Weisskopf and self-energy

Question

I am working my way through the 1934 paper by Weisskopf on the self-energy of the electron and is much helped by the English translation found here. I do have some difficulties with section 2 of this paper. Weisskopf starts from a Hamiltonian (in updated notation and SI units) $$ H=\beta mc^{2}+c\left(\boldsymbol{\alpha}\cdot\boldsymbol{p}\right)+ec\left(\boldsymbol{\alpha}\cdot\boldsymbol{A}\right)+\frac{1}{2\varepsilon_{0}}\int\left(E^{2}+c^{2}B^{2}\right)d^{3}\boldsymbol{r} $$ (no scalar potential). The self-energy is then given by the expectation value of $$ E=-\int\boldsymbol{j}\cdot\boldsymbol{A}d^{3}\boldsymbol{r}+\frac{1}{2}\varepsilon_{0}\int\left(E^{2}+c^{2}B^{2}\right)d^{3}\boldsymbol{r}, $$ where $\boldsymbol{j}=-ec\boldsymbol{\alpha}$ is the current-density and $\boldsymbol{A}$, $\boldsymbol{E}$ and $\boldsymbol{B}$ describe the field which is generated by the current density $\boldsymbol{j}$ and the charge density $\rho$. Weisskopf next divides the electromagnetic field into irrotational ($\parallel$) and solenoidal ($\perp$) parts $$ \begin{array}{lcl} \boldsymbol{A} & = & \boldsymbol{A}_{\parallel}+\boldsymbol{A}_{\perp}\\ \boldsymbol{E} & = & \left(-\nabla\phi-\partial_{t}\boldsymbol{A}_{\parallel}\right)-\partial_{t}\boldsymbol{A}_{\perp}\\ \boldsymbol{B} & = & \boldsymbol{B}_{\perp}, \end{array} $$ allowing a separation of the energy into electrostatic and electrodynamic parts (the spatial integral over cross-terms give zero).

The electrostatic part is given by $$ E^{S}=\int\left(-\boldsymbol{j}\cdot\boldsymbol{A}_{\parallel}+\frac{1}{2}\varepsilon_{0}\boldsymbol{E}_{\parallel}^{2}\right)d^{3}\boldsymbol{r} $$ that Weisskopf reworks into $$ E^{S}=\frac{1}{2}\int\phi^{\prime}(\boldsymbol{r})\rho(\boldsymbol{r}) d^{3}\boldsymbol{r},\quad \phi^{\prime}\left(\boldsymbol{r}\right)=\frac{1}{4\pi\varepsilon_{0}}\int\frac{\rho\left(\boldsymbol{r}^{\prime}\right)}{\left|\boldsymbol{r}-\boldsymbol{r}^{\prime}\right|}d^{3}\boldsymbol{r}^{\prime}. $$ This part I can follow.

However, I have problems with the electrodynamical part. The starting expression is $$ E^D=-\int\boldsymbol{j}_{\perp}\cdot\boldsymbol{A}_{\perp}d^{3}\boldsymbol{r}+\frac{1}{2}\varepsilon_{0}\int\left(E_{\perp}^{2}+c^{2}B_{\perp}^{2}\right)d^{3}\boldsymbol{r},\quad (*) $$ Weisskopf writes that this term can be "reduced to the time average" $$ E^D=-\frac{1}{2}\int\boldsymbol{j}_{\perp}\cdot\boldsymbol{A}_{\perp}d^{3}\boldsymbol{r} $$ using the following argument :

According to a well-known theorem from analytical mechanics, the change in energy through the adiabatic change of a parameter $a$ is given by $$ \Delta E=\overline{\frac{\delta H(p,q,a)}{\delta a}}\Delta a \quad (1) $$ where the bar denotes time average or, in quantum theory, the diagonal element. We will regard $\boldsymbol{j}_{\perp}$ as a parameter independent of the remaining variables, since only $\boldsymbol{j}_{\parallel}$ is determined by the continuity equation. If we set $\boldsymbol{j}_{\perp}=\lambda \boldsymbol{j}_{\perp}^0$ and if we allow the current $\boldsymbol{j}_{\perp}$ to increase from zero to its actual value $\boldsymbol{j}_{\perp}^0$ through a change in $\lambda$ and if we use as a basis the expansion $$ H^{\prime}=-\int\boldsymbol{j}_{\perp}\cdot\boldsymbol{A}_{\perp}d^{3}\boldsymbol{r} = -\lambda H^{\prime}_1 + \lambda^2 H^{\prime}_2+\ldots \quad (2) $$ where $H_0^{\prime}$ is proportional to the charge $e$, and $H_1^{\prime}$ is proportional to $e^2$, we obtain the expression: $$ E^D = \int_0^1 \overline{\frac{\partial H}{\partial \boldsymbol{j}_{\perp}^0}}\boldsymbol{j}_{\perp}^0 d\lambda = \overline{H}^{\prime}_1 + \frac{1}{2} \overline{H}^{\prime}_2+\ldots $$ The first term is proportional to the charge $e$ and must vanish because of the symmetry of the self-energy with respect to the sign of $e$; if we limit ourselves to terms that are quadratic in $e$, we thus obtain $E^D=\frac{1}{2} \overline{H}^{\prime}_2$.

My specific questions are:

1. Where can I find more information about the "well-known theorem from analytical mechanics" (1), and how does it translate into modern quantum mechanics ? I know that early quantum theory heavily used the notion of action-angle variables and adiabatic invariants.
2. Why does the second part of the starting expression $(*)$ for $E^D$ seems to disappear and how is the expansion (2) to be understood ?
3. I am also a bit intrigued by the remark that $\overline{H}^{\prime}_1$ vanishes due to charge conjugation symmetry. Any comments on that ?

Answer 1

I can provide an answer to the first question. The "well-known theorem from analytical mechanics" is discussed in §49 on adiabatic invariants in the third edition of "Mechanics" by L. D. Landau and E. M. Lifschitz, found here. We consider some system in periodic motion (libration/rotation) where appears an additional parameter $\lambda (t)$ which

varies slowly (adiabatically) with time as the result of some external action.

More precisely, we assume $$ T \frac{d\lambda}{dt} \ll \lambda . $$ A nice example of such a system is the Rayleigh–Lorentz pendulum; a detailed discussion if found in this paper by L. L. Sánchez-Soto and J. Zoido [ Am. J. Phys. 81, 57–62 (2013)]. Continuing with Landau and Lifschitz:

..since $\lambda$ is assumed to vary only slowly, the rate of change $\dot{E}$ of the energy will also be small. If this rate is averaged over the period $T$ and the "rapid" oscillations of its value are thereby smoothed out. the resulting value $\dot{E}$ determines the rate of steady slow variation of the energy of the system, and this rate will be proportional to the rate of change $\dot{\lambda}$ of the parameter

This leads to the expression $$ \overline{\frac{dE}{dt}}=\overline{\frac{\partial H}{\partial \lambda}}\frac{d\lambda}{dt};\quad \overline{\frac{\partial H}{\partial \lambda}}=\frac {1}{T}\int_0^T \frac{\partial H}{\partial \lambda} dt. $$ where one should not that the time derivative of $\lambda$ is taken out of the time average, since $\lambda$ is assumed to vary slowly.

In modern quantum-mechanical language we can connect this formula with the Hellmann–Feynman theorem $$ \frac{d E}{d \lambda} = \langle\psi(\lambda)|\frac{\partial H}{\partial \lambda}|\psi(\lambda\rangle $$ which we may rearrange to $$ \Delta E = \int _{E(0)}^{E(1)}dE = \int_0^1 \langle\psi(\lambda)|\frac{\partial H}{\partial \lambda}|\psi(\lambda)\rangle d\lambda $$