pivoxa15 said:
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:
<br />
\Delta H \approx \mu_{\textrm{Bohr}}B_0 \sigma_z<br />
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
<br />
\mu_{\textrm{Bohr}} \frac{d B_0}{dz}<br />
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
<br />
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}<br />
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.
