# Klein-Gordon Green's Function

I am having trouble understanding the derivation presented in chapter 2 of Overview of Quantum Field Theory, in which the authors show that the a particular function, denoted as $$D_R(x-y)$$ is the Green's function for the Klein-Gordon equation. Here is the derivation for $$D_R(x - y)$$, which is equation (2.54) in the book:

$$\langle0|[\phi(x), \phi(y)] |0\rangle = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}\big(e^{-ip\cdot (x - y)} - e^{ip\cdot (x - y)} \big) = \int \frac{d^3p}{(2\pi)^3}\int \frac{dp^0}{2\pi i}\frac{-1}{p^2 - m^2}e^{-ip\cdot (x - y)}$$

$$D_R(x - y) \equiv \theta(x^0 - y^0)\langle0|[\phi(x), \phi(y)] |0\rangle$$

The results of equation 2.56 (see below) are confusing me. I can derive the equations after the first equals sign (it is just the product rule); however, right after the second equals sign I see that the first term contains a $$\pi(x)$$. Why can we say this, because after all the derivative on the first term applies only to the step function $$\theta(x^0 - y^0)$$? Also after the first equal sign, I am confused again: why is the expression after the second equals sign equal to the four dimensional Dirac delta? I believe it has something to do with equation 2.54 but I am not sure how the math works; can someone explain?

$$(\partial^2 + m^2)D_R(x - y) = (\partial^2 \theta(x^0 - y^0))\langle0|[\phi(x), \phi(y)] |0\rangle + 2(\partial_\mu \theta(x^0 - y^0))(\partial^{\mu}\langle0|[\phi(x), \phi(y)] |0\rangle) + \theta(x^0 - y^0)(\partial^2 + m^2)\langle0|[\phi(x), \phi(y)] |0\rangle$$

$$= -\delta(x^0 - y^0)\langle0|[\pi(x), \phi(y)] |0\rangle + 2\delta(x^0 - y^0)\langle0|[\pi(x), \phi(y)] |0\rangle + 0$$

$$= - i\delta^4(x - y)$$

• Regarding your first question, see this. You can “move” a derivative from a delta function to what it is multiplying, if you also negate. May 17, 2021 at 3:05
• Regarding your second question, apply the equal-time canonical commutation relation. May 17, 2021 at 3:11
• @G.Smith Thank you!!
– user287576
May 17, 2021 at 10:12

For the first term note that the Lorentz index can only be $$\mu=0$$. Then use $$\partial \theta(x) = \delta(x)$$ in \begin{align} \int_{-\infty}^{+\infty} \frac{\partial ^2 \theta(x)}{\partial x^2} f(x) dx = &\, \int_{-\infty}^{+\infty} \frac{\partial \delta(x)}{\partial x} f(x) dx \nonumber\\ =&\, \delta(x) f(x) \Big|_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \delta(x) \frac{\partial f(x)}{\partial x} f \end{align} Hence \begin{align} [(\partial^0)^2\theta(x^0-y^0)] \langle 0| [\phi(x),\phi(y)]|0\rangle = &\,- \delta(x^0-y^0) \partial_0 \langle 0| [\phi(x),\phi(y)]|0\rangle\nonumber\\ = &\,- \delta(x^0-y^0) \langle 0| [\pi(x),\phi(y)]|0\rangle \end{align}