# Simple Stern-Gerlach question

1. Jul 12, 2011

I really did use the search function to try and figure this out for myself before posting , but this has been bugging me for approximately two days.

I've been trying to understand the Stern-Gerlach experiment, primarily from hyperphysics ( http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c6 ).

I understand all of the math presented except the very first step, which reads:

"The potential energy of the electron spin magnetic moment in a magnetic field applied in the z-direction is given by:

$U = -\mu \cdot B = -\mu_{B}\frac{g}{2}B_{z} = \pm\mu_{B}B_{z}$"

How does the dot product go to a regular scalar product? That seems to make the assumption that all of the magnetic moments are facing in the $\hat{z}$ direction. If you're just shooting silver atoms out of a furnace, I would expect those moments to be randomly oriented in space (but, of course, have the same magnitude).

What if the magnetic field inhomogeneity was in the $\hat{y}$ direction instead? Would the beam not split? That's the kind of absurdity that I can't make sense of in the linked article.

2. Jul 12, 2011

### Jacques_L

Your questions are the good ones...
Spin is a relationnal "thing" or property : it can be "all pro", or "all con", and nothing else.

Now you have to use Feynman diagrams of first order : two apex, one complete edge between the apex, and four half-edges.

First use (just hypothetical, however) : between the furnace and the big magnets.
The terminal apex is the magnetic reaction between the field and the spin of the atom.
The initial apex is the last collision of the silver atom in the furnace.
And we cannot tell anything experimental about the spin state during this first journey. Maybe the first journey cannot be distinguished from the second journey, where the magnets have imposed to the spin to chose "all pro", or "all con" the strong field.

Second use, always valid : after the magnets (and maybe before, from the furnace), and up to an eventual set of Stern&Gerlach magnets.
In the original Stern & Gerlach experiment, the final apex is the impact on the sensor, which dictates nothing more to the spin.
In sum : only the magnets have dictated the direction of the spin at least from the beginning of the flight between the two polar pieces.

3. Jul 12, 2011

### Staff: Mentor

What he's doing is defining the z-direction for each atom to be in the direction of that atom's magnetic moment. Each atom has its z-axis in a different direction.

This is different from what most books do. I just checked three "intro modern physics" books in my office. Ohanian and Tipler both define the z-direction as the direction of the magnetic field so

$$U = - \vec \mu \cdot \vec B = -\mu_z B$$

Beiser doesn't define a z-axis at all but instead uses magnitudes and angle:

$$U = - \vec \mu \cdot \vec B = -\mu B \cos \theta$$

4. Jul 12, 2011

### Bill_K

FirstYearGrad, The B field points in the z-direction, so naturally μ·B = μzBz. Why would you expect anything else??
The idea that spins can be "randomly oriented" is a classical idea. In quantum mechanics, given any spin system and any axis, the spin will be measured to have a component along that axis, and that is all! The spin projection along the axis will be random, either 'up' or 'down' in the case of spin ½. In the case you're describing, μz = gμB/2, and g = 2, so U = ±μBBz.

5. Jul 13, 2011

To Bill_K, it's not so much that I didn't expect $\mu \cdot B = \mu_{z}B_{z}$, it's that I didn't expect $\mu_{z} = \mu = \mu_{B}$. It seems like it's just something I need to struggle with for awhile.
Anyway, here is what I have gathered from the three responses above: somehow, regardless of the perspective from which you measure it, the spin angular momentum always measures out to be $\pm \hbar /2$, is that correct?