Question 3.10: Obtain the position operator in the momentum basis (Equation...

\left\langle \begin{matrix} p\begin{vmatrix} \hat{x} \\\end{vmatrix} S(t) \\ \end{matrix} \right \rangle=\left\langle p \begin{matrix} \begin {vmatrix} \hat{x}\int{dx} \\ \end{vmatrix} x \\ \end{matrix} \right\rangle \left\langle \begin{matrix} x|| S(t) \\ \end{matrix} \right\rangle 

=\int{\left\langle \begin{matrix} p|x|x \\ \end {matrix} \right\rangle }\left\langle \begin{matrix} x|S(t) \\ \end{matrix} \right\rangle dx

where I’ve used the fact that is an eigenstate of \hat{x} (\hat{x} |x〉 =x|x〉) ; x can then be pulled out of the inner product (it’s just a number) and

\left\langle \begin{matrix} p\begin{vmatrix} \hat{x} \\\end{vmatrix} S(t) \\ \end{matrix} \right \rangle=\int{x\left\langle \begin{matrix} p|x \\ \end{matrix} \right\rangle } \Psi (x,t)dx

=\int{x\frac{e^{-ipx/\hbar }}{\sqrt{2\pi \hbar } } } \Psi (x,t)dx

=i\hbar \frac{\partial}{\partial p} \int{\frac{e^{-ipx/\hbar }}{\sqrt{2\pi \hbar } } } \Psi (x,t)dx

Finally we recognize the integral as \Phi (p,t) (Equation 3.54 [\Phi (p,t)=\frac{1}{\sqrt{2\pi \hbar }}\int_{-\infty }^{\infty } e^{-ipx/\hbar }\Psi (x,t)dx].).