 # From a Hamiltonian for the Dirac Equation, we can add a potential term to it simply by adjusting the momentum operator so that p^(nu)->p^(nu)−A^(nu), where A^(nu) is the relevant potential. But how do you calculate A^(nu)? For example, what would A^(nu) be for an electron in an electromagnetic field given by the tensor F^(alpha beta)? Matias Aguirre 2022-07-23 Answered
From a Hamiltonian for the Dirac Equation, we can add a potential term to it simply by adjusting the momentum operator so that ${p}^{\mu }\to {p}^{\mu }-{A}^{\mu }$, where ${A}^{\mu }$ is the relevant potential. But how do you calculate ${A}^{\mu }$? For example, what would ${A}^{\mu }$ be for an electron in an electromagnetic field given by the tensor ${F}^{\alpha \beta }$?
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The field tensor can be derived from the vector potential like so:
${F}^{\mu \nu }={\mathrm{\partial }}^{\mu }{A}^{\nu }-{\mathrm{\partial }}^{\nu }{A}^{\mu }$
If $F$ is simple enough, you can usually construct an appropriate $A$ without too much difficulty. Otherwise you're stuck inverting this with a bunch of indefinite integrals.
Note that $A$ is not uniquely determined by this relation. If ${A}^{\mu }$ is a valid vector potential, then for any analytic function $\varphi$
${A}^{\prime \mu }={A}^{\mu }+{\mathrm{\partial }}^{\mu }\varphi$
will give equivalent results.