Suppose one has a position eigenstate $|x\rangle$. This can always be expressed in the momentum space representation via resolution of the identity:
$$|x\rangle = \int_{\mathbb{R}}dp \ \langle p| x\rangle|p\rangle = \dfrac{1}{\sqrt{2\pi \hbar}}\int_{\mathbb{R}}dp \ e^{-\frac{ipx}{\hbar}}|p\rangle$$
Now, in subsequent calculations I will be evolving a state like this according to a Hamiltonian typically involving functions of $\hat{x}$ and $\hat{p}$. In order to understand how to carry that out, I need to understand how the basic property $\hat{x}|x\rangle = x |x\rangle$ may be verified from the above by direct application of $\hat{x}$ in momentum space representation, as this will hopefully give insight into how my Hamiltonian can be applied to a state in this representation. In particular, I'm greatly confused about how to apply the derivative that arises in the momentum representation correctly here, and believe this is at the heart of my misunderstanding.
$$\hat{x}|x\rangle = i\hbar\frac{\partial}{\partial p}\left[\dfrac{1}{\sqrt{2\pi \hbar}}\int_{\mathbb{R}}dp' \ e^{-\frac{ip'x}{\hbar}}|p'\rangle\right]$$
It is tempting to bring the derivative under the integral sign, then apply some kind of product rule to the integrand? Overall, very confused as to how to apply the position operator to an object like this when in the momentum space representation. Any clarification on how this should work is greatly appreciated!
Best Answer
Your order of operations is incorrect. The correct starting point is \begin{align} {\hat x} | x \rangle &= {\hat x} \left( \frac{1}{\sqrt{2\pi \hbar}}\int dp e^{- \frac{i}{\hbar} p x } | p \rangle \right) \\ &= \frac{1}{\sqrt{2\pi \hbar}}\int dp e^{- \frac{i}{\hbar} p x } {\hat x} | p \rangle \end{align} Next, we use $$ {\hat x} | p \rangle = - i \hbar \frac{d}{dp} | p \rangle $$ Then, \begin{align} {\hat x} | x \rangle &= \frac{1}{\sqrt{2\pi \hbar}}\int dp e^{- \frac{i}{\hbar} p x } \left( - i \hbar \frac{d}{dp} | p \rangle \right) \\ &= - i \hbar \frac{1}{\sqrt{2\pi \hbar}}\int dp e^{- \frac{i}{\hbar} p x } \frac{d}{dp} | p \rangle \\ &= i \hbar \frac{1}{\sqrt{2\pi \hbar}}\int dp \frac{d}{dp} e^{- \frac{i}{\hbar} p x } | p \rangle \\ &= i \hbar \left( - \frac{i}{\hbar} x \right) \frac{1}{\sqrt{2\pi \hbar}}\int dp e^{- \frac{i}{\hbar} p x } | p \rangle \\ &= x | x \rangle. \end{align}