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The fermion fields are anti-commuting with each other,$$\{\Psi_{\alpha},\Psi_{\beta}\}=(\gamma^{0})_{\alpha\beta}~\delta^{3}(\vec{x}-\vec{y})\tag1$$ I was reading Srednicki's book quantum field theory, in section 40, eq.(40.43)&eq.(40.44), the charge conjugate of $~\bar{\Psi}A\Psi~$ transforms as $$C^{-1}\bar{\Psi}A\Psi C=\Psi^{T}CAC\bar\Psi^{T} \tag2$$ Since $\Psi^{T}CAC\bar\Psi^{T}$ is a $1\times1$ matrix, we can transpose it without changing its value, so $$C^{-1}\bar{\Psi}A\Psi C=(\Psi^{T}CAC\bar\Psi^{T})^{T}=-\bar\Psi C^{T}A^{T}C^{T}\Psi,\tag3$$ there is an extra minus sign because we have exchanged the two fermion fields $\Psi$,$\bar\Psi$ here.

My question is according to the anti-commuting relation $(1)$, we should have $$\Psi\bar\Psi \sim \delta^{3}(\vec{x}-\vec{y})-\bar\Psi\Psi,$$ rather than $\Psi\bar\Psi \sim -\bar\Psi\Psi,$ so why there is no term$\sim \delta^{3}(\vec{x}-\vec{y})$ in equation $(3)$? I think the fields here are operators rather than the classical fields.

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  • $\begingroup$ Keep in mind that those operators are in actual facts operator valued distributions, namely they must be smeared against test functions to become actual operators. There must indeed be a $\delta$ term but the integrations in different points of space should get rid of it, eventually (also one should know what the actual space-time dependence of those $A$ and $C$ are, to correctly integrate everything over). $\endgroup$ Commented Aug 7, 2016 at 0:19
  • $\begingroup$ @GennaroTedesco Thank you for your answer. Here in fact the $\delta$ term is $\delta^{3}(\vec{0})$, which is just a constant. Srednicki in this chapter want to illustrate the CPT theorem, which only need the Lagrangian to be invariant under CPT transformation. $\bar\Psi A \Psi $ here should be a common term in lagrangian. Although under CPT transformation, this term will create an extra term $\sim\delta^{3}(\vec{0})$ , it doesn't matter because we can tolerate the lagrangian adding a constant term. We still say the lagrangian is unchanged under CPT transformation and so the physics. $\endgroup$ Commented Aug 8, 2016 at 4:17

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My knowledge of fermionic field theory might need more than just a refresher, but by evaluating the transposition, you are not using the anticommutator, are you? The eigenvalues of the fermionic fields are not complex, but Grassmann numbers. If fermionic fields themselves can be considered as operators/"matrices" over the Grassmann numbers, then an additional minus-sign is needed to exchange the Grassmann-valued-coefficients. However, I really do not remember fermionic field theory very well anymore.

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You are right in one sense. However if you have a Fermi field $\Psi_a$ it is a pure operator. This acts on a Fock space of ladder states $\{|n\rangle\}$ to give the wave. For this fermion operator it to have a representation in space you must act on this with $|x\rangle\langle x|$ so the wave vector is $$ |\psi_a(x)\rangle~=~|x\rangle\langle x|\Psi_a|\sum_n|n\rangle $$ Peeling off the sum over Fock states we then have a representation of the fermion operator with position space. The commutator $\{\Psi_a(x),~\Psi_b(y)\}$ is then easily seen to involves $\langle x|y\rangle~=~\delta^3(x-y)$. The delta function emerges from representing the operator in space.

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    $\begingroup$ "However if you have a Fermi field Ψa it is a pure operator" is incorrect. Fields are operator valued distributions, namely they became operators only when smeared out against functions. Integrations of those functions takes care of the extra $\delta$ term. $\endgroup$ Commented Aug 7, 2016 at 0:01

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