1. Not finding help here? Sign up for a free 30min 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!

Lorentz Invariance of Propagator for Complex Scalar Field

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that

    [itex][\hat{\phi}(x_1),\hat{\phi}^\dagger(x_2)] = 0 [/itex]
    for [itex](x_1 - x_2)^2 < 0 [/itex]

    where [itex]\phi[/itex] is a complex scalar field

    2. Relevant equations

    [itex]\hat{\phi}=\int\frac{d^3 \mathbf{k}}{(2\pi)^3 \sqrt{2\omega}}[\hat{a}(k)e^{-ik\cdot x} + b^\dagger(k)e^{ik\cdot x}][/itex]

    with

    [itex][\hat{a}(k),\hat{a}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})[/itex]
    [itex][\hat{b}(k),\hat{b}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})[/itex]

    and all other commutation relations vanish.

    3. The attempt at a solution
    This is problem (7.3) in Aitchison and Hey's QFT book. I have a couple questions about the solutions that are posted online here. They get eventually to the difference of two integrals:

    [itex]\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{-ik\cdot (x_1-x_2)}-\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{ik\cdot (x_1-x_2)}[/itex]

    So far so good. Then they argue that for space-like intervals, x1-x2 can be transformed to -(x1-x2) by a Lorentz transformation, and thus the second term cancels the first.

    Now, I agree that each of the integrals must be Lorentz invariant, so you are allowed to evaluate each one in whichever frame you want, but I don't see how one is allowed to transform just the x1-x2 vector without also transforming the k.

    And a related question: In a different problem (5.6), when we come to a similar integral, they suggest transforming to a frame where t1=t2, and since the resulting integral is an odd function of k, it must be zero. And since it's Lorentz invariant, it must be zero for all space-like intervals. It seems to me that this argument works for this problem, too, so I'm not sure why they bring up this other argument about transforming x1-x2 to -(x1-x2).

    Anyone want to help explain what I'm missing?

    Thanks
     
  2. jcsd
  3. Sep 28, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You are integrating over [itex]\vec{k}[/itex]. Thanks to the invariance of [itex]\mathrm{d}^3 \vec{k}/\omega[/itex] where [itex]\omega=\sqrt{m^2+\vec{k}^2}[/itex] you know that the result is a Lorentz scalar [tex]F(x_1-x_2)[/tex]. That means that under Lorentz transformations
    [tex]F'(x')=F(x)=F(\Lambda^{-1} x).[/tex]
    Thus you only need to transform [itex]x=x_1-x_2[/itex].

    It's only crucial to show that for a spacelike [itex]x[/itex] you can always find a Lorentz transformation such that [itex]\Lambda x=-x[/itex]. Since you can orient your coordinate system always such that [itex]x=(0,\xi,0,0)[/itex] you just need to find a Lorentz transformation which makes out of this [itex]\Lambda x=(0,-\xi,0,0)[/itex].
     
  4. Sep 28, 2013 #3
    Thanks for the reply. I guess I'm missing the last part of that equality [itex]F(x) = F(\Lambda^{-1}x)[/itex]. As a test, if I define a function

    [itex]F(x)=k\cdot x[/itex]

    where [itex]k[/itex] is some arbitrary 4-vector, then F(x) is a Lorentz scalar. If I have some transformation such that [itex]x'=\Lambda x[/itex], then

    [itex]F'(x')=k'\cdot x'=k\cdot x=F(x)\not=F(\Lambda^{-1} x)[/itex]

    I understand that in the original problem we're integrating over [itex]\vec{k}[/itex], but it still seems to me we need to transform the [itex]\vec{k}[/itex].
     
  5. Sep 29, 2013 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Argh. That's of course a typo. The correct rule is
    [tex]F'(x')=F(x)=F(\Lambda^{-1} x'),[/tex]
    because
    [tex]x'=\Lambda x \; \Leftrightarrow \; x=\Lambda^{-1} x.[/tex]
     
  6. Sep 29, 2013 #5
    Right after I posted that, I figured that's what you meant. And it finally struck me why we don't have to worry about the [itex]\vec{k}[/itex]. Thanks for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Lorentz Invariance of Propagator for Complex Scalar Field
  1. Complex Scalar Field (Replies: 0)

Loading...