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


1 Answer 1


$$\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[p_b(x_b-x_N)-\epsilon H(p_b,x_b,t_b)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}e^{(i/\hbar)[-\epsilon H(p_b,x_b,t_b)]}e^{(i/\hbar)[p_b(x_b-x_N)]}$$ $$=\int_{-\infty}^\infty\frac{dp_b}{2\pi\hbar}(1-(i/\hbar)[-\epsilon H(p_b,x_b,t_b)])e^{(i/\hbar)[p_b(x_b-x_N)]}$$ (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)

  • $\begingroup$ Thank you for your help. I still have some doubts about your passage! Bear with me if I'm not missing something here or maybe wanting something a bit as rigorous. From the third line to the fourth line, you transformed p_b to its quantum operator analogue. And this is not clear to me how it works out. To which conditions, does this hold true? I only see epsilon being small as an argument to write the exponential as a sum of its first series terms, but not as a factor to change from a vector variable to the momentum operator (differential). $\endgroup$ Nov 7, 2022 at 21:35
  • $\begingroup$ $\partial_{x}\exp(px)=p\exp(px)$,$\partial_{x}^2\exp(px)=p^2\exp(px)$,$\partial_{x}^3\exp(px)=p^3\exp(px)$,...,$f(\partial_{x})\exp(px)=f(p)\exp(px)$ $\endgroup$
    – Dan Piponi
    Nov 7, 2022 at 21:51
  • $\begingroup$ Oh, right! This last expression (in your comment above) I came across long ago! We use in displacement quantum operator, for instance. Things got a lot more clear! Just let me know if I'm not getting too ahead of this result. Since what is in your comment is true, do we have to explicitly open the exponential into a few series power terms? Since f could be exactly T(p,t), by separating the Hamiltonian in the exponential (which is still classical) then the product of the exponential is the sum of its arguments, right? $\endgroup$ Nov 7, 2022 at 22:50
  • $\begingroup$ 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 $\endgroup$
    – Dan Piponi
    Nov 8, 2022 at 5:29
  • $\begingroup$ I have heard of it, but I'm not familiar. But I can see the analogy here, by the Feynman's double-slit example generally given to introduce path integrals. Thank you again! That helped me a lot. $\endgroup$ Nov 8, 2022 at 10:33

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