Dirac to Pauli

The Pauli Hamiltonian is not obtained by merely deleting the lower two components of a Dirac spinor. In the Dirac basis, those components are small in a positive-energy, low-momentum regime, but they mediate the magnetic spin coupling. Solving for them to leading order gives

$$H_{\mathrm{eff}} = q\Phi +\frac{\boldsymbol\pi^2}{2m} -\frac{q\hbar}{2m} \boldsymbol\sigma\cdot\mathbf B,$$

where $q$ is signed and $\boldsymbol\pi=-i\hbar\nabla-q\mathbf A$. The coefficient corresponds to the tree-level Dirac value $g=2$. This page derives that result and states the hierarchy needed to trust it.

Required background. The Covariant Dirac Equation, Gamma-Matrix Conventions, and Minimal Coupling supply the Hamiltonian, Dirac basis, signed charge, and kinetic-momentum commutator. Two-component Pauli algebra and controlled asymptotic expansions are assumed.

Separate the rest energy and spinor blocks

The minimally coupled Dirac equation in Hamiltonian form is

$$i\hbar\partial_t\psi = \left( c\boldsymbol\alpha\cdot\boldsymbol\pi +\beta mc^2 +q\Phi \right)\psi.$$

In the Dirac basis,

$$\boldsymbol\alpha\cdot\boldsymbol\pi = \begin{pmatrix} 0_2&\boldsymbol\sigma\cdot\boldsymbol\pi\\ \boldsymbol\sigma\cdot\boldsymbol\pi&0_2 \end{pmatrix}, \qquad \beta = \begin{pmatrix} I_2&0_2\\ 0_2&-I_2 \end{pmatrix}.$$

For a state concentrated in the positive-energy sector, remove the rest- energy phase and write

$$\psi(t,\mathbf x) = e^{-imc^2t/\hbar} \begin{pmatrix} \varphi(t,\mathbf x)\\ \chi(t,\mathbf x) \end{pmatrix}.$$

The two-component amplitudes $\varphi$ and $\chi$ are often called the large and small components in this regime. Those names are conclusions of the scale hierarchy, not permanent basis-independent labels.

Substitution gives the exact coupled equations

$$i\hbar\partial_t\varphi = q\Phi\varphi +c\boldsymbol\sigma\cdot\boldsymbol\pi\,\chi,$$

and

$$i\hbar\partial_t\chi = c\boldsymbol\sigma\cdot\boldsymbol\pi\,\varphi +(q\Phi-2mc^2)\chi.$$

The lower block lies a rest-energy gap $2mc^2$ away after the positive-branch phase has been removed. Losing this factor of two gives the wrong magnetic and kinetic coefficients.

After removing $mc^2$, the upper component is slow while the lower component is offset by $-2mc^2$. The off-diagonal operator $c\boldsymbol\sigma\cdot\boldsymbol\pi$ creates the leading small component; substituting it back produces both kinetic and magnetic terms.

Leading elimination of the small component

Rearrange the lower equation:

$$\left( 2mc^2+i\hbar\partial_t-q\Phi \right)\chi = c\boldsymbol\sigma\cdot\boldsymbol\pi\,\varphi.$$

The positive-energy nonrelativistic regime requires gauge-covariant residual energies and kinetic couplings small compared with the gap. An absolute scalar-potential offset is gauge dependent; physically relevant potential differences and transition frequencies must respect the expansion. A useful local hierarchy is

$$\left|i\hbar\partial_t-q\Phi\right| \ll2mc^2, \qquad c|\boldsymbol\pi|\ll2mc^2,$$

together with fields that vary slowly enough not to induce transitions across the gap. To leading order, replace the operator in parentheses by $2mc^2$:

$$\chi \simeq \frac{\boldsymbol\sigma\cdot\boldsymbol\pi}{2mc}\,\varphi.$$

For a free positive-energy packet, this estimates

$$\frac{\|\chi\|}{\|\varphi\|} \sim \frac{|\mathbf p|}{2mc} \sim \frac{v}{2c}.$$

Setting $\chi=0$ would discard the entire order needed to generate the Pauli spin term. The correct leading procedure is to solve for $\chi$ and substitute it into the upper equation.

For spatially or temporally varying potentials, the formal inverse

$$\chi = \left( 2mc^2+i\hbar\partial_t-q\Phi \right)^{-1} c\boldsymbol\sigma\cdot\boldsymbol\pi\,\varphi$$

is only a formal ordered expression until an inverse and its boundary or initial conditions have been chosen. For time evolution the solution also contains the homogeneous lower-component term fixed by the initial state; the inverse alone must not silently discard it. Stationary reduction and time-dependent elimination are distinguished in Effective Hamiltonians. Beyond leading order, time derivatives, potentials, and kinetic momenta must remain in their stated order.

The Pauli identity with noncommuting momenta

Use

$$\sigma_i\sigma_j = \delta_{ij}I_2+i\epsilon_{ijk}\sigma_k.$$

Then

$$\begin{aligned} (\boldsymbol\sigma\cdot\boldsymbol\pi)^2 &= \sigma_i\sigma_j\pi_i\pi_j \\ &= \boldsymbol\pi^2 +i\epsilon_{ijk}\sigma_k\pi_i\pi_j \\ &= \boldsymbol\pi^2 +\frac{i}{2}\epsilon_{ijk}\sigma_k[\pi_i,\pi_j]. \end{aligned}$$

Minimal coupling gives

$$[\pi_i,\pi_j] = iq\hbar\epsilon_{ij\ell}B_\ell.$$

Using $\epsilon_{ijk}\epsilon_{ij\ell}=2\delta_{k\ell}$,

$$\boxed{ (\boldsymbol\sigma\cdot\boldsymbol\pi)^2 = \boldsymbol\pi^2 -q\hbar\boldsymbol\sigma\cdot\mathbf B }.$$

The magnetic term is a direct consequence of the noncommuting kinetic momenta. If one replaces $\pi_i\pi_j$ by commuting numbers, the spin coupling vanishes incorrectly.

Effective Pauli Hamiltonian

Substitute the leading $\chi$ into the upper equation:

$$i\hbar\partial_t\varphi = q\Phi\varphi +\frac{1}{2m} (\boldsymbol\sigma\cdot\boldsymbol\pi)^2\varphi.$$

The Pauli identity gives

$$i\hbar\partial_t\varphi = \left[ q\Phi +\frac{\boldsymbol\pi^2}{2m} -\frac{q\hbar}{2m} \boldsymbol\sigma\cdot\mathbf B \right]\varphi.$$

With the physical spin operator

$$\mathbf S=\frac{\hbar}{2}\boldsymbol\sigma,$$

the magnetic term is

$$-\frac{q}{m}\mathbf S\cdot\mathbf B = -g\frac{q}{2m}\mathbf S\cdot\mathbf B \quad\text{with}\quad g=2.$$

For an electron, $q=-e$, so

$$-\frac{q\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B = +\frac{e\hbar}{2m}\boldsymbol\sigma\cdot\mathbf B.$$

This sign says that an electron’s magnetic moment is antiparallel to its spin. The lower-energy Zeeman state in a positive $z$ field therefore has spin antiparallel to $\mathbf B$.

From Pauli dynamics to relativistic corrections

The algebraic elimination above is a leading positive-energy approximation. It is not the full Foldy–Wouthuysen transformation. A systematic Foldy–Wouthuysen construction seeks a unitary block diagonalization of the Dirac Hamiltonian order by order, transforming states and observables as well as the Hamiltonian.

At higher order, a controlled calculation can generate:

• the free kinetic correction $-\mathbf p^4/(8m^3c^2)$;
• spin–orbit coupling;
• the Darwin term;
• field-gradient, time-dependence, and operator-ordering contributions.

Their form depends on whether the fields are static, which field orders are retained, and how the noncommuting operators are ordered. The higher-order calculation must preserve Hermiticity and transform the state normalization and observables consistently.

The Foldy–Wouthuysen Transformation explains the representation change and transformed observables. The static Foldy–Wouthuysen expansion provides those declared assumptions and derives the higher-order operators.

The leading Pauli result is reliable when the state stays far from the negative-energy sector, momenta are small compared with $mc$, external fields are weak and slowly varying on the Compton scale, and pair creation is negligible. It does not provide a Lorentz-covariant two-component theory or a description of antiparticles.

Exercises

1. Noncommuting Pauli identity

Derive $(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ without assuming that the components of $\boldsymbol\pi$ commute.

Solution

Only the antisymmetric part of $\pi_i\pi_j$ survives contraction with $\epsilon_{ijk}$:

$$i\epsilon_{ijk}\sigma_k\pi_i\pi_j = \frac{i}{2}\epsilon_{ijk}\sigma_k[\pi_i,\pi_j].$$

Substituting $[\pi_i,\pi_j]=iq\hbar\epsilon_{ij\ell}B_\ell$ and contracting the epsilon symbols gives $-q\hbar\boldsymbol\sigma\cdot\mathbf B$.

2. Electron sign

Set $q=-e$ with $e>0$ in the effective Hamiltonian. Which spin projection has lower energy in a uniform field $\mathbf B=B\hat{\mathbf z}$?

Solution

The spin term is

$$H_{\mathrm{spin}} = +\frac{e\hbar B}{2m}\sigma_z.$$

Its eigenvalues are $\pm e\hbar B/(2m)$. The $\sigma_z=-1$ state, whose spin is antiparallel to $\mathbf B$, has the lower energy.

3. Size of the small component

Estimate $\|\chi\|/\|\varphi\|$ for a free packet with speed $v=10^{-2}c$.

Solution

At leading order $|\mathbf p|\simeq mv$, so

$$\frac{\|\chi\|}{\|\varphi\|} \sim \frac{|\mathbf p|}{2mc} \simeq \frac{v}{2c} =5\times10^{-3}.$$

The probability weight of the lower component is smaller by roughly the square of this ratio.

References

• L. L. Foldy and S. A. Wouthuysen, “On the Dirac Theory of Spin 1/2 Particles and Its Non-Relativistic Limit,” Physical Review 78, 29–36, 1950, doi:10.1103/PhysRev.78.29.
• W. Greiner, Relativistic Quantum Mechanics: Wave Equations, 3rd ed., Springer, 2000, doi:10.1007/978-3-662-04275-5.
• J. J. Sakurai, Advanced Quantum Mechanics, Addison–Wesley, 1967.
• F. Schwabl, Advanced Quantum Mechanics, 3rd ed., Springer, 2005, doi:10.1007/3-540-28528-8.