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I'm currently studying Wick's theorem for fermions with Peskin's and Schroeder's Introduction to QFT (p.115 & p.116). Here, Wick contractions are defined as $$ \psi^\bullet (x)\bar{\psi}^\bullet(y) = \begin{cases}\ \ \{\psi^+(x), \bar \psi^-(y)\} & \text{if } x^0 > y^0\ , \\ -\{\bar \psi^+(y), \psi^-(x)\} & \text{if } x^0 < y^0\ \end{cases} \tag{4.105} $$ and one can show that this coincides with the Feynman propagator $S_F(x-y)$. So far so good.
Then, it goes on to state Wick's theorem for fermions as $$ T\left[\psi_1\bar{\psi}_2\psi_3 ...\right] = N\left[ \psi_1\bar{\psi}_2\psi_3 ... +\text{ all possible contractions} \right],\tag{4.111} $$ where $\psi_i \equiv \psi(x_i)$ and spinor indices where neglected. I'm a bit confused by what is meant with 'all possible contractions' as the contraction was only defined for the ordering $\psi^\bullet (x)\bar{\psi}^\bullet(y)$ and not for $\bar{\psi}^\bullet(x)\psi^\bullet(y)$. How would one define the latter or rather is it necessary to define it at all? Consider for instance the vacuum expectation value $$ \langle 0 \vert T \psi_1\bar{\psi}_2\psi_3\bar{\psi}_4 \vert 0 \rangle $$ such that only fully contracted expressions remain. How would one apply Wick's theorem in this case? My guess would be to anticommute $\bar{\psi}_2$ and $\psi_3$ and evaluate the resulting expression, i.e. $$ \langle 0 \vert T \psi_1\bar{\psi}_2\psi_3\bar{\psi}_4 \vert 0 \rangle = - \langle 0 \vert T \psi_1\psi_3\bar{\psi}_2\bar{\psi}_4 \vert 0 \rangle + \delta^3(\mathbf{x}_2 - \mathbf{x}_3)\delta^{ab} \langle 0 \vert T \psi_1\bar{\psi}_4 \vert 0 \rangle, $$ with $a,b$ being the spinor indices of $\bar{\psi}_2$ and $\psi_3$. In this case one could simply apply the contraction definition from above, minding that one gets additional factors of (-1) for contractions of not neighbouring operators. Could someone confirm this approach and/or elaborate more on this?

Qmechanic
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basan
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  • This is essentially explained in my Phys.SE answer here. – Qmechanic Jan 12 '21 at 16:16
  • Thank you. I will have a look at it and try to understand it. – basan Jan 13 '21 at 08:59
  • @Qmechanic If I‘m understanding your answer for the other question correctly, then one can infer directly from basic Algebra considerations that the contraction with interchanged Grassmann operators, i.e. $\bar{\psi}^\bullet_2 \psi^\bullet_3$, simply yields an additional factor of (-1) compared to $\psi_3^\bullet \bar{\psi}^\bullet_2$. Thus, I don‘t even need to use the equal-time anti-commutation relations. – basan Jan 15 '21 at 13:35
  • Yes, Wick thm is not nec. about equal-time. 2. The fermionic signs follow simple rules.
  • – Qmechanic Jan 15 '21 at 14:05
  • Thank you very much. That clarified a lot and I agree with the comment from the other post, that the physics mystifies the mathematical idea in this case. Can I somehow mark your comment as an answer to my question? – basan Jan 15 '21 at 14:36