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!

Klein-Gordon Causality calculation

  1. Jan 17, 2008 #1
    [SOLVED] Klein-Gordon Causality calculation

    1. The problem statement, all variables and given/known data
    In Peskin and Schroeder on page 27 it is stated that when we compute the Klien-Gordon propagator in terms of creation and annihilation operators the only term that survived the expansion is
    [tex]
    <0|a_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> \ \ (1).
    [/tex]
    I am unsure of why the term
    [tex]
    <0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0>
    [/tex]
    would vanish.

    2. Relevant equations
    The expansion of the field is given by
    [tex]
    \phi (x) = \int \frac{d^{3}p}{(2 \pi)^{3}} \frac{1}{\sqrt{2E_{\textbf{p}}}}(a_{\textbf{p}}}e^{-ip\cdot x} + a^{\dagger}_{\textbf{p}}e^{ip\cdot x})
    [/tex]
    and the normalization condition for states is
    [tex]
    <\textbf{p}|\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}-\textbf{q}).
    [/tex]


    3. The attempt at a solution
    Looking at the normalization condition given above I got,
    [tex]
    <0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q}}|0> = <0|\textbf{p}+\textbf{q}> = (2\pi)^{3}\delta^{3}(\textbf{p}+\textbf{q}).
    [/tex]
    However this mean that (1) is not the only surviving term, and from my calculations this also gives a factor of 2 that should not be there. I am unsure of how this term vanishes.
     
  2. jcsd
  3. Jan 17, 2008 #2

    StatusX

    User Avatar
    Homework Helper

    The quick answer is that ap kills the vacuum, so since:

    [tex]<0|a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>^* = <0|a_{\textbf{q}}a_{\textbf{p} }|0>=0[/tex]

    the term vanishes (alternatively, let the operators act to the left rather than the right).

    Also, note that [itex] a^{\dagger}_{\textbf{p}}a^{\dagger}_{\textbf{q} }|0>[/itex] is not the 1-particle state [itex]|\textbf{p}+\textbf{q}> [/itex], but is the two particle state consisting of one particle with momentum p and another with momentum q (it is true that the total momentum is then p+q, and maybe that's what you mean, but if so it's confusing notation). Furthermore, the inner product of the vacuum with any n-particle state is zero, be it the 2-particle state you should have used or even the one-particle state |p+q>, so your second equality is also wrong.
     
  4. Jan 17, 2008 #3
    Thank you very much that does help. Could you possibly point me in the direction as to why
    [tex]
    <0|\textbf{p};\textbf{q}> = 0
    [/tex]
    where [tex]|\textbf{p}, \textbf{q}>[/tex] is a two particle state. I remember this but can't recall why. Also how do I mark this as answered?
     
  5. Jan 17, 2008 #4

    olgranpappy

    User Avatar
    Homework Helper

    the state you call [itex]|0>[/itex] is *not* a 2-particle state of zero momentum, it is the vacuum--it has no particles. It is orthogonal to any state that has particles. It is a basic property of the creation and annihilation operators that
    [tex]
    a_p |0>=0
    [/tex]
    for all p

    and thus
    [tex]
    <0|a_p^\dagger = 0
    [/tex]
    for all p.

    and thus
    [tex]
    <0|a_p^\dagger a_q^\dagger = 0
    [/tex]

    and thus
    [tex]
    <0|a_p^\dagger a_q^\dagger|0>=0
    [/tex]
    for all p and q. And the above is certainly not equal to the delta function expression you gave in your first post. cheers,

    adam
     
  6. Jan 17, 2008 #5

    StatusX

    User Avatar
    Homework Helper

    There are a few ways to see it:

    1. ap annhilates the vacuum state, from which it follows by my last post.
    2. The states are different eigenstates of the Hermitian operator N giving the number of particles in the system.
    3. The inner product on a Fock space is defined so that only states in the same number N-particle subspace can have a non-vanishing inner product.

    Of course, these are all related to each other. Also, things get a little less clear in an interacting theory, and often we have to redefine things (renormalize) so that these statements remain true.
     
  7. Jan 17, 2008 #6
    That makes complete sense, thank you for your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Klein-Gordon Causality calculation
Loading...