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I am reflecting on the solution of the time-independent Schroedinger equation.

My reasoning is that the stationary state of the time-independent Schroedinger equation must be a real valued function because of their stationary condition they must not carry any flux.

Is my reasoning correct?

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For Hamiltonian operators of the form $$H = -\frac{\hbar^2}{2m} \Delta + V(x)$$ $V$ must be real to assure that $H$ is at least symmetric and eigenvalues $E$ are real. Therefore, if $0\neq \psi= \psi(x)$ is an eigenfunction, so that: $$-\frac{\hbar^2}{2m} \Delta \psi + V(x) \psi = E\psi \tag{1}$$ taking the complex conjugate of both sides you have: $$-\frac{\hbar^2}{2m} \Delta \overline{\psi} + V(x)\overline{\psi} = E\overline{\psi}\:.\tag{2}$$ (1) and (2) together imply that the real valued functions $\Psi= \psi + \overline{\psi}$ and $\Phi= i(\psi - \overline{\psi})$ are eigenfunctions with the same eigenvalue $E$. Notice that the pair $\Psi$, $\Phi$ encompass the same information as that of the pair $\psi$ and $\overline{\psi}$: They are linearly independent if and only if $\psi$ and $\overline{\psi}$ are and generate the same vector subspace (of the eigenspace associated to $E$).

If an eigenspace with energy $E$ has dimension $\geq 2$, you may have a nonvanishing flux of a generic eigenfunction with that energy.

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You can choose it to be real, obviously you can always multiply the wave function by a phase factor. You can easily prove this by taking the complex conjugate of the eigenvalue equation.

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  • $\begingroup$ Multiplying by a phase factor can lead to currents. For example, the macroscopic superconducting wavefunction can be considered as a "constant" sqrt(density) with a phase factor. That phase factor, when you compute "psi-star-del-psi" and its complex conjugate, can lead to a non-zero difference if the phase has spatial dependence. $\endgroup$ – eqb May 29 '14 at 19:28

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