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!

Explicit expressions for creation/annihilation operator of the free scalar field

  1. May 22, 2008 #1
    I've been trying to work my way through some of my lecture notes, and have hit this snag. (n.b. I use [itex]k_0 \equiv +\sqrt{\vec{k}^2 + m^2}[/itex])

    We have
    [tex]a(q) = \int d^3 x e^{iqx} \{ q_0 \phi(x) + i \pi(x) \} [/tex]
    [tex]a^{\dagger}(q) = \int d^3 x e^{-iqx} \{ q_0 \phi(x) - i \pi(x) \}[/tex]

    To calculate the commutation relation between these operators, we simply multiply them out as required, and substitute the canonical commutation relation between fields and their conjugate momenta.

    I work through the relatively tedious steps and get

    [tex][a(q),a(p)] = \int d^3 x d^3 y e^{i(qx-py)} \delta^3(\vec{x} - \vec{y}) (q_0 - p_0)[/tex]
    [tex]= \int d^3 x e^{i(q-p)x} (q_0 - p_0)[/tex]
    [tex]= \int d^3 x e^{i(q_0-p_0)x^0} e^{i(\vec{q}-\vec{p})\cdot\vec{x}} (q_0 - p_0)[/tex]

    In my notes, the next step is to replace [itex]\vec{q}[/itex] with [itex]\vec{p}[/itex] and so get 0. However, if we integrate over x, surely we are left with a loose delta function outside an integral, which would mean that [itex][a(q),a(p)] = 0 \Leftarrow q=p[/itex] which I know is wrong.

    Can anyone explain that last step? Any textbooks I've seen assume this is trivial and just go on to state the commutation relation between the creation/annihilation operators rather than calculating it.
     
    Last edited: May 22, 2008
  2. jcsd
  3. May 22, 2008 #2
    Never mind. After doing the integral, I got

    [tex]
    (q_0 - p_0) \delta^3(\vec{q}-\vec{p}) e^{i(q_0-p_0)t}
    [/tex]

    which conspires to be zero when the delta function is non-zero because of the first term in brackets, and is zero everywhere else because of the delta function.
     
  4. May 22, 2008 #3

    Avodyne

    User Avatar
    Science Advisor

    Correct!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?