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The probability of current in quantum mechanics when the is a magnetic potential, A, is defined as: $$\boldsymbol j=\frac{1}{2m}(\psi^*\hat{\boldsymbol p} \psi-\psi\hat{\boldsymbol p}\psi^* -2q{\boldsymbol A}|\psi|^2) \tag{1}$$

However, when dealing with plain waves A is ofter defined as a complex quantity: $${\boldsymbol A}={\boldsymbol A_0}e^{jwt-{\boldsymbol K}·{\boldsymbol x}}\tag{2}$$

This would lead to an imaginary current which as far as I know is something unphysical. My question is how to interpret a complex potential like this. Is it wrong? Is only the real part of it which is contributes to the probability of current? Or it is the absolute value?

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Complex-valued $\mathbf E$ or $\mathbf A$ appear often in calculations involving EM waves. It is understood there that such complex-valued quantity represents a real-valued quantity, this practice is because of easier manipulation of complex exponentials. This is possible and useful when all terms of the wave equation are at most linear functions of $\mathbf E$ or $\mathbf A$ (linear wave equation) but it does not work for situations where the equation has higher order terms.

Not so in calculations involving Schroedinger's equation - if we involve $\mathbf A$, it appears in second power already in the simplest of cases. Using complex exponential as opposed to real sine function would lead to very different evolution of $\psi$. In particular, it would make the Hamiltonian non-Hermitian so $\int|\psi|^2\,d\tau$ wouldn't be conserved in time.

This can be used in a cavalier way to describe particle capture but in the standard calculations of evolution of $\psi$ under action of external field, where no particles disappear/decay and the above integral is constant 1, the physical state of external EM field has to be described by a vector potential or electric field that is a real-valued quantity.

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When you work with plane waves you always have to take the real part when you want to talk of something physical. For example, if I compute the electric field from your potential, assuming there's no Coulomb potential $\phi$,

$$\vec E = -\frac{1}{c}\frac{\partial \vec A}{\partial t} - \vec \nabla \phi = i\frac{\omega}{c} \vec A_0 e^{i\omega t - i \vec k \cdot \vec x}$$

which is still complex, but the electric field is real. You use complex numbers because taking derivatives is simpler and it's also legit if you do only linear operations, but when you want to talk of physical objects you have to take the real part.

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As you know J is the probability current. The 1 dimensional one is defined as follows:

$$J(x,t) = \frac{i\hbar}{2m} (\Psi\frac{\partial \Psi^*}{\partial x}-\Psi^*\frac{\partial \Psi}{\partial x})$$

Notice that, based on this definition, the complex part of the probability current will vanish.

Let me explain myself. Imagine a wave function of the form:

$$\Psi (x,t) = f(x)e^{-iat}$$

Where $f(x)$ is a real function and $a$ a real constant.

Use the probability current definition and see how the imaginary part vanishes. I will work out one of the components explicitly:

$$\Psi\frac{\partial \Psi^*}{\partial x} = f(x)e^{-iat}\frac{df}{dx}e^{iat}= f(x)\frac{df}{dx}$$

Source: Introduction to Quantum mechanics; Second edition, David J. Grifiths.

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