I was reading Cohen-Tannoudji's Quantum Mechanics Vol. 1 on the section about the free particle, with the general Schroedinger equation given by

$$i\hbar\frac{\partial}{\partial t}\phi({\mathbf{r}},t) = -\frac{\hbar^{2}}{2m}\Delta \phi(\mathbf{r},t) + V(\mathbf{r},t)\phi(\mathbf{r},t)$$

with solution to the case of the one dimensional free particle being

$$\phi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(k)e^{i[kx-\omega(k)t]}dk$$

From which

$$\phi(x,0)=\frac{1}{\sqrt{2\pi}}\int g(k)e^{ikx}dk \Rightarrow g(k)=\frac{1}{\sqrt{2\pi}}\int\phi(x,0)e^{-ikx}dx$$

He then claims that as a consequence of the last statement $\phi(x,0)$ is a valid form for any potential rather than just a free particle. This isn't immediately obvious to me so can anybody spell it out for me why this is so?

  • $\begingroup$ Where in the book does it say that ? $\endgroup$
    – Adam
    Dec 5, 2017 at 6:55
  • $\begingroup$ pg 21-23 (ch 1. C. 1) $\endgroup$
    – A. Kriesz
    Dec 5, 2017 at 7:07

1 Answer 1


They are just saying that you can always write down a wave-function in Fourier space, in particular the initial wave-function.


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