Calculating the Stern Gerlach Angle in Magnetic Fields

[pivoxa15]

How can you calculate the angle of the split of the electrons after passing it thorugh a non uniform magnetic field?

I know how to calculate it by elementary means using the magnitude of S and Sz and doing trig. But is that angle always the same no matter what intensity magnetic field its passed through?

 

[olgranpappy]

How can you calculate the angle of the split of the electrons after passing it thorugh a non uniform magnetic field?

the actual magnitude of the "split" depends on the details of the apparatus and incoming beam.

 

[pivoxa15]

Does it depend on the magnetic quantum number by any chance? It definitely depends on the spin quantum number.

 

[olgranpappy]

Does it depend on the magnetic quantum number by any chance? It definitely depends on the spin quantum number.

Yeah, but often one considers atoms that have $$\ell=0$$, for example Ag atoms (as described in section 1.1 of Sakurai "Modern Quantum Mechanics"), because this type of atom makes for simpler examples.

Silver can be thought of as having a full d-band and so there is only one "valence" electron in the 5s state. Thus the total orbital angular momentum is zero, but the total spin angular momentum is \hbar/2 (and, of course, the total angular momentum is \hbar/2) and we consider the energy perturbation to be:

$$\Delta H \approx \mu_{\textrm{Bohr}}B_0 \sigma_z$$

where $$\mu_{\textrm{Bohr}}$$ is the Bohr Magneton and B_0 is the external field in the z-direction and \sigma_z is the Pauli matrix. So the atoms shooting out of an "oven" into the apparatus feel a force (only in the region where B_0 is changing--the "fringing" part) due to the changing B_0 field of either plus or minus
$$\mu_{\textrm{Bohr}} \frac{d B_0}{dz}$$
since the spin is quantized. And thus there appear two "spots" on the detecting screen.

I believe that you can figure out the approximate angular distance between the spots using
$$2\theta \approx 2\frac{\delta p}{p} = 2\frac{\int F dt}{p} \approx 2\frac{B_0\mu_{\textrm{Bohr}}/v}{mv}<br /> =\frac{B_0 \mu_{\textrm{Bohr}}}{E_0}$$
where E is the incident energy of the atom. The above is quite approximate indeed and should only hold for \mu B_0 << E_0.