Electromagnetism – Why Canonical Momentum for Dirac Equation Is Not Defined in Terms of Gauge Covariant Derivative

covarianceelectromagnetismgauge-invariancemomentumoperators

The canonical momentum is always used to add an EM field to the Schrödinger/Pauli/Dirac equations. Why does one not use the gauge covariant derivative? As far as I can see, the difference is a factor i in front of the vector potential. I know I'm combining two seemingly unrelated things, but they seem very similar, an the covariant form seems much "better" with respect to the inherent gauge freedom in the EM field. I can also see that with the canonical momentum form, the equations remain unchanged after an EM and a QM (phase) gauge transformation. Suffice to say my field theory knowledge is not that impressive.

Best Answer

The identification goes as follows:

$$ \text{Kin. Mom.}~=~ \text{Can. Mom.} ~-~\text{Charge} \times \text{Gauge Pot.} $$

$$ \updownarrow $$

$$ m\hat{v}_{\mu} ~=~ \hat{p}_{\mu} - qA_{\mu}(\hat{x})$$

$$ \updownarrow $$

$$ \frac{\hbar}{i} D_{\mu} ~=~ \frac{\hbar}{i}\partial_{\mu} - qA_{\mu}(x) $$

$$ \updownarrow $$

$$ D_{\mu} ~=~ \partial_{\mu} -\frac{i}{\hbar} qA_{\mu}(x) $$

$$ \updownarrow $$

$$ \text{Cov. Der.}~=~ \text{Par. Der.} ~-~\frac{i}{\hbar}\text{Charge} \times \text{Gauge Pot.} $$

The imaginary unit $i$ is needed, e.g. because the derivative is an anti-hermitian operator (recall the usual integration-by-part proof), while the momentum is required to be a hermitian operator in quantum mechanics.

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