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Quantization for fermionic field works exactly like bosonic field. You just replace regular functional derivative $\frac{\delta}{\delta \phi(x)}$ as in $$\pi(x)\to (-i)\frac{\delta}{\delta \phi( x)}$$ …

I am trying to compute the 2-point Green function $\tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta, \bar{\eta}]$ You have to take the $\eta=0$/$\bar{\eta}=0 $ limit a …

$$ (\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} = \psi_{a} \gamma^{0\ast}_{ab} \chi_{b}^{\ast} $$ is not correct. Rather $$ (\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} =\chi_{b}^{\ast} \gamma^{ …

Theoretically speaking, the expectation value of a Fermi field operator must be a Grassmann number. How do we observe a Grassmann number-value in experiments? Beats me! (Added note: Since here we are …

It is wrong to claim that: $$\{\psi_\alpha,\psi^\dagger_\beta\}=\delta_{\alpha \beta}$$ since Grassmann odd numbers should always anticommute with each other: $$\{\psi_\alpha,\psi^\dagger_\beta\}=0$$ …

when doing classical Dirac fields, sometimes they are treated as complex-valued spinors but sometimes they are treated as grassmann-valued spinors. Dirac fields should always be treated as grass …

The second sign $$ -\gamma^0_{ab}\gamma^2_{bc}\psi_c\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}=\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}\gamma^0_{ab}\gamma^2_{bc}\psi_c $$ comes from swapping the Grassmann-o …

The definition given by Nakahara is correct. Specifically: $$ \frac{\delta G[\psi(x)]}{\delta \psi(y)} \\ = \frac{1}{\epsilon}\{G[\psi(x) + \epsilon \delta(x-y)] - G[\psi(x)]\} \\ = \frac{1}{\epsilon …

Why is the Grassmann algebra part of the field $(v_1\wedge\cdots\wedge v_n)$ often omitted in physics textbooks when discussing Fermionic theories and is only brought up when the path integral is int …