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Rigorous Feynman pathintegral derivation

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nahsihorst
#1
Jun18-11, 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><p|q> = e^{-i \delta t (p^2 /2m)} |p><p|q>[/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 :)
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Bill_K
#2
Jun18-11, 06:20 PM
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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.
nahsihorst
#3
Jun18-11, 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?

Fredrik
#4
Jun18-11, 08:59 PM
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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).
Bill_K
#5
Jun19-11, 07:37 AM
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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
nahsihorst
#6
Jun19-11, 03:21 PM
P: 4
Quote Quote by Bill_K View Post
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.?
Fredrik
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Jun19-11, 07:10 PM
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Quote Quote by nahsihorst View Post
Can one also use this definition and is it equal to the taylor exp.?
I'm pretty sure it would work for bounded operators, but I haven't tried to prove it. I'm guessing that it doesn't work in general for unbounded operators. Actually, I think that applies to the power series definition of the exponential too; it works for bounded operators but not in general for unbounded operators.

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.
theZ
#8
Jun19-11, 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 Baker-Campbell-Hausdorff 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
nahsihorst
#9
Jun20-11, 06:37 AM
P: 4
Thanks at all. I think the Suzuki-Trotter expansion finally solved the problem for me.


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