Magnetic Moment

A magnetic moment is the coefficient of a particle’s linear spin response to a weak magnetic field. Minimal Dirac coupling fixes one contribution; a covariant Pauli interaction supplies an independent correction. This page matches that correction to a low-energy Hamiltonian and to electromagnetic form factors. The Pauli Equation owns the magnetic-moment convention, the minimal $g=2$ result, Zeeman splitting, and spin precession.

Required background. The Pauli Equation defines the signed-charge convention, and Minimal Coupling defines the field tensor. Helpful background. The Gordon identity on Dirac Current relates the covariant vertex to its spin response.

A covariant Pauli interaction

Take a massive charged Dirac particle with signed charge $q\ne0$. Parameterize its moment by

$$\boldsymbol\mu =g\frac{q}{2m}\mathbf S,\qquad g=2(1+a),\qquad \mathbf S=\frac{\hbar}{2}\boldsymbol\sigma.$$

For the charged lepton convention, $g$ is positive and the electron sign resides in $q=-e$. The parameter $a$ is the anomaly relative to minimal Dirac coupling. For general composites the coefficient can have either sign; no positive-$g$ assumption should be imposed on an independently matched moment.

The local Dirac–Pauli equation with a constant real coefficient $a$ is

$$\left[ i\hbar c\gamma^\mu D_\mu-mc^2 -\frac{a q\hbar}{4m} \sigma^{\mu\nu}F_{\mu\nu} \right]\psi=0,$$

where $\sigma^{\mu\nu}=i[\gamma^\mu,\gamma^\nu]/2$. Its field-strength interaction is gauge invariant: $F_{\mu\nu}$ is invariant and the spinor transforms by the same phase as in minimal coupling. It does not change the charge in $D_\mu$.

This is a specified effective interaction, not a derivation of $a$ from a wave equation. A measured coefficient, constituent model, or underlying quantum field theory must determine it. Taking a constant coefficient also omits resolved momentum dependence and further derivative operators.

Matching the magnetic coefficient

With the tensor convention of Minimal Coupling,

$$F_{0i}=\frac{E_i}{c},\qquad F_{ij}=-\epsilon_{ijk}B_k,$$

the contraction is

$$\sigma^{\mu\nu}F_{\mu\nu} =2\left( \frac{i}{c}\boldsymbol\alpha\cdot\mathbf E -\boldsymbol\Sigma\cdot\mathbf B \right).$$

Multiplication of the equation by $\beta=\gamma^0$ therefore gives the additional Hamiltonian

$$\delta H =-\frac{a q\hbar}{2m}\beta\boldsymbol\Sigma\cdot\mathbf B +\frac{i a q\hbar}{2mc}\beta\boldsymbol\alpha\cdot\mathbf E.$$

Both terms are Hermitian for real fields: $\beta\boldsymbol\Sigma$ is Hermitian and $\beta\boldsymbol\alpha$ is anti-Hermitian. The electric term is odd in the energy-block decomposition. It is part of the Lorentz-covariant magnetic interaction, not a rest-frame electric dipole energy proportional to $\mathbf S\cdot\mathbf E$.

At leading low-energy order the upper block yields

$$\delta H_{\rm spin} =-\frac{a q\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B.$$

Adding the minimal Dirac contribution gives

$$H_{\rm spin} =-\frac{(1+a)q\hbar}{2m} \boldsymbol\sigma\cdot\mathbf B,$$

which matches $g=2(1+a)$. This matching assumes weak, slowly varying fields and the positive-energy nonrelativistic regime. At higher order, the odd electric term and its commutators also contribute. Changing only the Zeeman coefficient in a relativistic Hamiltonian does not include all effects of the covariant Pauli operator.

Static form factors and the anomaly

For this paragraph set $\hbar=c=1$. For on-shell spinors of the same mass, define the momentum transfer $k=p'-p$ and write the parity-even electromagnetic vertex as

$$\Gamma^\mu(p',p) =q\left[ F_1(k^2)\gamma^\mu +\frac{iF_2(k^2)}{2m} \sigma^{\mu\nu}k_\nu \right].$$

The physical charge has been factored out, so $F_1(0)=1$. With this normalization,

$$a=F_2(0),\qquad g=2\bigl[F_1(0)+F_2(0)\bigr].$$

One way to see why both form factors enter the spin coefficient is the Gordon identity:

$$\bar u'\gamma^\mu u =\bar u'\left[ \frac{(p'+p)^\mu}{2m} +\frac{i\sigma^{\mu\nu}k_\nu}{2m} \right]u.$$

After this substitution the spin-dependent part of the vertex carries $F_1+F_2$, whereas the convection part carries $F_1$. The static, long-wavelength magnetic response selects the former combination. Grozin (2005) develops this form-factor interpretation and its perturbative calculation.

The local Pauli coefficient matches the zero-transfer value. At finite transfer, form factors describe additional structure and radiative response; a constant anomalous moment cannot reproduce that information. Schwinger’s leading QED correction is discussed on the Pauli owner page; its loop calculation requires quantized fields and is outside this one-particle matching argument.

Intrinsic spin response and orbital energy

Return to SI units. For a uniform field, choose the symmetric gauge $\mathbf A=\mathbf B\times\mathbf r/2$. The leading low-energy orbital term expands as

$$\frac{(\mathbf p-q\mathbf A)^2}{2m} =\frac{\mathbf p^2}{2m} -\frac{q}{2m}\mathbf B\cdot\mathbf L +\frac{q^2}{8m}|\mathbf B\times\mathbf r|^2.$$

Thus the energy linear in a weak field contains both orbital and intrinsic contributions,

$$H_{\rm linear} =-\frac{q}{2m} (\mathbf L+g\mathbf S)\cdot\mathbf B.$$

The quadratic term is the diamagnetic contribution. These are terms in a Hamiltonian with a specified orbital state and boundary conditions. The derivative $-\partial E/\partial B$ of a complete energy level can include orbital motion, state mixing, and relativistic binding corrections. It need not equal a free particle’s intrinsic spin moment.

For an isolated, differentiable nondegenerate eigenvalue, the Hellmann–Feynman theorem relates that derivative to $-\langle\partial H/\partial B\rangle$. At a degeneracy, first diagonalize the perturbation in the degenerate subspace. Atomic Landé factors and material-dependent effective spin Hamiltonians therefore require the appropriate state and environment; they are not alternative measurements of the same free Dirac coefficient.

Neutral particles and independently signed moments

The parameterization $a q/(2m)$ is inconvenient when $q=0$. Instead introduce a real coefficient $\mu_*$ with units of magnetic moment and use

$$\left[ i\hbar c\gamma^\mu D_\mu-mc^2 -\frac{\mu_*}{2}\sigma^{\mu\nu}F_{\mu\nu} \right]\psi=0.$$

For a neutral Dirac particle, $D_\mu=\partial_\mu$, but the Pauli term can remain nonzero. Its low-energy interaction is

$$H_{\rm spin}=-\mu_*\boldsymbol\sigma\cdot\mathbf B, \qquad \boldsymbol\mu=\frac{2\mu_*}{\hbar}\mathbf S.$$

For the additional moment of a charged particle, $\mu_*=a q\hbar/(2m)$ recovers the previous equations. For a neutral particle it is an independent signed coefficient. This observation does not assert that every neutral particle permits such a diagonal operator. For an identical anticommuting Majorana field, the tensor bilinear $\bar\Psi_{\rm M}\sigma^{\mu\nu}\Psi_{\rm M}$ vanishes, excluding this diagonal Pauli moment (Dreiner, Haber, and Martin, 2010). This is a field-statistics statement. Majorana Spinors explains why identities for anticommuting fields must be distinguished from those for commuting classical spinor components.

Exercises

Checking Hermiticity. Show that $i\beta\alpha_i$ is Hermitian. Why does the explicit $i$ in the electric interaction not make the Hamiltonian non-Hermitian?

Solution

Since $\alpha_i^\dagger=\alpha_i$, $\beta^\dagger=\beta$, and $\{\alpha_i,\beta\}=0$,

$$(i\beta\alpha_i)^\dagger =-i\alpha_i\beta=i\beta\alpha_i.$$

The matrix that multiplies $i$ was anti-Hermitian. For real multiplicative $E_i$, the product is Hermitian on the appropriate common domain.

Orbital contamination. In a weak field $B\hat{\mathbf z}$, suppose the perturbation is diagonal in a state with $L_z=\hbar m_\ell$ and $S_z=\hbar m_s$. Find its first-order slope.

Solution

The linear shift is

$$\Delta E =-\frac{q\hbar B}{2m}(m_\ell+g m_s).$$

The slope measures the total combination. Only after determining or removing the orbital contribution can it isolate the intrinsic $g$.

Neutral matching. A neutral spin-half particle has $\boldsymbol\mu=\gamma\mathbf S$ with signed gyromagnetic ratio $\gamma$. Determine $\mu_*$ and explain why assigning it a finite charge-normalized anomaly is unsuitable.

Solution

Matching gives $\mu_*=\gamma\hbar/2$. The product $a q\hbar/(2m)$ vanishes for $q=0$ at fixed finite $a$, so it cannot represent a nonzero neutral moment. The independently signed coefficient avoids a singular normalization.

References

• Dreiner, Herbi K., Howard E. Haber, and Stephen P. Martin. “Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry.” Physics Reports 494, 1–196 (2010). doi:10.1016/j.physrep.2010.05.002. Corrected manuscript.
• Grozin, Andrey. “Lectures on QED and QCD.” Lecture notes (2005). arXiv:hep-ph/0508242.
• Schwinger, Julian. “On Quantum-Electrodynamics and the Magnetic Moment of the Electron.” Physical Review 73, 416–417 (1948). doi:10.1103/PhysRev.73.416.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.