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Consider the general solution to the time-dependent Schrödinger equation for a free particle

\begin{align*} \Psi(x,t) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i\left(\hbar kx-\frac{1}{2}\frac{\hbar^2k^2}{m}t\right)/\hbar} dk\\ \phi(k)&= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \Psi(x,0) e^{-ikx}dx \end{align*}

The time-dependent Schrödinger equation for the free particle is \begin{equation} i\hbar \frac{\partial \Psi}{\partial t} = \frac{1}{2} \frac{\hbar^2k^2}{m} \frac{\partial^2 \Psi}{\partial x^2} \end{equation}

Presumably if $\Psi(x,t)$ is a solution, I should be able to plug it into both sides and show they are equal.

My Problem:

I don't know how to take the partial derivative of $\Psi(x,t)$. That expression looks very complicated and I have never (1) taken the derivative of a Fourier transform or (2) done so when I have an inverse Fourier transform nested inside it (ie the $\phi(k)$ term).

My Question:

Can someone explain how to perform these partial derivatives? If it's the same method for differentiating with respect $t$ and $x$, then you need only explain for one case.

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closed as off-topic by ACuriousMind, JamalS, BMS, Kyle Kanos, Jim Mar 24 '15 at 15:30

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    $\begingroup$ the partial differentiation acting on $\Psi(x,t)$ wrt to $x$ and $t$ only hits the exponential term inside the integral, your confusion is probably from the $x$ in the definition of $\phi(k)$ but that is a dummy integration variable which you could have as well denoted by any other letter. $\endgroup$ – Ali Moh Mar 20 '15 at 22:57
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    $\begingroup$ There is a $k^2$ too much on the RHS of your Schrödinger equation. $\endgroup$ – ACuriousMind Mar 20 '15 at 23:52
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$$\frac{\partial}{\partial t}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i\hbar kx}\left(\frac{\partial}{\partial t}e^{-\frac{1}{2}\frac{\hbar^2k^2}{m}t/\hbar}\right) dk$$

$$\frac{\partial}{\partial x}\Psi(x,t) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) \left(\frac{\partial}{\partial x}e^{i\hbar kx}\right)e^{-\frac{1}{2}\frac{\hbar^2k^2}{m}t/\hbar} dk$$

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