Why should I not use the Hermitian adjoint in this equation?

Question

This is superficially a question about translation operators but perhaps more to do with bra & ket basics. I've been working through Barton Zwiebach's lecture notes (chapter 6) for MIT Quantum Physics II and there is this line that I don't quite follow: $$\langle p \vert T_{x_0} \vert x_1 \rangle = \langle p \vert e^{\frac{-i}{\hbar}\hat{p}x_0}\vert x_1\rangle = e^{\frac{-i}{\hbar} p x_0} \frac{e^{\frac{-i}{\hbar} p x_1}}{\sqrt {2 \pi \hbar }}= \langle p \vert x_0 + x_1 \rangle$$ It's the second equality I have a problem with. Now, I get that $ \langle p \vert \hat{p}= p \langle p \vert $ and $ \langle p \vert x_1 \rangle = \frac{e^{\frac{-i}{\hbar} p x_1}}{\sqrt {2 \pi \hbar }}$ but something makes me want to take the Hermitian adjoint and put $$\langle p \vert e^{\frac{-i}{\hbar}\hat{p}x_0}\vert x_1\rangle = e^{\frac{i}{\hbar} p x_0} \frac{e^{\frac{-i}{\hbar} p x_1}}{\sqrt {2 \pi \hbar }} $$ which I know is wrong.

Answer 1

Perhaps it becomes clearer with a slightly different notation: If $\psi_p$ is a (generalised) eigenvector of the momentum operator $\hat{p}$ with (generalised) eigenvalue $p$, and $\phi$ another Hilbert space vector, then $$\langle \psi_p|\mathrm{e}^{-\mathrm{i}\hat{p}x_0}|\phi\rangle = \langle \mathrm{e}^{\mathrm{i}\hat{p}x_0}\psi_p|\phi\rangle = \langle \mathrm{e}^{\mathrm{i}p x_0}\psi_p|\phi\rangle = \mathrm{e}^{-\mathrm{i}px_0} \langle \psi_p|\phi\rangle.$$ Note that the scalar product is anti-linear in the first component.

Answer 2

I can't second-guess the voices you hear, but if you wish to act on the ket instead of the bra, recall $$ \langle p \vert \hat{p}\vert x_1 \rangle = \langle p \vert \hat{p}\int\!\! dp'~~|p'\rangle \langle p'\vert x_1 \rangle\\ =\int\!\! dp'~~p'\delta(p-p') \frac{e^{\frac{-i}{\hbar} p' x_1}}{\sqrt {2 \pi \hbar }}= p \frac{e^{\frac{-i}{\hbar} p x_1}}{\sqrt {2 \pi \hbar }} .$$

What holds for $\hat p$ morphs to $f(\hat p)$, so take $f(\hat p)=e^{\frac{-i}{\hbar} \hat p x_1}$ to get $$ \langle p \vert e^{\frac{-i}{\hbar}\hat{p}x_0}\vert x_1\rangle = e^{\frac{-i}{\hbar} p x_0} \frac{e^{\frac{-i}{\hbar} p x_1}}{\sqrt {2 \pi \hbar }}= \langle p \vert x_0 + x_1 \rangle.$$

• If, on the other hand, you are really asking whether $ \langle p \vert i\vert x_1 \rangle = i\langle p \vert x_1 \rangle $ is consistent, of course it is.