# Calculation in the book Path Integrals by Kleinert

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 $$$$(x_bt_b|x_N t_N ) ≈ \int^∞_{−∞} \frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N) - \epsilon H (p_b,x_b,t_b)]}.\tag{2.21}$$$$ so far, so good, however, then it says: "The momentum $$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 "

$$$$(x_bt_b|x_N t_N ) ≈ e^{−i\epsilon H(−ih∂_{x_b},x_b,t_b)/\hbar} \int^∞_{−∞}\frac{dp_b}{2\pi \hbar} e^{ip_b(x_b−x_N )/\hbar} =e^{−i\epsilon H(−i\hbar∂_{x_b},x_b,t_b)/\hbar} δ(x_b−x).\tag{2.22}$$$$ 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?

$$\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N)-\epsilon H(p_b,x_b,t_b)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[-\epsilon H(p_b,x_b,t_b)]}e^{(i/\hbar)[p_b(x_b-x_N)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}(1-(i/\hbar)[-\epsilon H(p_b,x_b,t_b)])e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (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)
• $\partial_{x}\exp(px)=p\exp(px)$,$\partial_{x}^2\exp(px)=p^2\exp(px)$,$\partial_{x}^3\exp(px)=p^3\exp(px)$,...,$f(\partial_{x})\exp(px)=f(p)\exp(px)$ Nov 7, 2022 at 21:51
• As you'll later take the limit as $\epsilon$ goes to zero you only need terms linear in $\epsilon$. BTW this derivation is a form of Huygens' principle, if you're familiar with that from elsewhere. The last term in my derivation is essentially the "ripples" you get get from a point source after an infinitesimal time Nov 8, 2022 at 5:29