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Rigorous Feynman pathintegral derivation

  1. Jun 18, 2011 #1
    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 :)
     
  2. jcsd
  3. Jun 18, 2011 #2

    Bill_K

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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.
     
  4. Jun 18, 2011 #3
    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?
     
  5. Jun 18, 2011 #4

    Fredrik

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    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).
     
  6. Jun 19, 2011 #5

    Bill_K

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    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
     
  7. Jun 19, 2011 #6
    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.?
     
  8. Jun 19, 2011 #7

    Fredrik

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    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.
     
  9. Jun 19, 2011 #8
    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–Campbell–Hausdorff_formula
     
  10. Jun 20, 2011 #9
    Thanks at all. I think the Suzuki-Trotter expansion finally solved the problem for me.
     
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