1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Free, massless boson propagaor

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Cosider a single, free, massless boson with action [itex]S=\int\mathcal{L}=\frac{1}{2\pi}\int\partial X \overline{\partial}X[/itex] in two dimensions [itex]\overline{\partial}X(z,\overline{z}) = \partial_{\overline{z}} X(z,\overline{z})[/itex]
    Show, that the propagator [itex]\langle X(z,\overline{z})X(w,\overline{w})\rangle=-\frac{1}{2}log|z-w|[/itex].
    Use [itex]z=\sigma^{1}+i\sigma^{0}[/itex] and the integration measure [itex]2i\, dz\wedge d\overline{z}=d\sigma^{1}\wedge d\sigma^{0}[/itex].
    [itex]\sigma^{0}, \sigma^{1}[/itex] are the real coordiates.

    2. Relevant equations
    [itex]\langle X(z,\overline{z})X(w,\overline{w})\rangle = \frac{\int_X exp(-S[X])X(z,\overline{z})X(w,\overline{w})}{\int_X exp(-S[X])}

    3. The attempt at a solution
    Unfortuntely, I don't really know how to start. I don't even know why the integration measure is [itex]2i\, dz\wedge d\overline{z}=d\sigma^{1}\wedge d\sigma^{0}[/itex].
    It would be very nice if someone could just give me an ansatz.
  2. jcsd
  3. Jun 5, 2012 #2
    I think you will find the following attachment useful

    Attached Files:

  4. Jun 5, 2012 #3
    Thanks, that is helpful, indeed. But I still have some trouble with the formulation in complex coordinates.

    That's how I far I got:
    The path integral of a total derivative vanishes. Therefore, I obtain:
    [itex]0=\int\mathcal{D}X\frac{\delta}{\delta X(z,\overline{z})}(e^{-S}X(w,\overline{w})) = \int\mathcal{D}X\, e^{-S}\left[-\frac{\delta S}{\delta X(z,\overline{z})}X(w,\overline{w})+\delta(w-\overline{w})\delta(z-\overline{z})\right][/itex]

    Now, I have to evaluate the functional derivative of the action:
    [itex] \frac{\delta S}{\delta X(z,\overline{z})}
    = \frac{1}{2\pi}\frac{\delta}{\delta X(z,\overline{z})}\int dzd\overline{z} \partial X \overline{\partial}X[/itex]
    [itex]= \frac{1}{2\pi} \int dzd\overline{z}\frac{\delta} {\delta X(z,\overline{z})}(\partial X)\overline{\partial}X+\partial X\frac{\delta}{\delta X(z,\overline{z})}(\overline{\partial}X)[/itex]
    [itex]=-\frac{1}{\pi}\partial\overline{\partial} X(z,\overline{z})[/itex]
    In the last step I have used integration by parts with vanishing boundary terms.

    Plugging this into the equation above I find:
    [itex]0=\int\mathcal{D}X\, e^{-S} \left[\frac{1}{\pi} \partial \overline{\partial} X(z,\overline{z})X(w,\overline{w})+\delta(w-\overline{w})\delta(z-\overline{z})\right][/itex]
    [itex]\Rightarrow \partial\overline{\partial} X(z,\overline{z})X(w,\overline{w}) = -\pi\delta(w-\overline{w})\delta(z-\overline{z})[/itex]

    Next, I must show that [itex]X(z,\overline{z})X(w,\overline{w})=-\frac{1}{2}ln|z-w|[/itex] solves this equation, but I am unable to do so:
    It would be suffiecient to show: [itex]-\frac{1}{2}\int dzd\overline{z} \,\partial\overline{\partial}ln|z-w| = \pi[/itex]
    But here I fail:
    [itex]\int dzd\overline{z}\, \partial\overline{\partial}ln|z-w|=\int dzd\overline{z}\,\overline{\partial}\frac{1}{z-w} = \ldots[/itex]

    Can someone tell me how to proceed?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook