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How can $\psi=Ae^{i(kx-\omega t +\phi)}$ represent a plane wave travelling in $+x$ direction?
I knew the equation $\psi = A\cos(kx - \omega t + \phi)$. I also know Euler's identity and that the second equation is just the real part of the first one.
The same question was asked here, but I couldn't understand Bobak's answer in there. He does not mention the need for a "2 component object".
Sorry if this seems redundant but QM is a very new topic for me.
Both plane waves have time dependence, so the question is which one satisfies the Time-Dependent Schrodinger Equation (TDSE) in one dimension,
$$i\hbar\frac{\partial}{\partial t}\psi(x,t)=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi(x,t).$$
The answer is that the complex one does (for $V=0$, since we’re talking about a free particle) and the real one does not. So the complex one is a valid wavefunction and the real one isn’t.
The TDSE is complex so it is not surprising that its solutions are complex.
Regardless of what it means, carrying the imaginary part and using Euler's formula actually makes the algebra much less cumbersome. Once the legwork is done, just extract the real part and you'll be back in familiar territory (a real plane wave). Later on, you'll learn more about how the imaginary part is otherwise useful. Hope this helps.
