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Consider the Green's function of fermion operators with imaginary time, $$\mathcal{G}(\nu, \nu', \tau) = - \langle T_\tau c_{\nu}(\tau) c_{\nu'}^\dagger(0)\rangle\tag{1}$$ To show it satisfies the periodicity, $$\mathcal{G}(\nu, \nu', \tau) = - \mathcal{G}(\nu, \nu', \tau+ \beta) \tag{2}$$ one needs to used the following identity, (see eq (71) and (72) on page 16 of http://folk.ntnu.no/johnof/green-2013.pdf) $${\rm Tr}(ABC \ldots XY Z) = {\rm Tr}(ZAB \ldots XY )\tag{71}$$

$$Tr(e^{-\beta H}c_{\nu'}^\dagger e^{H \tau} c_\nu e^{-H \tau} ) = Tr( e^{H \tau} c_\nu e^{-H \tau} e^{-\beta H}c_{\nu'}^\dagger )\tag{3}$$ which is important for the appearance of the minus sign in equation (2). However, since $c_\nu$ and $c_\nu^\dagger$ are fermions, I doubt equation (3) is not correct. My question is whether (3) is indeed correct or not? If (3) is not correct, how can (2) hold?

Qmechanic
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user34104
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2 Answers2

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I can see why you might be confused. If the trace were only over the spinor indices on a fermi field, $\psi_\alpha$ say, then there would be an additional minus sign. Here, however, the trace is over the entire many-body Hilbert space and the $c_\nu$'s are just like any other operator and so have a cyclic trace. This is what Ismasou's terse comment is saying.

mike stone
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  • Hello, I am still confused why if trace over the entire many-body Hilbert space, their don't have minus sign, is in this space fermion field don't have the anti-commutation property? Could you please elaborate more? – Daren Aug 31 '22 at 02:51
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  1. Consider the trace$^1$ $$ {\rm Tr}_{\cal H}(A)~:=~\sum_{i} A^i{}_i \tag{A}$$ of an operator $A:{\cal H}\to {\cal H}$ with matrix elements $$A^i{}_j~:=~\langle i | A |j\rangle\tag{B}$$ in some basis elements $|i\rangle$.

  2. We assume that we can assign a Grassmann-parity $|i|$ to the elements $|i\rangle$ and a Grassmann-parity $|A|$ to the operator $A$. The Grassmann parity of the matrix elements $A^i{}_j$ are then $$|A^i{}_j| = |i| + |A| + |j|. \tag{C}$$

  3. The cyclic trace property $$ {\rm Tr}(AB)~=~ \sum_{ij} A^i{}_j B^j{}_i~=~\sum_{ij} B^j{}_i A^i{}_j~=~{\rm Tr}(BA)\tag{D}$$ holds for a class of infinite-dimensional matrices where the only non-zero matrix elements of the operators $A$ and $B$ are Grassmann-even:
    $$\begin{align} A^i{}_j \text{ Grassmann-odd} \quad\Rightarrow& \quad A^i{}_j~=~0\cr B^i{}_j \text{ Grassmann-odd} \quad\Rightarrow& \quad B^i{}_j~=~0 \end{align}\tag{E} $$

  4. The links and references are implicitly assuming property (E) even for Grassmann-odd operators. This is readily fulfilled in practice, essentially because there are no Grassmann-odd numbers in Nature.

References:

  1. G.D. Mahan, Many-Particle Physics, 1990; p. 134 + eq. (3.2.5).

  2. G.D. Mahan, Many-Particle Physics, 2000; p. 110 + eq. (3.29).

  3. A. Das, Finite Temperature Field Theory, 1997; p. 3.

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$^1$ In contrast the supertrace $$ {\rm sTr}_{\cal H}(A)~:=~\sum_{i} (-1)^{|i|(1-|A|)} A^i{}_i \tag{F}$$ is super-cyclic $$\begin{align} {\rm sTr}(AB) ~=~& \sum_{ij} (-1)^{|i|(1-|A|-|B|)}A^i{}_j B^j{}_i\cr ~=~&(-1)^{|A||B|}\sum_{ij} (-1)^{|j|(1-|A|-|B|)}B^j{}_i A^i{}_j\cr ~=~&(-1)^{|A||B|}{\rm sTr}(BA). \end{align}\tag{G}$$

Qmechanic
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