2
$\begingroup$

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.)

$\endgroup$
1
  • 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
1
$\begingroup$

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): $$\int_{-\infty}^{+\infty}dxe^{i(k-\frac{p}{\hbar})x} = 2\pi\delta(k - \frac{p}{\hbar}).$$

Note
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.

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $p=hf=2\pi\hbar/\lambda=\hbar k$,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.