When Feynman was trying to formulate path integral of quantum mechanics, he was inspired by Dirac's remark which roughly states that $e^{i\frac{S}{\hbar}}$corresponds to the transition amplitude, where $S$ is the action.

I'm wondering what was Dirac's original argument behind this statement.


1 Answer 1


Dirac is indeed the forefather of the path integral approach to quantum mechanics. His reasoning was described in the 1933 paper "Lagrangian in/of quantum mechanics". See the full text:


Conceptually, he had the whole thing. He realized that the Poisson brackets have a counterpart in quantum mechanics, the commutators, and because the Lagrangian seems so natural and effective in classical physics, it should have a counterpart in quantum mechanics, too. He indeed derived the possibility to compute the transition amplitude as a sum over histories with the Feynmanian exponential inserted in.

He had formally derived this insight by inserting lots of completeness relations, as integrals over the position $Q$, inside the transition amplitude.

Dirac really didn't get too far in calculating this sum over histories, verifying that it's the right thing, dealing with subtleties that arise in the picture, and he didn't make it any useful. Those things were done by Feynman but all the formal features of the path-integral approach were already known to Dirac.

  • $\begingroup$ Do you happen to know what Dirac means with the term "contact transformations" when he writes in the beginning of the paper: "Lagrangian theory is closely connected with the theory of contact transformations" ? I have never heard the term "contact transformations" anywhere. $\endgroup$
    – asmaier
    May 23, 2016 at 20:45
  • $\begingroup$ I haven't heard it, either, but I have used the same thing in explaining the difference between class physics and QM - without reading this Dirac text. "Transformation" means the transformation of the phase space associated with the time evolution, in this case. The adjective "contact" means that the probability density that the point $(x,p)$ at one moment turns to $(x',p')$ at another moment is proportional to a (Dirac) delta-function - it's only nonzero if one evolves it in the right way. The word "contact" generally does refer to the delta-function, like in "contact interactions". $\endgroup$ May 24, 2016 at 4:00
  • $\begingroup$ On the other hand, the matrix elements - probability amplitudes - are continuous, different from 0 and 1, and basically nonzero for any pair of initial and final state. That's due to the uncertainty principle and this general form of the matrix elements is why Dirac talks about "transformations" in QM - it obviously means unitary transformations but it generalizes the delta-function-based classical contact ones. See motls.blogspot.com/2016/03/… for my text that was trying to convey the same idea, 2nd part. $\endgroup$ May 24, 2016 at 4:04
  • 3
    $\begingroup$ canonical transformations are sometimes called contact transformations, you can read more about this on Goldstein's first edition of classical mechanics. $\endgroup$ Jan 5, 2017 at 21:08

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.