Quantum current density in general relativity

FunkyDwarf
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Hi All,

As shown here

http://en.wikipedia.org/wiki/Klein–Gordon_equation#Gravitational_interaction

one can modify the derivative operator in the Minkowski Klein-Gordon equation to generate the GR case for an arbitrary metric.

My question is, what would the quantum current density j be in this case? Given here
http://en.wikipedia.org/wiki/Four-current
is the classical current density in the EM case but that's not really what I'm after.

Furthermore, if i generate a radial wave equation from the GR KG equation in some metric and transform it in a Schrodinger like form, can I just use the standard tools of non-rel QM and use \psi^* \frac{d}{dx} \psi for j?

Cheers!
 
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I think a lagrangian like
<br /> \mathcal{L}=\bar{\psi}\gamma^\mu \partial^\nu \psi g_{\mu\nu}<br />
would get you essentially
<br /> \bar{\psi}\gamma^\mu \partial^\nu \psi=\theta^{\mu\nu}<br />
when varied by the metric, where the current corresponds to the 0\nu part.

But I kinda made that up, so we should keep that tentative lol.
 
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