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I have some questions about the Green’s function of the Klein-Gordon operator and the Feynman propagator. The first is about retarded Green’s function:

\begin{eqnarray} \int_{-\infty}^\infty\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\mathbf{p}}\left(e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right)&=& \int_{-\infty}^\infty\frac{d^3p}{(2\pi)^3}\left\{\frac{1}{2E_\mathbf{p}}e^{-ip\cdot(x-y)}\Big|_{p^0=E_\mathbf{p}}+\frac{1}{-2E_\mathbf{p}}e^{-ip\cdot(x-y)}\Big|_{p^0=-E_\mathbf{p}}\right\}\tag{1}\\ &\underset{x^0>y^0}{=}&\int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{-ip\cdot(x-y)}.\tag{2} \end{eqnarray}

  1. Why is (2) to be performed along the contour on P&S, p. 30? (Fig. 1)

I think that the contour above is derived from this process; According to Cauchy’s integral formula (2) is the contour integration along the clockwise contour closed below (Fig. 2) and $\int_{\text{arc}}$ is vanished due to Jordan’s lemma, so only the contour in Fig. 1 remains. Is this correct?

  1. Why is (3) equal to (4), Feynman propagator, shifting poles infinitesimally from real to imaginary axis? Is it just because they are equal when $\epsilon\to0$?

\begin{equation} \int_{-\infty}^\infty\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2}e^{-ip\cdot(x-y)},\tag{3} \end{equation}

\begin{equation} \int_{-\infty}^\infty\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}.\tag{4} \end{equation}

Fig. 1

Fig. 2

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Let me provide with some details and try to answer your questions:

The Green's function of the Klein-Gordon operator can be found to be \begin{equation} G(x-y)\sim \int d^3\vec{p}\hspace{0.1cm}e^{i\vec{p}\cdot(\vec{x}-\vec{y})}\int_{-\infty}^{+\infty} dp_0\hspace{0.1cm}\frac{e^{-ip_0(x_0-y_0)}}{(|p_0|+E_{\vec{p}})(|p_0|-E_{\vec{p}})}\hspace{0.1cm}, \end{equation} where $E_{\vec{p}}=\sqrt{\vec{p}^2+m^2}$.

One way to compute this function is to perform the $p_0$-integration using complex analysis, i.e. supposing that our energy $p_0$ (which is of course a real parameter) is an imaginary parameter. Then, all you need to do is to shift the poles $\pm E_{\vec{p}}$ by arbitrarily small imaginary quantities $\epsilon$. There are 2 ways to shift each pole, namely $E_{\vec{p}}+i\epsilon$, $E_{\vec{p}}-i\epsilon$, $-E_{\vec{p}}+i\epsilon$ and $-E_{\vec{p}}-i\epsilon$.The notions of "retarded", "advanced" and "Feynman" propagators correspond to different choices in the shifting of these 2 poles.

The retarded Green's function is identified by shifting both poles by a negative amount, i.e. $$E_{\vec{p}}\to E_{\vec{p}}-i\epsilon\qquad\text{and}\qquad-E_{\vec{p}}\to -E_{\vec{p}}-i\epsilon.$$ Note that these shifts are exactly equivallent to the contour you mention in Peskin-Schroeder's textbook, since in that case you keep the poles fixed but shift the contour so that it passes above the poles. These being said, we choose the contour this way (or shift the poles this way) because that's how we define the "retarded" Green's function from the very beginning. What you mention regarding the Cauchy's integral formula is correct! However, it only explains why we have to close the contour in the lower half-plane. If we chose to close it in the upper half-plane, the contour wouldn't enclose any pole and we would get zero by virtue of the Residue Theorem.

Finally, I hope I could give you an answer to your second question as well.

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  • $\begingroup$ I have two questions about your answer. 1. In P&S, the retarded Green's function of the K.-G. field is defined as $\theta(x^0-y^0)\langle 0\vert\left[\phi(x),\phi(y)\right]\vert 0\rangle$, so I think this is different to the equation you mentioned first since it has poles at $\pm E_\mathbf{p}$ so that this integral does not converge. 2. How can we just shift the poles despite the fact that shifting poles change what is integrated? Is it just because $E_\mathbf{p}\pm i\epsilon\to E_\mathbf{p}$ when $\epsilon\to 0$? $\endgroup$ – Orient Aug 23 '18 at 3:56
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    $\begingroup$ @Orient 1. If you read carefully, P&S state that $\theta(x^0-y^0)\langle 0\vert\left[\phi(x),\phi(y)\right]\vert 0\rangle$ together with the prescription of going around the poles define the retarded Green's function. 2. Yes, the arbitrary $\epsilon$ will be taken to go to 0 at the end of the day so it doesn't affect anything really. $\endgroup$ – G K Aug 23 '18 at 10:40

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