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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!

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3 Answers 3

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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}

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  • $\begingroup$ Hi, can you clarify why my order of operations is incorrect? I see in your first line that you pass your operator through the exponential factor first despite it's dependence on $p$ and $x$. Why is that? Is it due to the fact that the operator itself is only applicable to states? How does one decide on which order of operations is the correct one? My second question is what happens to the boundary term during the integration by parts? How do you argue that it vanishes? $\endgroup$ Feb 22, 2023 at 12:50
  • $\begingroup$ No, the relationship ${\hat x} | p \rangle = - i \hbar \frac{d}{dp} | p \rangle$ is not quite right. The simplest way to work with this in a self-consistent and rigorous way is as in this answer of mine. $\endgroup$ Feb 22, 2023 at 12:51
  • $\begingroup$ @EmilioPisanty If I've understood correctly, you explain that $\langle x| \hat{x} = i\hbar \frac{\partial}{\partial p} \langle{x}|$. Writing $\langle x |$ in momentum representation results in a similar scenario symbolically: $\langle x| \hat{x} = \dfrac{1}{\sqrt{2\pi \hbar}} i\hbar \frac{\partial}{\partial p} \int_{\mathbb{R}}dp \ e^{-\frac{ipx}{\hbar}}\langle p| $. Now, applying this symbolic relation to an arbitrary state $|\Phi \rangle$ and bringing the derivative under the integral sign, there is a need for a product rule under the integral sign. Is that correct? $\endgroup$ Feb 22, 2023 at 13:07
  • $\begingroup$ @Emilio Pisanty The expression in the answer is right, after all. The signs from interchanging bras with kets and x with p cancel each other, after all. The answer is fine. The OP's original sin is not using the proper definition as commented above. $\endgroup$ Feb 22, 2023 at 14:41
  • $\begingroup$ @Theoreticalhelp - ${\hat x}$ is an operator that acts on Hilbert space states. It commutes with all numbers. In the integral, $x$ and $p$ are simply numbers. ${\hat x}$ does not care about those at all. That's why it commutes past all of them to directly at on the state. Basically ${\hat O} ( c |\Psi\rangle ) = c ( {\hat O}|\Psi\rangle)$ for all $c \in {\mathbb C}$. $\endgroup$
    – Prahar
    Feb 22, 2023 at 20:46
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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.

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  • $\begingroup$ Hi, it feels like this perspective doesn't quite answer my question in the spirit I was seeking. Firstly, I think there is an error in your latter integrals, which I believe should be over $x$, not $p$. I fully expected (and indeed took it as a definitive property apriori) that $\hat{x}|x\rangle = x |x\rangle$ but my question was about a direct verification of this fundamental fact by computation in the momentum basis - the particulars of that calculation are not fully captured by your example. $\endgroup$ Feb 22, 2023 at 13:34
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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 $π‘₯Μ‚=π‘–β„βˆ«\!\!𝑑𝑝 ~ |π‘βŸ©βˆ‚_π‘βŸ¨π‘| $ ? (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.

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  • $\begingroup$ This is what I was after, thank you. $\endgroup$ Feb 22, 2023 at 15:24

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