How would you evaluate \begin{equation}|iD\!\!\!\!/-m|\end{equation} Where $D_{\mu}=\partial_{\mu}-ieA_{\mu}$. I have an idea of how to do this without the gauge field, because it's essentially \begin{equation}|i\partial\!\!\!/-m|=|\partial^{2}-m^{2}|^{1/2}\end{equation} and A. Zee's book covers this is some detail. From what I can tell, you use $\ln|A|=\mathrm{tr}\ln A$. The trace is in momentum space \begin{equation}\frac{1}{2}\int \frac{d^{4}k}{(2\pi)^{4}} \ln (k^{2}-m^{2}-i\epsilon)+C\end{equation} How do you do this in the presence of a gauge field? I've gotten this far \begin{equation}|iD\!\!\!\!/-m|=e^{\frac{1}{2}\mathrm{tr}\ln \left(D^{2}-(e/2)\sigma^{\mu\nu}F_{\mu\nu}+m^{2}\right)}\end{equation} With $\sigma^{\mu\nu}=(i/2)[\gamma^{\mu},\gamma^{\nu}]$. But I have no idea how to proceed.
2 Answers
I take it you mean the determinant with the straight bars...?
If so, the only way to compute it for general background gauge field is, as nervxxx mentioned in the comments, to expand the determinant in $A_\mu$.
If you are considering a specific background gauge field (like a constant magnetic field) you should look whether you can find the eigenvalues of your operator, i.e. solve \begin{equation} (i D\!\!\!/ - m) \psi = \lambda \psi \;. \end{equation} The determinant will then be given as the product of all the eigenvalues. (In order to regularize the expression, it may be useful to rewrite it as the exponential of the sum of the logarithm of all eigenvalues, but that depends on the specifics of the problem.)
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"Without the gauge field", you reduce the problem from the spinor differential equation to a second order scalar differential equation. For what it's worth, you can do the same for the Dirac operator with the gauge field, but you will get a fourth order partial differential equation (in a general case) for one function. The procedure described in my articles https://arxiv.org/abs/1008.4828 (Journal of Mathematical Physics, vol. 52, p. 082303 (2011)) and https://arxiv.org/abs/1502.02351 requires minor modification in your case. I am not sure though this will help you get the final answer.
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