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Green's function of the Klein-Gordon operator

  1. Sep 9, 2013 #1
    Again, from the Peskin and Schroeder's book, I can't quite see how this computation goes:

    See file attached

    The thing I don't get is how the term with [itex](\partial^{2}+m^{2})\langle 0| [\phi(x),\phi(y)] | 0 \rangle[/itex] vanishes, and also why they only get a [itex]\langle 0 | [\pi(x),\phi(y)] | 0 \rangle[/itex] from the [itex]\partial_{t}\langle 0 | [\phi(x),\phi(y)] | 0 \rangle[/itex] and not also a [itex]\langle 0 | [\phi(x),\pi(y)] | 0 \rangle[/itex]
     

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  3. Sep 9, 2013 #2

    vanhees71

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    Me neither. Peskin/Schroeder is pretty unclear concerning propagators and its analytic properties.

    First of all, one has to specify which propagator one is talking about, and this depends on what you want to do with it. In the case of perturbation theory in vacuum qft you need the time-ordered propagator, which is defined as the vacuum-expectation value of free field operators (here for an uncharged Klein-Gordon field)
    [tex]\mathrm{i} D(x-y)=\langle 0|\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(y)|0\rangle.[/tex]
    Now you plug in the expansion of the field operator in terms of creation and annihilation operators
    [tex]\hat{\phi}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega(\vec{p})}} [\hat{a}(\vec{p}) \exp(-p \cdot x) + \hat{a}^{\dagger}(\vec{p}) \exp(+p \cdot x) ]_{p^0=\omega(\vec{p})}.[/tex]
    Then you can write the propgator after some algebra with vacuum expectation values of annihilation and creation operator products as
    [tex]\mathrm{i} D(x-y)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3 2 \omega(\vec{p})} \left [\Theta(\xi^0) \exp(-\mathrm{i} p \cdot \xi) + \Theta(-\xi) \exp(+\mathrm{i} p \cdot \xi) \right]_{p^0=\omega(\vec{p}),\xi=x-y}.[/tex]
    Now you take the Fourier transform of this wrt. [itex]\xi[/itex] with a regulating factor [itex]\exp(-\epsilon |\xi^0|)[/itex], which leads you to
    [tex]\tilde{D}(p)=\int_{\mathbb{R}^4} \mathrm{d} \xi D(\xi) \exp(+\mathrm{i} p \cdot \
    \xi)=\frac{1}{p^2-m^2+\mathrm{i} \epsilon}.[/tex]
    The [itex]\mathrm{i} \epsilon[/itex] has to be understood to be taken in the weak limit [itex]\epsilon \rightarrow 0^+[/itex].

    For a more detailed explanation, why one has to use this time-ordered propagator, and also this derivation, see my QFT manuscript,

    http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

    Chapter 3.
     
  4. Sep 10, 2013 #3
    It is because of the free field eqn. (∂2+m2)[itex]\phi[/itex]=0,the last term vanishes.Note also that ∂μ will act only on operators whose argument is x,not y.Also the first term will involve a by part to get the result.Rest is just simple calculation based on commutator relation and some property of step function.
     
  5. Sep 12, 2013 #4
    That was the key I was missing, probably missed due to writting it so abstractly :P. Thanks!
     
  6. Sep 12, 2013 #5

    vanhees71

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    That's not writing it "abstractly" but just bad notation. If there are more than one space-time argument involved, you always should write explicitly wrt. which one you take partial derivatives.

    Now, I guess what you want to show is that the Feynman propagator of free fields fulfills the differential equation of the Green's function of the free Klein-Gordon operator. For this you just take the derivatives of the expectation value of the time-ordered field-operator product, writing the time-ordering symbol out in terms of Heaviside unitstep functions. Then you use
    [tex]\partial_{t_1} \Theta(t_1-t_2)=-\partial_{t_2} \Theta(t_2-t_1)=\delta(t_1-t_2)[/tex]
    and the canonical equal-time commutation relations for the field operators.
     
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