# Should physical solution to Schrödinger eq. always be real in one dimencional space? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 1 at 4:56

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.