just a three-photon reaction (one high-energy photon
turning into two lower-energy photons) you would predict
(incorrectly) that the process would be relativistically
invariant.

For simplicity let's analyze this in the frame where the
crystal is at rest.

Since the applied static E-field is linearly polarized,
the only part of the incident beam that produces a grating
is the linearly-polarized part.  So it's simplest to analyze
the incident photon and the incident phonon in the linear
polarization basis.  The two incident objects have equal-and-
opposite momentum (hbar k and negative hbar k) for a total
of zero, and also zero total angular momentum.  Therefore
the two outgoing photons have zero total linear momentum
and zero total angular momentum.

It is amusing to analyze the emergent photons in the circular
polarization basis.  Each of them will have one unit of
angular momentum -- but by conservation we know that the
momenta will be equal and opposite, for a total of zero.

So this is not orthogonal at all.  Orthogonal wavefunctions
exhibit zero correlation.  These wavefunctions are 100%
correlated.  If you catch one of the photons, you can't
predict whether its angular momentum will be +1 or -1, but
you know that if you catch the other one it will be opposite
to the first.

BTW, expressing this result in terms of RCP and LCP is a
bit tricky, because it is conventional to apply the right
hand rule relative to the direction of propagation of the
photon, and the two photons are propagating in opposite
directions.  So the result is sometimes stated that 50%
of the time both are RCP and 50% of the time both are LCP,
but this tends to obscure the central physics, namely that
they have equal-and-opposite angular momentum.  Things
are much simpler when expressed in terms of the angular
momentum bivector.

Help stamp out cross products!  Help stamp out the right
hand rule!

For homework, to test your understanding of the physics,
analyze the following situation:  polarize the crystal
with a strong beam of one wavelength, then turn that off
and quickly (before the crystal has time to readjust)
send in a beam of a different wavelength.

For a discussion of tangentially-related symmetry issues, see
  http://www.monmouth.com/~jsd/physics/pierre-puzzle.htm

This posting is the position of the writer, not that of Edward
Leedskalnin, William Beckford, Sarah Winchester, or Edward James.

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.