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If we consider the trace of two fermion field ($\psi_A \ \psi_B$), and using the cyclic property of trace (which has Grassmann property in general I thought). $$ \text{Tr}(\psi_A \ \psi_B)=(-)\text{Tr}(\psi_B \ \psi_A). $$ I am told that(not directly, I deduced, maybe wrong):

(1) if fermion field in spinor space, above equation have a minus sign;

(2) if fermion field in many-body Hilbert space, above equation don't have minus sign.

Why? Is in Hilbert space the anti-commutation property don't satisfy?

Here is how I deduce above two cases:

  1. I am learning "Finite Temperature Field Theory" by Ashok Das, on page 3, the book derived the Kubo-Martin-Schwinger (KMS) relation; $$ \begin{aligned} \left\langle A_H(t) B_H\left(t^{\prime}\right)\right\rangle_\beta &=Z^{-1}(\beta) \operatorname{Tr} \rho(\beta) A_H(t) B_H\left(t^{\prime}\right) \\ &=Z^{-1}(\beta) \operatorname{Tr} e^{-\beta \mathcal{H}} A_H(t) B_H\left(t^{\prime}\right) \\ &=Z^{-1}(\beta) \operatorname{Tr} \mathrm{e}^{-\beta \mathcal{H}} A_H(t) \mathrm{e}^{\beta \mathcal{H}} \mathrm{e}^{-\beta \mathcal{H}} B_H\left(t^{\prime}\right) \\ &{\color{blue}=}Z^{-1}(\beta) \operatorname{Tr} A_H(t+i \beta) \mathrm{e}^{-\beta \mathcal{H}} B_H\left(t^{\prime}\right) \\ &{\color{blue}=}Z^{-1}(\beta) \operatorname{Tr} \mathrm{e}^{-\beta \mathcal{H}} B_H\left(t^{\prime}\right) A_H(t+i \beta) \\ &=\left\langle B_H\left(t^{\prime}\right) A_H(t+i \beta)\right\rangle_\beta \end{aligned} $$ And I am troubled for the step where I use "blue" color to mark. That step use the cyclic property of trace. And the book further said: "this relation holds independent of Grassmann parities of the operator $A$ and $B$, namely, for both bosons as well as fermionic operators."

  2. Then I searched this question inside Physics SE, and I find this similar question. You can check this question and the answer by "mike stone".

That's how I deduce my argument.

Daren
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1 Answers1

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Consider gamma matrices. We have $\gamma^1 \gamma^2 = -\gamma^2\gamma^1$ but the cyclic property of the trace is still true: ${\rm tr}\{\gamma^1 \gamma^2\}= {\rm tr}\{\gamma^2 \gamma^1\}$. Of course this tells us that ${\rm tr}\{\gamma^1 \gamma^2\}= -{\rm tr}\{\gamma^1 \gamma^2\}$, so in this case we must have${\rm tr}\{\gamma^1 \gamma^2\}=0$.

Your example is more complicated, but the same cyclic property holds.

mike stone
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