I'm trying to solve problem 3.12 in D.J. Griffiths's "Introduction to Quantum Mechanics 3rd ed."; it is as follows:

Find [the momentum space wave equation] $\Phi(p,t)$ for the free particle in terms of $\phi(k)$.

$\phi(k)$ is defined in the free particle's 1D position space wave equation

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi(k)e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}dk$$

as $$\phi(k) = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\Psi(x,0)e^{-ikx}dx$$

That is, if we use the definition of the Fourier transform where $\frac{1}{\sqrt{2\pi}}e^{\mp ikx}$ is used in the integrands for the Fourier and inverse Fourier transforms respectively (I was always taught $e^{\mp 2\pi isx}$, but I'll roll with Griffiths's scaling), then $\phi(k)$ is really just the Fourier transform for the initial state of the wave equation in position space.

Now, the problem I'm running up against is the following: if I use Griffiths's way of converting $\Psi(x,t)$ to $\Phi(p,t)$ (position space to momentum space wave equation), i.e.

$$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{+\infty}_{-\infty}\Psi(x,t)e^{-i\frac{p}{\hbar}x}dx$$

I get

$$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{+\infty}_{-\infty}\left(\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi(k)e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}dk\right)e^{-i\frac{p}{\hbar}x}dx$$

My intuition says that the two exponentials should just cancel out, and so the only way I see of simplifying the expression is if I assume that $p=\hbar k$ (I try to be very careful with this substitution, because it often causes trouble with constant factors). I get:

$$\Phi(p,t) = \frac{1}{\sqrt{\hbar}}\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\left(\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi\left(\frac{p}{\hbar}\right)e^{i\frac{p}{\hbar}x}e^{-i\frac{p^2}{2m\hbar}t}d\left(\frac{p}{\hbar}\right)\right)e^{-i\frac{p}{\hbar}x}dx$$

The inner integral performs an inverse Fourier transform, the outer a Fourier transform, so they cancel, to get:

$$\Phi(p,t) = \frac{1}{\sqrt{\hbar}} \phi\left(\frac{p}{\hbar}\right) e^{-i\frac{E}{\hbar}t}$$

This is nice and all, but I've read and been told before that $\phi(k)$ is the time-independent momentum space wave equation similar to $\psi(x)$, not $\frac{1}{\sqrt{\hbar}}\phi(k)$. What should be the scaling factor? I feel like $p=\hbar k$ is either not always applicable, or it may only be done when additional factors are added in front of the Fourier integrals (even if the integration variable is $x$ and the scaling is thus not really due to substitution of $dx$).

(I've looked here, but it doesn't give me any answers.)

  • 1
    $\begingroup$ This is incorrect. $k$ is an integration variable here whereas $p$ is a constant parameter. What you need to do is to perform the integration over $x$ first and then do the integral over $k$. $\endgroup$ – Prahar Mitra Mar 27 '20 at 13:04

As @Prahar has correctly pointed out, there is a purely mathematical error when equating the integration variable $k$ with the external variable $p$. Using two different symbols (e.g., $k$ and $k'$) would be the correct approach.

Further, the key to the solution is using the Fourier representation of the $\delta$-function (afterr changing the order of integration): $$\int_{-\infty}^{+\infty}dxe^{i(k-\frac{p}{\hbar})x} = 2\pi\delta(k - \frac{p}{\hbar}).$$

Griffiths scaling in Fourier transform is commonly used in physics, both in space ($k$) and in time ($\omega$) transforms. Also, don't be surprised to see differentials written right after the integration sign, before the integrand - as I have done - although this is more typical of quantum mechanics.

  • $\begingroup$ This answer helped me the most in realising where my error was. However, I'm still getting the same solution: $$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\frac{1}{\sqrt{2\pi}\hbar}\int^{+\infty}_{-\infty}\phi\left(\frac{k'}{\hbar}\right)e^{-i\frac{k'^2}{2m\hbar}t}\cdot\left[\int^{+\infty}_{-\infty}e^{i\frac{k'-p}{\hbar}x}dx\right]dk'$$ $$= \frac{1}{\sqrt{\hbar}}\int^{+\infty}_{-\infty}\phi(k)e^{-i\frac{\hbar k^2}{2m}t}\delta(k-\frac{p}{\hbar})dk$$ The remaining factor seems to match what @ZeroTheHero was saying, but I don't know how I could get a solution i.f.o. $p$ instead of $p/\hbar$. $\endgroup$ – Mew Mar 28 '20 at 9:46
  • $\begingroup$ @Mew this is about converting variable of a probability function. The probability of momentum in interval $[p, p+dp]$ is $$|\Phi(p,t)|^2dp = |\Phi(k,t)|^2dk = |\Phi(k,t)|^2d\frac{p}{\hbar},$$ that is $$|\Phi(p,t)|^2 = \frac{1}{\hbar}|\Phi(k,t)|^2,$$ and therefore $$\Phi(p,t) = \frac{1}{\sqrt{\hbar}}\Phi(k,t).$$ $\endgroup$ – Vadim Mar 28 '20 at 9:56

The problem is with your use of mixed variables $k$ and $p$. First, it is best to think of \begin{align} \langle x\vert p\rangle &=\frac{1}{\sqrt{2\pi\hbar}}e^{-i p x/\hbar}\, ,\qquad \langle p\vert x\rangle = \langle x\vert p\rangle^* =\frac{1}{\sqrt{2\pi\hbar}}e^{+i p x/\hbar} \end{align} which justifies the symmetric placement of the $\sqrt{2\pi \hbar}$ factor, but on the other hand \begin{align} \langle x\vert k\rangle &=\frac{1}{\sqrt{2\pi}}e^{-i k x }\, ,\qquad \langle k\vert x\rangle = \langle x\vert k\rangle^* =\frac{1}{\sqrt{2\pi}}e^{+i k x} \end{align} so that \begin{align} \Psi(p,t)&=\langle p\vert \psi\rangle = \int dx \langle p\vert x\rangle \langle x\vert \Psi(t)\rangle = \int dx \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\Psi(x,t)\, ,\\ \Psi(k,t)&=\langle k\vert \psi\rangle = \int dx \langle k\vert x\rangle \langle x\vert \Psi(t)\rangle = \int dx \frac{1}{\sqrt{2\pi}}e^{ikx}\Psi(x,t)\, ,\\ &= \sqrt{\hbar} \,\Psi(p,t)\, , \end{align} where the unit operator \begin{align} \hat 1=\int dx \vert x\rangle \langle x\vert \end{align} has been used.


Momentum $p$ and wave number $k$ are in fact related by $p=\hbar k$. Because they just differ by a constant, usually both are considered to describe the momentum of quantum systems (especially if you do your work with $\hbar=1$).

Note that this follows directly from the de Broglie relation $p=hf=2\pi\hbar/\lambda=\hbar k$,


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