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Problem with Path Integral Expressions in Peskin And Schroeder Section 9.1

  1. Aug 16, 2009 #1
    Hi again everyone,

    I have some doubts about the path integral expressions given in Section 9.1 of Peskin and Schroeder (pg 281 and 282).

    For a Weyl ordered Hamiltonian H, the propagator has the form given by equation 9.11, which reads

    [tex]U(q_{0},q_{N};T) = \left(\prod_{i,k}\int dq_{k}^{i}\int \frac{dp_{k}^{i}}{2\pi}\right)\exp{\left[i\sum_{k}\left(\sum_{i}p_{k}^{i}(q_{k+1}^{i}-q_{k}^{i})-\epsilon H\left(\frac{q_{k+1}+q_{k}}{2},p_{k}\right)\right)\right]}[/tex]

    Now, as the authors point out, for the case when [itex]H = \frac{p^2}{2m} + V(q)[/tex], we can do the momentum integral, which is (taking the potential term out)

    [tex]\int\frac{dp_{k}}{2\pi}\exp{\left(i\left[p_k(q_{k+1}-q_{k})-\epsilon\frac{p_{k}^2}{2m}\right]\right) = \frac{1}{C(\epsilon)}\exp{\left[\frac{im}{2\epsilon}(q_{k+1}-q_{k})^2\right]}[/tex]


    [tex]C(\epsilon) = \sqrt{\frac{2\pi\epsilon}{-im}}[/tex]

    Now, I do not understand how the distribution of factors [itex]C(\epsilon)[/itex] equation 9.13 (given below) comes up. To quote the authors:

    So, my question is: how did we get this term:


    PS -- Is it because the momentum index goes from 0 to N-1 and the coordinate index goes from 1 to N-1? The momentum integral product produces N terms, so to write the coordinate integral with the same indexing as before (in the final expression), i.e. k = 1 to N-1, we factor a [itex]C(\epsilon)[/itex] out?

    Also, in equation 9.12,

    [tex]U(q_{a},q_{b};T) = \left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)\exp{\left[i\int_{0}^{T}dt\left(\sum_{i}p^{i}\dot{q}^{i}-H(q,p)\right)\right]}[/tex]

    shouldn't we just write

    [tex]\left(\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]

    instead of

    [tex]\left(\prod_{i}\int \mathcal{D}q(t)\mathcal{D}p(t)\right)[/tex]

    since the [itex]\mathcal{D}[/itex] itself stands for [itex]\prod[/itex]?
    Last edited: Aug 16, 2009
  2. jcsd
  3. Aug 16, 2009 #2
    Another query I have concerns the derivation of equation 9.14, which is

    [tex]\langle \phi_{b}({\bf{x}})|e^{-iHT}|\phi_{a}({\bf{x}})\rangle = \int \mathcal{D}\phi\exp\left[i\int_{0}^{T}d^{4}x \mathcal{L}\right][/tex]


    [tex]\langle \phi_{b}({\bf{x}})|e^{-iHT}|\phi_{a}({\bf{x}})\rangle = \int\mathcal{D}\phi\int\mathcal{D}\pi \exp{\left[i\int_{0}^{T}d^{4}x\left(\pi\dot{\phi}-\frac{1}{2}\pi^2-\frac{1}{2}(\nabla\phi)^2-V(\phi)\right)\right]}[/tex]

    To quote the authors,

    Now, I am a bit confused about the meaning of [itex]\mathcal{D}\phi[/itex] in these two equations. In going from the second equation to the first, we evaluate a Gaussian integral over [itex]\pi[/itex]. What happens to the constant factor that originates from this step?


    [tex]\int \exp{(-ax^2 + bx + c)} = \sqrt{\frac{\pi}{a}}\exp{\left(\frac{b^2}{4a} +c\right)}[/tex]

    and so

    [tex]\int \exp{\int d^{4}x(-a\phi^2 + b\phi + c)} = \sqrt{\frac{\pi}{a}}\exp{\int d^{4}x\exp{\left(\frac{b^2}{4a} +c\right)}[/tex]
  4. Aug 18, 2009 #3
  5. Aug 18, 2009 #4


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    It gets absorbed into the definition of [itex]{\cal D}\phi[/itex], which changes (by that constant factor) from one equation to the next.
  6. Aug 18, 2009 #5
    I see, thanks Avodyne...I just wanted to confirm that.
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