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?


4 Answers 4


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


How is Larmor precession described in QM?

Let's say you have a spin ½ particle (a proton). Larmor precession means that if you intend to measure its angular momentum in a direction perpendicular to an applied magnetic field, the expectation value will oscillate periodically.

The behaviour of spin is derived from the commutator relations of the spin angular momentum measurement operators, e.g. $[S_y, S_z]=i\hbar S_x$ etc. (These could be motivated by the non-commuting of rotations in geometry, or just by analogy with orbital angular momentum.)

Choosing any axis, the proton's spin has two eigenstates $ |{↑}⟩$ and $|{↓}⟩$. It turns out that every possible quantum state of the spin (and every measurement operator) can be represented in the basis of these two eigenvectors. For example, $S_z=\frac\hbar2(|{↑}⟩⟨{↑}|-|{↓}⟩⟨{↓}|)$ and $S_x=\frac\hbar2(|{↑}⟩⟨{↓}|+|{↓}⟩⟨{↑}|)$.

If it is subjected to a static magnetic field $B$ (say along the $z$ axis), there will be Zeeman splitting such that the eigenstates for that axis have different energy levels (attributable to the torque on the particle's intrinsic magnetic dipole moment $\gamma S$). The time-independent Schrödinger equation then gives the phase evolution of each eigenstate. Consider the state $\vec\chi(0) = \cos\frac\theta2\ \ |{↑}⟩\ + e^{i\phi}\sin\frac\theta2\ \ |{↓}⟩$. This state will evolve in time, becoming:

$$\vec\chi(t) = \cos\frac\theta2\ \exp\frac{iB\gamma t}2\ |{↑}⟩\ + e^{i\phi}\sin\frac\theta2\ \exp\frac{-iB\gamma t}2\ |{↓}⟩$$

Thus, the expectation value $⟨S_z⟩=\frac\hbar2\cos\theta$ is time-invariant, whereas other expectation values such as $⟨S_x⟩=\frac\hbar2\sin\theta\cos(\phi-B\gamma t)$ will oscillate with a frequency proportional to the external applied field. (Note, $\vec\chi(0)$ was the $+\frac\hbar2$ eigenvector for a measurement along an axis with inclination $\theta$ and azimuth $\phi$.)

So if a proton is in the $|{↑}⟩$ or $|{↓}⟩$ eigenstate it will not precess. However, if it was prepared in any other state (for example, by first measuring its spin along the $x$ axis, or alternatively after nutating it with some perturbation) then it will precess at the Larmor frequency, meaning that the outcome of a later measurement along any transverse axis will depend on how long you waited before performing the measurement. (For example, if you measure along the $x$ axis twice, separated by a delay of an integer number of Larmor periods, you will get the same result both times. If the delay is half a period, you will get the opposite measurement result. If the delay is a quarter period then you could get either result because at that moment the spin along that axis was uncertain.)

[See Griffiths Introduction to Quantum Mechanics for more detail.]

What happens to precession in the classical limit?

Classically, a spinning object with electric charge, if situated in a uniform magnetic field, will precess with the Larmor frequency. (The external field exerts a torque on the object's magnetic dipole moment, and the object's angular momentum results in a gyroscopic effect.)

$$\gamma\vec L \times \hat zB = \frac{d\vec L}{dt} \implies \vec L = a\cos B\gamma t\ \hat x + a\sin B\gamma t\ \hat y + b\hat z$$

The basis for MRI (an imaging technology commonly used in modern hospitals) is that a parcel of tissue has a bulk magnetic moment which precesses at the Larmor frequency. This rotating magnetic field induces a periodic current in a detector coil.

Note this type of macroscopic measurement will be too weak of an interaction to collapse individual atoms into specific quantum states. Thus, the observed bulk magnetisation will not be quantised.

How does temperature affect this?

If the particle interacts with a thermal bath then there is typically (for common temperatures and field strengths) ample energy to excite the particle between spin energy levels. Generally, magnetic susceptibility reduces with increasing temperature, as microscopic alignment becomes more chaotic. This will contribute to the relaxation processes in NMR/MRI.


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.


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.)


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