Understanding action of position operator on momentum space representation of a position eigenstate

Question

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!

Answer 1

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}

Answer 2

Writing objects like $\hat{x} | x \rangle$ refers to states and operators in a geometrical way (that is basis independent). You have defined the state $|x\rangle$ as being the eigenstate of the position operator, i.e. $\hat{x} | x \rangle = x | x \rangle$. This is true regardless of if you write $|x\rangle = \dfrac{1}{\sqrt{2\pi \hbar}}\int_{\mathbb{R}}dp \ e^{-\frac{ipx}{\hbar}}|p\rangle$ or not.

To understand how the position operator looks in the momentum basis, we expand the equation in the momentum basis by $\langle p|\hat{x} | x \rangle = x \langle p| x \rangle$. However this object is horribly singular, its better to consider $\langle p | \hat{x} | \Psi \rangle$ for $|\Psi \rangle \in \mathcal{H}$.

We can invert the expression relating position and momentum eigen-kets

$$|p\rangle = \dfrac{1}{\sqrt{2\pi \hbar}}\int_{\mathbb{R}}dx \ e^{\frac{ipx}{\hbar}}|x\rangle$$

Substituting this into our expression $\langle p | \hat{x} |\Psi \rangle$ and using the fact that $\hat{x}$ is self adjoint, i.e. $\langle x|\hat{x} = \langle x|x$, then

$$\langle p | \hat{x} | \Psi \rangle = \dfrac{1}{\sqrt{2\pi \hbar}}\int_{\mathbb{R}}dx \ e^{\frac{-ipx}{\hbar}}x \langle x|\Psi\rangle = i\hbar \frac{\partial}{\partial_p}\Psi(p)$$

We see that the action of the position operator written in the momentum basis takes the form of a differential operator w.r.t. momentum.

Answer 3

I am providing a parallel proof/insight to the impeccable answer of @Prahar, as you appear conflicted about dummy variables. Your last equation is not even wrong: it's gobbledygook. A derivative w.r.t. p cannot act nontrivially on something which is not a function of p, like your square bracket.

Why don't you simply use the definition $̂=ℏ∫\!\! ~ |⟩∂_⟨| $ ? (Recall this is similar to the expression interchanging x with p, but for a minus sign.) $$ \hat x |x\rangle= i\hbar \int \!\!dp dp'~ |p\rangle \partial_p\langle p| p'\rangle e^{-ixp'/ \hbar } /\sqrt{2\pi \hbar}\\ - i\hbar \int \!\!dp dp'~ (\partial_p |p\rangle )\langle p| p'\rangle e^{-ixp'/ \hbar } /\sqrt{2\pi \hbar}= i\hbar \int \!\!dp ~|p\rangle (\partial_p e^{-ixp/ \hbar }) /\sqrt{2\pi \hbar}= x|x\rangle, $$ where, in the penultimate step, the p-p' delta function is collapsed before the second integration by parts.