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In David Tong's Quantum Field Theory Lecture Notes, Page 115 Eq. 5.34, the Feynman propagator was defined to be

$$ S_F(x-y)=\langle 0|T\psi(x)\bar\psi (y)|0\rangle \newcommand{\normord}[1]{:\mathrel{#1}:\;}. \tag{5.34} $$

However, in Eq. 5.36,

$$\mathrm{contraction}(\psi(x)\bar\psi(y))=T(\psi(x)\bar\psi(y)) \ -\normord{\psi(x)\bar\psi(y)} =S_F(x-y), \tag{5.36}$$

which seemed to indicate that $$\normord{\psi(x)\bar\psi(y)}=0.$$

What happened here? How did Wick's theorem work for Feynman propagator in Dirac Equation?

Qmechanic
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ShoutOutAndCalculate
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2 Answers2

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Note that the LHS of eq. (5.56) is an operator. This implies that there is an implicitly written identity operator ${\bf 1}$ on the RHS of eq. (5.56). By sandwiching eq. (5.56) with the vacuum state, we only get $$\langle 0 | :\psi(x)\bar\psi(y):|0\rangle~=0;$$ not OP's last equation.

There is an analogous situation for bosons. For more details, see e.g. this related Phys.SE post.

Qmechanic
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It's not true $:\psi(x)\bar\psi(y):=0$, but it is true that $\langle0|:\psi(x)\bar\psi(y):|0\rangle=0$.

I just looked up Tong's notes and it looks like he works out a specific example in detail for the bosonic case at eq 3.33 that may be helpful to you.

octonion
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