I am trying to reconcile two different ways of producing the Schrödinger equation in momentum space starting from the Schrödinger equation in abstract notation. The time dependent Schrödinger equation of course reads: $$i\hbar\frac{\mathrm{d} }{\mathrm{d} t}{|\psi(t)\rangle}=\hat H{|\psi(t)\rangle}=(\frac{\hat P^2}{2m}+ V(\hat X)){|\psi(t)\rangle}$$

The basic technique of projecting it onto momentum space is the following: $$i\hbar\frac{\mathrm{d} }{\mathrm{d} t}{\langle p|\psi(t)\rangle}=\langle p|(\frac{\hat P^2}{2m}+ V(\hat X)){|\psi(t)\rangle} =\frac{ p^2}{2m}\langle p|\psi(t)\rangle+\langle p| V(\hat X){|\psi(t)\rangle}$$ where we have used the fact that the continuum of eigenbasis vectors of $\hat P$ are labelled their respective eigenvalues $p$ satisfying: $\hat P | p\rangle= p| p\rangle$ and now using the definition $\langle p| \psi(t)\rangle=: \widetilde{\psi}(p,t)$ we can write the above equation:$$\begin{equation} i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\widetilde{\psi}(p,t)=\frac{p^2}{2m}\widetilde{\psi}(p,t)+\langle p| V(\hat x){|\psi(t)\rangle}\end{equation}\tag{P.S.E.}\label{P.S.E.}$$ Now comes the tricky part of dealing with the $\langle p|V(\hat X){|\psi(t)\rangle}$ term.

  1. The rigorous way of evaluating it is to first insert the identity operator in the momentum basis to have: $$\langle p|V(\hat X){|\psi(t)\rangle}= \int dp'\langle p|V(\hat X)|p'\rangle\langle p'{|\psi(t)\rangle}=\int dp'\langle p|V(\hat X)|p'\rangle \widetilde{\psi}(p',t)$$ Now we can insert the identity operator in the position basis twice to evaluate $\langle p|V(\hat X)|p'\rangle=\widetilde{V}(p-p')$ where $\widetilde{V}(p)$ is the Fourier Transform of $V(x)$ w.r.t. $p$. Bringing everything together, the Schrödinger equation in the momentum basis reads: $$\begin{equation} i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\widetilde{\psi}(p,t)=\frac{ p^2}{2m}\widetilde{\psi}(p,t)+\widetilde{V}(p) *\end{equation}\widetilde{\psi}(p,t)\tag{1}\label{1}$$ where the $*$ sign implies convolution.

  2. As can be seen from above, equation $\ref{1}$ is messy and hard to deal with for any practical calculation. On the other hand, we know that the operator $\hat X$ is represented by $i\hbar\frac{\partial }{\partial p}$ in the momentum basis (In abstract notation: $\langle p|\hat X|\psi(t)\rangle = i\hbar\frac{d}{dp}\langle p|\psi(t)\rangle$). Keeping that in mind, is it legal to use this for writing the momentum space Schrödinger equation $\ref{P.S.E.}$ as $$\begin{equation} i\hbar\frac{\mathrm{d} }{\mathrm{d} t}\widetilde{\psi}(p,t)=\frac{ p^2}{2m}\widetilde{\psi}(p,t)+V(i\hbar\frac{d}{dp})\widetilde{\psi} (p,t)\end{equation}\tag{2}\label{2}$$ where I have just replaced the $\hat X$ operator with $i\hbar\frac{\partial }{\partial p}$ ?

I ask this specifically because this is the kind of argument used to say that the S.E. is exactly the same in the position and momentum basis when we are considering the simple harmonic oscillator problem (where $x$ and $p$ are at equal footing in the Hamiltonian).

If the answer to my question is yes (which is my guess as one can almost always expand $V(x)$ in a power series of $x$ and then replace $x$ with $i\hbar\frac{\partial }{\partial p}$ everywhere), then what is the point of going through this whole rigmarole I describe in 1? If the answer is no in general, are there specific cases (where the functional dependence of the potential is polynomial?) when we can apply this trick and why?

Apologies for the long question and any help is appreciated.


2 Answers 2


I would interpret 1 (plus a few additional steps and an assumption about $V$) as a derivation of 2.

In particular

\begin{eqnarray} \langle p | V(\hat{X}) | \psi(t) \rangle &=& \int {\rm d} x \int {\rm d} p' \langle p | x \rangle \langle x | V(\hat{X}) | p' \rangle \langle p'| \psi(t) \rangle \\ &=& \int {\rm d} x \int {\rm d} p' e^{i p x} V(x) e^{-ip'x} \langle p' | \psi(t) \rangle \\ &=& \int {\rm d} x \int {\rm d} p' e^{-ip'x} V\left(-i\frac{\partial}{\partial p}\right)e^{i p x} \langle p' | \psi(t) \rangle \\ &=& V\left(-i\frac{\partial}{\partial p}\right) \int {\rm d} x \int {\rm d} p' e^{i x(p-p')} \langle p' | \psi(t) \rangle \\ &=& V\left(-i\frac{\partial}{\partial p}\right) \int {\rm d} p' \delta(p-p')\langle p' | \psi(t) \rangle \\ &=& V\left(-i\frac{\partial}{\partial p}\right) \langle p | \psi(t) \rangle \\ &=& V\left(-i\frac{\partial}{\partial p}\right) \tilde{\psi}(p,t) \end{eqnarray}

To go from line 2 to 3, I assumed I could Taylor expand $V(x)$, and replace each term like $x^n e^{i p x}$ with $(-i \partial_p)^n e^{i px}$. Then to go from line 3 to 4, I pulled $V(-i \partial_p)$ outside the integral. I think everything else is straightforward, but I'm happy to add additional clarification if needed.

  • $\begingroup$ Yeah, as I mentioned in my question already, I would expect this to happen as soon as you can taylor expand V(x) in power series of x. If I can always do this, method 1 seems like all work for nothing. Maybe there are non-trivial examples when it becomes essential that we start from 1? $\endgroup$
    – Arnab
    Commented Nov 14, 2020 at 15:01
  • $\begingroup$ How would you know you could do method 2 without this argument? $\endgroup$
    – Andrew
    Commented Nov 14, 2020 at 15:03
  • $\begingroup$ Also in step 3, x should be replaced with $i\hbar \partial_p$ without a negative sign. $\endgroup$
    – Arnab
    Commented Nov 14, 2020 at 15:04
  • $\begingroup$ What I wrote is fine, since $-i \partial_p e^{ipx}=x$ (I'm setting $\hbar=1$). $\endgroup$
    – Andrew
    Commented Nov 14, 2020 at 15:05
  • 1
    $\begingroup$ Oh, I see what you mean. Well, ultimately that comes from the fundamental commutation relation between x and p, which you have also implicitly used in your derivation for 2. At any rate, thanks for taking the time to answer my question. Appreciate it! $\endgroup$
    – Arnab
    Commented Nov 14, 2020 at 15:21

I would like to complete @Andew's answer. Method 1 is more general, and therefore, it's best to see method 2 as a consequence of 1. This can be seen by applying multiple times integration by parts to the integral of the convolution product. In general, from $V = x^n$, you deduce $\tilde V = 2\pi i^n\delta^{(n)}(p)$, so $$ \tilde V*\tilde \psi = (2\pi i^n\delta^{(n)}(p))*\tilde \psi \\ =(2\pi\delta(p))*((-i)^n\tilde \psi^{(n)}) \\ =(-i)^n\tilde \psi^{(n)} $$ using the measure $\frac{dp}{2\pi}$ in the integration as always, and the second step uses integration by parts.

Method 2 is more useful for a full analytical resolution. For example, take a particle under a uniform force field: $$ H = \frac{p^2}{2m}-Fx $$ in momentum space, this becomes: $$ \frac{p^2}{2m}\tilde\psi-Fi\hbar\tilde \psi' = E\tilde\psi $$ which you can easily solve: $$ \tilde \psi = C\exp\left(\frac{1}{Fi\hbar}\left(\frac{p^3}{6m}-Ep\right)\right) $$ and Fourier transforming back to position space, you obtain the integral representation of the Airy function.

Method 1 is typically used in perturbative expansions. Look at the Born expansion for example. Generally, when writing down Feynman diagrams, the reason why you integrate out the internal momentum lines is because you are calculating a convolution product. Also, mathematically, the method 1 has nicer properties, like when you convert a ODE to an integral equation on which you can apply fixed point theorems etc. To prove general mathematical results, it will therefore be the better starting points.

Hope this helps and tell me if something's not clear.


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