Foldy–Wouthuysen Transformation

The Foldy–Wouthuysen transformation is a unitary change of representation that makes the free Dirac Hamiltonian block diagonal in its energy sectors. In external fields, a controlled version of the method organizes relativistic corrections to two-component dynamics. Transforming a Hamiltonian also requires transforming its observables; changing representation does not by itself remove a physical interference effect.

Required background. Hamiltonian Form supplies the free spectral structure and Gamma-Matrix Conventions fixes the matrix basis. Helpful background. Dirac to Pauli gives the leading limit, while Zitterbewegung explains the position-operator question.

Even and odd operators

In the Dirac basis, call an operator even if $[\beta,\mathcal E]=0$ and odd if $\{\beta,\mathcal O\}=0$. Even operators are block diagonal and odd operators are block off diagonal relative to $\beta$. These labels describe matrix blocks, not spatial parity.

A minimally coupled Hamiltonian has the form

$$H=\beta mc^2+\mathcal E+\mathcal O,\qquad \mathcal E=q\Phi,\qquad \mathcal O=c\boldsymbol\alpha\cdot\boldsymbol\pi.$$

The aim is to suppress the odd part to a declared order, or exactly when the algebra permits it. Evenness alone does not specify a unique representation: further unitary transformations within the two blocks can change the component form without reintroducing odd terms.

The exact free transformation

For the free massive operator, put $M=mc^2>0$, $\mathcal O=c\boldsymbol\alpha\cdot\mathbf p$, and $E_{\mathbf p}=\sqrt{M^2+c^2\mathbf p^2}$. With the convention $\psi_{\rm FW}=U\psi_D$, define

$$U(\mathbf p)= \frac{E_{\mathbf p}+M+\beta\mathcal O} {\sqrt{2E_{\mathbf p}(E_{\mathbf p}+M)}}.$$

The sign of $\beta\mathcal O$ is tied to this direction of the transformation. Reversing $U$ and $U^\dagger$ without reversing that sign would give the wrong block diagonalization.

Since $(\beta\mathcal O)^\dagger=-\beta\mathcal O$, $(\beta\mathcal O)^2=-\mathcal O^2$, and $\mathcal O^2=E_{\mathbf p}^2-M^2$, one obtains

$$UU^\dagger =\frac{(E_{\mathbf p}+M)^2+\mathcal O^2} {2E_{\mathbf p}(E_{\mathbf p}+M)} =I.$$

The numerator identity $U(\beta M+\mathcal O)=\beta E_{\mathbf p}U$ then gives

$$H_{\rm FW}=U H_D U^\dagger=\beta E_{\mathbf p}.$$

The upper block evolves with $+E_{\mathbf p}$, the lower block with $-E_{\mathbf p}$. The transformation is a bounded unitary Fourier multiplier; it preserves the free $H^1$ domain. It is exact for every free momentum, not merely a small-momentum expansion.

An operator rotation aligns the energy projectors

For nonzero $\mathbf p$, the same map can be written

$$U=\cos\theta+ \beta\boldsymbol\alpha\cdot\widehat{\mathbf p}\,\sin\theta, \qquad \tan(2\theta)=\frac{c|\mathbf p|}{M},$$

with $0\leq\theta<\pi/4$. The continuous value at zero momentum is $U=I$. This is a rotation in matrix space, not a spatial rotation of the particle.

The exact free projectors become constant block projectors:

$$U P_\pm(\mathbf p)U^\dagger=\frac{I\pm\beta}{2}.$$

Thus an upper-only state in the FW representation is in the positive-energy sector. That statement was false for a generic upper-only state in the original Dirac representation.

For the covariantly normalized free mode $u_s(p)$, direct substitution gives $Uu_s(p)=\sqrt{2E_{\mathbf p}}(\chi_s,0)^T$. This connects the matrix transformation to the explicit normalization on Free Dirac Spinors.

Position and spin must be transformed too

For any observable $A_D$, the same physical observable in the new representation is $A_{\rm FW}=UA_DU^\dagger$. Its expectation is unchanged when the state is transformed by $U$.

In momentum representation, the original Dirac position is $\mathbf X_D=i\hbar\nabla_{\mathbf p}$. Its image is

$$U\mathbf X_DU^\dagger =i\hbar\nabla_{\mathbf p} +i\hbar U(\nabla_{\mathbf p}U^\dagger).$$

The second term is present because $U$ depends on momentum. One cannot simply erase it and call the resulting operator the unchanged Dirac position.

There is also a useful mean-position choice: take $i\hbar\nabla_{\mathbf p}$ itself as the position in the FW representation. Its pullback to the Dirac representation is $U^\dagger i\hbar\nabla_{\mathbf p}U$. Under $H_{\rm FW}=\beta E_{\mathbf p}$, this chosen position has velocity $\beta c^2\mathbf p/E_{\mathbf p}$ with no free intersector oscillatory term.

This is consistent with the zitterbewegung calculation: the original position and the mean-position operator are different observables. A unitary representation change preserves the predictions of either one when it is transformed consistently.

Likewise, the simple FW spin $\hbar\boldsymbol\Sigma/2$ commutes with $H_{\rm FW}$. Its Dirac-representation version is $U^\dagger(\hbar\boldsymbol\Sigma/2)U$, generally not the original component expression $\hbar\boldsymbol\Sigma/2$.

Low momentum and external fields

For $|\mathbf p|\ll mc$,

$$U=I+\frac{\beta\mathcal O}{2mc^2} +O\!\left(\frac{\mathbf p^2}{m^2c^2}\right).$$

The leading anti-Hermitian generator $G=\beta\mathcal O/(2mc^2)$ has $[G,\beta mc^2]=-\mathcal O$, which cancels the initial odd term. Repeated commutators generate the kinetic and field-gradient corrections.

In a general external field, the free momentum formula cannot simply be substituted and declared exact. Kinetic momenta can fail to commute, and the scalar potential fails to commute with the momentum-dependent map. If $U$ also depends on time, then

$$H_{\rm FW} =UHU^\dagger+i\hbar\dot U U^\dagger.$$

The derivative term is as necessary here as in gauge covariance. A systematic calculation must specify the field strengths, gradients, time scales, and momentum range retained in its expansion. Small momentum alone does not justify ignoring rapid driving or strong intersector transitions.

Block diagonalizing the free equation does not eliminate its negative sector, turn it into antiparticle creation operators, or establish that a time-dependent interacting problem preserves particle number.

Exercises

1. Show that $G=\beta\mathcal O/(2M)$ is anti-Hermitian and cancels $\mathcal O$ at first order in $e^GH e^{-G}$.

Solution

Hermiticity of $\beta,\mathcal O$ and $\{\beta,\mathcal O\}=0$ give $G^\dagger=-G$. Also $[G,\beta M]=(\beta\mathcal O\beta-\mathcal O)/2 =-\mathcal O$. This cancels the original odd term in the first commutator expansion.

2. What free positive-energy Hamiltonian remains after subtracting rest energy in the FW representation?

Solution

It is $E_{\mathbf p}-mc^2$. Expanding gives $\mathbf p^2/(2m)-\mathbf p^4/(8m^3c^2)+\cdots$. The full square root is exact; the truncation is controlled only in a specified low-momentum regime.

3. Does the absence of an odd block in $H_{\rm FW}$ make every transformed observable block diagonal?

Solution

No. The transformed original position contains a momentum-derivative connection term and can connect the energy blocks. Diagonalizing one operator does not simultaneously diagonalize every other observable.

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 — the original transformation and interpretation.
• R. G. Littlejohn, “The Foldy-Wouthuysen Transformation,” Physics 221B, Notes 49, University of California, Berkeley, 2021–22, lecture notes — ordered block diagonalization in external fields.
• B. Thaller, The Dirac Equation, Springer, 1992 — free transformations and position and spin operators.