Suppose $\psi = \psi_{real} + i \psi_{imag}$ be the wave function, then both $\psi_{real}$ and $\psi_{imag}$ can be used to solve the Schrodinger's equation

This can be demonstrated by plugging $\psi$ into the Schrodinger's equation and equating the real and complex part to get two separate schrodinger's equation

What is the practical implication of this finding?

Does it mean in the lab if $\psi_{real}$ is measured then it can be used to find the other parameters without knowing the full wave function?

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    $\begingroup$ Could you tell us how you would measure $\psi$ at all, no matter whether real or imaginary? The wavefunction is not an observable! $\endgroup$ – ACuriousMind Dec 27 '14 at 13:54

I think you're confusing a few things. If f and g are solutions of a linear homogeneous differential equation, so is f+g. But if f+g is a solution f and g are not necessarily solutions!


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