Magnetic Moments from Orbital Motion

A charged particle with orbital angular momentum carries a magnetic moment. For one particle of charge $q$ and mass $m$, the orbital contribution is

$$\boldsymbol\mu_L = \frac{q}{2m}\mathbf L.$$

For an electron one usually writes $q=-e$ with $e>0$ and defines the Bohr magneton

$$\mu_B = \frac{e\hbar}{2m_e}.$$

Then

$$\boldsymbol\mu_L = -\mu_B\frac{\mathbf L}{\hbar}.$$

The minus sign is physical: the electron’s orbital magnetic moment points opposite to its orbital angular momentum. This page is the symmetry-side account of that relation and its immediate Zeeman and Larmor consequences. The broader historical and experimental role of magnetic moments is treated in Magnetic Moments, spin and $g$-factor conventions are collected in Magnetic Moments and g-Factors, and the spin-only magnetic-field Hamiltonian is treated in Spin in Magnetic Fields.

Classical Origin

The formula is easiest to remember from a current loop. A charge $q$ moving in a circular orbit of angular frequency $\omega$ forms a current

$$I_{\rm loop} = \frac{q\omega}{2\pi}.$$

For radius $r$, the magnetic moment magnitude is current times area:

$$\mu = I_{\rm loop}\pi r^2 = \frac{q\omega r^2}{2}.$$

The orbital angular momentum magnitude is

$$L = m r^2\omega,$$

so

$$\mu = \frac{q}{2m}L.$$

The vector relation is

$$\boldsymbol\mu_L = \frac{q}{2m}\mathbf L.$$

Quantum mechanics promotes this to an operator statement for the orbital part of the magnetic moment. The proportionality is not a new angular-momentum algebra; it is the magnetic response associated with charged orbital motion.

Electron Convention and Bohr Magneton

For the electron, the charge is negative. With $e>0$ and $q=-e$,

$$\boldsymbol\mu_L = -\frac{e}{2m_e}\mathbf L.$$

The Bohr magneton rewrites the coefficient in natural atomic units:

$$\mu_B = \frac{e\hbar}{2m_e}, \qquad \boldsymbol\mu_L = -\mu_B\frac{\mathbf L}{\hbar}.$$

The orbital $g$ factor is therefore

$$g_L=1$$

in the convention

$$\boldsymbol\mu_L = -g_L\mu_B\frac{\mathbf L}{\hbar}$$

for an electron. This should be contrasted with the leading electron spin relation

$$\boldsymbol\mu_S = -g_s\mu_B\frac{\mathbf S}{\hbar}, \qquad g_s\simeq2.$$

The factor of approximately two is a spin fact, not an orbital fact. Applying $g_s\simeq2$ to an ordinary orbital moment is a common source of wrong Zeeman splittings.

Magnetic Coupling

The energy of a magnetic moment in a magnetic field is

$$H_Z = -\boldsymbol\mu\cdot\mathbf B.$$

For the orbital moment of a charge $q$,

$$H_Z^{(L)} = -\frac{q}{2m}\mathbf L\cdot\mathbf B.$$

For an electron,

$$H_Z^{(L)} = \mu_B \frac{\mathbf L\cdot\mathbf B}{\hbar}.$$

If the field is uniform and points along $z$,

$$\mathbf B=B\hat{\mathbf z},$$

then

$$H_Z^{(L)} = \mu_B B\frac{L_z}{\hbar}.$$

On a state with

$$L_z|\ell m\rangle = \hbar m|\ell m\rangle,$$

the first-order orbital Zeeman shift is

$$\Delta E_{\ell m}^{(L)} = \mu_B Bm.$$

The sign is worth reading carefully. For an electron with $B>0$, positive $m$ means $L_z>0$, but the magnetic moment points in the negative $z$ direction. The energy $-\boldsymbol\mu\cdot\mathbf B$ is therefore higher.

Relation to Minimal Coupling

The same term appears from the charged-particle Hamiltonian in a magnetic field. For a particle in a scalar potential $V(r)$ and a vector potential $\mathbf A$,

$$H = \frac{1}{2m} \left( \mathbf P-q\mathbf A \right)^2 + V(r).$$

For a uniform magnetic field, choose the symmetric gauge

$$\mathbf A = \frac12\mathbf B\times\mathbf R.$$

For constant $\mathbf B$, this gives

$$\begin{aligned} H &= \frac{\mathbf P^2}{2m} +V(r) -\frac{q}{2m}\mathbf B\cdot\mathbf L \\ &\quad +\frac{q^2}{8m} \left\lvert \mathbf B\times\mathbf R \right\rvert^2 . \end{aligned}$$

The middle term is exactly $-\boldsymbol\mu_L\cdot\mathbf B$. The final term is quadratic in the field and is often called the diamagnetic term in this atomic context. Weak-field orbital Zeeman discussions usually keep the linear term first, but the full minimally coupled Hamiltonian contains both.

This derivation also explains a limitation of the simple formula. In a magnetic field, canonical momentum and kinetic momentum differ:

$$\boldsymbol\pi = \mathbf P-q\mathbf A.$$

The simple orbital Zeeman term is a perturbative statement around the zero-field orbital angular momentum $\mathbf L=\mathbf R\times\mathbf P$. Fully gauge-covariant magnetic-field dynamics belongs to Minimal Coupling in Wave Mechanics.

Symmetry and Zeeman Splitting

Without a magnetic field, a central Hamiltonian has full rotational symmetry:

$$[H_0,L_i]=0, \qquad i=x,y,z.$$

A fixed external field

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

selects an axis. In the linear orbital Zeeman approximation,

$$H = H_0+\mu_B B\frac{L_z}{\hbar}$$

for an electron in a central potential. This Hamiltonian still commutes with $L_z$, so $m$ remains a good label. Since the perturbation is proportional to $L_z$, it splits the formerly degenerate $m$ values:

$$\Delta E_m = \mu_B Bm.$$

This is the simplest normal-Zeeman pattern. It is not the general atomic Zeeman effect. Real atomic levels can include electron spin, spin–orbit coupling, total angular momentum, nuclear effects, and field-strength regimes where different labels become appropriate. The historical and spectroscopic context is in Zeeman Effect Revisited; the perturbative basis hierarchy and quadratic correction are in Zeeman Effect as a Perturbation Example.

The full uniform-field Hamiltonian including the quadratic diamagnetic term has only axial rotational symmetry in general. It preserves $L_z$ for a central problem, but it need not preserve $L^2$.

Larmor Precession Preview

Write the orbital moment as

$$\boldsymbol\mu_L = \gamma_L\mathbf L, \qquad \gamma_L=\frac{q}{2m}.$$

The orbital Zeeman Hamiltonian is

$$H_Z^{(L)} = -\gamma_L\mathbf L\cdot\mathbf B.$$

Using the angular-momentum commutators,

$$[L_i,L_j] = i\hbar\sum_k\epsilon_{ijk}L_k,$$

the Heisenberg equation gives

$$\frac{d\mathbf L}{dt} = \gamma_L\,\mathbf L\times\mathbf B.$$

Thus the transverse orbital angular momentum precesses around the magnetic field. The frequency magnitude is

$$\omega_L = \left\lvert\gamma_L\right\rvert B = \frac{\lvert q\rvert B}{2m}.$$

For an electron this is

$$\omega_L = \frac{eB}{2m_e} = \frac{\mu_BB}{\hbar}.$$

The spin precession page uses the same Hamiltonian logic with $\boldsymbol\mu=\gamma\mathbf S$, but the spin gyromagnetic ratio contains the spin $g$ factor. See Larmor Precession.

Physical Interpretation

The orbital magnetic moment is not a statement that the electron travels on a little classical circle. It is a statement about how a charged orbital wavefunction responds to rotations and magnetic fields. The current-loop picture gives the coefficient, while quantum angular momentum gives the allowed projections.

For an orbital eigenstate,

$$L_z=\hbar m$$

means the magnetic moment component for an electron is

$$\mu_{L,z} = -\mu_B m.$$

A magnetic field can distinguish different $m$ values because it supplies a physical axis. Without that axis, the $2\ell+1$ states in a central potential are related by rotations and are degenerate.

The Bohr magneton sets the natural scale of electron magnetic energies in atomic physics. Nuclear magnetic moments are much smaller because they involve nuclear masses rather than the electron mass. Effective magnetic moments in solids, molecules, and bands can be very different because the relevant degrees of freedom and effective Hamiltonians are different.

Common Mistakes

• Forgetting the electron sign: for $q=-e$, $\boldsymbol\mu_L$ is antiparallel to $\mathbf L$.
• Using the spin value $g_s\simeq2$ for an orbital magnetic moment; ordinary orbital motion has $g_L=1$.
• Treating $m$ as a literal classical orbit rather than an angular-momentum projection.
• Writing a Zeeman shift without specifying the charge convention and the magnetic moment being used.
• Ignoring the quadratic diamagnetic term when the field is not weak enough for the linear approximation.
• Confusing canonical angular momentum $\mathbf R\times\mathbf P$ with fully gauge-covariant mechanical motion in a magnetic field.
• Assuming a uniform magnetic field spatially separates a beam; Stern–Gerlach separation requires a field gradient.

Cross-Links

• Magnetic Moments in Matter carries this orbital operator into projected material subspaces and compares it with spin, ordered, and fluctuating moment measures without treating bulk orbital magnetization as a sum of site-local angular momenta.
• Orbital Angular Momentum
• Angular Momentum Algebra
• Central Potentials and Rotational Symmetry
• Hydrogen Atom Angular Structure
• Spin in Magnetic Fields
• Magnetic Moments and g-Factors
• Larmor Precession
• Spin–Orbit Coupling
• Magnetic Moments
• Zeeman Effect Revisited
• Zeeman Effect as a Perturbation Example
• Zeeman Effect
• Minimal Coupling in Wave Mechanics
• Charged Particle in a 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.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• C. J. Foot, Atomic Physics, Oxford University Press, 2005.

Exercises

1. Derive the electron orbital Zeeman Hamiltonian for $\mathbf B=B\hat{\mathbf z}$ from $\boldsymbol\mu_L=-\mu_B\mathbf L/\hbar$.

Solution

The magnetic interaction is

$$H_Z = -\boldsymbol\mu_L\cdot\mathbf B.$$

For an electron,

$$\boldsymbol\mu_L = -\mu_B\frac{\mathbf L}{\hbar}.$$

With $\mathbf B=B\hat{\mathbf z}$,

$$H_Z^{(L)} = -\left( -\mu_B\frac{\mathbf L}{\hbar} \right)\cdot B\hat{\mathbf z} = \mu_BB\frac{L_z}{\hbar}.$$

2. Find the orbital Zeeman shifts for an electron with $\ell=1$ and $m=-1,0,1$.

Solution

The linear orbital shift is

$$\Delta E_m = \mu_BBm.$$

For $m=-1,0,1$, the shifts are

$$-\mu_BB, \qquad 0, \qquad \mu_BB.$$

3. Show that the orbital Larmor frequency for a charge $q$ has magnitude $\lvert q\rvert B/(2m)$.

Solution

Write

$$\boldsymbol\mu_L = \gamma_L\mathbf L, \qquad \gamma_L=\frac{q}{2m}.$$

The Hamiltonian is

$$H = -\gamma_L\mathbf L\cdot\mathbf B.$$

The Heisenberg equation gives

$$\frac{d\mathbf L}{dt} = \gamma_L\mathbf L\times\mathbf B.$$

For a uniform field of magnitude $B$, the transverse components rotate with angular frequency magnitude

$$\omega_L = \lvert\gamma_L\rvert B = \frac{\lvert q\rvert B}{2m}.$$

4. Explain why the simple orbital Zeeman term is not the whole magnetic-field Hamiltonian.

Solution

Minimal coupling gives

$$H = \frac{1}{2m} \left( \mathbf P-q\mathbf A \right)^2 +V.$$

For a uniform magnetic field in symmetric gauge, this expands to

$$H = H_0 -\frac{q}{2m}\mathbf B\cdot\mathbf L +\frac{q^2}{8m} \left\lvert \mathbf B\times\mathbf R \right\rvert^2.$$

The middle term is the orbital Zeeman coupling. The final term is quadratic in $B$ and is not included in the simple linear Zeeman shift. Spin magnetic moments, relativistic corrections, and many-body effects can add further terms in realistic atoms and materials.