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.

  • $\begingroup$ Did you try putting into into the time-dependent Schrodinger equation for $V=0$? $\endgroup$
    – G. Smith
    Aug 20, 2019 at 18:34
  • $\begingroup$ I don't think I fully understand your question. $\psi=Ae^{i(kx-\omega t +\phi)}$ is a plane wave since it has the same value for all $y$ and $z$ values. It travels in the x-direction because the $-\omega t$ term shifts the function along the positive x-axis as time increases. These ideas are not unique to QM. Are you actually trying to ask how this is a valid wave function specifically in QM? $\endgroup$ Aug 20, 2019 at 18:48
  • $\begingroup$ Also you should give more detail as to why the question you link to should not be considered a duplicate. Not understanding an answer does not mean your question is not a duplicate. You actually have to explain what about your question is different. What are you looking for that is unique to your question compared to the potential duplicate? $\endgroup$ Aug 20, 2019 at 18:50
  • $\begingroup$ @AaronStevens My question is how can both of these equations represent the same things? $\endgroup$ Aug 20, 2019 at 18:52
  • $\begingroup$ @harshit54 They don't though $\endgroup$ Aug 20, 2019 at 18:54

2 Answers 2


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.

  • $\begingroup$ Thank you for answering. My discussion with Aaron and G.Smith cleared my doubts already. $\endgroup$ Aug 20, 2019 at 19:20

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