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I Klein Gordon propagator

  1. Jun 12, 2017 #1
    Hello! I am reading about Klein Gordon operator from Peskin book and he reaches at a point the integral ##\int_0^\infty \frac{1}{p^2-m^2}e^{-ip(x-y)}dp^0##. He then explains the different approaches of doing this integral, depending on how you pick the contour around the 2 poles. Why does the contour matter? One should get the same result in the end, no matter how you twist your contour around the poles (some ways are easier to solve but the result should be the same in the end). So what is the point of all this? Thank you!
     
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  3. Jun 12, 2017 #2

    vanhees71

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    The correct choice of the contour is the essence of the entire calculation. In vacuum QFT you need the time-oredered propagator (which in this case is the same as the Feynman propagator). A much better derivation than the handwaving one given in Peskin&Schroeder is to simply evaluate the expectation value, defining the free KG propagator, using the mode decomposition of the field operator:
    $$\mathrm{i} \Delta(x)=\langle \Omega|\mathcal{T} \hat{\phi}(x) \hat{\phi}(0) \omega \rangle.$$
    The mode decomposition is
    $$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} \left [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x)+\hat{a}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=\omega_{\vec{p}}},$$
    where
    $$\omega_{\vec{p}}=+\sqrt{\vec{p}^2+m^2}.$$
    Using this convention the commutation relations for the annihilation and creation operators read
    $$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').$$
    Now first evaluate the Mills representation Wightman function ("fixed order correlation function")
    $$\mathrm{i} \Delta^{21}(x)=\langle \hat{\phi}(x) \hat{\phi}(0),$$
    i.e.,
    $$\mathrm{i} \Delta_{M}^{21}(t,\vec{p})=\int_{\mathbb{R}^3} \langle \hat{\phi}(x) \hat{\phi}(0) \rangle \exp(-\mathrm{i} \vec{x} \cdot \vec{p}).$$
    Then you get the Mills representation of the time-ordered function as
    $$\Delta_{M}(t,\vec{p})=\Theta(t) \Delta_{M}^{21}(t,\vec{p}) + \Theta(-t) \Delta_{M}^{21}(-t ,\vec{p}).$$
    Then to finally get the momentum-space Green function do the final Fourier transformation with respect to ##t##. The final result, of course should be
    $$\tilde{\Delta}(p)=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$
     
  4. Jun 12, 2017 #3
    Thank you, I will take a look on your derivation. But letting the physics meaning of it aside, mathematically speaking, the contour integral of a holomoprhic function should be the same no matter the contour (if the end points are the same, and here they are ##\pm \infty##). So, in the end, why would one get different results, for different contours?
     
  5. Jun 12, 2017 #4

    vanhees71

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    There are poles on the real ##p^0## axis. So the integral as it stands is not defined, and you have to find the "detour" of the complex integration path around the poles which gives you the Green's function you want. There are infinitely many Green's functions of the KG operator. Besides the time-ordered one we need here, there's the retarded and advanced Green's function as well. In general given any Green's function you can add any solution of the homogeneous KG equation you like and still have a Green's function. The physics question decides which specific one you need. For perturbation theory in vacuum QFT you need the time-ordered one, as is derived a little later in Peskin&Schroeder.
     
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