Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Path integral formulation

  1. Sep 27, 2010 #1
    I'm not quite satisfied by the derivation I've found in Sakurai (Modern Quantum Mechanics) and was trying to 'derive' it myself. I'd like some help to seal the deal. I've described below what I've done. Please tell me where to go from there.

    I know the solution to the Schrodinger equation can be given in terms of a unitary operator,

    [tex] \left| \psi(t) \right\rangle = U(t,t_{0})\left| \psi(t_{0}) \right\rangle [/tex]

    [tex] \left\langle x \left| \psi(t) \right\rangle = \left\langle x \right| U(t,t_{0})\left| \psi(t_{0}) \rangle [/tex]

    which describes time evolution.

    I know that I can introduce intermediary time intervals, thus splitting the above term

    [tex] \langle x \left| \psi(t) \right\rangle = \int dx_{0} \left\langle x \right| U(t,t_{0}) \left| x_{0} \rangle \left\langle x_{0} \right| \psi(t_{0}) \rangle[/tex]

    I here consider

    [tex] \left\langle x \right| U(t,t_{0}) \right| x_{0} \rangle = \left\langle x,t \right| x_{0},t_{0} \rangle [/tex]

    as the Green's function of this operation.

    Now, splitting the interval between x and x_{0} by introducing n points [tex]x_{1}, \dots , x_{n} [/tex] such that

    [tex] \left\langle x,t \right| x_{0},t_{0} \rangle = \int dx_{1} \int dx_{2} \dots \int dx_{n} \left\langle x_{1},t_{1} \right| x_{0},t_{0} \rangle \left\langle x_{2},t_{2} \right| x_{1},t_{1} \rangle \dots \left\langle x,t \right| x_{n},t_{n} \rangle [/tex]

    So far so good. Now comes the crunch. How does everyone seem to get


    as the integrand as they let

    [tex]n \rightarrow \infty [/tex].
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 27, 2010 #2
    You'll need to use the known form of the time-evolution operator at some point to get the Hamiltonian to appear. The point of the large number of time steps is that you can use a linear approximation for time evolution in each timestep.
    Or see chap3 here for example: http://www.tcm.phy.cam.ac.uk/~bds10/tp3.html
  4. Sep 28, 2010 #3
    I see. Suppose for simplicity's sake, I substitute the time evolution operator with some time independent Hamiltonian [tex]\hat{H}[/tex].

    [tex]U = exp(\frac{-\hat{H}t}{\hbar})[/tex]

    For some very small time interval, say [tex] \epsilon [/tex]

    [tex]U = exp(\frac{-i\hat{H}\epsilon}{\hbar})[/tex]

    From here, we use [tex]\delta \hat{S} = \hat{H}\epsilon[/tex]?

    so [tex]exp(\frac{-i\delta \hat{S} }{\hbar}) = exp(\frac{-i\hat{H}}{\hbar})[/tex]

    Something like that? Now each of these exponentials, when multiplied, adds up the [tex] \delta S [/tex] in the exponent to give you S over the entire path?
    Last edited: Sep 28, 2010
  5. Sep 28, 2010 #4
    I think it's more subtle than that. The action in the Hamiltonian formulation is
    [tex]S = \int p\dot q - H(p,q)[/tex] and the first step you should arrive at is a phase-space path integral, with integrals over momentum and position. To reach the Lagrangian formulation, which is an integral over all classical paths in configuration space, you'll need to integrate out the quadratic momentum terms. You might consider splitting the Hamiltonian into a kinetic and potential part.
  6. Sep 29, 2010 #5
    I just used the fact that
    [tex]\frac{\delta S}{\delta t} = H[/tex]

    So we get, for really small [tex] \delta t [/tex]
    [tex] \delta S = H\delta t [/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook