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Computation of propagation amplitudes for KG field

  1. Dec 22, 2014 #1
    Note: I'm posting this in the Quantum Physics forum since it doesn't really apply to HEP or particle physics (just scalar QFT). Hopefully this is the right forum.

    In Peskin and Schroeder, one reaches the following equation for the spacetime Klein-Gordon field:
    $$\phi(x,t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}\Big(a(p)e^{-ip\cdot x}+a^\dagger(p)e^{ip\cdot x}\Big)$$
    Then they say that the propagation amplitude for a particle to go from a spacetime point x to a spacetime point y is [itex]\langle 0\vert\phi(x)\phi(y)\vert 0\rangle[/itex] where the ket |0> is the vacuum. I understand this up to here. But then they start computing it as follows (equation 2.50):
    $$\langle 0\vert\phi(x)\phi(y)\vert 0\rangle = \int\frac{d^3p\,d^3q}{(2\pi)^6}\frac{1}{2\sqrt{E_{p}E_{p'}}}\Big(a(p)e^{-ipx}+a^\dagger(p)e^{ipx}\Big)\Big(a(q)e^{-iqy}+a^\dagger(q)e^{iqy}\Big)$$

    Clearly there will now be four terms when you expand the parentheses, and the book claims that all of these vanish except for the term with [itex]\langle 0\vert a(p)a^\dagger(q)\vert 0\rangle[/itex]. Two questions:

    a) Wouldn't this term vanish also since a(p) kills the vacuum bra, producing a zero?

    b) Why doesn't the term with [itex]a^\dagger(p)a^\dagger(q)[/itex] stay? In that term there are no annihilation operators to kill the vacuum, so surely the term should vanish.
     
  2. jcsd
  3. Dec 22, 2014 #2

    WannabeNewton

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    No. ##a_p## annihilates ##|0\rangle## which clearly means ##a^{\dagger}_p## annihilates ##\langle 0|##, not ##a_p##.

    See above.
     
  4. Dec 22, 2014 #3

    vanhees71

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    You forgot to "sandwich" the right-hand side of your equation between ##\langle 0|## and ##|0 \rangle##. Then it becomes immediately clear that all expressions with a annihilation operator acting to the right and a creation operator acting to the left (it becomes an annihilation operator when letting it act on the left argument of a scalar product) to the vacuum state, gives 0. The only non-vanishing term is thus indeed
    $$\langle 0|a(p) a^{\dagger}(q) 0 \rangle =\langle a^{\dagger}(p) 0|a^{\dagger}(q)0 \rangle=(2 \pi)^3 \delta^{(3)}(\vec{q}-\vec{p}).$$
     
    Last edited: Dec 22, 2014
  5. Dec 22, 2014 #4

    dextercioby

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    Again, it's all due to the damn bra-ket notation which is confusing. The equation should read:

    ## \langle 0, a (p) a^{\dagger} (q) 0\rangle = \langle a^{\dagger} (p) 0, a^{\dagger}(q)0\rangle ##
     
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