Alternative way to compute expectation value of momentum?

Question

This might be ridiculously incorrect, but is it possible to find the expectation value of momentum like this?

In the position space: $$\langle x | \psi \rangle = \psi(x)$$ $$\langle \hat{A} \rangle_{x\ basis} = \langle \psi | \hat{A} | \psi \rangle \equiv \int dx \ \psi^{*}(x) \ \hat{A} \ \psi(x)$$ Can we then say that in momentum space: $$\langle p | \psi \rangle = \psi(p) $$ $$\langle \hat{A} \rangle_{p\ basis}=\int dp\ \psi^{*}(p) \ \hat{A} \ \psi(p)$$ If so, can we use the relationship, $\langle p | \psi \rangle = \int \langle x | \psi \rangle \ e^{-ipx/\hbar}\ dx$ (discussed in this post: https://physics.stackexchange.com/a/462977/369620) to solve for $\langle p | \psi \rangle$ given $\langle x | \psi \rangle$, and thus, from $\langle x | \psi \rangle$ alone, solve for the expectation value of an operator in momentum space?

Answer 1

Yes, you can work out the expectation value of an operator in that way. The key thing to notice is that the expectation value does not depend on which basis you choose to work with. For example, it holds that (assuming I'm not missing any normalization constants in the Fourier transform, and I could be) $$\langle \hat{p} \rangle = \int \psi^*(p) p \psi(p) \mathrm{d}p = - i \hbar \int \psi^*(x) \frac{\mathrm{d}}{\mathrm{d}x} \psi(x) \mathrm{d}x.$$ It is up to you to choose which option is more convenient for the problem at hand.