Position mean in momentum-space (QM)

[WWCY]

Homework Statement

Find $\langle x \rangle$ in the momentum representation

I am having trouble understanding some of the steps needed to get to the expression, assistance is greatly appreciated!

Homework Equations

3. The Attempt at a Solution [/B]
$$\langle x \rangle = \langle \psi | x | \psi \rangle = \int dp \ dp' \langle \psi | p \rangle \langle p | \hat{x} | p'\rangle \langle p' | \psi \rangle$$
$$ \langle p | \hat{x} | p'\rangle = \int dx \langle p | x \rangle \langle x | \hat{x} | p' \rangle = \int dx \ x \langle p | x \rangle \langle x |p' \rangle$$
Since
$$\langle x | p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{i}{\hbar} p x}$$
I rewrote line 2 as
$$ \langle p | \hat{x} | p'\rangle = -\frac{i\hbar}{2 \pi \hbar} \partial _{p'}\int dx \ \exp {\big [ \frac{ix}{\hbar} (p'-p) \big ]} = -i \hbar \ \partial _{p'} \delta (p' - p) = -i \hbar \ \partial _{p'} \delta (p - p')$$
Here's where things went wrong, if I substitute the result above back into line 1, I have
$$\langle x \rangle = -i\ \hbar \int dp \ dp'\ \tilde{\psi }(p)\ \partial _{p'} \delta (p - p')\ \tilde{\psi }(p')$$
If I choose to integrate over $p'$ right away, I get a sign error. If I integrate over $p'$ by parts, I get the right answer. It seems that the partial derivative acting on the delta is what's causing problems while performing the integral. But even so, integration by parts "passes" the partial derivative over to $\tilde{\psi} (p')$ and yet it gets me the right answer. So what's wrong (and right)?

Many thanks in advance!

 

[anathema]

Thank you for posting your question. It seems like you are on the right track, but there are a few things that need to be clarified. First, the expression for $\langle x \rangle$ in the momentum representation is given by
$$\langle x \rangle = \int dp \ dp' \langle \psi | p \rangle \langle p | \hat{x} | p'\rangle \langle p' | \psi \rangle$$
which is the same as what you have written. However, the next step should be
$$\langle p | \hat{x} | p'\rangle = \int dx \langle p | x \rangle \langle x | \hat{x} | p' \rangle = \int dx \ x \langle p | x \rangle \langle x |p' \rangle$$
Notice that in the second line, the integral is over $x$, not $p'$. This is where you made a mistake. The next step, using the expression for $\langle x | p \rangle$, should be
$$\langle p | \hat{x} | p'\rangle = \int dx \ x \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{i}{\hbar} p x} \frac{1}{\sqrt{2\pi \hbar}} e^{-\frac{i}{\hbar} p' x} = \frac{1}{2\pi \hbar} \int dx \ x e^{\frac{i}{\hbar} (p-p') x}$$
From here, you can use the fact that
$$\int dx \ x e^{ikx} = i \hbar \frac{\partial}{\partial k} \delta (k)$$
to get the correct expression for $\langle p | \hat{x} | p'\rangle$. This will then give you the correct expression for $\langle x \rangle$ in the momentum representation.

I hope this helps and clears up any confusion you may have had. Let me know if you have any further questions.