Relation between Zeeman effect and Larmor precession

I'm wondering if there's some intrinsic reason that the precessional frequency, $\omega$, of a spin about an external magnetic field matches the frequency of a photon emitted (or absorbed) in transitioning between the levels.

I'm trying to understand the magnetic resonance in MRI, so I'm looking at spin $\frac{1}{2}$ protons that split into two levels in an external field with Zeeman splitting given by $\Delta E = \gamma \hbar B$ and so a photon of energy given by $E = \hbar \omega$ will cause a resonant transition between the levels, giving the resonant frequency $\omega = \gamma B$. The main thing I'm wondering is why this $\omega$ happens to match the Larmor frequency of precession. Is it just coincidental or is there some quantum mechanical reason underlying this?

• In fact, Larmor derived his expression $\omega=\gamma B$ trying to explain the Zeeman effect! This means it is not a coincidence at all: Larmor managed to explain the Zeeman effect in the context of classical mechanics (way before the birth of quantum mechanics). See, for example, this post of mine in History of Science and Mathematics: How did gyromagnetic ratio come up before quantum mechanics, and who introduced it? Feb 27, 2016 at 11:39

Consider an electron bounded to an hydrogen like atom and take a strong external field $$B$$ along the $$\hat{z}$$ direction, it has orbital magnetic moment $$\vec{M_{L}}$$ and intrinsic magnetic moment $$\vec{M_{S}}$$ given by

$$\vec{M_{L}}=\frac{\mu_{B}\vec{L}}{\hbar} \hspace{1.5 cm} \vec{M_{S}}=g\frac{\mu_{B}\vec{S}}{\hbar}$$

and the hamiltonian will be given by

$$H=-\frac{\hbar^2}{2m}\nabla^2-\frac{Ze^2}{4\pi\epsilon_{0}r}+\frac{\mu_{B}B}{\hbar}(L_{z}+gS_{z})$$

the determination of the gyromagnetic factor $$g$$ is crucial since the splitting of the energy levels strictly depends on it:

$$E=E_{n}+\mu_{B}B(m_{l}+gm_{s})$$

In order to determine the $$g$$ factor we can for example make a Stern-Gerlach experiment along $$\vec{z}$$ direction with Ag atoms prepared in a $$|S_{x}\rangle$$ (or also $$|S_{y}\rangle$$). If you put another Stern-Gerlach device with magnetic field along $$\hat{x}$$ (or $$\hat{y}$$) at the end of the one along $$\vec{z}$$ direction, you can take note of the number of electrons found in one of these two states $$|S_{x}\rangle$$ state and $$|S_{y}\rangle$$ state. At a given time $$t$$ the probabilities are

$$P(|S_{y}\rangle)=\cos^2\left(\frac{\omega_L t}{2}\right) \hspace{1.5 cm} P(|S_{y}\rangle)=\sin^2\left(\frac{\omega_L t}{2}\right)$$

with $$\omega_{L}$$ (Larmor frequency) given by $$g \frac{\mu_{B}}{\hbar} B$$. You can make a fit for the frequency and determine $$g$$. By this type of experiment we know that $$g=2$$.

If we now return to the problem of the electron in a strong field and we consider the selection rules for dipole transition ($$\Delta m_{l}=0,-1,+1$$ and $$\Delta m_{s}=0$$), we note that the spectral line corresponding to a transition from n to n' energy levels is split into three components, each of those separated by the Larmor frequency, the same frequency we found in the spin's precession.