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Rigorous Feynman pathintegral derivation 
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#1
Jun1811, 04:38 PM

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><p) [itex]e^{i \delta t (\hat p^2 /2m)} q> = e^{i \delta t (\hat p^2 /2m)} p><pq> = e^{i \delta t (p^2 /2m)} p><pq>[/itex] ([itex] \hat p [/itex]is an operator) which is obviously true. But in my case I have [itex]e^{i \delta t (\hat p^2 /2m + V(\hat q))} q> \neq e^{i \delta t( \hat p^2 /2m + V(q))} q>[/itex] as the commutator of [itex]\hat p[/itex] and [itex]\hat q[/itex] 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[itex] \hat q = q [/itex]and [itex]\hat p = p [/itex]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 :) 


#2
Jun1811, 06:20 PM

Sci Advisor
Thanks
P: 4,160

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.



#3
Jun1811, 06:33 PM

P: 4

True, writing them as exponentials is reasonable. But I don't like to reverse this in the end, writing
[itex]1+\delta x = e^x[/itex] Could you explain why writing them as exponentials is not correct? 


#4
Jun1811, 08:59 PM

Emeritus
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PF Gold
P: 9,529

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 [tex]e^x=1+x+\mathcal O(x^2).[/tex] 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 "[itex]\lim_{x\rightarrow 0}[/itex]" 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). 


#5
Jun1911, 07:37 AM

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Thanks
P: 4,160




#6
Jun1911, 03:21 PM

P: 4

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.? 


#7
Jun1911, 07:10 PM

Emeritus
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PF Gold
P: 9,529

It seems to work for unbounded operators in some special cases. For example, define D by Df(x)=f'(x) for all smooth f. Then the power series definition combined with Taylor's formula tells us that f(x)=(exp(D)f)(0) for all x. In those cases when the power series definition doesn't work, the exponential is defined by Stone's theorem. 


#8
Jun1911, 09:05 PM

P: 4

I recommend you look up the "Trotter Expansion" of a matrix exponential. The path integral can be understood as a Trotter expansion in which resolutions of the identity (alternating between Q and P) are inserted. The Trotter expansion can be derived from the BakerCampbellHausdorff formula, where operators 'X' and 'Y' will refer to the kinetic and potential energy in a small time slices.
http://en.wikipedia.org/wiki/Baker–C...sdorff_formula 


#9
Jun2011, 06:37 AM

P: 4

Thanks at all. I think the SuzukiTrotter expansion finally solved the problem for me.



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