Understanding derivation of expectation value of power of momentum operator

Question

I have been reading through the book Fourier Series and Integrals by Dym and McKean. In one of their derivations, they have the following line $$ \int \psi^* B^n \psi = \int \psi^* [(2\pi i)^{-1} D]^n \psi = \int \hat\psi^* \gamma^n \hat\psi = \int \gamma^n |\hat\psi|^2 $$ Where they define the momentum operator $B$ as $B = (2\pi i)^{-1} D$, with $D$ representing the differentiation operator. However, I'm not sure how the second equality follows. We have the identity $(2\pi i \gamma)^p D^q \hat f = \left[D^p (-2\pi i x)^qf\right]\hat{}$, but I don't see how this gets us what we want and in particular how $\psi^*$ turns into $\hat\psi^*$. Could I get some guidance?

Answer 1

The second equality is explained as follows. Using the Parseval Theorem we have \begin{align*} \int \psi^*(x) \left[ (2\pi i)^{-1} D\right]^n \psi(x) \mathrm{d}x &= \int \frac{1}{(2\pi i)^n}\widehat{\psi^*}(\gamma) \widehat{D^n \psi}(\gamma) \mathrm{d}\gamma \\ &= \frac{1}{(2\pi i)^n} \int \widehat{\psi^*}(\gamma)(2\pi i)^n\gamma^n \widehat{\psi}(\gamma)\mathrm{d}\gamma, \end{align*} where we used the differentiation rule $\widehat{f^{(n)}}(\gamma) = (2\pi i \gamma)^n \widehat{f}(\gamma).$