Spin precession

The previous section’s SG apparatus answers “which way does the spin point right now”. This section asks the next natural question: between two measurements, what is the spin doing?

This is not mere theoretical fussiness. Every hydrogen nucleus in your body is a spin $\tfrac12$, and a hospital’s magnetic resonance imaging (MRI) machine spends its whole day talking to them — the physics it relies on is precisely the answer to “how does spin evolve between measurements”.

The Hamiltonian: one term is all that survives

Place a particle with magnetic moment $\vec\mu$ in a uniform field $\vec B=B\hat z$. Uniform means no force (last section’s force came from the gradient), so the particle does not move; the position ledger is perfectly quiet, and only the spin ledger evolves. The energy is just the classical orientation energy:

$$\hat H=-\vec\mu\cdot\vec B=-\gamma B\,\hat S_z\tag{5.5.1}$$

Here $\vec\mu=\gamma\hat{\vec S}$, and the proportionality constant $\gamma$ is the gyromagnetic ratio — one fixed value per particle species. For the electron $|\gamma_e|=g_e\mu_B/\hbar$ (with $g_e\approx2.002$); the proton’s is about 660 times smaller (larger mass, smaller moment).

This is arguably the simplest non-trivial Hamiltonian in quantum mechanics: a $2\times2$ diagonal matrix. The eigenstates are just $\ket{\uparrow},\ket{\downarrow}$, with energies

$$E_\uparrow=-\frac{\gamma\hbar B}{2},\qquad E_\downarrow=+\frac{\gamma\hbar B}{2}\tag{5.5.2}$$

(We narrate with $\gamma$ positive; the electron’s $\gamma$ is actually negative, which merely reverses the sense of rotation.) The gap between the two levels, $\Delta E=\gamma\hbar B$, opens linearly with the field — this is the Zeeman splitting in its simplest two-level form. Converting that gap to a frequency, write

$$\omega\equiv\frac{\Delta E}{\hbar}=\gamma B\tag{5.5.3}$$

This frequency is about to become the star of the show.

Solving the evolution: the spin circles the field

Evolution of an arbitrary initial state and the precession of ⟨S⟩basic~9 minexpand

Step 1: the initial state. Let the spin at $t=0$ point “up” along some direction $(\theta,\phi=0)$ (using the general formula of section 5.3; taking $\phi=0$ loses no generality):

$$\ket{\chi(0)}=\cos\frac\theta2\ket{\uparrow}+\sin\frac\theta2\ket{\downarrow}\tag{5.5.4}$$

Step 2: apply the stationary-state expansion. The eigenstates of $\hat H$ are ready-made; just hang a phase clock $\ee^{-\ii E t/\hbar}$ on each component (see section 3.9):

$$\ket{\chi(t)}=\cos\frac\theta2\,\ee^{+\ii\omega t/2}\ket{\uparrow} +\sin\frac\theta2\,\ee^{-\ii\omega t/2}\ket{\downarrow}\tag{5.5.5}$$

(substituting $E_\uparrow=-\hbar\omega/2$, $E_\downarrow=+\hbar\omega/2$). Pull out the overall phase $\ee^{+\ii\omega t/2}$ and discard it:

$$\ket{\chi(t)}\ \sim\ \cos\frac\theta2\ket{\uparrow} +\ee^{-\ii\omega t}\sin\frac\theta2\ket{\downarrow}\tag{5.5.6}$$

Step 3: recognise who this is. Compare with the general spin state $\cos\tfrac\theta2\ket{\uparrow}+\ee^{\ii\phi}\sin\tfrac\theta2\ket{\downarrow}$ — this is exactly the “up” state along the direction $(\theta,\ \phi(t)=-\omega t)$! The polar angle $\theta$ does not budge; the azimuthal angle rotates uniformly. Evolution = circling the z axis — a line of latitude on the Bloch sphere.

Step 4: check the expectation values. Compute each (sandwich with the Pauli matrices):

$$\langle\hat S_z\rangle=\frac\hbar2\cos\theta=\text{const}\tag{5.5.7}$$$$\langle\hat S_x\rangle=\frac\hbar2\sin\theta\cos\omega t,\qquad \langle\hat S_y\rangle=-\frac\hbar2\sin\theta\sin\omega t\tag{5.5.8}$$

$\langle\vec S\rangle$ is a vector of constant length rotating uniformly about the field direction at angular velocity $\omega=\gamma B$ — exactly like a classical gyroscope precessing under gravity. This rotation is called Larmor precession, and $\omega$ the Larmor frequency.

Resonance: how to flip a spin

Watching a spin go round in circles is not much use by itself; we also want to control it — say, flip an up spin to down. The level gap is $\hbar\omega$, so intuition says: apply a perturbation at exactly the frequency $\omega$.

Concretely: on top of the large static field $B_0\hat z$, add a small field $B_1$ rotating in the xy plane ($B_1\ll B_0$) at frequency $\omega_{\text{rf}}$. The qualitative results (the full solution uses the rotating-frame trick of section 3.9; here we make the physics clear):

• Off resonance ($\omega_{\text{rf}}$ away from $\omega_0=\gamma B_0$): the small field rotates “out of step” with the spin; its push helps one moment and hinders the next, averaging to almost nothing.
• On resonance ($\omega_{\text{rf}}=\omega_0$): view things from a frame rotating along with the spin. The effect of the big field $B_0$ is exactly cancelled by the rotation, leaving only a stationary small field $B_1$ — so the spin turns slowly about $B_1$ at the gentle rate $\omega_1=\gamma B_1$, arcing from the north pole down to the south and back. The probability swings back and forth between $\ket{\uparrow}$ and $\ket{\downarrow}$:

$$P_{\downarrow}(t)=\sin^2\frac{\omega_1 t}{2}\quad(\text{on resonance})\tag{5.5.9}$$

This swing is the Rabi oscillation. Time the shutdown of the small field precisely and you can park the spin anywhere you like: run for $t=\pi/\omega_1$ (a ”$\pi$ pulse”) and the flip is complete; run half as long (a ”$\pi/2$ pulse”) and the spin lies flat on the equator — the key move NMR uses below.

NMR and MRI: turning this section into a machine

The workflow of nuclear magnetic resonance (NMR), in the language of this section, takes three sentences:

1. Polarise. Put the sample in a strong field $B_0$. In thermal equilibrium slightly more spins point up than down (the lower-energy state is a touch more populated — only about ten parts per million at room temperature and 3 T, but with $10^{23}$ nuclei in the sample the net magnetisation is plenty), giving the sample a net magnetic moment along z.
2. Flip. Fire a radio-frequency $\pi/2$ pulse tuned to $\omega_0=\gamma B_0$, laying the net moment flat into the xy plane.
3. Listen. The flattened moment precesses at the Larmor frequency like a spinning bar magnet, inducing an oscillating voltage at $\omega_0$ in a pickup coil. Measure that signal’s frequency and decay.

Why is it worth listening to? Because the $B$ in $\omega_0=\gamma B$ is the actual field at the nucleus, slightly shielded by the surrounding electron cloud — the same nucleus in different chemical environments shifts in frequency by a few parts per million (the “chemical shift”). From one spectrum a chemist reads off molecular structure; that is NMR spectroscopy.

MRI runs the trick in reverse: deliberately make $B_0$ a gradient field that varies with position, so that frequency becomes a barcode for location — receive a signal at 127.68 MHz and you know it came from the slice of tissue where the field matches that frequency. Add the fact that signals decay at different rates in different tissues (water and fat have different relaxation times), and a three-dimensional image of soft tissue can be reconstructed. The hammering noise you hear inside an MRI machine is the gradient coils rapidly switching that “barcode”.

∎Key formulas

• Hamiltonian and splitting

  $$\hat H=-\gamma B\hat S_z,\qquad \Delta E=\gamma\hbar B=\hbar\omega_0$$

  Two-level Zeeman splitting; ω₀ is independent of the initial direction

• Larmor precession

  $$\langle\hat S_x\rangle=\tfrac\hbar2\sin\theta\cos\omega_0 t,\quad \langle\hat S_z\rangle=\text{const}$$

  Polar angle fixed, azimuth rotating uniformly: a line of latitude on the Bloch sphere

• Larmor frequency

  $$\omega_0=\gamma B$$

  Proton 42.6 MHz/T, electron 28.0 GHz/T — one channel per particle species

• Rabi oscillation

  $$P_{\downarrow}(t)=\sin^2\tfrac{\omega_1 t}{2},\qquad \omega_1=\gamma B_1$$

  On resonance the populations swing at full amplitude; a π pulse flips completely

?Self-check4 questions

1. 1.

   During Larmor precession, which of the following quantities changes with time?

   • ⟨S_x⟩ and ⟨S_y⟩
   • ⟨S_z⟩
   • The precession frequency ω₀
   • The magnitude |S|

   Submit
2. 2.

   Why does a Rabi flip require the radio-frequency field to be tuned to ω₀ = γB₀?

   • On resonance, in a frame rotating with the spin, the small field B₁ appears stationary and can keep pushing the spin the same way
   • Only photons of this frequency can be absorbed by the magnetic field
   • Because B₁ must exceed B₀
   • Off resonance, energy would not be conserved

   Submit
3. 3.

   Which statements about NMR/MRI are correct? (Select all that apply.)

   Select all that apply

   • MRI uses gradient fields to turn the Larmor frequency into an encoding of position
   • The chemical shift comes from the electron cloud slightly shielding the external field
   • A single proton induces a smooth sinusoidal voltage in the coil
   • The job of the π/2 pulse is to lay the net magnetisation flat into the plane perpendicular to B₀

   Submit
4. 4.

   The proton gyromagnetic ratio is γ = 2.675×10⁸ rad·s⁻¹·T⁻¹. In a 3.0 T MRI magnet, what is the proton Larmor frequency f = γB/2π in MHz? (One decimal place.)

   MHz50% relative tolerance

   Submit

What comes next

Up to now we have been playing with a single spin. But in real systems angular momenta rarely come alone: the electron in hydrogen has both orbital angular momentum $\hat{\vec L}$ and spin $\hat{\vec S}$; helium has the spins of two electrons; protons and neutrons assemble into nuclei.

Put two angular momenta side by side — what is the total? The classical answer, “add the vectors”, runs into immediate trouble in quantum mechanics: not even one vector’s three components can be simultaneously sharp, so what exactly would you add? The next section gives the correct addition, and its answer will also explain “singlet” and “triplet” — words that will keep returning in chemical bonds and entangled states.