Matrix elements of $p$ in $x$ basis? I am trying to self study QM and did not understand the following part of Shankar’s QM page 153. How does he go from the first to the second equation?

 A: *

*Start from $\langle x | U(t) | x^\prime \rangle =\langle x | e^{-iHt/\hbar} |x^\prime \rangle$, where $H= P^2/2m$ ($P$= momentum operator).


*Insert the unit operator in the form $\mathbf{1}= \int\limits_{-\infty}^\infty \! dp \,|p\rangle\langle p|$, where $| p \rangle$ are the momentum eigen-"states" satisfying the eigenvalue equation $P |p \rangle = p |p\rangle$, which implies $f(P) |p\rangle = f(p) |p\rangle $ for a function $f(P)$ of the momentum operator:
$\qquad \langle x |e^{-iP^2t/2m\hbar} |x^\prime\rangle =\int\limits_{-\infty}^\infty \! dp \, \langle x | e^{-iP^2t/2m\hbar}|p \rangle \langle p|x^\prime\rangle= \int\limits_{-\infty}^\infty \! dp \, e^{-ip^2 t/2m\hbar} \langle x |p\rangle \langle p | x^\prime \rangle$.


*Remembering that $\langle x | p \rangle = \frac{e^{ipx/\hbar}}{\sqrt{2 \pi \hbar}}$ and $\langle p| x^\prime \rangle = \langle x^\prime | p \rangle^\ast =\frac{e^{-ipx^\prime/\hbar}}{\sqrt{ 2 \pi \hbar}}$, you obtain the Fresnel integral

$\qquad \langle x |U(t) | x^\prime \rangle = \frac{1}{2 \pi \hbar} \int\limits_{-\infty}^\infty \! dp \, e^{-ip^2 t/2m \hbar} e^{ip(x-x^\prime)/\hbar}$.


*The remaining integration can carried out by observing that the exponent

$\qquad -\frac{ip^2 t}{2m \hbar}+\frac{ip(x-x^\prime)}{\hbar} = - \frac{it}{2 m \hbar} \left(p^2- 2\frac{m (x-x^\prime)}{t} p \right)=-\frac{it}{2 m \hbar} \left[ \left(p-\frac{m(x-x^\prime)}{t}\right)^2 - \frac{m^2 (x-x^\prime)^2}{t^2}         \right] $
suggests the variable transformation $p^\prime = p-m(x-x^\prime)/t$ yielding the final result
$\qquad \langle x |U(t) | x^\prime \rangle =\sqrt{\frac{m}{2 \pi i \hbar t}} \, e^{im(x-x^\prime)^2/2  t \hbar}$.
