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More Path Integral questions

  1. Aug 14, 2009 #1
    Hi everyone,

    In chapter 5 of Lewis Ryder's book on QFT, the expression for the propagator as a path integral is derived. Equation 5.7, which is the expression for the propagator over a small path [itex](q_{j+1} t_{j+1};q_{j}t_{j})[/itex], reads

    [tex]\langle q_{j+1} t_{j+1} |q_{j}t_{j}\rangle = \frac{1}{2\pi\hbar}\int dp \exp{\left[\frac{i}{\hbar}p(q_{j+1}-q_j)\right]} - \frac{i\tau}{\hbar}\langle q_{j+1}|H|q_{j}\rangle[/tex]

    where [itex]\tau = t_{j+1}-t_{j}[/itex]. This expression holds quite generally, but equation 5.13, which reads

    [tex]\langle q_{f} t_{f} |q_{i}t_{i}\rangle = \int \frac{\mathcal{D}q\mathcal{D}p}{h}\exp{\frac{i}{\hbar}\left[\int dt p\dot{q}-H(p,q)\right]}[/tex]

    is derived under the assumption that H is of the form

    [tex]H = \frac{p^2}{2m} + V(q)[/tex]

    This allows us to express the propagator as a function of the action S[q(t)] in the above expression.

    But what if H is not of this form? What does the propagator look like there? I suppose it depends on the specific case (the author points out one example of a Lagrangian [itex]L = f(q)\dot{q}^2/2[/itex] which requires the introduction of an effective action different from [itex]\int L dt[/itex]), but are there any general rules or classes of systems where one can write the above expression, but which do not have the canonical form of H given above?

    The author also states that Feynman began with the above expression for the propagator, which is not a very rigorous thing to do, given the counterexample in the previous paragraph.

  2. jcsd
  3. Aug 14, 2009 #2
    I got a partial answer on page 281 of Peskin and Schroeder.
  4. Aug 14, 2009 #3


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    In general the formula is correct if the classical H(p,q) is the Weyl symbol of the quantum hamiltonian, defined as

    [tex]H(p,q)\equiv\int ds\,e^{ips/\hbar}\langle q{-}{\textstyle{1\over2}}s|\hat H|q{+}{\textstyle{1\over2}}s\rangle.[/tex]
  5. Aug 15, 2009 #4
    I think Avodyne is correct, as p^2/(2m)+V(q) is not the most general Hamiltonian you can have. You should be able to have H(p,q)=ap^2+bp+pf(q)+V(q) for constants a, b and arbitrary function f(q). This ought to be the most general Hamiltonian that allows one to safely pass into the Lagrangian scheme. The only example of such a cross term pf(q) I can think of are in 3-gluon vertices in non-Abelian gauge theories (or just boson 3-vertices in general).
    Last edited: Aug 15, 2009
  6. Aug 15, 2009 #5
    Thanks RedX and Avodyne.

    Where can I read more about this Weyl symbol, and specially this integral representation?
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