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

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.

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.
• Thanks - I think I see where I was going wrong. Does it make sense to say this?: $\langle p \vert \alpha \hat{p} \vert x \rangle = \langle \alpha ^* \hat{p} p \vert x \rangle = \alpha p \langle p \vert x \rangle$ Commented Mar 12, 2023 at 18:31
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.