About the behavior of the position and momentum operators

Question

Following my book I came to know the following expressions for the position and momentum operators ($\hat{x},\hat{p}$): \begin{align}&\langle x|\hat{x}|\psi\rangle=x\psi(x) \ \ \ \ \ &(1)\\[1.5ex] &\langle x|\hat{p}|\psi \rangle=-i\hbar\frac{d}{dx}\psi(x) \ \ \ \ \ &(2)\\[1.5ex] &\langle p|\hat{x}|\psi\rangle=i\hbar\frac{d}{dp}\psi(p) \ \ \ \ \ &(3)\\[1.5ex] &\langle p|\hat{p}|\psi\rangle=p\psi(p) \ \ \ \ \ &(4)\end{align} To prevent misunderstandings: $|x\rangle,|p\rangle$ are the eigenvectors of position and momentum respectively, $|\psi\rangle$ is a general state, $x,p$ are the values of position and momentum (that can be seen as eigenvalues of the respective operators) and $\psi(x)=\langle x|\psi\rangle,\psi(p)=\langle p | \psi\rangle$ are simply the wavefunctions of the state $\psi$.

However we also know that the following is true: \begin{align} &\hat{x}|x\rangle=x|x\rangle \ \ \ \ \ &(5)\\[1.5ex] &\hat{p}|p\rangle=p|p\rangle \ \ \ \ \ &(6) \end{align} My question is: is there a way to write down explicitly the action of the position operator on the eigenvectors of momentum and the action of the momentum operator on the eigenvector of position, similarly to what we do in equations (5) and (6)? So in practice what I am asking is: $$\hat{x}|p\rangle=?$$ $$\hat{p}|x\rangle=?$$ Trying to answer my own question I came up with the following brutal line of reasoning: we can use equations (2) and (3): $$\langle x|\hat{p}|\psi \rangle=-i\hbar\frac{d}{dx}\langle x|\psi\rangle$$ $$\langle p|\hat{x}|\psi\rangle=i\hbar\frac{d}{dp}\langle p |\psi\rangle$$ and using the fact that $\hat{x},\hat{p}$ are both hermitian operators we can think to write: $$\hat{x}|p\rangle=i\hbar\frac{d}{dp}|p\rangle \ \ \ \ \ (7)$$ $$\hat{p}|x\rangle=-i\hbar\frac{d}{dx}|x\rangle \ \ \ \ \ (8)$$ however I really don't trust my own reasoning here, firstly because it is not mathematically formal at all; and secondly because deriving a vector representing a state just like it was a function make no sense in my mind ($d/dp|p\rangle=???$).

What is going on? Is my reasoning at least partially correct? Can we find an explicit form for $\hat{x}|p\rangle,\hat{p}|x\rangle$?

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Notice that I have edited my question to correct a mistake present into equation (3), that had nothing to do with the question itself. Some of the answers may refer to this mistake edited out.

Answer 1

The key results to remember are \begin{align} \langle x\vert p\rangle &= \frac{1}{\sqrt{2\pi\hbar}}e^{i p x/\hbar}\, \\ \hat p\langle x\vert p\rangle &=-i\hbar\frac{d}{dx}\langle x\vert p\rangle \, ,\\ \psi(x)&=\langle x\vert\psi\rangle\, ,\\ \psi(p)&=\langle p\vert\psi\rangle. \end{align} Thus, \begin{align} \hat x \langle p\vert x\rangle : =\langle p\vert \hat x\vert x\rangle = x \frac{1}{\sqrt{2\pi\hbar}}e^{-i p x/\hbar}=+i\hbar \frac{d}{dp}\langle \hat p\vert x\rangle \end{align} and thus \begin{align} \hat x\psi(p)&=\langle p\vert\hat x\vert\psi\rangle\, ,\\ &=\int dx \langle p\vert \hat x\vert x\rangle \langle x\vert \psi\rangle\, ,\\ &= \int dx i\hbar\frac{d}{dp} \langle p\vert x\rangle\langle x\vert \psi\rangle\, ,\\ &= i\hbar\frac{d}{dp}\int dx \langle p\vert x\rangle\langle x\vert \psi\rangle = i\hbar\frac{d}{dp}\psi(p)\, . \end{align} The other cases are done in a similar way, remembering that $\langle p\vert x\rangle = \langle x\vert p\rangle^*$. Note that derivatives must act on functions, not on kets.

The change of sign in the derivative is similar to the change in sign of $x$ and $p$ when one makes a canonical transformation $x\to P,p\to -Q$.

Answer 2

You should be able to confirm all the given expressions, but not your derived ones, (7), (8), by the formal representations, (setting $\hbar=1$, i.e. nondimensionalizing, to avoid predictable mistakes; think of it as absorbing $\hbar$ into all derivative operators, or else its inverse square root into x and p), $$ \bbox[yellow,5px]{\hat p = \int\! dp~ |p\rangle p\langle p|= \int\! dx~ |x\rangle (-i\partial_x) \langle x| \\ \hat x = \int\! dp~ |p\rangle i\partial_p \langle p|= \int\! dx~ |x\rangle x \langle x| } ~ . $$ That should allow you to catch your sign errors in your last two equations, (7) and (8).

Your curiosity objective, (7) and (8) with their signs corrected, $$\hat p|x\rangle= i\partial_x |x\rangle, \quad \hat x |p\rangle= -i\partial_p|p\rangle,$$ is not as useful as operators acting on bras, as you ought to find out. Try complex conjugating (2) and (3).

Do you now at last see $$ \langle \phi| \hat p| \psi \rangle= -i \int \! dx ~ \langle \phi| x\rangle \partial_x \langle x| \psi \rangle\\ = -i \int \! dx ~ \phi(x)^* \partial_x \psi(x) = i \int \! dx ~ \partial_x \phi(x)^* ~ \psi(x) ~~? $$

Answer 3

Here's a more abstract derivation of the action of the momentum operator on position eigenstates. To start, we note that momentum is the generator of spacial translation in classical mechanics. Therefore, it stands to reason that we should identify momentum in quantum mechanics as the generator of spacial translations. Let $U(x)$ be defined so that $$U(x)|x'\rangle=|x'+x\rangle.$$ Then $U(x)$ is the translation operator (I'm working in one dimension for simplicity). We require that $U(0)=1$ (translation by zero doesn't change anything). We also want $U(x)$ to be unitary ($U^\dagger=U^{-1}$) so that it doesn't mess up normalization of the vectors it acts on. These requirements means that, for small $\epsilon$, we can expand (think Taylor expansion) $U(x)$ as $$U(\epsilon)=U(0)+\epsilon K\equiv 1-\frac{i\epsilon}{\hbar}P,$$ where $K$ is anti-Hermitian (and so $P$ is Hermitian). The anti-Hermitian requirement on $K$ comes from the fact that $U(\epsilon)$ needs to be unitary to order $\epsilon$. Then we define $P=i\hbar K$ (the factor of $i$ makes $P$ Hermitian, and the factor of $\hbar$ is necessary for dimensional analysis if we identify $P$ as momentum). $P$ is the generator of spacial translations. Any finite translation can be performed by simply performing a bunch of infinitesimal ones using $P$. Then we have $$\left(1-\frac{i\epsilon}{\hbar}P\right)|x\rangle=|x+\epsilon\rangle\implies P|x\rangle=i\hbar\frac{|{x+\epsilon}\rangle-|x\rangle}{\epsilon}.$$ Taking the limit as $\epsilon\xrightarrow{}0$, we get $$P|x\rangle=i\hbar\frac{\partial}{\partial x}|x\rangle.$$