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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 $x^0 = y^0$

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

... 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: $$ \frac{1}{2} \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle + \frac{1}{2} \hspace{1mm} \langle 0|\phi(x) \phi(y)|0 \rangle = \langle 0|\phi(x) \phi(y)|0 \rangle \tag{4B} $$ $$ =\langle 0|\phi(t,\vec x) \phi(t, \vec y)|0 \rangle =\langle 0|\phi(t, \vec y) \phi(t, \vec x)|0 \rangle $$

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