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

Complex scalar field and contraction

  1. Aug 20, 2011 #1
    Hi guys,

    If I use the definition of the scalar complex field as the combination of two scalar real fields, I can get

    [tex]\phi (x) = \int \frac{d^3 p}{(2\pi )^3} \frac{1}{\sqrt{2p_0}} [ \hat a _{\vec{p}} e^{-ip.x} + \hat b _{\vec{p}}^{\dagger } e^{ip.x}][/tex]

    which I can rewrite in terms of (like in Peskin & Schroeder)

    [tex]\phi (x) = \phi ^{+} (x) + \phi ^{-} (x)[/tex]

    where [itex]\langle 0|\phi ^{-} = 0[/itex] and [itex]\phi ^{+} |0\rangle = 0[/itex].


    My problem is: when you try to calculate the contraction of the field with itself

    [tex]\text{\contraction}\{\phi (x)\}\{\phi (y)\} = \begin{cases} [\phi ^{+} (x), \phi ^{-} (y)] , & \text{if } x_0 > y_0 \\ [\phi ^{+} (y), \phi ^{-} (x)] , & \text{if } x_0 < y_0 \end{cases}[/tex]

    which is supposed to be the Feynman Propagator, you obtain it for a scalar real field, but for a scalar complex field as defined above, you obtain terms with [itex]\hat a _{\vec{p}} \hat b _{\vec{p \prime}}^{\dagger }[/itex]. The operators commute, so the vacuum expectation of these terms would be 0.


    I guess I'm wrong, but can someone see where? :)
     
  2. jcsd
  3. Aug 21, 2011 #2
    Hmm, I think the problem is that you are trying to contract the complex scalar field with itself, when I think you can actually only contract it with it's conjugate field (similarly to spin-1/2 fields, i.e. see page 116 of Peskin and Schroeder). I haven't been able to find a reference to back me up on this but I think it must be the case, for the very reason you have discovered.

    I.e. in your definition of the contraction there should be a dagger on the second scalar field in the first commutators and on the first field in the second commutator -if you want them to be equal to the Feynman propagator- and the commutators you have written down are indeed zero.

    It makes perfect sense now that I think about it more. The positive and negative frequency components of a complex scalar field are totally separate fields in some sense, so of course their commutators should vanish, in a free theory anyway...
     
    Last edited: Aug 21, 2011
  4. Aug 22, 2011 #3
    Now that you say it, it seems pretty logical. Thanks for explanation. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex scalar field and contraction
Loading...