**Appendix B. Miscellaneous Derivations for S-matrix Evaluation**

For use in deriving the following, we begin with the fermion anticommutator,

$$\{b\_i(p), b\_j^\dagger(p')\} = \{d\_i(p), d\_j^\dagger(p')\} = (2\pi)^3 \frac{E}{m} \delta\_{l\bar{l}} \delta^3(\mathbf{p} - \mathbf{p'}),\tag{A7}$$

and since there is only one electron and no positron in both the initial and final states for the rest frame and *E<sup>i</sup>* = *E<sup>f</sup>* = *p<sup>i</sup>* = *p<sup>f</sup>* = *m* = *ω* = 2*π*/*t* with the spins being summed over, we have

$$\begin{split} \psi(x) \mid i \rangle &= \sqrt{\frac{m}{E\_i r^3}} \psi(x) b\_i^\dagger(p\_i) \mid 0 \rangle \\ &= \sqrt{\frac{1}{r^3}} \int \frac{d^3 p}{(2\pi)^3} \sum\_{j=1}^2 b\_j(m) u^j(m) e^{-iEt} b\_i^\dagger(m) \mid 0 \rangle \\ &= \sqrt{\frac{1}{r^3}} \int \frac{d^3 p}{(2\pi)^3} \sum\_{j=1}^2 u^j(m) e^{-iEt} \{b\_j(m), b\_i^\dagger(m)\} \mid 0 \rangle \\ &= \sqrt{\frac{1}{r^3}} u^i(m) e^{-iEt} \, . \end{split}$$

and similarly

$$
\begin{aligned}
\langle f \mid \bar{\Psi}(\mathbf{x}) &= \sqrt{\frac{1}{r^3}} \,\bar{u}^f(m) e^{+iEt}, \\
\langle \psi(\mathbf{x}) \mid i \rangle &= \sqrt{\frac{1}{r^3}} \,\bar{v}^i(m) e^{+iEt}, \\
\langle f \mid \bar{\Psi}(\mathbf{x}) &= \sqrt{\frac{1}{r^3}} \,\bar{v}^f(m) e^{-iEt},
\end{aligned} \tag{A8}$$

in which the fermion anticommutator Equation (A7) is used for simplification [21]. Note that the spatial components, <sup>√</sup> 1/*r* 3 , of the spin vector are factored out in these plane-wave equations.

For the derivation of Equation (20), we begin with Equation (18) for the electrostatic self-interaction contribution,

$$\mathbf{h}\_I = e \,\,\overline{\psi}(\mathbf{x}) \gamma^\mu A\_\mu \psi(\mathbf{x}) \,\,\tag{A9}$$

and substitute it into the first order S-Matrix,

$$S\_{fi}^{(1)}(E1) = -i \int d^4 \mathbf{x} \,\langle f \mid e \,\Psi(\mathbf{x}) \gamma^\mu A\_\mu \psi(\mathbf{x}) \mid i \rangle . \tag{A10}$$

Making substitutions using Equation (A8) from above, and then going to the rest frame, along with using *A*<sup>0</sup> = *e*/4*πr*, *α* = *e* <sup>2</sup>/4*π* and taking *r* to be constant, the S-matrix expression works out to give

$$\begin{aligned} \left(^{11}\_{fi}(\mathbf{E}1) &= -i\frac{e^2}{4\pi r} \int d^4x \left(\sqrt{\frac{1}{r^3}} \dot{u}^f(m) e^{+imt} \gamma^0 \sqrt{\frac{1}{r^3}} u^i(m) e^{-imt}\right) \\ &= -i\frac{\alpha}{r^4} \left(\bar{u}^f(m) \gamma^0 u^i(m)\right) \int d^4x \\ &= -i\frac{\alpha}{r^4} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ &= -i \, t \frac{\alpha}{r} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \tag{A11} \end{aligned} \tag{A12}$$

where we have taken R *d* <sup>4</sup>*x* = *r* 3 *t*.

*S*

The S-matrix evaluation for Equation (21) in the rest frame is as follows. We begin with Equation (19),

$$\mathbf{h}\_I(\mathbf{x}) = -\frac{3\kappa}{8} (\bar{\psi}(\mathbf{x})\gamma^5\gamma\_k\psi(\mathbf{x})) (\bar{\psi}(\mathbf{x})\gamma^5\gamma^k\psi(\mathbf{x}))\_\prime \tag{A12}$$

and substitute it into the first order S-Matrix:

$$S\_{fi}^{(1)}(E2) = i\frac{3\kappa}{8} \int d^4x \left< f \mid (\bar{\psi}(\mathbf{x})\gamma^5\gamma\_k\psi(\mathbf{x})) (\bar{\psi}(\mathbf{x})\gamma^5\gamma^k\psi(\mathbf{x})) \mid i \right> . \tag{A13}$$

Then, because the spatial components, <sup>√</sup> 1/*r* 3 , of the spin vectors are factored out in the plane-wave equations that we derived in Equation (A8), by substituting Equation (A8) into Equation (A13) and then going to the rest frame, the S-matrix expression works out to give

$$\begin{split} S\_{fi}^{(1)}(E2) &= i\frac{3\kappa}{8} \int d^4x \, \frac{1}{\tau^3} \left( \bar{u}^f(m)e^{i\int m\bar{r}} \, \gamma^5 \gamma\_0 \, \bar{v}^i(m)e^{i\int m\bar{r}} \right) \frac{1}{\tau^3} \left( \sigma^f(m)e^{-i\int m\bar{r}} \, \gamma^5 \gamma^0 \, \bar{u}^i(m)e^{-i\int m\bar{r}} \right) \\ &= i\frac{3\kappa}{8\tau^6} \int d^4x \, \left( \bar{u}^f(m)\gamma^5 \gamma\_0 \, \bar{v}^i(m) \, \bar{v}^f(m) \, \gamma^5 \gamma^0 \, \bar{u}^i(m) \right) \\ &= i\frac{3\kappa}{8\tau^6} \left( \bar{u}^f(m)\gamma^5 \gamma\_0 \, \left( -\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \gamma^5 \gamma^0 \, \bar{u}^i(m) \right) \int d^4x \\ &= i \frac{3\kappa \, \gamma^5 t}{8\tau^6} \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) \\ &= i \frac{3\kappa}{8\tau^3} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right) . \tag{A14} \tag{A15} \end{split}$$

where we have again taken R *d* <sup>4</sup>*x* = *r* 3 *t*. Similarly, the S-matrix expression for the anti-fermion works out to give

$$\begin{split} S\_{fi}^{(1)}(E4) &= i\frac{3\kappa}{8} \int d^4x \, \frac{1}{r^3} \left( \boldsymbol{\sigma}^f(m) e^{-int} \boldsymbol{\gamma}^5 \boldsymbol{\gamma}\_0 \, \boldsymbol{u}^i(m) e^{-int} \right) \, \frac{1}{r^3} \left( \boldsymbol{\pi}^f(m) e^{+int} \boldsymbol{\gamma}^5 \boldsymbol{\gamma}^0 \boldsymbol{v}^i(m) e^{+int} \right) \\ &= i \frac{3\kappa}{8r^6} \left( \boldsymbol{\sigma}^f(m) \, \boldsymbol{\gamma}^5 \boldsymbol{\gamma}\_0 \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \, \boldsymbol{\gamma}^5 \boldsymbol{\gamma}^0 \boldsymbol{v}^i(m) \right) \int d^4x \\ &= i \frac{3\kappa \boldsymbol{\gamma}^3 \boldsymbol{t}}{8r^6} (-1) \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) \\ &= -i \, i \frac{3\kappa}{8r^3} \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right) . \tag{A15} \end{split} \tag{A16}$$

From the last two equations it is easy to see that the fermion and anti-fermion S-matrix equations are coupled. In other words, one might initially think that in the rest frame *ψ*¯(*t*)*γ* 5*γ* <sup>0</sup>*ψ*(*t*) = 0, which would result in the entire term being zero, but that is not the case because of the coupling of mixed states between fermions and anti-fermions is an intermediate step in the self-interaction.

### **References**


© 2020 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).
