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I am reading a book on Schroedinger's equation and it says that

"The relation between $\psi(x, 0)$ and $\phi(p)$ [where the latter is the amplitude in the $\psi(x,t)$ integral] is obtained by noting that [the Schroedinger equation] is a Fourier integral and thus can be inverted."

May I ask what the bold bit means? Thanks.

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    Could you write the expression for $\psi(x, t)$ down?2011-08-22
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    Sure. $$\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}dp\phi(p)e^{\frac{i(px-Et)}{\hbar}}$$2011-08-22

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The bold bit is referring to the the Fourier Inversion Theorem. Of course, the author is probably glossing over the requirements for convergence of the inverse fourier transform, but often the inverses exist as distributions.

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    Thanks. So that bit just means that we can make the function in the integral the subject function... Just out of curiosity are these ***"Fourier integrals"*** common?2011-08-22
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    Fourier integrals are just "Fourier transforms".2011-08-22
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    @H Taylor: as Jonas Teuwen says, Fourier Integrals are simply Fourier Transforms; therefore, they are quite common.2011-08-22