I am trying to derive an equation which is a standard result in physics (the momentum space Schrödinger equation).

(Background: The wavefunction is a complex valued function of position coordinates and time $\psi:\mathbb{R^3}\times\mathbb{R^+}\rightarrow\mathbb{C}$. It is square integrable ($L^2$) and is generally assumed to behave such that as ${r\rightarrow\pm\infty}$ then $\psi$ and $\nabla \psi$ fall off to zero sufficiently fast. Starting from the fact the position-space ($\psi$) and momentum space ($\phi$) wave functions are fourier transforms of each other):- $$\phi(\mathbf{p},t) = \int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r'$$ (Integral runs over all space) Taking the partial time derivative $$\partial_t\phi(\mathbf{p},t) =\partial_t\left(\int \psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r' \right) = \int \partial_t\psi(\mathbf{r'},t) e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}d^3r'$$ Because $t$ is a parameter while $p$ and $r$ are dynamical variables, i.e have no explicit $t$ dependence. Now using a postulated differential equation of $\psi$ $$i\hbar \partial_t\psi(\mathbf{r},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t) + V(\mathbf{r})\psi(\mathbf{r},t) $$ I get $$i\hbar \partial_t\phi(\mathbf{p},t)= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r' $$

First Question: I moved my exponential term to the left, so that the operator $\hat{H} =-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbb{r}) $ acts only on $\psi(\mathbb{r},t)$ and not on $\psi(\mathbb{r},t)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}$. What is the justification for this exchange mathematically?

Evaluating the first integral by using integration by parts and identifying the resulting integral as the fourier transform I get $$ -\frac{\hbar^2}{2m}\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\nabla^2\psi(\mathbf{r'},t)d^3r' = \frac{p^2}{2m}\phi(\mathbf{p},t)$$

Second Question: I am having trouble understanding the second integral perhaps for the same reason as my first question: after fourier transforming $\psi$ $$ \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})\psi(\mathbf{r'},t)d^3r'= \int \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'})e^{\frac{i}{\hbar}\mathbf{p'}\cdot\mathbf{r'}}\phi(\mathbf{p'},t)d^3r'd^3p' $$ I thought I had done it but when I looked it up, there was a caveat that $V(\mathbb{r})$ must be analytic and the following must be true:$$\color{blue}{e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}V(\mathbf{r'}) = V(i\hbar\nabla_p)e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}}$$

Under this assumption I have the correct equation in momentum representation. $$i\hbar \partial_t \phi(\mathbf{p},t) = \frac{p^2}{2m}\phi(\mathbf{p},t)+V(i\hbar\nabla_p)\phi(\mathbf{p},t)$$

I am not getting where the blue thing came from. Can someone explain what conditions hold for $\hat{f}(x)\hat{g}(y)=\hat{g}(y)\hat{f}(x)$ particularly in this context.

Third Question: Any other less clumsy way to derive this?

  • $\begingroup$ Oh hey, I didn't know you can use color commands in MathJax. $\endgroup$ Apr 14, 2011 at 10:20
  • $\begingroup$ @Willie Wong '\color{blue}{text-to-color}' $\endgroup$
    – Please Delete Account
    Apr 14, 2011 at 11:06

1 Answer 1


First question: You didn't move anything on the right-hand side, you just Fourier-transformed both sides of the previous equation and then commuted the time derivative with the Fourier transform on the left-hand side.

Second question: You seem to be missing $\phi(\mathbf p,t)$ in the double integral, I believe the sign in the exponential with $\mathbf p'\cdot \mathbf r'$ is wrong, there's a supernumerary closing parenthesis in the equation after that, and you didn't introduce $\hat{f}(x)$ and $\hat{g}(y)$; the question may either resolve itself or become easier to answer if you address all these issues.

[Edit in response to the comment] To make the answer to the first question more explicit: Starting from the time-dependent Schrödinger equation,

$$i\hbar \partial_t\psi(\mathbf{r},t)=-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t) + V(\mathbf{r})\psi(\mathbf{r},t)\;,$$

you Fourier-transform both sides:

$$\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}i\hbar \partial_t\psi(\mathbf{r'},t)d^3r'= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$

Then you exchange the time derivative and the integral:

$$\partial_t\int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}i\hbar \psi(\mathbf{r'},t)d^3r'= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$

Mathematically one might be interested in the conditions under which this is valid, but physicists tend not to worry about that sort of thing; in any case, this is certainly much less problematic than exchanging the Laplacian and the phase factor would be. Now you just have to substitute the definition of $\phi(\mathbf{p},t)$ to obtain

$$i\hbar\partial_t \phi(\mathbf{p},t)= \int e^{-\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r'}}\left[ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r'},t) + V(\mathbf{r'})\psi(\mathbf{r'},t)\right]d^3r'\;.$$

  • $\begingroup$ @Joriki edited. the $\hat{f}$ and $\hat{g}$ refer to two different operators. $\endgroup$
    – Please Delete Account
    Apr 14, 2011 at 7:50
  • $\begingroup$ regarding first question: I did not see it as a fourier transform of my previous equation.. i was plugging in a known DE for $\psi$, but seeing that both are equivalent, i think my trouble is understanding the meaning of "then commuted the time derivative with the Fourier transform on the left-hand side." can you explain a bit because I am quite shaky on these issues. $\endgroup$
    – Please Delete Account
    Apr 14, 2011 at 7:54
  • $\begingroup$ @joriki thanks for answering .. one more question.. " this is certainly much less problematic than exchanging the Laplacian and the phase factor would be. " you are referring to the "first integral" in my answer, I never exchanged it.. I used the integration by parts technique twice and never exchanged the phase factor with the laplacian throughout the process. But I am still interested in knowing why it would be problematic, what exactly is the issue with the exchanges. I want to know the machinery that runs behind it. $\endgroup$
    – Please Delete Account
    Apr 14, 2011 at 8:29
  • $\begingroup$ @Approximist: I think that's a misunderstanding. I was refering to your sentence "I moved my exponential term to the left ...". So I wasn't saying that what you did after that is problematic, only that you don't need to do anything as problematic as moving the exponential term to the left of the Laplacian. $\endgroup$
    – joriki
    Apr 14, 2011 at 8:35
  • $\begingroup$ @joriki some more question. You have taken the postulated DE (schrodinger eqn) Fourier transformed it, then taken the partial time derivative. (I took the Fourier transform, took the partial time derivative, then put in the postulated DE). The only difference I see two delicate points that allow us to arrive at the common solution. In my case I did not understand why it was justified to commute the time derivative with the phase and in your case you have not justified contd.. $\endgroup$
    – Please Delete Account
    Apr 14, 2011 at 17:41

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