Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis

Question

I am trying to calculate $\langle x\ |\ \hat{x}\ |\ p\rangle$. I can work in the $x$-basis like so:

$$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle=\int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle=x\langle x\ |\ p\rangle=\frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$

Which steps of this derivation are not correct, if any?

Second, if I try to do the same thing by using the $p$-basis representation of $\hat{x}$, I get into more trouble:

$$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'i\hbar\frac{\partial}{\partial p'}\langle x\ |\ p'\rangle\langle p'\ |\ p\rangle$$ $$=\int dp' (-\frac{xe^{ip'x/\hbar}}{\sqrt{2\pi\hbar}})\delta(p-p')=-\frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$

Hmmm - that doesn't look right - shouldn't the result be the same, no matter which basis or representation I use for $\hat{x}$? Where are the mistakes? Also, I'm not too confident with my own usage of the primes ' all over. Am I using those correctly?

Answer 1

This question (v6) [concerning the overall minus sign in OP's calculation] is essentially a Fourier transformed version of e.g. this Phys.SE post, see Emilio Pisanty's answer and my answer.

The main point is again that the derivative in the momentum Schrödinger representation $$\hat{x}~=~i\hbar\frac{\partial}{\partial p}, \qquad \hat{p}~=~p,$$ acts on the bra (rather than the ket):

$$\langle p | \hat{x}~=~i\hbar\frac{\partial\langle p |}{\partial p} ,\qquad \langle p | \hat{p}~=~p\langle p |.$$

If one insists to act on kets (as opposed to bras), then the momentum Schrödinger representation comes with the opposite sign:

$$\hat{x}|p\rangle~=~-i\hbar\frac{\partial|p\rangle}{\partial p} ,\qquad \hat{p}|p\rangle~=~|p\rangle p.$$

The overlap is given as $$\langle p | x\rangle~=~\frac{e^{px/i\hbar}}{\sqrt{2\pi\hbar}}. $$

Answer 2

You want to use $$ \hat x= i\hbar\frac{\partial}{\partial p} $$ in the momentum basis. This means that $$ <p|\hat x|\psi>= i\hbar\frac{\partial}{\partial p} <p|\psi> $$ Thus, by hermiticity of $\hat x$, we evaluate
$$ <x|\hat x|p> = (<p|\hat x|x>)^* $$ $$ =(i\hbar\frac{\partial}{\partial p} <p|x>)^* $$ $$ =(i\hbar\frac{\partial}{\partial p} e^{-ipx/\hbar})^* $$ $$ xe^{ipx/\hbar} $$ which is what you want.

Answer 3

I think you have the answer for your second question. For your first question let me clarify:

$$\langle x\ |\ \hat{x}\ |\ p\rangle \overset{(1)}{=} \int dx'\langle x\ |\ \hat{x}\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(2)}{=} \int dx'x'\langle x\ |\ x'\rangle\langle x'\ |\ p\rangle \overset{(3)}{=} x\langle x\ |\ p\rangle =\frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}$$

In the first equality you use the fact that $ |x \rangle $, where $x \in \mathbb{R}$ is complete. That is $\int dx' |\ x'\rangle\langle x'| = \mathbb{1}$ In the second equality you just act on the ket. In the third equality you use the fact that $ |x \rangle $ is orthonormal ie $\langle x\ |\ x'\rangle = \delta(x-x') $, which leaves you with the following:

$$\int \mathrm{d}x'\delta(x-x') \cdot x' \langle x'\ |\ p\rangle = x\langle x\ |\ p\rangle = \frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}$$

So what you have done is not bogus at all.

Answer 4

Second, if I try to do the same thing by using the $p$-basis representation of $\hat{x}$, I get into more trouble:

$$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'i\hbar\frac{\partial}{\partial p'}\langle x\ |\ p'\rangle\langle p'\ |\ p\rangle$$ $$=\int dp' (-\frac{xe^{ip'x/\hbar}}{\sqrt{2\pi\hbar}})\delta(p-p')=-\frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$

This is wrong. You "moved" the partial derivative to the left through the bra-ket, which is not allowed. You should have "moved" the partial derivative to the right, and then integrated by parts like this:

$$\langle x\ |\ \hat{x}\ |\ p\rangle=\int dp'\langle x\ |\ \hat{x}\ |\ p'\rangle\langle p'\ |\ p\rangle=\int dp'\langle x\ |\ p'\rangle i\hbar\frac{\partial}{\partial p'} \langle p'\ |\ p\rangle$$ $$=\int dp' (\frac{e^{ip'x/\hbar}}{\sqrt{2\pi\hbar}})i\hbar\frac{\partial}{\partial p'}\delta(p-p') =-\int dp' \delta(p-p')i\hbar\frac{\partial}{\partial p'}(\frac{e^{ip'x/\hbar}}{\sqrt{2\pi\hbar}}) =+\frac{xe^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$