I have a statement that the path integrals formalism is eqivalent to operator formalism in quantum mechanics. Is it a correct statement?

I understand that each of these two formalisms has its advantages. Sometimes a problem can be easily solved in the operator formalism than using path integrals, and for certain others, it would be the opposite. But, are they equivalent?

The problem is the following. Suppose there exists a particle which interacts with a harmonic oscilator. $$ H=\frac{P^2}{2M} + V(X) + \frac{p^2}{2m}+ \frac{x^2}{2m} +\alpha x X $$ where $x$,$p$ is a coordinate and momentum of harmonic oscilator, $X$,$P$ is a coordinate and momentum of a particle.

If I want to investigate only a particle with coordinate $X$ and don't care what will be with a harmonic oscillator, I can integrate out over harmonic oscillator coordinate $x$ in path integral formalism and obtain some action, which will depend on $X$ and $\dot X$.

enter image description here

enter image description here

The above is an image of Feynman's book, "Quantum Mechanics and Path Integrals." In this picture, $X$ and $x$ correspond to harmonic oscillator coordinate, thus opposite to my convention. I want to obtain the same or equivalent quantity but using operators. How can I obtain the same thing in the operator formalism?

  • 1
    $\begingroup$ Quick comment : In Heisenberg picture one can solve for $X(t)$ and $P(t)$ and use them to get closed equations for $x(t)$ and $p(t)$. $\endgroup$ – Sunyam Nov 18 '18 at 10:56
  • $\begingroup$ Don't you just need to sum all the 1PI diagrams? $\endgroup$ – Eric David Kramer Nov 18 '18 at 11:51
  • $\begingroup$ Sunyam. Can you obtain something intresting? I can write equations, but I can not solve it. $\endgroup$ – Peter Nov 18 '18 at 14:50
  • $\begingroup$ Eric. What do your mean? What is 1PI diagram? The sum it is not a answer. I want to obtain some close form, like in path integral formalism. $\endgroup$ – Peter Nov 18 '18 at 14:51
  • $\begingroup$ If X is a slow mode , you just complete the square of the fast x part, so it provides an overall correction to the slow part, $(-\alpha^2 X^2 +(n+1/2)) /(2m)$. Of course the two formalisms are equivalent, provided you translate between them correctly. $\endgroup$ – Cosmas Zachos Nov 18 '18 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.