Wavefunction in external electromagnetic fields

Question

If we consider a spin in an external magnetic field, it starts to precess around the vector of the magnetic field. The same should be true for any angular momentum. But what does this mean for the wavefunction of the electron? To be more precise: What happens to the wavefunction if we apply an external magnetic field? Is there an easy way to picture it? If I assume a 2P3/2 state (hence L=1, s=1/2 and j=3/2) and an external magnetic field, then if we just considered this angular momentum we would see precession around B. But does this also mean that the orbital is going to precess?

If so, what would be the consequence? Would this mean that the electric dipole would also be rotating (for asymmetric orbitals) since the probability distribution would be rotating as well?

As requested, more information: I am talking about atoms. When I ask about what happens to the wavefunction, I mostly care about what any change of the wave function means for the electric dipole since the dipole moment is given by $<d>= -e <r>$. Probably one can even answer this question, when spin is completely neglected and only the evolution of $L$ in a magnetic field is considered, but I am not sure.

Answer 1

The principled way to answer this question is to write down the Schrödinger equation for a particle in magnetic field and try to solve it. From the context of the question it seems that you are talking about an electron with spin-1/2 - the case that is treated extensively in textbooks, including its motion in magnetic field. However, it is not clear whether you are talking about a free electron, an electron bound in an atom or some other situations. I will therefore suggest a few directions to look into:

• Free electron in magnetic field. In this case the orbital motion is described in terms of the Landau levels. Let me note in this context that the orbital angular momentum is a good quantum number only in the case of spherical symmetry. Once we apply a magnetic field this symmetry is broken, and the angular momentum is not a good quantum number anymore, except for its projection on the magnetic field diretcion - the states with definite angular momentum get mixed with time-dependent coefficients, which is what we mean by precession.Landau levels could be solved, in principle, in terms of angular momentum - this is an interesting exercise. But yes, the orbital states significantly change in this case.
• Electron bound in atom Atoms have spherical symmetry and therefore it is natural to treat them in terms of angular momentum even when this symmetry is broken. The problem of an electron in a central-symmetric potential with a in magnetic field is solved in many textbooks. In this case description in terms of precession of the full magnetic momentum indeed comes in handy, as it accounts fully for the changes to the orbital motion in the magnetic field.
• Electrons in solids In this case the band structure has to be solved with magnetic field included in the Hamiltonian. This turns out to be hard, since the translational symmetry becomes dependent on the gauge choice for the magnetic field. For realistic magnetic fields one can however often limit oneself to Peierls substitution - plugging the magnetic field into the tight-binding Hamiltonian as if it were an exact one. It results in a number of interesting effects, grounded in changes of orbital behavior, such as Shubnikov-de Haas oscillations, integer quantum Hall effect, and others. (Fractional quantum Hall yet is another beast, which is usually not treated in terms of a Hamiltonian.)

Answer 2

For a spin in an not excessively strong external magnetic field we can ignore the spatial part of the wave function and use an effective spin Hamiltonian. For a single electron spin this is the $$H=g\mu_B \vec B \cdot \vec S$$, where g=-2. The solutions are simply $|+\rangle$ and $|-\rangle$, with eigenvalues $\pm \frac{1}{2} g\mu_B$. Whether these wave functions represent precession around the direction of $\vec B$ as for a classical dipole is at present just a convenient interpretation, as we have no kinematic model of electron spin.

It does not stop here. Because of radiative effects for a free electron, g=-2.002 319 304 362 56(35). In an atom without angular momentum nuclear spin the g-value will have slightly different values because of higher order angular momentum effects. If the atom has angular momentum, which has g=+1, then the combined g-value will be very different from -2. The description applies to any fermion, such as the proton with g=+5.5856946893(16). For neutral atomic hydrogen the hamiltonian gets additional terms: $$H=g\mu_B \vec B \cdot \vec S +g\mu_N \vec B \cdot \vec I + A \vec S \cdot \vec I$$ where $\mu_N$ is the nuclear magneton. The thrird term describes proton-electron momentum hyperfine interaction. For sufficiently high spins higher order spin hamiltonian terms are possible, such as quadrupole interaction. See also https://en.wikipedia.org/wiki/Zeeman_effect. Related subjects are MRI and 21 cm astronomy.