Exploring Quantum Physics - Week 1 Bonus Question 1

Question:

Which is the position operator in momentum space?

Solution:

By watching the lecture video alone, I have no idea how to approach this. Fortunately, I have got from the library a copy of "Quantum Mechanics Demystified" which give me an idea what is the momentum space.

So basically, we have $\psi(x)$ in position space, then we have $\phi(p)$ in momentum space as Fourier transform pair

$\begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\phi(p) e^{\frac{ i p x}{\hbar}} dp} \\ \phi(p) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx} \end{eqnarray*}$

So for the position operator, in position space, it should have representation $x \psi(x)$. Now in momentum space, it should have representation $\frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) x e^{\frac{-i p x}{\hbar}}} dx$. Note that we already have $\phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx}$, we just need to get one more x in the integrand, and here is the Eureka moment! Differentiate the integrand by $p$, that will yield:

$\begin{eqnarray*} \frac{\partial}{\partial p}\phi(p) &=& \frac{\partial}{\partial p}\frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) e^{\frac{-i p x}{\hbar}} dx} \\ &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) \frac{\partial}{\partial p}e^{\frac{-i p x}{\hbar}} dx} \\ &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) \frac{-i x}{\hbar} e^{\frac{-i p x}{\hbar}} dx} \\ i\hbar \frac{\partial}{\partial p}\phi(p) &=& \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{-\infty}^{\infty}{\psi(x) x e^{\frac{-i p x}{\hbar}} dx} \end{eqnarray*}$

So $i\hbar \frac{\partial}{\partial p}\phi(p)$ is exactly what we will need :) Sometimes you need another book, and sometimes, you need an Eureka moment!