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!
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!
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