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I know that a vector potential has to be taken into account for the schrödinger probability current: $$\vec{j}=\frac{1}{2m} \left[ \Psi^*\hat{\vec{p}}\Psi-\Psi\hat{\vec{p}}\Psi^* - 2q\vec{A} |\Psi|^2 \right]$$

Does a scalar potential also affect the probability current? Does it matter if the scalar potential depends on time?

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Take the probability density:

$$ \rho = \Psi^*\Psi $$

and diferentiate it wtr. time, then substiture your srhödinger equation, does the scalar potential appear?

$$ i\hbar\partial_t\rho = i\hbar\partial_t\Psi^*\Psi + \Psi^* i\hbar\partial_t\Psi \quad \quad (1) $$

And from Schöringer's equations $i\hbar \partial_t\Psi = \hat{H}\Psi = \frac{1}{2m}(\hat{p}-\frac{q}{c}\hat{A})^2\Psi + e\hat{\phi}\Psi$, noting thaht the hamiltonian is hermitic, we can directly subtitute in $(1)$

$$ i\hbar\partial_t\rho = \left[-\frac{1}{2m}(\hat{p}-\frac{q}{c}\hat{A})^2\Psi^*\right] - e\hat{\phi}\Psi^*\Psi + \Psi^* \left[\frac{1}{2m}(\hat{p}-\frac{q}{c}\hat{A})^2\Psi\right] + e\hat{\phi}\Psi^*\Psi $$

from you can readily see that the scalar potetial cancels out. The rest of the calculation I leave it as an execrise to the reader.

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  • $\begingroup$ I should also say that no, it doesn't depend on time. But as the wave function $\Psi$ has to obey Schrödinger's equation, wich has the time dependig scalar potential, the courrent will depend on time, put it isn't in its definition. $\endgroup$
    – iiqof
    Apr 15, 2013 at 7:37

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