# Self-energy of an electron

In his 1947 paper, Bethe states that the self-energy of an electron in a quantum state $m$, due to its interaction with transverse electromagnetic waves is

$$W = -\frac{2e^2}{3\pi\hbar c^3}\sum_{n}\int_0^K\mathrm{d}k\,\frac{k\,|v_{mn}|^2}{E_n-E_m+k}$$

where $v_{mn}$ is the matrix element of the velocity operator.

He says this is a result from ordinary radiation theory. Can someone post (or point me to) a derivation of this result?

[Note: I will set $c=\hbar=\varepsilon_0=1$ in this answer, because I cannot for the life of me remember where they go and I'm doing this off the top of my head.]

This is a result in perturbation theory. Consider a Hamiltonian $H_0$ for which we can exactly know the spectrum (one such Hamiltonian could be the Hamiltonian for a hydrogen-like atom, which is the case here), and another Hamiltonian $H$ given by

$$H=H_0+\varepsilon V,$$

where $V$ is considered to be some kind of interaction Hamiltonian and $\varepsilon$ is a "small" parameter characterizing the strength of this interaction/perturbation.

Now, we know the eigenstates and eigenvalues of $H_0$, let's call them $|n^{(0)}\rangle$ and $E_n^{(0)}$ for some index $n$. Furthermore, let's denote by $|n\rangle$ and $E_n$ the exact eigenstates and eigenvalues of the full Hamiltonian $H$. Finally, since $H\to H_0$ as $\varepsilon\to 0$, let's Taylor expand $|n\rangle$ and $E_n$ in powers of $\varepsilon$ as

$$|n\rangle=\sum_{k=0}^{\infty}\varepsilon^k|n^{(k)}\rangle\sim |n^{(0)}\rangle+\varepsilon|n^{(1)}\rangle+\cdots\\ E_n=\sum_{k=0}^{\infty}\varepsilon^kE_n^{(k)}\sim E_n^{(0)}+\varepsilon E_n^{(1)}+\varepsilon^2 E_n^{(2)}+\cdots.$$

This is essentially the idea of perturbation theory. We can derive explicite formulae for the higher corrections of the eigenstates and energies by solving the equation $H|n\rangle=E_n|n\rangle$ at each order in $\varepsilon.$ For instance, at first order we have

$$H_0|n^{(0)}\rangle+\varepsilon V|n^{(0)}\rangle+\varepsilon H_0|n^{(1)}\rangle+\mathcal{O}(\varepsilon^2)=E_n^{(0)}|n^{(0)}\rangle+\varepsilon E_n^{(1)}|n^{(0)}\rangle+\varepsilon E_n^{(0)}|n^{(1)}\rangle+\mathcal{O}(\varepsilon^2).$$

Noting that $H_0|n^{(0)}\rangle=E_n^{(0)}|n^{(0)}\rangle$ by definition, and acting on both sides with $\langle m^{(0)}|$ gives

$$\varepsilon\langle m^{(0)}|V|n^{(0)}\rangle+\mathcal{O}(\varepsilon^2)=\varepsilon E_{n}^{(0)}\delta_{m,n}+\mathcal{O}(\varepsilon^2).$$

Finally, demanding that terms of the same order in $\varepsilon$ are equal (since Taylor series are unique) gives us $E_{n}^{(1)}=\langle n^{(0)}|V|n^{(0)}\rangle$. That is, the first order correction to the energies are given by the diagonal terms in $V$.

If we continued this process and found the second order correction, we would find

$$E_{n}^{(2)}=\sum_{m\neq n}\frac{|\langle m^{(0)}|V|n^{(0)}\rangle|^2}{E_n^{(0)}-E_m^{(0)}}.$$

This is essentially exactly what Bethe did. He used perturbation theory on a system where $H_0$ is the sum of the hydrogen-like Hamiltonian and the free electromagnetic field, while $V$ is an operator representing the interaction between the electron and the electromagnetic field. These Hamiltonians are

$$H_0=\frac{\textbf{p}^2}{2m}-\frac{e^2}{4\pi r}+\frac{1}{2}\int\mathrm{d}^3\textbf{r}\,\left(|\textbf{E}(\textbf{r})|^2+|\textbf{B}(\textbf{r})|^2\right)\\ V=\frac{e}{m}\left(\textbf{p}\cdot\textbf{A}(\textbf{r})+\textbf{A}(\textbf{r})\cdot\textbf{p}\right)+\frac{e^2}{2m}|\textbf{A}(\textbf{r})|^2-e\phi(\textbf{r}),$$

where $\textbf{A}$ is the vector potential and $\phi$ is the electrostatic potential (the interaction is given by the usual replacement $\textbf{p}\to\textbf{p}-q\textbf{A}$ in the free Hamiltonian $H_0$). We can now pick a gauge, so let's pick the Coulomb gauge ($\boldsymbol{\nabla}\cdot\textbf{A}=0$, $\phi=0$). If we Fourier transform the vector potential as

$$\textbf{A}(\textbf{r})=\frac{1}{\sqrt{V}}\sum_{\textbf{k}}\sum_{\alpha=1}^{2}\left[a^{(\alpha)}_{\textbf{k}}\boldsymbol{\epsilon}^{(\alpha)}(\textbf{k})e^{i\textbf{k}\cdot\textbf{r}}+\text{h.c.}\right],$$

where $V$ is the volume of space, $\boldsymbol{\epsilon}^{(\alpha)}(\textbf{k})$ are polarization vectors, and $\textbf{k}$ is the Fourier wavenumber, then, noting that the Coulomb gauge requires $\textbf{k}\cdot\boldsymbol{\epsilon}^{\alpha}(\textbf{k})=0$, the electromagnetic part of the Hamiltonian becomes

$$H_0=\cdots +\sum_{\textbf{k}}\sum_{\alpha=1}^{2}\frac{\omega_{\textbf{k}}}{2}\left(a^{(\alpha)\dagger}_{\textbf{k}}a^{(\alpha)}_{\textbf{k}}+a^{(\alpha)}_{\textbf{k}}a^{(\alpha)\dagger}_{\textbf{k}}\right).$$

(Note that the $\cdots$ just represents the electron part which we don't care about right now.) Here, $\omega_{\textbf{k}}\equiv|\textbf{k}|$. This Hamiltonian is just an infinite collection of Harmonic oscillators labelled by momenta $\textbf{k}$ and polarizations $\alpha$. The eigenstates of a given wavenumber $\textbf{k}$ and polarization $\alpha$ have equal energy spacing with energy $\omega_{\textbf{k}}$. These states are interpreted as photons.

Welcome to Quantum Field Theory!

The eigenstates of $H_0$ are labeled by $|n,m,\ell\rangle\otimes|\textbf{k},\alpha\rangle$, where $|n,m,\ell\rangle$ is the typical state of an electron in an orbital in a hydrogen-like atom and $|\textbf{k},\alpha\rangle$ represents a photon of wavenumber (momentum) $\textbf{k}$ and polarization $\alpha$. The result Bethe gave is simply obtained by using the second-order perturbation expansion on the state $|n,m,\ell\rangle$ (an electron with no photons) with the interaction Hamiltonian given above. Furthermore, if you want to calculate this yourself, you may find the following replacements useful:

$$\delta_{\textbf{k},\textbf{k}'}\to\frac{(2\pi)^3}{V}\delta(\textbf{k}-\textbf{k}')\\ \sum_{\textbf{k}}f(\textbf{k})\to\frac{V}{(2\pi)^3}\int\mathrm{d}^3\textbf{k}\,f(\textbf{k}).$$

These are the standard rules for turning sums into integrals when considering momenta.

Finally, the upper limit $K$ in the integral is what is known as a UV cutoff. Basically, the idea is that the full integral with $K\to\infty$ is divergent, and so there is a need to cut the integral off at some point. Typically, this point is where your theory stops making sense or can't make predictions, but Bethe simply took the point to be $1/r_e$, where $r_e$ is the classical electron radius. Today, we treat these kinds of infinite integrals with a systematic treatment known as renormalization, which aims to make sense of these infinities by redefining various quantities (in this case, the mass of the electron gets changed by the self-interaction). Surprisingly, however, Bethe's result was pretty accurate.

(Please feel free to correct anything that I got wrong or any typos in my answer. I most likely got something wrong.)

I hope this helped!

• Thanks for the detailed post. Taking $\phi(r) = 0$ and noting that $\langle n',l',m'| |A(r)|^2|n,l,m\rangle =0, \nabla A(r) = 0$, we get $V_{n'l'm', nlm}= 2e \langle n',l',m'| A(r) . v |n,l,m\rangle = 2e A(r) . \vec v_{n'l'm', nlm}$, where $\vec v$ is the velocity operator. I don't see how one gets the shift in the denominator ($E_n - E_m \rightarrow E_n - E_m +k$). Can you clarify? – Coriolis1 Jul 4 '17 at 13:40
• The shift comes from the fact that the intermediate state has energy $E_n+k$, where the $k$ comes from the energy of the photon in the intermediate state. – Bob Knighton Jul 4 '17 at 16:12