Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Typically when I think of a spin moment in an external field, I visualize it classically in terms of a magnetic moment precessing around the external field vector.

Now how is this described quantum mechanically? My guess would be that the external field causes transitions between the up-state and down-state at the Larmor frequency. However if the Zeeman energy was much greater than the thermal energy, wouldn't the spin need to stay in the low energy state? Does that mean the spin ceases precession in the classical picture?

share|cite|improve this question

Larmor precession is quantum mechanically described by the interaction of the spinor with the magnetic field. Without magnetic field the spin-up eigenstates states in the x and y direction are given by:

$S_x^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ 1\end{array}\right) \exp(-iEt)$

$S_y^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}1 \\ i\end{array}\right)\exp(-iEt)$

The direction of the spin in the xy-plane is given by the angle between the two spinor components in the complex plane.

The two components get an extra energy factor from the magnetic field B but with a different sign:

$S^\uparrow ~~=~~ \frac{1}{\sqrt{2}}\left(\begin{array}{r}\exp(-iE_Bt) \\ \exp(+iE_Bt)\end{array}\right)\exp(-iEt)$

The result is that the spin direction rotates (precesses) in time.

Regards, Hans

share|cite|improve this answer

Normally this is best explained by NMR. U have a population of spins in an thermal equity. How the energy states of this ensemble are populated is defined by Maxwell-Boltzmann statistics. In case of deuterium atmoms, the nucleus of a single deuterium atom is either in spin down or spin up state, ratio of spin up to down is set by MB statistics. There is no vanishing of precession, the spin is a fundamental property like mass, it doesnt cease! This would mean zero energy, in QM u have the ground state energy != 0. And the energy states are discrete as the spin is discrete! It can add to total spin/angular momentum = 0, but this is another topic when looking on several electron atoms and molecules. Actually there is no classical match to QM Spin. The correspondence principle doesnt work here like for impulse operator/observable.

Back to NMR: if u put the deuterium ensemble into an external rf field with rf Larmor-frequency of the deuterium atom, u excite the ensemble out of thermal equity and relax back after a while. Look at NMR Spectroscopy, there is especially the need to implement very high magnetic fields (Zeeman-Energy) to trick out the Boltzmann-Statistics and yield a good signal-noise ratio.

share|cite|improve this answer

It is a bit confusing in the wave-function (spinor) picture. However one can just as easily work with the density matrix. This is the outer product of the wave function with itself. For a spin-1/2 (two component) wave-function, the pure part of the density matrix is a spin-1 object, a little vector pointing in some direction. This vector precesses exactly as in the classical case. (The non-pure part is the spin-0 irreducible rep, totally invariant under rotations. This shouldn't be too surprising -- if you don't know what direction it is, rotating it shouldn't change anything.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.