Spin State Energy Levels

Question

When a spin-1/2 particle is placed in a magnetic field that is strong enough and varies slowly enough in space and time, it will become polarized and its spin will either align or anti-align with the magnetic field direction. Let the magnetic field be denoted by $\mathbf{B}$ and the spin by $\mathbf{s}$ . Then, under this condition, the potential energy becomes

$$ U = \pm\gamma|\mathbf{s}||\mathbf{B}| $$

where $\gamma$ is the gyromagnetic ratio (which can be negative) and the upper and lower signs correspond to the spin-up and spin-down states, respectively.

I believe all of this is standard quantum mechanics. When the magnetic field exhibits a spatial gradient (which isn't too large), then the field will induce a force on the spin-1/2 particle. This is the basis of the Stern-Gerlach experiment. And, in particular, the SG experiment indicates that the spin-up and spin-down states are (at least roughly) equally likely when the particle source is initially unpolarized, or polarized in a direction perpendicular to $\mathbf{B}$ as observed in sequential SG experiments.

However, based on the potential energy function given above, it is clear that the anti-aligned state has a lower energy for a given value of $|\mathbf{B}|$ (assuming $\gamma > 0$). So should I be surprised that two states with different energy levels should appear to be equally likely? Is there any preference for the aligned state to eventually transition to the lower energy anti-aligned state within a constant magnetic field? I've never seen this issue addressed in QM textbooks or elsewhere.

Answer 1

The energy difference along with the larger thermodynamical likelihood for occupation of the lower level is real. There is an application, nuclear magnetic resonance (NMR) spectroscopy/imagining/quantum computing. But due to the very small energy difference for technically achievable magnetic fields, the effect is usually negligible at roomtemperature. NMR makes do with the tiny imballance present at room temperature.

Answer 2

If the particle source is "unpolarized", that literally means it is equally likely to find particles from this source in either energy eigenstate - that's the definition of "unpolarized", so you shouldn't be surprised about that.

When the spin of a particle is "perpendicular to the magnetic field", that's another way of saying that the particle is in an equal superposition of the two energy eigenstates, that is,

$$\lvert \psi \rangle = a\lvert \uparrow \rangle + b\lvert \downarrow \rangle$$

where $|a| = |b|$. This also means that the probability of finding the particle in either state is equally likely:

$$\langle \uparrow \rvert \lvert \psi \rangle = |a|^2 = |b|^2$$ $$\langle \downarrow \rvert \lvert \psi \rangle = |b|^2 = |a|^2$$

You are however correct that, in thermal equilibrium, under the influence of a magnetic field, you should expect more of the lower-energy eigenstate than of the higher energy, the relative probabilities being proportional as follows: $$p(\pm E) = {\rm exp}{\frac{\mp \gamma |s| |B|}{k T}}$$ But that is not the case during the SG experiment.