Friday, December 23, 2011

The Quantization of Spin Revisited

I started this series on the Stern Gerlach experiment two weeks ago because I wanted to make a certain point about just where and how the wave function supposedly "collapses". Stern Gerlach is a nice discussion topic because people say the particle jumps into one of two possible spin states...but just where does it jump? When it passes through the magnetic field, or when it hits the screen?

Then I got sidetracked by a fascinating Master's Thesis by a guy named Jared Stenson, who argues that we should really look at what the silver atoms do when they pass through a pure quadrupole field, instead of the traditional Stern Gerlach. Stenson claims, with some justification, that the beam of atoms then spreads into a donut. This is very different from the traditional Stern Gerlach where the beam simply splits in two.

Stenson did his analysis for an unpolarized beam, and last week I refined his analysis for the case of the polarized beam. I got a beautiful pattern of spin distributed about the donut. For an initial spin-up beam configuarion, this is the pattern at the detector screen:

It's very different from the pattern we get for the traditional Stern Gerlach setup...OR IS IT????

Stenson points out that the original experiment was done, not with a pencil-shaped beam, but with a fan-shaped source. There never were two disrete dots representing "spin-up" and "spin-down". There were in fact two bands, representing the splitting of the fan-shaped beam. No, that's not eve right: the two bands came together at the ends, making an elongated ellipse. And isn't an elongated ellipse just another name for an elongated circle?

What if Stern and Gerlach had used a pencil beam source instead of a fan-shaped beam? Would they have gotten Stenson's donut? Well, let's see: Stenson calculates his donut for the case of the pure quadrupole field, and Stern-Gerlach has a sharp magnetic pole in proximity with a wide flat pole. If we calculate the Stern-Gerlach field, to the second order approximation we get a constant field plus...a pure quadrupole.

And what is the effect of the constant field on the beam of silver atoms? AB-SO-LUTE-LY nothing! Ok, to be fair, if the spin is lined up cross-wise to the field, there should be precession; but because the field is constant, any precession is also constant across the beam profile; so when it leaves the field, there is no shifting of phases which would tend to make the beam curve or bend. It just keeps going straight.

Any diffraction or bending of the beam can only come from the quadrupole component; and that's what we've calculated. The silver atoms don't suddenly jump into one of two possible states at all...they spread themselves around the axis of the beam in a symmetric pattern, showing all possible valuse of spin orientation. In the picture, I've shown the result for a polarized beam, where the circular distribution is skewed in the direction of the spin polarization. For the unpolarized beam, you get the uniform circular donut pattern.

Yes, there is a way you can set up the experiment so the spin appears to jump into one of two possible values: instead of using a pencil-shaped beam, you spread the beam out into a fan-shape. Then you get diffraction effects and interference so that the deposition pattern on the screen spreads out into an elongated ellipse, which, ignoring the endpoints, looks like two discreet bands. But that's not the Stern-Gerlach experiment the way the whole world talks about it. The real story is right here on this blog, and I just wonder if you can find this analysis ANYWHERE else.

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