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Using Path Integral to calculate propagator

  1. Nov 16, 2012 #1
    Hi, It's great to find this forum.
    I'm teaching myself QM using Shankar, it's a great book, I've covered 8 chapters so far.
    My question is about the notion of using Path Integral method to calculate the propagator. The recipe given by Shankar says the propagator is
    [itex]
    U(x,t;x')=A\int \exp{\frac{iS[x(t)]}{\hbar}}\mathcal{D}[x(t)]
    [/itex]
    where the integral sums over all possible paths connecting (x',0) and (x, t)

    He went on to prove that if the potential is in the form of
    [itex]
    V=a+bx+cx^2+d\dot{x}+ex\dot{x}
    [/itex]
    then the propagator has a simpler form:
    [itex]
    U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t)
    [/itex]

    In his proof for this, he simply assumed that given x, t and x', the there exists one and only one classical path - at least the action S is well defined. But I found this part confusing, two counter examples:

    1. In a harmonic oscillator of angular frequency ω, let x=x'=0, and t=2π/ω, apparently there are infinite classical paths connecting (x', 0) to (x, t) - no matter what your initial velocity, you always come back to the same spot after one full period.

    2. Similarly, if x=x'=a > 0, and t=π/ω, there will be no classical paths (there's no way to come back to the same non trivial spot after half period!

    In case 1, luckily all such classical paths have the same action S=0, so we can still find the propagator following [itex] U(x,t;x')=e^{iS_{cl}/\hbar}\cdot A(t) [/itex]

    But in case 2, there is no classical path at all, what does this imply for the propagator, does the propagator not exist? In QM, the propagator connects one state to the other, so it should always be well-defined.

    I'm sure this is treated more rigorously in non-elementary Path Integral texts, but before diving into those, maybe someone here can explain in simple words what this means.
    Thanks
     
  2. jcsd
  3. Nov 16, 2012 #2

    RUTA

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    If there is no classical path connecting the points, then his derivation doesn't obtain. In those cases, you have to use eqn 8.6.1 directly -- you don't have his shortcut. Even though there is no way for the particle to get from x to x' in time t' classically (potential barrier for example), there may be a non-zero probability of finding the particle there.
     
  4. Nov 16, 2012 #3
    Thanks for the reply, that makes sense. What about the case where there are multiple classical paths connecting (x', t') and (x, t)? I have a hunch that this problem will go away if they all yield the same action S, but I don't know if this is true.
     
  5. Nov 16, 2012 #4

    RUTA

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    I think his derivation would still hold, you would just pick a particular classical path and proceed. Multiple classical paths for the same potential require unusual circumstances (e.g., curved spacetime or wormholes). Your oscillator example produces multiple classical paths for the same potential (same frequency), because you supplied boundary conditions that are automatically satisfied per periodicity and therefore you don't have a unique solution to the second-order, classical differential equation of motion.
     
  6. Nov 16, 2012 #5

    atyy

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    Is the particle on a ring related to what you are asking?
    http://felix.physics.sunysb.edu/~abanov/Teaching/Spring2009/Notes/abanov-cp02-upload.pdf: "This vector potential is not observable in classical mechanics but affects the quantum spectrum because of multiple-connectedness of the ring (there are many non-equivalent ways to propagate from the point 1 to the point 2 on a ring)."
     
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