Larmor Precession

Larmor precession is the rotation of a magnetic moment, spin expectation value, or Bloch vector around a static magnetic field. For a spin with

$$\boldsymbol\mu=\gamma\mathbf S,$$

the Zeeman Hamiltonian is

$$H = -\boldsymbol\mu\cdot\mathbf B = -\gamma\mathbf S\cdot\mathbf B.$$

In a uniform static field of magnitude $B_0$, the precession frequency magnitude is

$$\omega_L = |\gamma|B_0.$$

This page derives the operator and spinor forms of the precession. The static Hamiltonian, energy splitting, and sign conventions are introduced in Spin in Magnetic Fields.

Setup

Take a constant magnetic field

$$\mathbf B=B_0\hat{\mathbf z}.$$

The Hamiltonian is

$$H = -\gamma B_0S_z.$$

For spin-$1/2$,

$$S_z=\frac{\hbar}{2}\sigma_z,$$

so

$$H = -\frac{\gamma\hbar B_0}{2}\sigma_z.$$

The signed angular frequency

$$\omega=\gamma B_0$$

is useful in algebra. The positive precession frequency is $|\omega|$.

Heisenberg Derivation

The Heisenberg equation is

$$\frac{dA}{dt} = \frac{i}{\hbar}[H,A]$$

for an operator with no explicit time dependence. Using

$$[S_i,S_j] = i\hbar\sum_k\epsilon_{ijk}S_k,$$

and

$$H=-\gamma\mathbf B\cdot\mathbf S,$$

one obtains

$$\frac{d\mathbf S}{dt} = \gamma\,\mathbf S\times\mathbf B.$$

For $\mathbf B=B_0\hat{\mathbf z}$, this gives

$$\frac{dS_x}{dt} = \omega S_y,$$

and

$$\frac{dS_y}{dt} = -\omega S_x,$$

while

$$\frac{dS_z}{dt}=0.$$

Thus the component along the field is conserved and the transverse components rotate.

Operator Solution

The coupled equations imply

$$\frac{d^2S_x}{dt^2} = -\omega^2S_x, \qquad \frac{d^2S_y}{dt^2} = -\omega^2S_y.$$

With operators evaluated at $t=0$, the solution is

$$S_x(t) = S_x(0)\cos\omega t + S_y(0)\sin\omega t,$$

and

$$S_y(t) = S_y(0)\cos\omega t - S_x(0)\sin\omega t.$$

Also

$$S_z(t)=S_z(0).$$

The sign of $\omega=\gamma B_0$ determines the sense of rotation. Many authors quote only the magnitude $|\omega|$ and handle the sense of precession by convention.

Spinor Phase Derivation

The same result appears in the Schrödinger picture. Write an initial spinor in the $S_z$ basis:

$$|\psi(0)\rangle = \alpha|+z\rangle+\beta|-z\rangle.$$

The energies are

$$E_{+z} = -\frac{\hbar\omega}{2}, \qquad E_{-z} = \frac{\hbar\omega}{2}.$$

Therefore

$$|\psi(t)\rangle = \alpha e^{i\omega t/2}|+z\rangle + \beta e^{-i\omega t/2}|-z\rangle.$$

The global phase is irrelevant, but the relative phase changes as

$$\frac{\beta(t)}{\alpha(t)} = \frac{\beta}{\alpha}e^{-i\omega t}.$$

If the initial state is written on the Bloch sphere as

$$|\psi(0)\rangle = \cos\frac{\theta}{2}|+z\rangle + e^{i\phi_0}\sin\frac{\theta}{2}|-z\rangle,$$

then

$$|\psi(t)\rangle \sim \cos\frac{\theta}{2}|+z\rangle + e^{i(\phi_0-\omega t)} \sin\frac{\theta}{2}|-z\rangle.$$

The polar angle $\theta$ is constant, while the azimuth changes:

$$\phi(t)=\phi_0-\omega t.$$

The Bloch vector therefore precesses around the $z$ axis.

Bloch-Vector Equation

For a pure or mixed spin-$1/2$ state,

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right).$$

Under

$$H = -\frac{\hbar\gamma}{2}\mathbf B\cdot\boldsymbol\sigma,$$

the Bloch vector obeys

$$\frac{d\mathbf r}{dt} = \gamma\,\mathbf r\times\mathbf B.$$

This is the same precession equation as for $\langle\mathbf S\rangle$, since

$$\langle\mathbf S\rangle = \frac{\hbar}{2}\mathbf r.$$

The trajectory is a circle of fixed polar angle around the field direction when the field is static and the system is closed.

Energy Eigenstates Do Not Precess

If the spin is prepared in $|+z\rangle$ or $|-z\rangle$ for a field along $z$, the state only accumulates a phase. Its Bloch vector sits at the north or south pole, so there is no visible precession of the expectation value.

Precession requires transverse coherence. For example,

$$|+x\rangle = \frac{1}{\sqrt2} \left( |+z\rangle+|-z\rangle \right)$$

has a Bloch vector in the equatorial plane. The relative phase between the two $S_z$ components changes in time, so the vector rotates around the $z$ axis.

Relation to Magnetic Resonance

The Larmor frequency is also the transition frequency between the two spin projections in a static field:

$$\Delta E=\hbar\omega_L.$$

An oscillating transverse field near this frequency can drive transitions. That driven problem is not merely free precession; it introduces an explicitly time-dependent Hamiltonian and, after approximations, Rabi oscillations. The historical technique entry is Magnetic Resonance, the spectroscopy-facing measurement map is Magnetic Resonance Overview, and the two-level driven model is Rabi Oscillations: First Encounter. Magnetometry places the Larmor law inside a calibrated field measurement with finite bandwidth, spatial weighting, noise, and systematic effects.

Sign Conventions

Three signs are easy to mix:

• the sign of the particle charge inside $\gamma$;
• the sign convention for $\boldsymbol\mu=\gamma\mathbf S$;
• the orientation chosen for the positive magnetic field axis.

For an electron, using $e>0$ and

$$\boldsymbol\mu_S = -g_s\mu_B\frac{\mathbf S}{\hbar},$$

one has $\gamma<0$. The spin expectation therefore precesses in the opposite sense compared with a positive-$\gamma$ spin in the same $\mathbf B$. The frequency magnitude is still $|\gamma|B_0$.

When comparing books, check whether they define the Larmor frequency as $\omega_L=\gamma B_0$, $\omega_L=-\gamma B_0$, or only as the positive magnitude.

Common Mistakes

• Forgetting that the sign of $\gamma$ changes the sense of precession.
• Saying that an energy eigenstate “precesses” when only its global phase changes.
• Confusing Larmor precession in a static field with Rabi oscillations driven by a transverse field.
• Treating the Bloch vector as a literal classical spinning object rather than an expectation-value representation.
• Dropping the distinction between angular frequency $\omega$ and ordinary frequency $f=\omega/(2\pi)$.
• Assuming damping or relaxation is present in the closed-system Larmor equation; relaxation requires environmental or open-system physics.

Cross-Links

• Spin in Magnetic Fields
• Magnetic Moments from Orbital Motion
• Bloch Sphere
• Spin Rotations
• Pauli Matrices
• Heisenberg Equations of Motion
• Spin-1/2 as a Canonical System
• Magnetic Resonance
• Magnetic Resonance Overview
• Magnetometry
• Rabi Oscillations: First Encounter
• Spin in Magnetic Field Hamiltonian

References

• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. P. Slichter, Principles of Magnetic Resonance, 3rd ed., Springer, 1990.
• A. Abragam, The Principles of Nuclear Magnetism, Oxford University Press, 1961.

Exercises

1. Starting from $H=-\gamma B_0S_z$, derive $dS_x/dt=\gamma B_0S_y$.

Solution

Use the Heisenberg equation:

$$\frac{dS_x}{dt} = \frac{i}{\hbar}[H,S_x].$$

With $H=-\gamma B_0S_z$,

$$\frac{dS_x}{dt} = -\frac{i\gamma B_0}{\hbar}[S_z,S_x].$$

Since

$$[S_z,S_x]=i\hbar S_y,$$

one obtains

$$\frac{dS_x}{dt} = \gamma B_0S_y.$$

2. A spin starts in $|+x\rangle$ in a field $\mathbf B=B_0\hat{\mathbf z}$. Write the state at time $t$ up to global phase.

Solution

Use

$$|+x\rangle = \frac{1}{\sqrt2} \left( |+z\rangle+|-z\rangle \right).$$

Under the Hamiltonian $H=-(\hbar\omega/2)\sigma_z$, the state becomes

$$|\psi(t)\rangle = \frac{1}{\sqrt2} \left( e^{i\omega t/2}|+z\rangle + e^{-i\omega t/2}|-z\rangle \right).$$

Removing the global phase $e^{i\omega t/2}$ gives

$$|\psi(t)\rangle \sim \frac{1}{\sqrt2} \left( |+z\rangle + e^{-i\omega t}|-z\rangle \right).$$

3. For $\gamma<0$ and $B_0>0$, does the Bloch azimuth $\phi(t)$ increase or decrease in the convention of this page?

Solution

This page uses

$$\phi(t)=\phi_0-\omega t, \qquad \omega=\gamma B_0.$$

If $\gamma<0$ and $B_0>0$, then $\omega<0$. Therefore

$$\phi(t)=\phi_0+|\omega|t,$$

so the azimuth increases.

4. Why does an eigenstate of $S_z$ not show transverse Larmor precession in a field along $z$?

Solution

An $S_z$ eigenstate is also an energy eigenstate of $H=-\gamma B_0S_z$. Time evolution only multiplies it by a phase. Since pure states are rays, a global phase does not change the Bloch vector or any spin expectation value. The state has no transverse coherence between $|+z\rangle$ and $|-z\rangle$, so there is no transverse vector to rotate.