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 Thanks P: 3,860 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?
Emeritus
PF Gold
P: 9,014

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).
Thanks
P: 3,860
 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.?
Emeritus