# Why commutator of positive and negative parts of scalar field is equal to the Feynman propagator?

Peskin & Schroeder state that the contraction of two fields, defined as the commutator: $$[\phi^+(x),\phi^-(y)]\qquad \text{assuming}\ x^0>y^0$$ is equal to the Feynman propagator $$D_F(x-y)$$. But, why is that true? If we take the definition of positive and negative frequency parts of the field: \begin{aligned} \phi^+(x) &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}a_{\mathbf{p}}e^{-ip\cdot x}\\ \phi^-(x) &= \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}a^\dagger_{\mathbf{p}}e^{+ip\cdot x} \end{aligned} And apply their commutator, we have: $$\begin{multline} [\phi^+(x),\phi^-(y)] = \int\frac{d^3p}{(2\pi)^3}\frac{d^3p'}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\frac{1}{\sqrt{2E_{\mathbf{p'}}}}e^{-ip\cdot x}e^{+ip'\cdot y}\underbrace{[a_{\mathbf{p}},a^\dagger_{\mathbf{p'}}]}_{=(2\pi)^3\delta^3(\mathbf{p}-\mathbf{p'})}\\ =\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}e^{-ip\cdot(x-y)} \end{multline}$$ From here, I don't know how to go about getting the Feynman propagator. Any suggestion?

• Isn't that explained in P&S p. 27-31? Nov 29, 2020 at 20:04
• So, if i understand it well, it is just a definition: $$\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\mathbf{p}}}e^{-ip\cdot(x-y)} = D_F(x-y)$$ But, as he says on page 83, why is that suppose to be equal to $\int \frac{d^{4} p}{(2 \pi)^{4}} \frac{i e^{-i p \cdot(x-y)}}{p^{2}-m^{2}+i \epsilon}$ ?
– Dani
Nov 29, 2020 at 20:16

Just evaluate the $$p^0$$ part of the integral. For $$x^0>y^0$$, we can close the contour below and enclose the pole at $$p^0=E_\mathbf{p}$$ (there's a pole at this point because $$p$$ is on-shell, so $$p^2-m^2=0$$). Then we evaluate the integral using Cauchy's integral formula, (I'm playing fast and loose with the $$i\epsilon$$) $$\int \frac{d p^0}{2 \pi } \frac{ie^{-i p \cdot(x-y)}}{p^{2}-m^{2}+i \epsilon}=\int\frac{d p^0}{2 \pi i} \frac{-e^{-ip \cdot(x-y)}}{(p^0+\sqrt{m^{2}+\mathbf{p}^2})(p^0-\sqrt{m^{2}+\mathbf{p}^2})+i \epsilon}$$ $$=-\int\frac{d p^0}{2 \pi i} \frac{e^{-ip \cdot(x-y)}}{(p^0+E_\mathbf{p})(p^0-E_\mathbf{p})+i \epsilon}=\frac{e^{-ip \cdot(x-y)}}{2E_\mathbf{p}+i \epsilon},$$ where it's implicitly understood that $$p^0=E_\mathbf{p}$$ in the final $$e^{-ip \cdot(x-y)}$$. Note that a sign flip occurred because we're integrating clockwise instead of counterclockwise as assumed in Cauchy's integral formula. Using the above, we get (for $$x^0>y^0$$ only) that $$\int \frac{d^{4} p}{(2 \pi)^{4}} \frac{i e^{-i p \cdot(x-y)}}{p^{2}-m^{2}+i \epsilon}=\int \frac{d^{3} p}{(2 \pi)^{3}} \frac{e^{-i p \cdot(x-y)}}{2E_\mathbf{p}+i \epsilon}.$$