# Feynman propagator at equal times

In Timo Weigand's lecture notes on page 36, Equation $$(1.165)$$, he defines the Feynman propagator (free scalar field theory):

$$D_F(x-y)$$ := $$\langle 0|T\phi(x) \phi(y)|0 \rangle \tag{1}$$

This means $$(1)$$ should either be $$\langle 0|\phi(x) \phi(y)|0 \rangle \hspace{2mm}\text{or}\hspace{2mm} \langle 0|\phi(y) \phi(x)|0 \rangle \tag{2}$$

In the next lines, he $$D_F(x-y)$$ as:

$$D_F(x-y) = \Theta(x^0 - y^0) \langle 0|\phi(x) \phi(y)|0 \rangle + \Theta(y^0 - x^0) \langle 0|\phi(y) \phi(x)|0 \rangle \tag{3}$$

To preserve causality in QFT, $$\phi(x)$$ and $$\phi(y)$$ should commute if $$x$$ and $$y$$ are spacelike separated. Now, consider spacetime points (events) $$x$$ and $$y$$ such that $$x^0 = y^0$$. This means that these events are obviously spacelike separated. Therefore $$(3)$$ becomes:

$$\Theta(0) \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle + \Theta(0) \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle = 2\langle 0|\phi(x) \phi(y)|0 \rangle \tag{4}$$

So the only way for $$(2)$$ and $$(4)$$ to be agreed with each other is when:

$$\langle 0|\phi(x) \phi(y)|0 \rangle = \langle 0|\phi(y) \phi(x)|0 \rangle = 0$$

Even if the $$\Theta(0)$$ is defined to be $$0$$, then too the above equality should hold to make $$(2)$$ and $$(4)$$ agree with each other.

With the above observations, is the following statement correct?: The Feynman propagator, $$D_F(x-y)$$ vanishes when $$x^0 = y^0$$

• The value of the Heaviside function at zero is not necessarily 1. It is common to choose $\Theta(0) = 1/2$ or even $\Theta(0) = 0$, or to conceive of it purely as a distribution with no value at zero at all (like the Dirac delta). Nov 16, 2022 at 16:34
• Nov 16, 2022 at 17:15
• When $x^0=y^0$, the propagator need not be zero (and in fact is not zero) when $x^i=y^i$. Usually the propagator is singular there. Nov 16, 2022 at 18:22

$$D_F(x-y) = \Theta(x^0 - y^0) \langle 0|\phi(x) \phi(y)|0 \rangle + \Theta(y^0 - x^0) \langle 0|\phi(y) \phi(x)|0 \rangle \tag{3}$$
... this means that these events are obviously spacelike separated. Therefore $$(3)$$ becomes:
$$\Theta(0)\hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle + \Theta(0) \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle = 2\langle 0|\phi(x) \phi(y)|0 \rangle \tag{4A}$$
No, using $$\Theta(0)=1/2$$, it becomes: $$\frac{1}{2} \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle + \frac{1}{2} \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle = \langle 0|\phi(x) \phi(y)|0 \rangle \tag{4B}$$ $$=\langle 0|\phi(t,\vec x) \phi(t, \vec y)|0 \rangle =\langle 0|\phi(t, \vec y) \phi(t, \vec x)|0 \rangle$$