I've been reviewing some fundamental calculations for path integral and I came across a passage to which I cannot figure out how it is supposed to be done. On page 92, the author started off, by splitting the integral into products of integrals
$$
(x_bt_b|x_at_a) ≈
\int^∞_{−∞}
dx_N (x_bt_b|x_N t_N ) (x_N t_N |x_at_a),\tag{2.20}
$$
to
\begin{equation}
(x_bt_b|x_N t_N ) ≈ \int^∞_{−∞} \frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N) – \epsilon H (p_b,x_b,t_b)]}.\tag{2.21}
\end{equation}
so far, so good, however, then it says: "The momentum $p_b$ inside the integral can be generated by a differential operator $\hat{p}_b ≡ −ih∂_{x_b}$ outside of it. The same is true for any function of $p_b$, so that the Hamiltonian can be moved before the momentum integral, yielding "
\begin{equation}
(x_bt_b|x_N t_N ) ≈ e^{−i\epsilon H(−ih∂_{x_b},x_b,t_b)/\hbar}
\int^∞_{−∞}\frac{dp_b}{2\pi \hbar}
e^{ip_b(x_b−x_N )/\hbar} =e^{−i\epsilon H(−i\hbar∂_{x_b},x_b,t_b)/\hbar}
δ(x_b−x).\tag{2.22}
\end{equation}
And then it follows a few more steps to end up in the Schrödinger equation. I have tried a few mathematical tricks, being unsuccessful to achieve this expression. I leave a link with this part of the book to give more room for context, even though my problem seems to be pure technical, I assume. Can anyone help me?
Link: http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pthic02.pdf
Best Answer
$$\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N)-\epsilon H(p_b,x_b,t_b)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[-\epsilon H(p_b,x_b,t_b)]}e^{(i/\hbar)[p_b(x_b-x_N)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}(1-(i/\hbar)[-\epsilon H(p_b,x_b,t_b)])e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (because $\epsilon$ is small) $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}(1-(i/\hbar)[-\epsilon H(-i\hbar\partial_{x_b},x_b,t_b)])e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (assuming $H(x,p,t)=V(x,t)+T(p,t)$ so the $x_b$ and $\partial_{x_b}$ in $H$ don't interfere) $$=(1-(i/\hbar)[-\epsilon H(-i\hbar\partial_{x_b},x_b,t_b)])\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (as there's no dependency on $p_b$ in the bit factored out front) $$=e^{(i/\hbar)[-\epsilon H(-i\hbar\partial{x_b},x_b,t_b)]}\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (again using $\epsilon$ small)