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Itzykson-Zuber path integral

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Itzykson-Zuber ch. 9-1-1:
    If
    [tex]H=\frac{P^2}{2m}-QF(t) [/tex]
    then
    [tex]\frac{\delta}{i\delta F(t)}\langle f\mid i\rangle_F=\langle f\mid Q(t)\mid i \rangle_F[/tex]

    Ok, I understand that.
    But then it states: if
    [tex]H=\frac{P^2}{2m}+V(Q)[/tex]
    then
    [tex]\int\mathcal{D}(q)\exp\left\{i\int dt\left[\frac{m\dot q^2}{2}-V(q)\right]\right\}=\exp\left\{-i\int dt V\left[\frac{\delta}{i\delta F(t)}\right]\right\}\langle f\mid i\rangle_F\big|_{F=0}[/tex]

    I can't understand this formula....
    3. The attempt at a solution
    I tried everything all week long
     
  2. jcsd
  3. Jun 24, 2012 #2

    Mute

    User Avatar
    Homework Helper

    It looks like the propagator [itex]\langle f | i\rangle_F [/itex] is defined as

    [tex]\langle f | i \rangle_F = \int \mathcal D[q \dot{q}] \exp\left[i \int dt~\left\{ \frac{1}{2}m\dot{q}^2 - q(t) F(t) \right\} \right]. [/tex]

    Suppose you can evaluate this path integral, but the path integral that you really want to evaluate is

    [tex]Z = \int \mathcal D[q \dot{q}] \exp\left[i \int dt~\left\{ \frac{1}{2}m\dot{q}^2 - V(q)\right\} \right]. [/tex]

    Is there a way to calculate this from [itex]\langle f | i \rangle_F[/itex]? Formally, yes. You've noted that

    [tex]\frac{\delta}{i\delta F(t)} \langle f | i \rangle_F = \langle f | q(t) | i \rangle_F,[/tex]

    which I will assume you understand the meaning of. Now, what if you differentiated twice? You'd get

    [tex]\frac{\delta}{i\delta F(t_1)} \frac{\delta}{i\delta F(t_2)} \langle f | i \rangle_F = \langle f | q(t_2) q(t_1) | i \rangle_F.[/tex]

    Ok, so what does this have to do with the formula? Well, suppose you replace q with the operator [itex]\frac{\delta}{i\delta F(t)}[/itex] in the potential V(q). What does

    [tex]\exp\left[i\int dt~V\left[\frac{\delta}{i\delta F(t)}\right] \right] \langle f | i \rangle_F [/tex]

    mean? Well, you have to understand it in a power series sense: expand [itex]V\left[\frac{\delta}{i\delta F(t)}\right][/itex] in a power series, and then expand the exponential in a power series. The derivatives will act on [itex]\langle f | i \rangle_F[/itex] to give a factor of q(t) for each derivative [itex]\langle f | i \rangle_F[/itex]. If you then re-sum up both the exponential series and the series for V, you will end up with a term V(q) in your exponential which isn't there in the original propagator [itex]\langle f | i \rangle_F[/itex]. So, what you find is that

    [tex]\exp\left[i\int dt~V\left[\frac{\delta}{i\delta F(t)}\right] \right] \langle f | i \rangle_F = \int \mathcal D[q\dot{q}] \exp\left[i\int dt~\left\{ \frac{1}{2}m\dot{q}^2 - V(q) - q(t)F(t)\right\}\right][/tex]

    But there's still that qF(t) term there, so set F(t) to zero after all of the derivatives have acted to get rid of it and have the path integral that you really want.

    To summarize this, the formula gives you a way of systematically calculating the path integral [itex]Z[/itex] given the propagator [itex]\langle f | i \rangle_F [/itex]. (Although it may be a rather inefficient way!)
     
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