I spent the last week putting together all the analytical
machinery for this beautiful problem of the polarized beam in a quadrupole
field, and today I’m going to solve it. Ironically, I’m not going to use any of
my intiricate machinery after all. I’m going to solve the problem by guessing
the answer, and then showing that it’s the only answer that makes sense. It’s
another way of doing things.

As you will recall this is a modified Stern Gerlach
experiment proposed by a grad student at Brigham Young by the name of Stenson.
He wants us to analyze the splitting of the beam not in the traditional Stern
Gerlach filter, but in a pure quadropole magnetic field. It’s a really good
problem, and Stenson gives us the answer for the case of the unpolarized beam.
Now I’m going to give the answer for a polarized beam.
Stenson starts off by observing that the solution for the
ordinary case is almost universally described as the beam splitting into two
paths, so you get two dots of silver on the screen. He points out that this is
wrong: that in the original experiment, the beam was fan-shaped, and the
deposit was in the shape of an ellipse. He goes on to doubt whether you would
even get the famous two-dot result if you used a pencil-shaped beam, and I
agree with him.

He then presented the solution for the pencil-shaped beam
going through a quadrupole field, and he shows it as a donut pattern. This
makes physical sense. But he only does the case of the randomly polarized beam.
What about the polarized beam?
Of course now it’s complicated because the polarization can
be in any direction with respect to the quadrupole. But then we get back to the
old idea of basis states: if you can analyze it for the two orthogonal basis
state polarizations, namely spin-up and spin-down, you should get all other
cases as superpositions of these.

Furthermore, there is a four-fold symmetry to the field. If
you solve the spin-up case, then rotate 90 degrees, you are once again aligned
in the same way (except the relative spins are opposite) so your solution
should have the same form. Ditto for 180 and 270 degrees. Four-fold symmetry.
Finally, whatever solution you find, when you turn it upside
down and add your two solutions together, you have to get the uniform donut
pattern shown by Stenson. Things have to add up.

Putting all these conditions together, lets take a wild
guess at the solution. That’s what I did, and I came up with the theory that
the intensity of the pattern has to vary sinusoidally about the periphery of
the ring. I’ve drawn it here:
The nice thing about this solution is that it approximates
the traditional Stern-Gerlach to the extent that the spin-up particle is drawn
(more or less) towards the upwards-pointing field; and at the same time,to the
extent that the field points sideways relative to the particle, the trajectory
splits left and right. It’s a bit like you had two traditional S-G’s one after
another.

Furthermore, the solution is nice because if you then take a
spin-down particle, you should get the opposite pattern. If you combine these
two patterns as you would in the case of a random beam, you just add the
intensities together, and the result is a uniform density distribution about
the ring. So we’ve replicated Stenson’s solution.
But there is one rather nasty problem with this
distribution. It’s a density function, which is of course the square of the
wave function, and you really haven’t solved the problem until you’ve
determined the wave function itself. The obvious thing is to take the square
root of the density function, but then you get cosine of the half-angle. The
problem is when you come around full circle, you’ve reveresed polarity. The two
branches of the function don’t line up.

What I figured out is you can add a *phase*to the cosine so that it comes full circle: you just multiply it by an exponential, which preserves the amplitude but reverses the sign after one full revolution. The function looks like this:

It turns out this function can be rewritten in terms of sums
instead of products, and it is actually quite simple:

There’s a factor of ½ which I haven’t bothered to keep track of,
but no matter. You can see that the amplitude of the function peaks at the top
of the circle, goes to zero as you move around the circle clockwise, and comes
back up to the peak when you complete the circle.

I have to admit I thought I had nailed it with this, and then I
noticed that it doesn’t quite do everything I think it ought to do. The beam
enters the magnetic field, and it is spread out into a donut shape: but you
look at the silver atoms as they land, they are still all pointing upwards, and
I don’t think that can be right. I the atoms to be trying to line themselves up
with the quadrupole field as they pass through it, so they land on the screen
pointing all different ways. Incredibly, I can make them do this without too
much trouble. We just have to remember two things: first, the formula for breaking
up exponentials into sines and cosines; and secondly, that the function psi is
really a spinor function and has to have two components. Let’s just try
assigning the real part to the up component and the imaginary part to the down
component and see if it works:

A careful inspection shows this function preserves all the
properties we wanted: it peaks at top dead center, goes to zero at the bottom,
and the intensity (square of the amplitude) varies as cosine-squared. And oh yes,
the endopoints match up when you come full circle.

What’s more, we get an interesting distribution of spin
orientations as we go around the circle. I’ve sketched them out here:If you remember what the quadrupole field looked like (I posted the sketch about a week ago!) you can see that whatever the orientation of the local field, the silver atoms try to line themselves up parallel to minimize their energy. For example when theta is 90 degrees, for instane, the spinor evaluates (1,

*i) which is pointing sideways. Which is what the quadrupole field is doing for theta = 90. So it’s a physically satisfying picture.*

The big challenge is yet to come: first, we have to write out the wave function for a beam that enters the magnets with spin down. And then we have to examine combinations of those two wave functions, representing beams that enter the field with arbitrary spin orientations, e.g. spin sideways, and we have to see if the function does things that make sense. In particular, keeping in mind the four-fold symmetry of the field, we have to see if a beam entering with sideways spin polarization does the same thing as the form we’ve already calculated, only rotated by 90 degrees.

Hold on to your pants.

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