Darwin Term

The Darwin term is a spin-independent relativistic correction proportional to the Laplacian of an electrostatic potential energy. For a point Coulomb source it becomes a contact operator, so its leading nonrelativistic matrix element samples the wavefunction at the source. It contributes even to S states, where the spin–orbit operator vanishes. Its precise coefficient follows from the Foldy–Wouthuysen Expansion, without assigning a literal jittering trajectory to the electron.

Required background. The Foldy–Wouthuysen Expansion derives the electric-gradient operator. Helpful background. Free Dirac Spinors supports the matrix-element interpretation; The Hydrogen Atom supplies the nonrelativistic Coulomb eigenstates.

The electric-field-gradient operator

For a massive, minimally coupled Dirac particle in a static electrostatic potential energy $V=q\Phi$, the positive-energy FW Hamiltonian contains

$$H_{\rm D} =\frac{\hbar^2}{8m^2c^2}\nabla^2V =-\frac{q\hbar^2}{8m^2c^2}\nabla\cdot\mathbf E.$$

This expression uses the signed charge $q$. It is a multiplicative operator for a smooth $V$, and a distributional operator when the source is idealized as a point. It accompanies the kinetic and spin–orbit corrections at the same FW order; it is not an extra term to append to the exact Dirac Hamiltonian.

The coefficient on the FW owner page comes from the scalar part of $-[\mathcal O,[\mathcal O,V]]/(8m^2c^4)$. In electrostatics, Gauss’s law also gives

$$H_{\rm D} =-\frac{q\hbar^2}{8m^2c^2\epsilon_0}\rho_{\rm ext}.$$

Here $\rho_{\rm ext}$ is the prescribed external source density, not the probability density of the Dirac particle. A positive nuclear charge density produces a positive Darwin energy for an electron with $q=-e$.

The FW assumptions remain essential. These formulas describe a low-energy expansion and do not assert that an arbitrarily sharp or strong field is uniformly approximated by a finite number of terms.

Coulomb contact and the origin

Let

$$V(r)=-\frac{\kappa_{\rm C}}{r},\qquad \kappa_{\rm C}>0.$$

The distributional identity

$$\nabla^2\frac1r=-4\pi\delta^{(3)}(\mathbf r)$$

gives

$$H_{\rm D} =\frac{\pi\kappa_{\rm C}\hbar^2}{2m^2c^2} \delta^{(3)}(\mathbf r).$$

The Laplacian does vanish at each ordinary point $r>0$. That observation misses the entire contact term. For example, integrating the Laplacian over a ball and using the flux of $\nabla(1/r)$ through its boundary gives $-4\pi$, independent of the ball radius. A zero distribution could not have that flux.

For a normalized nonrelativistic two-component wavefunction $\varphi$ that is continuous at the origin, first-order perturbation theory gives

$$\Delta E_{\rm D} =\frac{\pi\kappa_{\rm C}\hbar^2}{2m^2c^2} \varphi^\dagger(0)\varphi(0).$$

Regular central-potential states with $\ell>0$ vanish at the origin, so this contact matrix element is zero. S states generally do not vanish. This is a statement about the unperturbed nonrelativistic states used in the expansion. It is not a prescription to insert a singular point-Coulomb Dirac spinor into a delta-function expectation value.

For a fixed, infinitely heavy point nucleus of charge $Ze$, write $a=Z\alpha$ and $\kappa_{\rm C}=a\hbar c$. The Schrödinger Coulomb length is $b=\hbar/(mc a)$. Since $|\psi_{n00}(0)|^2=1/(\pi n^3b^3)$,

$$\Delta E_{\rm D}(nS) =\frac{mc^2a^4}{2n^3}.$$

This positive contribution is only one part of the fine-structure shift. The complete correction also contains a negative kinetic contribution; one must combine operators before comparing with a spectral line or an exact Dirac energy.

Resolving the source before taking a limit

A finite source makes the contact limit transparent. As an example, spread the source uniformly over a ball of radius $R$. Its potential energy is

$$V_R(r)= \begin{cases} -\dfrac{\kappa_{\rm C}}{2R} \left(3-\dfrac{r^2}{R^2}\right),&r<R,\\[4pt] -\dfrac{\kappa_{\rm C}}{r},&r\geq R. \end{cases}$$

Both $V_R$ and its first radial derivative are continuous at $R$, so there is no surface delta function in its Laplacian. Instead,

$$\nabla^2V_R =\frac{3\kappa_{\rm C}}{R^3}\, \mathbf1_{r<R}.$$

The corresponding matrix element is

$$\Delta E_{{\rm D},R} =\frac{3\kappa_{\rm C}\hbar^2}{8m^2c^2R^3} \int_{r<R}d^3x\,\varphi^\dagger\varphi.$$

For a fixed wavefunction whose density is continuous at the origin, the integral approaches $4\pi R^3\varphi^\dagger(0)\varphi(0)/3$. The contact formula follows. For a regular state with density proportional to $r^{2\ell}$ near the origin, the finite-source Darwin matrix element scales as $R^{2\ell}$ when $\ell>0$.

This limit identifies a distribution and its matrix element. It does not prove uniform accuracy of the full FW truncation as $R\to0$, nor does it include the change of the unperturbed state caused by a finite nuclear radius. Finite-size spectroscopy requires those effects to be organized consistently.

What positive-energy spinor overlaps reveal

There is a second way to understand why a local potential acquires derivative corrections. Consider a weak static potential and its first-order matrix element between free positive-energy states. Normalize each spinor to $w^\dagger w=1$:

$$w(\mathbf p,\chi) =C_{\mathbf p} \begin{pmatrix} \chi\\[2pt] \dfrac{c\boldsymbol\sigma\cdot\mathbf p} {E_{\mathbf p}+mc^2}\chi \end{pmatrix}.$$

Here $\chi^\dagger\chi=1$ and $C_{\mathbf p}=\sqrt{(E_{\mathbf p}+mc^2)/(2E_{\mathbf p})}$. Expanding both the normalization factor and the lower component for $|\mathbf p|,|\mathbf p'|\ll mc$ gives

$$\begin{aligned} w^\dagger(\mathbf p',\chi')w(\mathbf p,\chi) ={}&\chi'^\dagger\Bigl[ 1-\frac{|\mathbf p'-\mathbf p|^2}{8m^2c^2}\\ &+\frac{i\boldsymbol\sigma\cdot(\mathbf p'\times\mathbf p)} {4m^2c^2} \Bigr]\chi+\cdots . \end{aligned}$$

In deriving this expression, use $(\boldsymbol\sigma\cdot\mathbf p') (\boldsymbol\sigma\cdot\mathbf p) =\mathbf p'\cdot\mathbf p +i\boldsymbol\sigma\cdot(\mathbf p'\times\mathbf p)$. The separate normalization terms are necessary; keeping only the lower-component product gives the wrong scalar answer.

Let $\mathbf k=\mathbf p'-\mathbf p$ and define $\widetilde V(\mathbf k)= \int d^3x\,e^{-i\mathbf k\cdot\mathbf x/\hbar}V(\mathbf x)$. The scalar overlap correction multiplies this transform by $-\mathbf k^2/(8m^2c^2)$. Since a Laplacian transforms to $-\mathbf k^2/\hbar^2$, this is precisely the matrix element of $\hbar^2\nabla^2V/(8m^2c^2)$. The spin-dependent overlap similarly matches the spin–orbit operator.

This check concerns the weak-potential, low-momentum matrix element. It explains the derivative interaction through the momentum dependence of the positive-energy spinors. A heuristic Compton-scale positional smearing may suggest its size, but it does not determine the coefficient or replace this calculation. The distinction between position observables and representations is discussed on Zitterbewegung and the Foldy–Wouthuysen Transformation.

Exercises

A source-free region. Can a particle have spin–orbit coupling where the Darwin operator vanishes? Give a Coulomb example.

Solution

Yes. Outside a point Coulomb source, $\nabla^2V=0$ but $\nabla V\ne0$. The Darwin operator is supported at the origin, whereas $H_{\rm SO}=\kappa_{\rm C}\mathbf L\cdot\mathbf S/ (2m^2c^2r^3)$ acts away from the origin. The two operators encode different parts of the electric-gradient correction.

The finite ball. For a density $\varphi^\dagger\varphi=C r^{2\ell}+o(r^{2\ell})$ near the origin, find the leading finite-ball Darwin shift.

Solution

The probability in the ball is $4\pi C R^{2\ell+3}/(2\ell+3) +o(R^{2\ell+3})$. Thus

$$\Delta E_{{\rm D},R} =\frac{3\pi\kappa_{\rm C}\hbar^2 C} {2m^2c^2(2\ell+3)}R^{2\ell} +o(R^{2\ell}).$$

For $\ell=0$ this tends to the contact expression. For $\ell>0$ it vanishes.

Zero momentum transfer. Set $\mathbf p'=\mathbf p$ in the spinor-overlap expansion. Explain the result and its implication for a constant potential energy.

Solution

Both displayed corrections vanish, leaving $\chi'^\dagger\chi$, as required by the exact normalization at equal momentum. A constant potential has no gradient or Laplacian, so it generates neither Darwin nor spin–orbit terms. It only shifts the energy origin.

References

• Bjorken, James D., and Sidney D. Drell. Relativistic Quantum Mechanics. McGraw–Hill (1964).
• 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.