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I Klein-Gordon propagator

  1. Jun 28, 2017 #1
    Hello! I am reading Peskin's book on QFT and in the first chapter (pg. 30) he introduces this: ##<0|[\phi(x),\phi(y)]|0> = \int\frac{d^3p}{(2\pi)^3}\int\frac{dp^0}{2\pi i}\frac{-1}{p^2-m^2}e^{-ip(x-y)}## and then he spends 2 pages explaining the importance of choosing the right contour integral (i.e. the right prescription of going around the poles at ##\pm E_0##). However I asked a question here just to make sure and I have been told that the way you go around the poles has no influence on the final value of the integral. So now I am confused. If the way you do the integral give the same result in the end, why is Peskin talking about, exactly? Am I missing something? Thank you!
     
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  3. Jun 28, 2017 #2

    Orodruin

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    It is not a question of how you go around the poles, but of if you go around the poles (depending on how you. Lose the contour).
     
  4. Jun 28, 2017 #3
    I am not sure I understand what you mean. You have to go around the poles otherwise you encounter a singularity point.
     
  5. Jun 28, 2017 #4

    Orodruin

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    You are reading it wrong. You do not have to go around the poles if your contour closes in the other direction. Only the enclosed poles affect the integral.

    Reminds me of the riddle "What is the contour integral around western Europe?"
    Zero. The Poles are in eastern Europe!
     
  6. Jun 28, 2017 #5
    Ok the joke is funny but still. I understand that you don't need to enclose the poles necessary, but my point is, whether you enclose them or not, you get the same result in the end (the same numerical value for the integral).
     
  7. Jun 28, 2017 #6

    Orodruin

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    No you don't. That is the point.
     
  8. Jun 28, 2017 #7
    OK, sorry I am confused. From what I know from Complex Analysis, for a holomorphic function, the path you choose shouldn't matter, just the end points. Why is it now different?
     
  9. Jun 28, 2017 #8

    Orodruin

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    The integrand is not holomorphic. It has poles.
     
  10. Jun 28, 2017 #9
    So the way you go around the poles gives you different results? But in the end you have a real integal, which should have a definite, fixed value. Using this complex integration, and going around the poles is just an easier way to solve an integral hard to solve just in real numbers. But in the end, as any definite integral on the real line, it must have only one value. So I am still a bit confused about why you get multiple values, based on the path.
     
  11. Jun 28, 2017 #10

    Orodruin

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    A priori, your integral is not well defined. It has poles on the real line. That is why you need the pole prescription. Different choices will give you different Green's functions.
     
  12. Jul 2, 2017 #11

    vanhees71

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    Now you are talking! Contrary to your previous claims, indeed how you avoid the poles is crucial to get the correct Green's function needed for your problem in question. Since the confusion is thanks to Peskin&Schroeder, I suppose what you want is the time-ordered Green's function, defined (for a neutral scalar field) as
    $$\Delta(x)=-\mathrm{i} \langle \Omega|\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(0)|\Omega \rangle,$$
    where
    $$\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(0)=\Theta(x^0) \hat{\phi}(x) \hat{\phi}(0) + \Theta(-x^0) \hat{\phi}(0) \hat{\phi}(x).$$
    In vacuum it's identical with the Feynman propagator.

    If, however, you have to calculate something in linear-response theory, you need the retarded propagator, defined as
    $$\Delta_{\text{ret}}(x) =-\mathrm{i} \langle \Omega|\Theta(x^0) [\hat{\phi}(x),\hat{\phi}(0)] |\Omega \rangle.$$
    Thanks to the ##\Theta## unit-step functions, the way to circumvent the poles are uniquely determined.

    A pedagogically much better way to introduce the free propagators in QFT is to use the plane-wave mode decomposition with creation and annihilation operators and first evaluate the Mills representation, i.e., the Green's functions Fourier transformed with respect to ##\vec{x}## only. Then you first keep the time and can do the final Fourier transform wrt. to time at the end!
     
  13. Jul 2, 2017 #12

    Orodruin

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    When did I claim that it does not matter how you avoid the poles? What I am saying in #2 is that it does not matter what the contour looks like, only if the pole prescription places the poles inside the contour or not. If you look at the thread linked in the OP, the confusion is whether or not it matters what the contour that encloses the pole looks like - it does not as long as it encloses the pole. You can deform the contour as you like as long as you do not change what poles are inside it.

    Edit: I am never suggesting integrating through the pole. By "going around" I mean that the pole is enclosed by the contour and by "not going around" I mean that the pole is outside of the contour.
     
  14. Jul 2, 2017 #13

    vanhees71

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    Ok, then I misunderstood your statement in #2.
     
  15. Jul 2, 2017 #14

    Orodruin

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    It seems the OP did too, so perhaps I could have formulated it better. I guess this is a problem of me using "going around" in the sense of enclosing and a possible meaning of "not going around" could be perceived as "going through". The misspelling in the parenthesis probably indicates I was on my phone at the time and perhaps not paying too much attention to formulation...

    I think this was also the source of the OPs confusion in the first place with the other thread using "going around" in the same meaning as I did (which may be why I chose that nomenclature) whereas "way of going around" as used in Peskin might refer to the pole prescription (no access to Peskin now).
     
  16. Jul 3, 2017 #15

    vanhees71

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    Peskin&Schroeder do the usual thing in going right away to the 4D Fourier transform, i.e., in momentum space, where you have the naive formula
    $$(p^2-m^2) \Delta(p)=1,$$
    which of course doesn't make sense without deforming the original path of the Fourier integral along the real ##p^0## axis. That it is the temporal component comes from the fact that the different propagators are distinguished by the ##\Theta## functions of time.

    Let's do the calculation, so that we have it as a record here in PF. Everything can be derived from the fixed-ordered correlation function (two-point Wightman function),
    $$\mathrm{i} \Delta^{>}(x)=\langle \Omega|\hat{\phi}(x) \hat{\phi}(0)|\Omega \rangle.$$
    To do so, we use the mode decomposition of the field operator
    $$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 E}}[\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{a}^{\dagger}(\vec{p}) \exp(\mathrm{i} p \cdot x)],$$
    where the normalization is such that
    $$[\hat{a}(\vec{p}_1) ,\hat{a}(\vec{p}_2)]=\delta^{(3)}(\vec{p}_1-\vec{p}_2)$$
    and ##p^0=E=\sqrt{\vec{p}^2+m^2}##.
    For the Wightman function we thus have
    $$\mathrm{i} \Delta^{>}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_2 \frac{1}{\sqrt{(2 \pi)^3 2 E_1}} \frac{1}{\sqrt{(2 \pi)^3 2 E_2}} \exp(-\mathrm{i} x \cdot p_1) \langle \Omega|\hat{a}(\vec{p}_1) \hat{a}^{\dagger}(\vec{p}_2)|\Omega \rangle.$$
    Now using
    $$\Omega|\hat{a}(\vec{p}_1) \hat{a}^{\dagger}(\vec{p}_2)|\Omega \rangle=\Omega|[\hat{a}(\vec{p}_1) ,\hat{a}^{\dagger}(\vec{p}_2)]|\Omega \rangle = \delta^{(3)}(\vec{p}_1-\vec{p}_2),$$
    we get
    $$\mathrm{i} \Delta^{>}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \frac{1}{(2 \pi)^3 2 E_1} \exp(-\mathrm{i} x \cdot p_1)|_{p_1^0=E_1}.$$
    From translation invariance of the vacuum state we get for the other Wightman function
    $$\mathrm{i} \Delta^{<}(x)=\langle \Omega |\phi(0) \phi(x) |\Omega \rangle = \langle \Omega |\phi(-x) \phi(0)|\Omega = \mathrm{i} \Delta^{>}(-x) \\ = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}_1 \frac{1}{(2 \pi)^3 2 E_1} \exp(\mathrm{i} x \cdot p_1)|_{p_1^0=E_1}.$$
    Now it's easy to do the Fourier transform for these two functions
    $$\mathrm{i} \tilde{\Delta}^{>}(p)=\int_{\mathbb{R}^4} \mathrm{d}^4 x \exp(\mathrm{i} x \dot p) \mathrm{i} \Delta^{>}(x)=\frac{1}{2E} (2 \pi) \delta(p^0-E)$$
    and
    $$\mathrm{i} \tilde{\Delta}^{<}(p)=\frac{1}{2E} (2 \pi) \delta(p^0+E),$$
    where ##E=\sqrt{\vec{p}^2+m^2}##.

    To get the time-ordered propagator, we only need the Fourier transformation of the ##\Theta## unit-step functions
    $$\tilde{\Theta}(p^0)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} p^0 t) \Theta(t).$$
    Of course, this integral doesn't exist, and we have to regularize it. The only way is to make ##p^0## complex. Obviously the integral converges if ##\mathrm{Im} p^0>0##. So for real ##p^0## we write ##p^0 \rightarrow p^0+\mathrm{i} \epsilon##. This gives
    $$\tilde{\Theta}(p^0)=\frac{\mathrm{i}}{p^0+\mathrm{i} \epsilon}.$$
    In the same way we find
    $$\tilde{\Theta}^{-}(p^0)=\int_{\mathbb{R}} \mathrm{d} t \exp(\mathrm{i} t p^0) \Theta(-t)=-\frac{\mathrm{i}}{p^0-\mathrm{i} \epsilon}.$$
    Now we can use the convolution theorem for Fourier transformations to get the time-ordered Green's function
    $$\Delta(x)=\Theta(x^0) \Delta^{>}(x) + \Theta(-x^0) \Delta^{<}(x),$$
    i.e.
    $$\mathrm{i} \tilde{\Delta}(p) = \int_{\mathbb{R}} \frac{\mathrm{d} k^0}{2 \pi} [\tilde{\Theta}(k^0) \mathrm{i} \tilde{\Delta}^{>}(p^0-k^0,\vec{p}) + \tilde{\Theta}^{-}(k^0) \mathrm{i} \Delta^{<}(p^0-k^0,\vec{p})].$$
    Plugging in the above expressions one finds by evaluating the ##\delta## distributions in the Wightman functions,
    $$\mathrm{i} \tilde{\Delta}(p)=\frac{\mathrm{i}}{(p^0)^2-(E-\mathrm{i} \epsilon)^2} = \frac{\mathrm{i}}{p^2-m^2+\mathrm{i} \epsilon},$$
    where one has to read ##\epsilon=0^+##, i.e., an "infinitesimal small positive real number".
     
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