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Shen712
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- In the Dirac equation for an electron, the covariant four-potential is generated by the electron itself. How can this four-potential interact with the electron itself?
Under the entry "Quantum electrodynamics" in Wikipedia, the Dirac equation for an electron is given by
$$ i\gamma^{\mu}\partial_{\mu}\psi - e\gamma^{\mu}\left( A_{\mu} + B_{\mu} \right) \psi - m\psi = 0 ,\tag 1 $$
or
$$
i\gamma^{\mu}\partial_{\mu}\psi - m\psi = e\gamma^{\mu}\left( A_{\mu} + B_{\mu} \right) \psi .\tag 2
$$
where ##A_{\mu}## is the covariant four-potential of the electrodynamic field generated by the electron itself, and ##B_{\mu}## is the external field imposed by external source.Under another entry "Electromagnetic four-potential" in Wikipedia, the components of the covariant four-potential ##A_{\mu} = (\phi, \mathbf{A})## are given by
$$
\phi \left( \mathbf{r}, t\right) = \frac{1}{4\pi\epsilon_{0}} \int d^{3}x' \frac{\rho \left( \mathbf{r'}, t_{r}\right)}{|\mathbf{r} - \mathbf{r'}|} ,\tag 3
$$
$$
\mathbf{A} \left( \mathbf{r}, t\right) = \frac{\mu_{0}}{4\pi} \int d^{3}x' \frac{j\left( \mathbf{r'}, t_{r}\right)}{|\mathbf{r} - \mathbf{r'}|} ,\tag 4
$$
where ##\rho\left( \mathbf{r}, t\right)## and ##\mathbf{j}\left( \mathbf{r}, t\right)## are charge density and current density respectively, and
$$
t_{r} = t - \frac{|\mathbf{r} - \mathbf{r'}|}{c} \tag 5
$$
is the retarded time.
When I try to solve the Dirac equation (2), I have problem dealing with the first term on the right-hand side, ##e\gamma^{\mu} A_{\mu} \psi##. The four-potential ##A_{\mu} = (\phi, \mathbf{A})## is generated by the electron itself, how can the electron interact with ##A_{\mu}##? Specifically, in this case, the charge density ##\rho## and current density ##\mathbf{j}## belong to the electron itself, and ##\mathbf{r} = \mathbf{r'}##. How can I calculate the term ##e\gamma^{\mu} A_{\mu} \psi##?
$$ i\gamma^{\mu}\partial_{\mu}\psi - e\gamma^{\mu}\left( A_{\mu} + B_{\mu} \right) \psi - m\psi = 0 ,\tag 1 $$
or
$$
i\gamma^{\mu}\partial_{\mu}\psi - m\psi = e\gamma^{\mu}\left( A_{\mu} + B_{\mu} \right) \psi .\tag 2
$$
where ##A_{\mu}## is the covariant four-potential of the electrodynamic field generated by the electron itself, and ##B_{\mu}## is the external field imposed by external source.Under another entry "Electromagnetic four-potential" in Wikipedia, the components of the covariant four-potential ##A_{\mu} = (\phi, \mathbf{A})## are given by
$$
\phi \left( \mathbf{r}, t\right) = \frac{1}{4\pi\epsilon_{0}} \int d^{3}x' \frac{\rho \left( \mathbf{r'}, t_{r}\right)}{|\mathbf{r} - \mathbf{r'}|} ,\tag 3
$$
$$
\mathbf{A} \left( \mathbf{r}, t\right) = \frac{\mu_{0}}{4\pi} \int d^{3}x' \frac{j\left( \mathbf{r'}, t_{r}\right)}{|\mathbf{r} - \mathbf{r'}|} ,\tag 4
$$
where ##\rho\left( \mathbf{r}, t\right)## and ##\mathbf{j}\left( \mathbf{r}, t\right)## are charge density and current density respectively, and
$$
t_{r} = t - \frac{|\mathbf{r} - \mathbf{r'}|}{c} \tag 5
$$
is the retarded time.
When I try to solve the Dirac equation (2), I have problem dealing with the first term on the right-hand side, ##e\gamma^{\mu} A_{\mu} \psi##. The four-potential ##A_{\mu} = (\phi, \mathbf{A})## is generated by the electron itself, how can the electron interact with ##A_{\mu}##? Specifically, in this case, the charge density ##\rho## and current density ##\mathbf{j}## belong to the electron itself, and ##\mathbf{r} = \mathbf{r'}##. How can I calculate the term ##e\gamma^{\mu} A_{\mu} \psi##?
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