In Griffiths Introduction to Quantum Mechanics, it is stated that the expectation value of any observable can be calculated in the momentum space Fourier space) in the following way.
In momentum space, then, the position operator is $i\hbar\partial/\partial p$. More generally,
$$
\langle Q\left(x,\,p\right)\rangle=\cases{\int\psi^\star\hat{Q}\left(x,\,\frac{\hbar}{i}\frac{\partial}{\partial x}\right)\psi\,{\rm d}x & \text{in position space;} \\ \int\Phi^\star\hat{Q}\left(-\frac{\hbar}{i}\frac{\partial}{\partial p},\,p\right)\Phi\,{\rm d}p & \text{in momentum space.}}\tag{3.58}
$$
In principle, you can do all the calculations in momentum space just as well (though not always as easily) as in position space.
Can anyone show the proof of this statement?
Best Answer
We can write the state in the position space as
$$\psi(x)=\frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)e^{ikx}dk$$
Therefore, for example
$$\int\psi^{\ast}(x)x\psi(x)dx=\int\left[\frac{1}{\sqrt{2\pi}}\int\tilde{\psi}^{\ast}(k_{1})e^{-ik_{1}x}dk_{1}\right]x\left[\frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k_{2})e^{ik_{2}x}dk_{2}\right]dx=$$
$$=\int\tilde{\psi}^{\ast}(k_{1})\tilde{\psi}(k_{2})\left[\frac{1}{2\pi}\int xe^{i(k_{2}-k_{1})x}dx\right]dk_{1}dk_{2}=$$
$$=\int\tilde{\psi}^{\ast}(k_{1})\tilde{\psi}(k_{2})\frac{1}{i}\frac{\partial}{\partial k_{2}}\left[\frac{1}{2\pi}\int e^{i(k_{2}-k_{1})x}dx\right]dk_{1}dk_{2}=$$
$$=\int\tilde{\psi}^{\ast}(k_{1})\tilde{\psi}(k_{2})\frac{1}{i}\frac{\partial}{\partial k_{2}}\delta(k_{2}-k_{1}) dk_{1}dk_{2}=\left[{\rm integration\:by\:parts}\right]=$$
$$=\int\delta(k_{2}-k_{1})\tilde{\psi}^{\ast}(k_{1})\left(-\frac{1}{i}\frac{\partial}{\partial k_{2}}\right)\tilde{\psi}(k_{2})dk_{1}dk_{2}=$$
$$=\int\tilde{\psi}^{\ast}(k)\left(-\frac{1}{i}\frac{\partial}{\partial k}\right)\tilde{\psi}(k)dk=\int\tilde{\psi}^{\ast}(p)\left(-\frac{\hbar}{i}\frac{\partial}{\partial p}\right)\tilde{\psi}(p)dp$$
Can you generalize this result?