# Transforming the Schrodinger equation from momentum-space to position-space [closed]

I am working on a problem right now that states the following:

A particle of mass $m$ is subjected to a force $\mathbf{F}(\mathbf{r}) = - \nabla V(\mathbf{r})$ such that the wave function $\varphi(\mathbf{p},t)$ satisfies the momentum-space Schrodinger equation:

$$(\mathbf{p}^2/2m - a \nabla_p^2)\varphi(\mathbf{p},t) = i \partial\varphi(\mathbf{p},t)/\partial t$$

where $\hbar = 1$, $a$ is some real constant and

$$\nabla_p^2 = \partial^2/\partial p_x^2 + \partial^2/\partial p_y^2 + \partial^2/\partial p_z^2.$$

Find the force $F(\mathbf{r})$.

Apparently I am suppose to be able to rewrite the momentum-space Schrodinger equation as the position-space Schrodinger equation as

$\left(\frac{-\nabla^2}{2m} + ar^2\right)\varphi(\mathbf{r},t) = i \frac{\partial \varphi(\mathbf{r},t)}{\partial t}$

In which case the potential can immediately be spotted as

$V(\mathbf{r} = ar^2)$,

and so the force is obviously

$F(\mathbf{r}) = -\nabla V(\mathbf{r}) = -\frac{\mathbf{r}}{r}\frac{d}{dr}V=-2ar$

I am not clear on how to rewrite the wave function in position-space however; it seems that I am suppose to use the fact that

$\psi(\mathbf{r},t) = (\frac{1}{2\pi})^{3/2} \int \varphi(\mathbf{k},t) e^{i\mathbf{k}\cdot\mathbf{r}} d\mathbf{k} \\ \varphi(\mathbf{k},t) = (\frac{1}{2\pi})^{3/2}\int\psi(\mathbf{r},t)e^{-i\mathbf{k}\cdot\mathbf{r}} d\mathbf{r}$

which yield the following relations

$\mathbf{p}^2\varphi(\mathbf{p},t) \to - \nabla^2 \psi(\mathbf{r},t)\\ \nabla_p^2(\mathbf{p},t) \to - r^2\psi(\mathbf{r},t)$

in order to transform from momentum-space to position-space.

However, I am at a loss at how to derive the latter relationships using the fact that $\psi(\mathbf{r},t)$ and $\varphi(\mathbf{p},t)$ are the fourier transforms of one another, and also, even if I were able to derive these relationships I can not see how I can just "place them" into the original momentum-space Schrodinger equation and obtain the position-space Schrodinger equation as I have written above. Can anyone please explain the mathematical intricacies that are being used to use the fourier transform relations to derive the new operators such that momentum-space Schrodinger equation can be rewritten? I guess I mean to ask, given the transformation of the operators (which I am not clear on) how does one just "rewrite" the momentum-space Schrodinger equation into the position-space Schrodinger equation?

## closed as off-topic by ACuriousMind♦, Kyle Kanos, John Duffield, user36790, Sebastian RieseNov 9 '15 at 13:40

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• If you look at Schrödinger equation as the equality of two functions of $p$, you can take the Fourier transform of both sides. Then, switching around derivatives and integrals, you can see that the desired form appears for each term of the equation. – Steven Mathey Nov 5 '15 at 8:34