In Mahan eq. 1.193 I see an expression for the second quantized current operator of the form:


However, in other online sources I instead see expressions like


Are these expressions equivalent? It seems to me that the $\psi(r)$ and $\nabla\psi^\dagger(r)$ do not commute, so I am surprised to see both definitions in different places online.

  • 1
    $\begingroup$ The right definition is the one that makes $j(r)$ hermitian. That should not be too difficult to check. $\endgroup$ – Hans Moleman Sep 7 at 22:47

The correct definition is the first one, you can easily check that it gives a physically correct hermitian current density $j(r) = j(r)^\dagger$. As hinted by Hans Moleman in the comment to your question.

Online resources might be quoting the second expression as a special case, only valid in certain situations. $\psi(r)$ and $\nabla\psi(r)$ do commute, for instance, for $\psi(r) = \mathrm{e}^{\mathrm{i}(kr-\omega t)}$.
In general, any spin-$0$ particle/field, described by a scalar, will obey the commutation relation. And usually spin-$0$ are the first things books do when explaining QFT and second quantisation, so I would not be surprised if that's the case for the resources you are looking at.


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