Spin precession · Quantum Mechanics

Intuition

Put a spin in a magnetic field along $z$ . Its energy depends only on $\hat{S}_{z}$ , so the probabilities of up and down along $z$ never change — but the transverse components do: the average spin vector turns about the field at a steady rate, the Larmor frequency, like a gyroscope under gravity. For an electron in one tesla that is about 28 GHz; for a proton about 43 MHz. Magnetic resonance, from chemistry labs to hospital scanners, listens to this precession.

A spinning top tilted in gravity does not fall over but swings its axis round in a circle. A spin in a magnetic field does the same with its average direction, about the direction of the field.

A spin starting as $\ket{+x}$ in a field along $z$ , with $\hat{H}=\omega\hat{S}_{z}$ , over two periods of : and , drawn in units of . Meanwhile stays zero: the average spin turns in the plane at right angles to the field.

Larmor precession

A spin with magnetic moment $\vec{\mu}=\gamma\hat{\mathbf{S}}$ in a field $B$ along has . Writing , the average spin turns about at the rate .

\begin{aligned}\hat{H}&=\omega\hat{S}_{z},\qquad \omega=-\gamma B\\ \frac{d}{dt}\langle\hat{\mathbf{S}}\rangle&=\vec{\omega}\times\langle\hat{\mathbf{S}}\rangle,\qquad \vec{\omega}=\omega\hat{\mathbf{z}}\end{aligned}

Properties

$\langle S_{z}\rangle$ is constant, since $\hat{S}_{z}$ commutes with ; and turn at the rate .

The spin precesses about the field

\hat{H}=\omega\hat{S}_{z}\ \Longrightarrow\ \frac{d\hat{S}_{x}}{dt}=-\omega\hat{S}_{y},\quad \frac{d\hat{S}_{y}}{dt}=\omega\hat{S}_{x}

Heisenberg’s equation needs the commutators of $\hat{S}_{z}$ with the transverse components, which the angular momentum algebra supplies. The two equations are those of a vector turning about $z$ , solved by a cosine and a sine.

Proof steps

\frac{d\hat{S}_{x}}{dt}=\frac{i}{\hbar}[\hat{H},\hat{S}_{x}]=\frac{i\omega}{\hbar}[\hat{S}_{z},\hat{S}_{x}]

Heisenberg’s equation of motion.

[\hat{S}_{z},\hat{S}_{x}]=i\hbar\hat{S}_{y}\ \Rightarrow\ \frac{d\hat{S}_{x}}{dt}=-\omega\hat{S}_{y}

The cyclic commutator.

[\hat{S}_{z},\hat{S}_{y}]=-i\hbar\hat{S}_{x}\ \Rightarrow\ \frac{d\hat{S}_{y}}{dt}=\omega\hat{S}_{x}

The same for $\hat{S}_{y}$ .

\begin{aligned}\hat{S}_{x}(t)&=\hat{S}_{x}\cos\omega t-\hat{S}_{y}\sin\omega t\\ \hat{S}_{y}(t)&=\hat{S}_{y}\cos\omega t+\hat{S}_{x}\sin\omega t\end{aligned}

A rotation by $\omega t$ about $z$ solves both.

\frac{d\hat{S}_{z}}{dt}=\frac{i\omega}{\hbar}[\hat{S}_{z},\hat{S}_{z}]=0

The component along the field is conserved.

Applications

Practice

The Larmor Frequency

In a field $B$ along $z$ , the average spin turns about $z$ at the Larmor frequency, the energy splitting divided by $\hbar$ . For an electron that is about 28 gigahertz per tesla.

Try it

At about 28 GHz per tesla, at what frequency does an electron spin precess in a field of 0.35 T, in GHz? Give one decimal place.