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. Below is a screenshot of two pages. I can very well follow the derivation of the electrostatic part of the self-energy, given by eq. (1). However, I am struggling to understand the derivation of the electrodynamical part, given in eq (2), where Weisskopf uses perturbation theory and notions of adiabatic change. Can anybody provide hints to understanding?

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 ?

I am working my way through the 1934 paper by Weisskopf on the self-energy of the electron [https://doi.org/10.1007/BF01333228] and is much helped by the English translation found here. Below is a screenshot of two pages. I can very well follow the derivation of the electrostatic part of the self-energy, given by eq. (1). However, I am struggling to understand the derivation of the electrodynamical part, given in eq (2), where Weisskopf uses perturbation theory and notions of adiabatic change. Can anybody provide hints to understanding  ?  

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. Below is a screenshot of two pages. I can very well follow the derivation of the electrostatic part of the self-energy, given by eq. (1). However, I am struggling to understand the derivation of the electrodynamical part, given in eq (2), where Weisskopf uses perturbation theory and notions of adiabatic change. Can anybody provide hints to understanding?