This question already has an answer here:

one dimensional Schrödinger equation: $$ \left[-\frac{\hbar^{2}}{2 m}\frac{\partial^2\psi(x)}{\partial{x}^2} +V(x)\right] \psi(x)=E \psi(x)$$

I know that to calculate the eigenfunctions $ \psi(x) $ depends on the potential $V(x)$, but in general, which are the characteristic of $\psi(x)$? it can be a complex function or a real function and how proof that?

What means that $\phi(x)$ is a physical solution if we also care about the probability?

Can unphysical wavefunctions give a right probability?


marked as duplicate by Gert, Bill N, Ben Crowell, Qmechanic quantum-mechanics May 1 at 4:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


It’s not super-clear what you’re asking but if $\psi(x)$ is a solution then so is $i\psi(x)$ and more generally so is $e^{i\varphi}\psi(x)$. There is nothing unphysical about either of these solutions.

Indeed in general the time-dependent solutions $\Psi(x,t)$ will be fully complex functions, yet they are certainly physical.

The predictions of the theory, such as average values of observables etc, typically depend on $x^2\vert\Psi(x,t)\vert^2$ or $\vert \Psi’(x,t)\vert^2$, or such combinations, which are real quantities.

Unphysical wavefunctions - for instance a function that is not continuous - do not necessarily give the right probabilities.


Not the answer you're looking for? Browse other questions tagged or ask your own question.