Momentum-Space Wavefunction - Expectation Values: Quantum Mechanics (Griffiths)

In D. J. Griffiths' "Introduction to Quantum Mechanics: 1st Edition," Problem 3.51 on page 117 is a 3-star problem. In the second edition, it's problem 3.12 on page 109, and is not a starred problem at all. I attribute this to being given more information about $\Psi_{x,t}$ in terms of $\Phi_{p,t}$ in the second edition. In the first edition, we are working from knowledge of the form of $\Phi$ alone. Since one is a Fourier transform of the other and vice versa, we can convert the integral used to evaluate $< x>$  from one with an integrand function containing $\Psi$ and x to one containing $\Phi$ and the x operator.

$$< x> = \int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

In the first edition, we're given

$$\Phi(p,t)=\frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}e^{-ipx/\hbar}\Psi(x,t)dx        (1)$$

We know $<x>=\int_{-\infty}^{\infty}x|\Psi(x,t)|^{2}dx$, so we just need to show that

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp=\int x|\Psi(x,t)|^{2}dx$$

Easier said than done, but here we go. It's not too bad, actually...

$$\int \Phi^{*}\left ( -\frac{\hbar}{i}\frac{\partial }{\partial p} \right )\Phi dp$$

Using (1) gives

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left ( -\frac{\hbar}{i} \frac{\partial }{\partial p}\right )\left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{-ipx/\hbar}\Psi dx \right ]dp$$

Simplifying a little...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}\Psi dx \right ]dp        (2)$$

Consider the integrand function in the second set of brackets. In particular,  $i\hbar\frac{\partial }{\partial p} e^{-ipx/\hbar}$.

Using the fact that $\frac{\mathrm{d} }{\mathrm{d} p}\left ( e^{-ipx/\hbar} \right )= -\frac{ix}{\hbar} e^{-ipx/\hbar}$ we can rewrite (2) using $x e^{-ipx/\hbar}$ in place of $i\hbar \frac{\partial }{\partial p}\left ( e^{-ipx/\hbar} \right )$ ...

$$\int \left [ \frac{1}{\sqrt{2\pi \hbar}}\int e^{ipx^{'}/\hbar}\Psi^{*}dx^{'} \right ]\left [ \frac{1}{\sqrt{2\pi \hbar}}\int x e^{-ipx/\hbar} \Psi dx \right ]dp$$

Simplifying further and using Fubini's Theorem gives

$$\frac{1}{2\pi \hbar}\int \int \int \left [ e^{ipx^{'}/\hbar}\Psi^{*} x e^{-ipx/\hbar}\Psi \right ]dx^{'}dxdp     (3)$$

Letting $y = \frac{p}{\hbar}$ and simplifying the integrand further gives

$$\frac{1}{2\pi }\int \int \int \left [ e^{-i(x-x^{'})/y} \Psi^{*} x \Psi \right ]dx^{'}dxdy     (4)$$

Evaluating the y-integral first...

$$\frac{1}{2\pi }\int e^{-i(x-x^{'})/y}dy = \delta (x-x^{'})$$

If you're not comfortable with this, check it for yourself by using Plancherel's Theorem to determine the Fourier transform of $\delta(x)$.

Now we have

$$\int \int \delta(x-x^{'})  \Psi^{*}(x^{'},t) x \Psi (x,t) dx^{'}dx     (5)$$

But $\int \delta(x-x^{'}) \Psi^{*}(x^{'},t) = \Psi^{*} (x,t)$  (see Eq. 3.129 in the first edition and Eq. 3.52 in the second edition), so we have

$$\int \Psi^{*}x \Psi dx = \int x |\Psi (x,t)|^{2}dx$$

QED