Foldy–Wouthuysen Expansion

The Foldy–Wouthuysen expansion turns the Dirac equation into an effective two-component theory with a specified accuracy. For static electromagnetic backgrounds, the first corrections contain the relativistic kinetic energy, spin–orbit coupling, and the Darwin term. Magnetic corrections occur at the same formal order unless an additional field-strength hierarchy suppresses them. This page derives the ordered Hamiltonian; the Foldy–Wouthuysen Transformation owns the representation change and the transformation of observables.

Required background. The Foldy–Wouthuysen Transformation introduces block diagonalization, and Minimal Coupling fixes the electromagnetic conventions. Helpful background. Dirac to Pauli derives the leading two-component limit.

Static Dirac Hamiltonian and expansion scales

Take $m>0$, real time-independent potentials, and the signed charge $q$. Write

$$\begin{aligned} H&=\beta M+V+\mathcal O,\qquad M=mc^2,\\ V&=q\Phi,\\ \mathcal O&=c\boldsymbol\alpha\cdot\boldsymbol\pi,\\ \boldsymbol\pi&=\mathbf p-q\mathbf A. \end{aligned}$$

Here $V$ is even and $\mathcal O$ is odd with respect to $\beta$. Initially use smooth fields and a common domain on which the displayed products and commutators make sense. The calculation is a low-energy asymptotic expansion, not an operator-norm Taylor series on the full unbounded Dirac Hilbert space.

For the formal counting, hold $m,q,\hbar,\Phi,\mathbf A$ and spatial derivatives fixed as $c$ increases. Then $M=O(c^2)$, $\mathcal O=O(c)$, and $V=O(1)$. For an actual state, kinetic momenta must be small compared with $mc$, field variation must not resolve arbitrarily short Compton-scale structure, and electric-potential differences sampled by the state must not invalidate the separation of energy sectors. A constant offset of $\Phi$ does not control this approximation: it only changes the energy origin. Strong or singular backgrounds need their own domain and scale analysis.

Removing the odd part with nested commutators

Use $\psi'=e^G\psi$, with the anti-Hermitian generator

$$G=\frac{\beta\mathcal O}{2M}.$$

The Baker–Campbell–Hausdorff series is

$$\begin{aligned} e^GHe^{-G} ={}&H+[G,H]\\ &+\frac12[G,[G,H]]+\cdots . \end{aligned}$$

Because $\beta\mathcal O=-\mathcal O\beta$,

$$[G,\beta M]=-\mathcal O,\qquad [G,\mathcal O]=\frac{\beta\mathcal O^2}{M}.$$

The first equation cancels the original odd term. Combining the mass and odd-operator series gives

$$\begin{aligned} e^G(\beta M+\mathcal O)e^{-G} ={}&\beta M+\frac{\beta\mathcal O^2}{2M} -\frac{\mathcal O^3}{3M^2}\\ &-\frac{\beta\mathcal O^4}{8M^3} +\cdots . \end{aligned}$$

The potential contributes

$$\begin{aligned} e^GVe^{-G} ={}&V+\frac{\beta[\mathcal O,V]}{2M}\\ &-\frac{[\mathcal O,[\mathcal O,V]]}{8M^2} +\cdots . \end{aligned}$$

For example, the sign of the double commutator follows from $[\beta\mathcal O,\beta[\mathcal O,V]] =-[\mathcal O,[\mathcal O,V]]$. The remaining leading odd operator is

$$\mathcal O_1 =\frac{\beta[\mathcal O,V]}{2M} -\frac{\mathcal O^3}{3M^2}, \qquad \mathcal O_1=O(c^{-1}).$$

A second transformation with $G_1=\beta\mathcal O_1/(2M)$ cancels it. Its new even terms start at $O(c^{-4})$ in the stated counting; remaining odd terms can be removed iteratively. The even Hamiltonian through $O(c^{-2})$ is therefore

$$\begin{aligned} H_{\rm FW} ={}&\beta M+V +\frac{\beta\mathcal O^2}{2M} \\ &-\frac{\beta\mathcal O^4}{8M^3} -\frac{[\mathcal O,[\mathcal O,V]]}{8M^2}\\ +\text{higher-order terms}. \end{aligned}$$

This is the static version of the expansion of Foldy and Wouthuysen (1950); Littlejohn’s notes give a useful complementary derivation with atomic power counting. For a time-dependent transformation, the additional $i\hbar(\partial_tU)U^\dagger$ term must also be expanded. Substituting time-dependent fields into this static formula does not perform that calculation.

Magnetic square and the electric commutator

The kinetic-momentum algebra gives

$$\mathcal O^2=c^2F,\qquad F=\boldsymbol\pi^2-q\hbar\boldsymbol\Sigma\cdot\mathbf B.$$

Consequently $\mathcal O^4=c^4F^2$ as an ordered operator. In particular,

$$\begin{aligned} F^2={}&\boldsymbol\pi^4 -q\hbar\{\boldsymbol\pi^2, \boldsymbol\Sigma\cdot\mathbf B\}\\ &+q^2\hbar^2\mathbf B^2. \end{aligned}$$

The anticommutator retains gradients of an inhomogeneous magnetic field. The last equality uses $(\boldsymbol\Sigma\cdot\mathbf B)^2=\mathbf B^2$ for an ordinary classical field. Replacing $F^2$ by $\boldsymbol\pi^4$ would discard terms of the same formal order for fixed nonzero $\mathbf B$. Such a replacement needs an additional weak-field counting.

For the electric term, first observe

$$[\boldsymbol\alpha\cdot\boldsymbol\pi,V] =-i\hbar\boldsymbol\alpha\cdot\nabla V.$$

Using $\alpha_i\alpha_j=\delta_{ij} +i\epsilon_{ijk}\Sigma_k$ and allowing each momentum to differentiate the field yields

$$\begin{aligned} &[\boldsymbol\alpha\cdot\boldsymbol\pi, [\boldsymbol\alpha\cdot\boldsymbol\pi,V]]\\ &\quad=-\hbar^2\nabla^2V -\hbar\boldsymbol\Sigma\cdot \bigl(\nabla V\times\boldsymbol\pi -\boldsymbol\pi\times\nabla V\bigr). \end{aligned}$$

Since $\nabla V=-q\mathbf E$ for these static potentials, the equivalent field form is

$$\begin{aligned} &[\boldsymbol\alpha\cdot\boldsymbol\pi, [\boldsymbol\alpha\cdot\boldsymbol\pi,q\Phi]]\\ &\quad=q\hbar^2\nabla\cdot\mathbf E +q\hbar\boldsymbol\Sigma\cdot (\mathbf E\times\boldsymbol\pi -\boldsymbol\pi\times\mathbf E). \end{aligned}$$

The cross-product difference is explicitly Hermitian. Here $\nabla\times\mathbf E=0$, so it also equals $2\mathbf E\times\boldsymbol\pi$. Keeping the difference makes the operator ordering visible.

Positive-energy effective Hamiltonian

In the upper FW block, set $\beta=1$ and $\boldsymbol\Sigma=\boldsymbol\sigma$. Define $F_{\rm P}=\boldsymbol\pi^2-q\hbar\boldsymbol\sigma\cdot\mathbf B$. The result is

$$\begin{aligned} H_+={}&mc^2+q\Phi +\frac{F_{\rm P}}{2m} -\frac{F_{\rm P}^2}{8m^3c^2}\\ &-\frac{q\hbar^2}{8m^2c^2}\nabla\cdot\mathbf E\\ &-\frac{q\hbar}{8m^2c^2}\boldsymbol\sigma\cdot (\mathbf E\times\boldsymbol\pi -\boldsymbol\pi\times\mathbf E) +\cdots . \end{aligned}$$

The leading terms reproduce the Pauli Equation. The lower block is still part of the transformed one-particle Dirac equation; interpreting it as positive-energy antiparticles requires the field-theory reorganization explained in Negative-Energy Solutions.

For a purely electrostatic problem, choose $\mathbf A=0$ and write $V=q\Phi$. Then

$$\begin{aligned} H_+={}&mc^2+V+\frac{\mathbf p^2}{2m}\\ &-\frac{\mathbf p^4}{8m^3c^2} +\frac{\hbar^2}{8m^2c^2}\nabla^2V\\ &+\frac{\hbar}{4m^2c^2} \boldsymbol\sigma\cdot(\nabla V\times\mathbf p) +\cdots . \end{aligned}$$

The second line contains the Darwin and spin–orbit operators. For a central potential, $\nabla V\times\mathbf p=V'(r)\mathbf L/r$ and $\mathbf S=\hbar\boldsymbol\sigma/2$, giving

$$H_{\rm SO} =\frac{V'(r)}{2m^2c^2r}\,\mathbf L\cdot\mathbf S.$$

For a Coulomb potential the Laplacian is a distribution. The contact term is interpreted through matrix elements or a specified finite-source limit, not by setting $\nabla^2(1/r)=0$ everywhere.

An exact magnetic check and its remainder

If $V$ is constant, it commutes with $\mathcal O$. The same algebra as in the exact free transformation diagonalizes the Hamiltonian through functional calculus, with the positive block

$$H_+=V+\sqrt{m^2c^4+c^2F_{\rm P}}.$$

On a compatible self-adjoint realization, $F_{\rm P}=(\boldsymbol\sigma\cdot\boldsymbol\pi)^2$ is nonnegative even though its separate spin term can have either sign. For a spectral value $f\geq0$, Taylor’s theorem gives

$$\begin{aligned} 0\leq{}& \sqrt{m^2c^4+c^2f} -mc^2-\frac{f}{2m} +\frac{f^2}{8m^3c^2}\\ \leq{}&\frac{f^3}{16m^5c^4}. \end{aligned}$$

Restricting to the spectral subspace $0\leq F_{\rm P}\leq\Lambda^2$ therefore bounds the energy-operator remainder by $\Lambda^6/(16m^5c^4)$. This is a genuine cutoff estimate for the magnetic square-root problem. It does not supply a remainder bound for arbitrary noncommuting electric potentials.

Exercises

The first even kinetic term. Combine the terms $[G,\mathcal O]$ and $[G,[G,\beta M]]/2$ in the first transformation. Why is the coefficient not $\beta\mathcal O^2/M$?

Solution

The two contributions are respectively $\beta\mathcal O^2/M$ and $-\beta\mathcal O^2/(2M)$. Their sum is $\beta\mathcal O^2/(2M)$. Keeping only the first commutator misses a contribution of exactly the same order.

An electric sign check. Let $V(\mathbf x)=k r^2/2$ with real $k$ and $\mathbf A=0$. Find both electric corrections.

Solution

Since $\nabla^2V=3k$ and $\nabla V=k\mathbf x$,

$$H_{\rm D}=\frac{3k\hbar^2}{8m^2c^2},\qquad H_{\rm SO}=\frac{k}{2m^2c^2}\mathbf L\cdot\mathbf S.$$

Both signs follow from the potential energy $V$, without guessing the sign of an electron’s charge.

Uniform magnetic-field counting. For $\mathbf B=B\hat{\mathbf z}$, $B>0$, use a simultaneous eigenstate of $\boldsymbol\pi^2$ and $\sigma_z$ with eigenvalues $P^2$ and $s=\pm1$. Compare the quartic correction with $-P^4/(8m^3c^2)$ alone.

Solution

The correct correction is

$$\begin{aligned} &-\frac{(P^2-q\hbar Bs)^2}{8m^3c^2}\\ &\quad=-\frac{P^4-2q\hbar BsP^2+q^2\hbar^2B^2} {8m^3c^2}. \end{aligned}$$

The omitted terms are not higher powers of $1/c$ under fixed-field counting. The exact square-root spectrum expands in $P^2-q\hbar Bs$, which is nonnegative on the allowed magnetic eigenstates.

References

• Foldy, Leslie L., and Siegfried 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.
• Littlejohn, Robert G. “The Foldy-Wouthuysen Transformation.” Physics 221B, Notes 49, University of California, Berkeley, academic year 2021–22 (2021). Lecture notes.
• Thaller, Bernd. The Dirac Equation. Springer (1992). doi:10.1007/978-3-662-02753-0.