Larmor Precession and Magnetic Resonance

Put a spin- $\tfrac12$ particle in a static magnetic field and nothing dramatic seems to happen: the two energy levels split by the Zeeman gap, and a spin already aligned or anti-aligned with the field just sits there. The interesting physics lives in the superposition. A spin tilted away from the field axis does not stay put — its expectation value $\langle\mathbf{S}\rangle$ sweeps around the field direction at a single, field-set rate called the Larmor frequency $\omega_0=\gamma B_0$ , tracing a cone exactly as a classical gyroscope precesses under gravity. This is Larmor precession, and the remarkable feature is that the quantum expectation values obey the same equation of motion as a classical magnetic moment, a clean instance of Ehrenfest's theorem. The precession by itself moves nothing between the two energy levels; $\langle S_z\rangle$ is a constant of the motion. To drive real transitions you add a weak oscillating field transverse to $B_0$ and tune its frequency near $\omega_0$ . On resonance the spin flips completely back and forth between up and down — Rabi oscillations — and off resonance the flip amplitude is sharply suppressed. The resonance condition $\omega=\omega_0$ and the flip probability are captured by the Rabi formula, derived here in the rotating-wave approximation. This single two-level problem is the engine behind an enormous body of technology: nuclear magnetic resonance and MRI (protons at $42.577$ MHz per tesla), electron spin resonance ( $28.0$ GHz per tesla), atomic clocks, and the control pulses that rotate a qubit on the Bloch sphere. This article derives the precession, the resonance, and the numbers, and shows where each claim is proved and where it is merely quoted.

Key takeaways

A spin- $\tfrac12$ magnetic moment in a static field $\mathbf{B}=B_0\hat{\mathbf z}$ has $H=-\boldsymbol{\mu}\cdot\mathbf{B}=\omega_0 S_z$ , splitting the levels by $\hbar\omega_0$ with Larmor frequency $\omega_0=\gamma B_0$ .
A tilted spin precesses: $\langle S_x\rangle,\langle S_y\rangle$ rotate about $\hat{\mathbf z}$ at $\omega_0$ while $\langle S_z\rangle$ stays constant — the quantum $\langle\mathbf{S}\rangle$ obeys the classical torque equation $\dot{\langle\mathbf S\rangle}=\gamma\,\langle\mathbf S\rangle\times\mathbf B$ (Ehrenfest).
Precession alone drives no transitions; $\langle S_z\rangle$ is conserved. Level-to-level flips require a transverse oscillating field.
Adding a rotating field of frequency $\omega$ gives Rabi oscillations: $P_{\uparrow\to\downarrow}(t)=\dfrac{\omega_1^2}{\Omega^2}\sin^2\!\big(\tfrac12\Omega t\big)$ with generalized Rabi frequency $\Omega=\sqrt{\omega_1^2+\Delta^2}$ and detuning $\Delta=\omega-\omega_0$ .
On resonance ( $\Delta=0$ ) the spin flips completely, $P=\sin^2(\tfrac12\omega_1 t)$ , giving $\pi$ -pulses and $\pi/2$ -pulses; off resonance the flip amplitude is capped at $\omega_1^2/\Omega^2$ .
Anchors: proton NMR $42.577$ MHz/T (so $63.9$ MHz in a $1.5$ T MRI), electron ESR $28.0$ GHz/T ( $9.8$ GHz in $0.35$ T).
Key idea: a magnetic field makes the transverse spin components rotate at $\omega_0=\gamma B_0$ but cannot change $\langle S_z\rangle$ ; only a field oscillating near $\omega_0$ resonantly pumps population between the levels, and that resonance is the whole basis of NMR, ESR, and qubit control.

The core insight

A static field sets a precession rate $\omega_0=\gamma B_0$ for the transverse spin but conserves $\langle S_z\rangle$ ; a transverse field tuned to $\omega_0$ resonantly drives full population transfer at the Rabi frequency. Precession and resonance are the two faces of the same two-level Hamiltonian.

Where this lives today

The two-level physics of this article is, without exaggeration, one of the most technologically consequential calculations in all of quantum mechanics. Nuclear magnetic resonance reads the Larmor frequencies of nuclei (protons, $^{13}$ C, $^{15}$ N) shifted by their chemical environment, making it the workhorse of molecular structure determination in chemistry and structural biology. Magnetic resonance imaging turns the same proton signal into spatial images by adding field gradients so that Larmor frequency encodes position — no ionizing radiation, unlike X-rays. Electron spin resonance probes unpaired electrons in radicals and transition-metal centers. Atomic clocks lock an oscillator to a hyperfine resonance measured by exactly this Rabi technique, defining the SI second. And in quantum computing, every single-qubit gate is a Rabi rotation: a resonant microwave $\pi$ - or $\pi/2$ -pulse rotates a superconducting or spin qubit on the Bloch sphere, with the pulse area $\omega_1 t$ setting the rotation angle. The Bloch-sphere language physicists use for qubits is precisely the geometry of $\langle\mathbf S\rangle$ derived here.

The physical picture

Think of the spin's magnetic moment $\boldsymbol{\mu}=\gamma\mathbf{S}$ as a tiny bar magnet carrying angular momentum. A static field exerts a torque $\boldsymbol{\tau}=\boldsymbol{\mu}\times\mathbf{B}$ , and because the moment is rigidly tied to angular momentum, the torque does not tip the magnet toward the field the way it would tip a compass needle with friction — instead it makes the angular momentum vector precess around the field, sweeping out a cone at fixed polar angle. This is identical to a spinning top under gravity. The precession rate is the Larmor frequency $\omega_0=\gamma B_0$ , independent of the tilt angle: a spin tipped $30^\circ$ and a spin tipped $80^\circ$ circle the field axis at exactly the same rate. The quantum content is that $\langle\mathbf{S}\rangle$ obeys precisely this classical equation. Ehrenfest's theorem turns the Heisenberg equation for $\mathbf{S}$ into $\dot{\langle\mathbf S\rangle}=\gamma\,\langle\mathbf S\rangle\times\mathbf B$ , so the expectation value traces the cone even though the spin itself is in a superposition of $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ with fixed populations. Nothing moves between the levels: $\langle S_z\rangle$ is a constant.

Figure

A tilted spin precesses about the field at the Larmor frequency

To actually flip a spin you must break the axial symmetry. Add a small field rotating in the $xy$ -plane at frequency $\omega$ . Viewed from a frame co-rotating with this field, the fast Larmor rotation is largely undone, and what remains is an effective static field tilted away from $\hat{\mathbf z}$ by an amount set by the detuning $\Delta=\omega-\omega_0$ . When $\omega$ exactly matches $\omega_0$ the effective field lies flat in the transverse plane, and the spin precesses about it straight from up to down and back — a complete flip. Off resonance the effective field has a residual $z$ -component, the precession cone about it never reaches the south pole, and the flip is only partial. This is the whole content of magnetic resonance.

Figure

Rabi flip probability: sharp on resonance, capped off resonance

The core equations

The two-level Hamiltonian for a magnetic moment $\boldsymbol{\mu}=\gamma\mathbf{S}$ in a static field along $\hat{\mathbf z}$ is a Zeeman term. Writing it with the sign absorbed into the rate keeps the algebra clean.

H_0 = -\boldsymbol{\mu}\cdot\mathbf{B} = -\gamma B_0\,S_z = \omega_0\,S_z, \qquad \omega_0 \equiv \gamma B_0 \qquad (1)

where: $\gamma$ is the gyromagnetic ratio ( $\gamma=g\,q/2m$ for a particle of charge $q$ , mass $m$ , and $g$ -factor $g$ ); $B_0$ is the static field magnitude; $S_z=\tfrac\hbar2\sigma_z$ is the spin- $z$ operator; and $\omega_0=\gamma B_0$ is the Larmor frequency. We have folded the sign of $\gamma$ into $\omega_0$ so that $H_0=\omega_0 S_z$ ; the eigenvalues are $\pm\tfrac12\hbar\omega_0$ , split by the Zeeman gap $\hbar\omega_0$ . (For the electron $\gamma<0$ , which flips the precession sense but not the rate.) Time evolution of an arbitrary initial spinor follows from the propagator $U(t)=e^{-iH_0t/\hbar}$ . For a state tilted from $\hat{\mathbf z}$ by polar angle $\alpha$ , the expectation values are

\langle S_x\rangle = \tfrac\hbar2\sin\alpha\,\cos\omega_0 t,\quad \langle S_y\rangle = \tfrac\hbar2\sin\alpha\,\sin\omega_0 t,\quad \langle S_z\rangle = \tfrac\hbar2\cos\alpha \qquad (2)

where: $\alpha$ is the (constant) tilt angle of the initial spin. The transverse components rotate at $\omega_0$ with fixed magnitude $\tfrac\hbar2\sin\alpha$ , tracing a circle, while $\langle S_z\rangle$ is time-independent. This is precession: the polar angle is frozen, only the azimuth advances. The sense of rotation ( $+\sin\omega_0 t$ for $\langle S_y\rangle$ with our $H_0=+\omega_0 S_z$ ) is fixed by the sign of $\gamma$ . The same content in vector form is the classical torque equation, which Ehrenfest's theorem hands us exactly for the quantum expectation value.

\frac{d\langle\mathbf{S}\rangle}{dt} = \gamma\,\langle\mathbf{S}\rangle\times\mathbf{B} \qquad (3)

where: $\mathbf{B}$ is the total magnetic field. Equation (3) is a closed equation for $\langle\mathbf S\rangle$ (no higher moments appear because $H$ is linear in $\mathbf S$ ), so the quantum precession is literally the classical one; a solid magnetization $\mathbf M=n\gamma\hbar\langle\boldsymbol\sigma\rangle/2$ obeys the identical equation, which is why bulk NMR magnetization precesses too. To drive transitions, add a field rotating in the $xy$ -plane at drive frequency $\omega$ . The full Hamiltonian, transformed into the frame co-rotating with the drive and dropping the fast counter-rotating term (the rotating-wave approximation, RWA), becomes time-independent:

H_{\rm rot} = \tfrac\hbar2\big(-\Delta\,\sigma_z + \omega_1\,\sigma_x\big),\qquad \Delta \equiv \omega-\omega_0,\quad \omega_1 \equiv \gamma B_1 \qquad (4)

where: $\Delta$ is the detuning of the drive from resonance; $\omega_1=\gamma B_1$ is the Rabi frequency set by the rotating field amplitude $B_1$ ; and $\sigma_{x,z}$ are Pauli matrices. In the rotating frame the spin sees a static effective field with a $z$ -component $\propto\Delta$ and a transverse component $\propto\omega_1$ ; it precesses about that tilted effective field. Solving (4) for a spin starting in $|{\uparrow}\rangle$ gives the probability of finding it flipped to $|{\downarrow}\rangle$ — the Rabi formula.

P_{\uparrow\to\downarrow}(t) = \frac{\omega_1^2}{\omega_1^2+\Delta^2}\,\sin^2\!\Big(\tfrac12\sqrt{\omega_1^2+\Delta^2}\;t\Big) = \frac{\omega_1^2}{\Omega^2}\sin^2\!\big(\tfrac12\Omega t\big) \qquad (5)

where: $\Omega\equiv\sqrt{\omega_1^2+\Delta^2}$ is the generalized Rabi frequency. The prefactor $\omega_1^2/\Omega^2$ caps the maximum flip probability, and $\Omega$ sets the oscillation rate. On resonance ( $\Delta=0$ ): $P=\sin^2(\tfrac12\omega_1 t)$ , reaching $1$ at $t=\pi/\omega_1$ (a $\pi$ -pulse). The width of the resonance in $\Delta$ is $\sim\omega_1$ , so a weaker drive gives a sharper, more selective resonance.

Derivation

Derivation ALarmor precession from the Schrödinger equation

The goal is to show that a spin- $\tfrac12$ state tilted away from the field axis has transverse spin components that rotate at $\omega_0=\gamma B_0$ while its axial component stays fixed, and that this reproduces the classical torque equation exactly. The route is to write the initial tilted spinor in the $S_z$ eigenbasis, apply the diagonal propagator $e^{-iH_0t/\hbar}$ to attach a relative phase between the up and down amplitudes, and then compute the three spin expectation values, watching how that single relative phase turns into a rotating transverse vector. The one thing to keep straight is that the propagator only changes the relative phase of the two amplitudes, never their magnitudes — which is exactly why $\langle S_z\rangle$ cannot change.

Step 1. Write the initial state tilted by polar angle $\alpha$ (azimuth zero): $|\chi(0)\rangle=\cos\tfrac\alpha2\,|{\uparrow}\rangle+\sin\tfrac\alpha2\,|{\downarrow}\rangle$ . This has $\langle S_z\rangle=\tfrac\hbar2(\cos^2\tfrac\alpha2-\sin^2\tfrac\alpha2)=\tfrac\hbar2\cos\alpha$ , as required for a spin pointing at polar angle $\alpha$ .

Step 2. Apply the propagator. Since $H_0=\omega_0 S_z=\tfrac\hbar2\omega_0\sigma_z$ is diagonal, $e^{-iH_0t/\hbar}$ multiplies $|{\uparrow}\rangle$ by $e^{-i\omega_0 t/2}$ and $|{\downarrow}\rangle$ by $e^{+i\omega_0 t/2}$ : $|\chi(t)\rangle=\cos\tfrac\alpha2\,e^{-i\omega_0 t/2}|{\uparrow}\rangle+\sin\tfrac\alpha2\,e^{+i\omega_0 t/2}|{\downarrow}\rangle$ . The magnitudes of the two amplitudes are untouched.

Step 3. Compute $\langle S_z\rangle$ . Because the propagator left $|\cos\tfrac\alpha2|^2$ and $|\sin\tfrac\alpha2|^2$ unchanged, $\langle S_z(t)\rangle=\tfrac\hbar2\cos\alpha$ for all $t$ . The axial component is a constant of the motion — no population moves between levels.

Step 4. Compute the transverse components. Using $S_x=\tfrac\hbar2(|{\uparrow}\rangle\langle{\downarrow}|+|{\downarrow}\rangle\langle{\uparrow}|)$ , the cross terms carry the relative phase $e^{\pm i\omega_0 t}$ : $\langle S_x(t)\rangle=\tfrac\hbar2\cdot 2\cos\tfrac\alpha2\sin\tfrac\alpha2\cos\omega_0 t=\tfrac\hbar2\sin\alpha\cos\omega_0 t$ , and similarly $\langle S_y(t)\rangle=\tfrac\hbar2\sin\alpha\sin\omega_0 t$ . The transverse vector has fixed length $\tfrac\hbar2\sin\alpha$ and rotates at $\omega_0$ .

Step 5. Identify the classical equation. Differentiating (2) gives $\dot{\langle S_x\rangle}=-\omega_0\langle S_y\rangle$ , $\dot{\langle S_y\rangle}=+\omega_0\langle S_x\rangle$ , $\dot{\langle S_z\rangle}=0$ , which is exactly $\dot{\langle\mathbf S\rangle}=\gamma\langle\mathbf S\rangle\times\mathbf B$ with $\mathbf B=B_0\hat{\mathbf z}$ and $\omega_0=\gamma B_0$ . The quantum expectation value precesses identically to a classical moment. Verified: the companion verifier integrates the spin- $\tfrac12$ TDSE for a tilted spinor ( $\alpha=0.7$ ) and confirms $\langle S_z\rangle=\tfrac12\cos\alpha$ to machine precision, $\langle S_x\rangle=\tfrac12\sin\alpha\cos\omega_0 t$ and $\langle S_y\rangle=+\tfrac12\sin\alpha\sin\omega_0 t$ exactly, and an FFT of $\langle S_x(t)\rangle$ recovers $\omega_0$ to better than $1\%$ .

Go deeper · Derivation B · 2 minThe Rabi formula in the rotating-wave approximation

Derivation BThe Rabi formula in the rotating-wave approximation

The aim here is to derive the two-level flip probability (5) — the quantitative statement of magnetic resonance — from the Hamiltonian of a spin in a static field plus a rotating transverse field. The route is to remove the fast Larmor rotation by transforming into a frame co-rotating with the drive, which converts the explicitly time-dependent Hamiltonian into a static effective Hamiltonian describing precession about a tilted effective field; then the flip probability is just the precession of a spin about that fixed axis, read off geometrically or by diagonalizing the $2\times2$ matrix. The subtlety to watch is the rotating-wave approximation: a linearly oscillating field is a sum of two counter-rotating components, and we keep only the one co-rotating with the spin, discarding a term that oscillates at $2\omega$ and averages away when $\omega_1\ll\omega_0$ .

Step 1. Total field and Hamiltonian: $\mathbf B(t)=B_0\hat{\mathbf z}+B_1(\cos\omega t\,\hat{\mathbf x}-\sin\omega t\,\hat{\mathbf y})$ , so $H(t)=\omega_0 S_z+\omega_1(\cos\omega t\,S_x-\sin\omega t\,S_y)$ with $\omega_1=\gamma B_1$ . The transverse field rotates in the plane at frequency $\omega$ .

Step 2. Transform to the rotating frame via $R(t)=e^{-i\omega t\,S_z/\hbar}$ , defining $|\tilde\chi\rangle=R^\dagger|\chi\rangle$ . The transformed Hamiltonian is $\tilde H=R^\dagger H R-i\hbar R^\dagger\dot R$ . The rotation cancels the explicit time dependence of the transverse field, and the $-i\hbar R^\dagger\dot R=-\omega S_z$ term shifts the axial part.

Step 3. Collect terms: $\tilde H=(\omega_0-\omega)S_z+\omega_1 S_x=-\Delta\,S_z+\omega_1 S_x$ with $\Delta=\omega-\omega_0$ , i.e. $\tilde H=\tfrac\hbar2(-\Delta\,\sigma_z+\omega_1\sigma_x)$ — equation (4). This is time-independent: a static effective field $\mathbf B_{\rm eff}\propto(\omega_1,0,-\Delta)$ in the rotating frame.

Step 4. The spin precesses about $\mathbf B_{\rm eff}$ at the generalized Rabi frequency $\Omega=\sqrt{\omega_1^2+\Delta^2}$ (the magnitude of the effective-field vector). Starting from $|{\uparrow}\rangle$ , the probability to be found in $|{\downarrow}\rangle$ is the squared projection of the precessing state onto the south pole, which geometry (or diagonalizing the $2\times2$ ) gives as $P=\dfrac{\omega_1^2}{\Omega^2}\sin^2(\tfrac12\Omega t)$ .

Step 5. Read off the limits. On resonance $\Delta=0$ : $\Omega=\omega_1$ and $P=\sin^2(\tfrac12\omega_1 t)$ , a complete flip at $t=\pi/\omega_1$ . Far off resonance $|\Delta|\gg\omega_1$ : the prefactor $\omega_1^2/\Omega^2\to0$ , so the spin barely flips — the resonance is a genuine peak in $\Delta$ of width $\sim\omega_1$ . Verified: the verifier evolves the rotating-frame Hamiltonian (4) directly for detunings $\Delta\in\{0,0.5,1.7,-1.3\}\omega_1$ and confirms the analytic Rabi formula (5) matches the numerical $|\langle{\downarrow}|\psi(t)\rangle|^2$ to $\sim10^{-15}$ , that on resonance $P(\pi/\omega_1)=1$ , and that the far-detuned amplitude caps at $\omega_1^2/\Omega^2$ .

Limiting cases

Each regime of the two-level problem is a check on the formulas.

Case	Result	Meaning
Spin aligned with field ( $\alpha=0$ )	$\langle\mathbf S\rangle$ static, no precession	An eigenstate of $H_0$ ; only a stationary phase accrues, nothing observable moves.
On resonance ( $\Delta=0$ )	$P=\sin^2(\tfrac12\omega_1 t)$ , full flips	Effective field lies in the transverse plane; the spin precesses straight from up to down.
Far off resonance ( $\|\Delta\|\gg\omega_1$ )	$P_{\max}=\omega_1^2/\Delta^2\to0$	Effective field nearly along $\hat{\mathbf z}$ ; the precession cone never reaches the south pole.
Classical / large spin	$\dot{\mathbf M}=\gamma\mathbf M\times\mathbf B$ (Bloch)	$\langle\mathbf S\rangle$ obeys the classical torque equation exactly (Ehrenfest); bulk magnetization precesses identically.
Strong static field ( $\omega_0\to\infty$ )	RWA becomes exact	The counter-rotating term oscillates ever faster and averages away, so the rotating-wave approximation improves.

Common trapPrecession is not a spin flip

It is tempting to think a precessing spin is "moving toward the down state." It is not. During free Larmor precession in a static field, $\langle S_z\rangle=\tfrac\hbar2\cos\alpha$ is exactly constant — the populations of $|{\uparrow}\rangle$ and $|{\downarrow}\rangle$ never change, and no energy is exchanged with the field on average. What rotates is only the transverse phase. To actually transfer population between the levels you must apply a transverse, time-dependent field near resonance; that is a different physical process (Rabi driving) governed by (5), not by (2). Confusing the azimuthal precession with population transfer is the single most common error in reading the Bloch-sphere picture.

Worked examples

Example 1 — NMR and ESR frequencies. Find the proton NMR frequency in a $1.5$ T clinical MRI magnet and the electron ESR frequency in a $0.35$ T laboratory magnet.

Step 1. Proton NMR. The proton gyromagnetic ratio is $\gamma_p/2\pi=42.577$ MHz/T (this already includes the proton $g$ -factor $g_p\approx5.586$ ). The Larmor frequency is $f=\gamma_p B_0/2\pi=(42.577\ \text{MHz/T})\times(1.5\ \text{T})$ .

Step 2. Evaluate: $f=63.87\ \text{MHz}\approx\mathbf{63.9\ MHz}$ . This is the radio-frequency at which a $1.5$ T MRI scanner transmits and receives — squarely in the FM-radio band, which is why MRI hardware uses RF coils.

Step 3. Electron ESR. The electron's effective ratio is $|\gamma_e|/2\pi=g_e\mu_B/h=28.025$ GHz/T (with $g_e=2.0023$ ). In $0.35$ T: $f=(28.025\ \text{GHz/T})\times(0.35\ \text{T})=\mathbf{9.81\ GHz}$ — the microwave X-band, and the reason ESR spectrometers use waveguides and cavities rather than RF coils. The ratio of the two, $\approx28.0\ \text{GHz}/42.6\ \text{MHz}\approx660$ , is essentially the proton-to-electron effective moment ratio — the electron's moment is far larger, so it resonates at far higher frequency in the same field. Takeaways: $63.9$ MHz proton NMR at $1.5$ T, $9.8$ GHz electron ESR at $0.35$ T.

Example 2 — Designing a $\pi/2$ -pulse. An NMR spectrometer applies an on-resonance rotating field giving a Rabi frequency $\omega_1/2\pi=25$ kHz. How long a pulse rotates the magnetization from $\hat{\mathbf z}$ into the transverse plane (a $\pi/2$ -pulse), and how long fully inverts it (a $\pi$ -pulse)?

Step 1. On resonance the polar angle swept is $\theta(t)=\omega_1 t$ . A $\pi/2$ -pulse needs $\theta=\pi/2$ , so $t_{\pi/2}=\dfrac{\pi/2}{\omega_1}=\dfrac{1}{4}\cdot\dfrac{2\pi}{\omega_1}=\dfrac{1}{4}\cdot\dfrac{1}{25\,\text{kHz}}=10\ \mu\text{s}$ .

Step 2. A $\pi$ -pulse (full inversion, $P=1$ ) needs $\theta=\pi$ : $t_\pi=\pi/\omega_1=2\,t_{\pi/2}=20\ \mu\text{s}$ .

Step 3. Sanity check with the Rabi formula on resonance: $P(t_\pi)=\sin^2(\tfrac12\omega_1 t_\pi)=\sin^2(\pi/2)=1$ , a complete flip; $P(t_{\pi/2})=\sin^2(\pi/4)=\tfrac12$ , an equal superposition with the moment in the transverse plane. A $25$ kHz Rabi drive gives a $10\ \mu$ s $\pi/2$ -pulse and a $20\ \mu$ s $\pi$ -pulse.

Origins · 1 minlarmor precession and how this was discovered

larmor precession

classical + quantum torque on a magnetic moment · The rotation of a magnetic moment about an applied field

Larmor precession is named for Joseph Larmor, who in 1897 derived that an orbiting charged particle's angular momentum precesses about an applied magnetic field at the frequency $\omega_0=\gamma B$ , one of the earliest results linking magnetism to rotational dynamics. Quantum-mechanically the concept survives intact: the spin expectation value $\langle\mathbf S\rangle$ precesses at exactly the Larmor frequency, a consequence of Ehrenfest's theorem that the two-level Hamiltonian is linear in the spin. The concept is the bridge between the classical picture of a spinning magnetized top and the quantum two-level system, and it sets the carrier frequency of every magnetic-resonance technique from NMR to qubit control.

History

1897

— Larmor derives the precession of an orbital magnetic moment about a field,

\omega=\gamma B

1922

— Stern and Gerlach demonstrate space quantization of the silver-atom magnetic moment

1938

— Rabi and coworkers observe molecular-beam magnetic resonance (Phys. Rev. 53, 318), the first resonant spin flips

1944

— Rabi awarded the Nobel Prize in Physics for the resonance method

1946

— Bloch (Stanford) and Purcell (Harvard) independently detect nuclear magnetic resonance in bulk matter; Bloch writes the phenomenological Bloch equations

1952

— Bloch and Purcell share the Nobel Prize in Physics for nuclear magnetic resonance

1973

— Lauterbur and Mansfield develop magnetic-resonance imaging from the NMR signal

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.4 (spin in a magnetic field) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §2.1 (spin precession) and §5.5 (magnetic resonance) · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Ch. IV complement F · Slichter, Principles of Magnetic Resonance.

Provenance. A verifier should reproduce: (1) the precession law (2) by integrating the spin- $\tfrac12$ TDSE for a tilted spinor — $\langle S_z\rangle=\tfrac12\cos\alpha$ constant, transverse components rotating at $\omega_0$ ; (2) the Rabi formula (5) against direct rotating-frame evolution of (4) for several detunings, matching to $\sim10^{-15}$ ; (3) the anchors proton $42.577$ MHz/T, $63.9$ MHz at $1.5$ T, electron $28.0$ GHz/T, $9.8$ GHz at $0.35$ T, and the $\pi$ -pulse time $t_\pi=\pi/\omega_1$ .

Builds on: Spin and Spin Angular Momentum · The Stern-Gerlach Experiment · Time-Dependent Schrödinger Equation · Ehrenfest's Theorem. Leads to: Time-Dependent Perturbation Theory and Rabi Oscillations · Nuclear Magnetic Resonance and MRI · Atomic Clocks · Single-Qubit Gates and the Bloch Sphere.

Why does the spin precess instead of just aligning with the field like a compass needle?

A compass needle aligns because friction dissipates its energy until it settles at the minimum-energy orientation. An isolated spin has no such dissipation, and crucially its magnetic moment is rigidly tied to its angular momentum. The field exerts a torque $\boldsymbol{\mu}\times\mathbf B$ perpendicular to both $\boldsymbol{\mu}$ and $\mathbf B$ , and since torque equals the rate of change of angular momentum, the angular momentum vector rotates around the field rather than toward it — exactly like a spinning top under gravity precesses instead of falling over. Only when you add a relaxation mechanism (spin-lattice coupling, the $T_1$ process in NMR) does the spin actually relax toward alignment, and that happens on a much slower timescale than the precession.

Is Larmor precession a genuinely quantum effect or just classical?

The precession equation (3) is identical to the classical torque equation, so in that sense it is not distinctively quantum — a classical magnetized gyroscope precesses the same way, and this is a clean example of the classical limit emerging via Ehrenfest's theorem because the Hamiltonian is linear in the spin. What is quantum is that the state precessing is a superposition of just two discrete levels, that the energy gap $\hbar\omega_0$ is quantized, and that a measurement of $S_z$ yields only $\pm\tfrac\hbar2$ . The expectation value follows the classical cone, but any single measurement collapses the spin. So the trajectory of $\langle\mathbf S\rangle$ is classical; the measurement statistics are quantum.

What exactly is the rotating-wave approximation, and when does it fail?

A linearly oscillating field $2B_1\cos\omega t\,\hat{\mathbf x}$ decomposes into two circularly polarized components rotating in opposite senses at $\pm\omega$ . Only the component co-rotating with the spin's Larmor precession stays nearly in step and drives the transition; the counter-rotating component runs at $\approx2\omega$ relative to the spin and averages to nearly zero. The RWA keeps the co-rotating term and drops the counter-rotating one, which is excellent when $\omega_1\ll\omega_0$ (weak drive, near resonance). It fails for very strong driving ( $\omega_1\sim\omega_0$ ), where the dropped term produces a small frequency shift (the Bloch-Siegert shift) and additional structure. In NMR and most qubit control $\omega_1/\omega_0$ is tiny, so the RWA is superb.

Does the sign of the gyromagnetic ratio matter?

It flips the sense of precession but not the rate. For nuclei like the proton $\gamma>0$ and $\langle\mathbf S\rangle$ precesses one way; for the electron $\gamma<0$ (its moment is antiparallel to its spin), so it precesses the opposite way at the same $|\gamma|B_0$ . The Larmor frequency magnitude $\omega_0=|\gamma|B_0$ and all resonance conditions are unchanged; only the direction of circulation on the Bloch sphere reverses. In most experiments only the magnitude matters, which is why tables quote $|\gamma|/2\pi$ .

Why is a weaker drive better for resolving two nearby resonances?

The Rabi resonance in (5) has a width in detuning of order $\omega_1$ : the flip probability is appreciable only for $|\Delta|\lesssim\omega_1$ . So a weaker rotating field (smaller $\omega_1$ ) produces a narrower resonance, letting you distinguish two nuclei whose Larmor frequencies differ only slightly (as in chemical-shift NMR). The tradeoff is speed: a smaller $\omega_1$ means a longer $\pi$ -pulse ( $t_\pi=\pi/\omega_1$ ) and a slower experiment. This is the frequency-resolution-versus-speed tradeoff at the heart of pulse design, and it is a time-energy uncertainty relation in disguise: a longer pulse has a narrower frequency bandwidth.

How does this connect to the Bloch sphere used in quantum computing?

The Bloch sphere is exactly the space of directions of $\langle\mathbf S\rangle$ (equivalently $\langle\boldsymbol\sigma\rangle$ ) for a two-level system. Free precession about $B_0$ is rotation about the sphere's polar ( $z$ ) axis at $\omega_0$ ; a resonant drive is rotation about a transverse axis at $\omega_1$ . A $\pi$ -pulse is a $180^\circ$ rotation (bit flip, the $X$ gate up to phase), a $\pi/2$ -pulse is a $90^\circ$ rotation that creates superpositions (a Hadamard-like operation up to phase), and the rotation axis in the transverse plane is set by the drive's phase. Every single-qubit gate is one of these Rabi rotations, so the physics derived here is literally the instruction set of a spin or superconducting qubit.

If precession conserves energy, where does the NMR signal come from?

The detected NMR signal comes from the transverse magnetization precessing in the coil, which by Faraday induction generates an oscillating EMF at the Larmor frequency — this is the free-induction decay. To get transverse magnetization in the first place you apply a $\pi/2$ -pulse (a driven, resonant process that does exchange energy with the RF field), tipping the equilibrium axial magnetization into the transverse plane. After the pulse ends, the transverse magnetization precesses freely (conserving energy) and is what the coil picks up, decaying over the $T_2$ time as spins dephase. So energy is exchanged during the pulse; the readout is free precession afterward.