Rigorous Feynman pathintegral derivation

by nahsihorst
Tags: path integral, quantum field theory, quantum mechanics
 P: 4 Hey, I'm trying to do exercise I.2.1. from Zee's QFT in a nutshell but I ran into a problem. The exercise is to derive the QM path integral with a Hamiltonian of the form 1/2 m p^2 + V(q). In the textbook he shows the proof for a free hamiltonian. He gets to a point where he has (I left out the integral for |p> = e^{-i \delta t (\hat p^2 /2m)} |p> = e^{-i \delta t (p^2 /2m)} |p>[/itex] ($\hat p$is an operator) which is obviously true. But in my case I have $e^{-i \delta t (\hat p^2 /2m + V(\hat q))} |q> \neq e^{-i \delta t( \hat p^2 /2m + V(q))} |q>$ as the commutator of $\hat p$ and $\hat q$ does not vanish. Thus I have no idea of how to prove this in general. In some QFT lecture notes I found they expand the exponential to first order, substitute$\hat q = q$and $\hat p = p$and write it again as an exponential. But I don't like this last step and want to do it more rigorous. Any hints? Thanks :)
 Sci Advisor P: 3,457 Well that actually sounds reasonable like a reasonable step. Since δt is infinitesimal, writing them as exponentials is somewhat of a lie in the first place.
 P: 4 True, writing them as exponentials is reasonable. But I don't like to reverse this in the end, writing $1+\delta x = e^x$ Could you explain why writing them as exponentials is not correct?
PF Patron
Emeritus
P: 8,837

Rigorous Feynman pathintegral derivation

When a physics book uses the term "infinitesimal", it means nothing more than that the expression that follows contains only a finite number of terms from a Taylor expansion around 0. For example, "for infinitesimal x, we have exp x=1+x" means that $$e^x=1+x+\mathcal O(x^2).$$ If it's OK to replace exp x with 1+x, then it's also OK to replace 1+x with exp x. The idea is that the neglected terms go to zero in the limit x→0. (You are working with an expression that follows a "$\lim_{x\rightarrow 0}$" in the actual calculation, right?)

(I didn't look at the details of this specific problem. I'm just making a comment about what appears to be the main issue).
P: 3,457
 But I don't like to reverse this in the end, writing 1+δx=ex
At that point he's got an infinite product, and to get back to an exponential I guess he's using the identity lim N-> ∞ (1 + a/N)N = ea
P: 4
 Quote by Bill_K At that point he's got an infinite product, and to get back to an exponential I guess he's using the identity lim N-> ∞ (1 + a/N)N = ea
Thanks a lot. I totally forgot about that relation. Just one last question:
Usually one defines the exponential of an operator with the help of the taylor series. Can one also use this definition and is it equal to the taylor exp.?
PF Patron