0K boundary condition for temperature Green's function

Question

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1. What the problem is asking

In the Matsubara (imaginary‑time) formalism the thermal Green function of a set of operators
[ A_{1}(\tau 1),A{2}(\tau 2),\dots ,A{n}(\tau _n) \qquad (0<\tau_i\le\beta ) ] is defined as

[ G(\tau 1,\dots ,\tau _n)=\frac{1}{Z}\, \operatorname{Tr}!\Big[ e^{-\beta H}\, T\tau\big{A_{1}(\tau 1)\cdots A{n}(\tau _n)\big}\Big] , \qquad Z=\operatorname{Tr}e^{-\beta H}. ]

Rickayzen states that if one of the times is equal to the end‑point of the interval, [ \tau_i =0\quad\text{or}\quad \tau_i=\beta , ] then the Green function satisfies the boundary condition

[ \boxed{ \; G(\tau _1,\dots ,0,\dots ,\tau _n)= -\;\epsilon\; G(\tau _1,\dots ,\beta ,\dots ,\tau _n)\; } \tag{1} ]

where

(\epsilon =+1) for bosonic operators (they commute)
(\epsilon =-1) for fermionic operators (they anticommute).

The question is: Why does (1) hold for any number of operators?
The asker tried a concrete example with three operators and seemed to obtain a contradiction. The resolution lies in a careful use of the cyclic property of the trace together with the (anti)‑periodicity of Heisenberg operators in imaginary time.

2. Step‑by‑step derivation

2.1 Definition of the thermal Green function

Write the Green function explicitly:

[ G(\tau 1,\dots ,\tau _n)=\frac{1}{Z} \operatorname{Tr}!\Big[ e^{-\beta H}\, T\tau\big{A_{1}(\tau 1)\cdots A{n}(\tau _n)\big}\Big]. \tag{2} ]

The time‑ordering operator (T_\tau) orders the factors from later to earlier imaginary times: [ T_\tau{A_{i_1}(\tau_{i_1})\cdots A_{i_n}(\tau_{i_n})} = \epsilon^{p}\, A_{j_1}(\tau_{j_1})\cdots A_{j_n}(\tau_{j_n}), ] where (\tau_{j_1}\ge \tau_{j_2}\ge\cdots\ge\tau_{j_n}) and
(p) is the number of pairwise interchanges of fermionic operators that are needed to bring the arguments into that order.
Thus each interchange of two fermionic operators contributes a factor (-1); interchanges of bosonic operators do nothing.
In compact form we write the overall sign as (\epsilon^{p}) with

[ \epsilon = \begin{cases} +1 & \text{bosons (commuting)}\[2pt] -1 & \text{fermions (anticommuting)}. \end{cases} \tag{3} ]

2.2 The Heisenberg picture in imaginary time

For any operator (A) in the Heisenberg picture

[ A(\tau)=e^{\tau H} A\, e^{-\tau H},\qquad 0\le\tau\le\beta . \tag{4} ]

Consequently

[ A(\beta)=e^{\beta H} A e^{-\beta H}. \tag{5} ]

Insert (5) into the trace:

[ e^{-\beta H} A(0)=e^{-\beta H}A = A(\beta) e^{-\beta H}. \tag{6} ]

No extra sign appears in (6); the sign will come from the cyclic property of the trace.

2.3 Cyclic property of the trace for fermionic/bosonic operators

For an ordinary (bosonic) operator (B),

[ \operatorname{Tr}(XY)=\operatorname{Tr}(YX). ]

If one of the operators, say (X), is fermionic (i.e. it contains an odd number of fermionic creation/annihilation operators) then moving it through another fermionic operator picks up a minus sign:

[ \operatorname{Tr}\big( X Y \big)= -\,\operatorname{Tr}\big( Y X \big) \qquad\text{if both (X) and (Y) are fermionic}. \tag{7} ]

More generally, when we move a single fermionic operator past a product of operators that contains an odd number of fermionic factors, we acquire a factor (\epsilon=-1); if the product contains an even number of fermionic factors the sign is (+1).
This is precisely the factor (\epsilon) appearing in (1).

2.4 Moving the operator at (\tau_i=0) to the back of the trace

Assume that the (k^{\text{th}}) argument of the Green function is the one we want to replace: [ \tau_k =0 . ]

Write the definition (2) with the operators explicitly ordered before the time‑ordering symbol:

[ G(\ldots ,0,\ldots)=\frac{1}{Z} \operatorname{Tr}!\Big[ e^{-\beta H}\, T_\tau{ \ldots A_k(0) \ldots}\Big]. \tag{8} ]

Using (6) we replace (A_k(0)) by (A_k(\beta)) to the left of the trace:

[ e^{-\beta H}A_k(0)=A_k(\beta) e^{-\beta H}. ]

Hence

[ G(\ldots ,0,\ldots)=\frac{1}{Z} \operatorname{Tr}!\Big[ A_k(\beta)\, e^{-\beta H}\, T_\tau{ \ldots}\Big]. \tag{9} ]

Now we cyclically permute the factor (A_k(\beta)) from the leftmost position to the rightmost position of the trace.
During this permutation we have to move (A_k(\beta)) past all the other operators that are inside the time‑ordering symbol. The number of fermionic interchanges performed is exactly the parity of the total number of fermionic operators that appear outside of (A_k). Denote this parity by (\epsilon) (the same symbol as in (1)). The cyclic move therefore yields

[ \operatorname{Tr}!\Big[ A_k(\beta)\, e^{-\beta H}\, T_\tau{ \ldots}\Big] = \epsilon\, \operatorname{Tr}!\Big[ e^{-\beta H}\, T_\tau{ \ldots A_k(\beta) }\Big]. \tag{10} ]

(If all operators are bosonic, (\epsilon=+1); if an odd number of fermionic operators are present, (\epsilon=-1).)

2.5 Restoring the time‑ordering

Inside the trace on the right‑hand side of (10) the operator (A_k(\beta)) now appears at the right end of the ordered product.
The time‑ordering operator (T_\tau) will place it according to its time argument (\beta).
Since (\beta) is the largest imaginary time, (A_k(\beta)) must be moved to the leftmost position of the ordered product.
Moving it past the other (n-1) operators involves exactly (n-1) permutations.

For bosons each permutation contributes no sign.
For fermions each permutation contributes a factor (-1).

Thus the total sign from re‑ordering (A_k(\beta)) to its proper position is ((-1)^{n-1}) for fermions and (+1) for bosons.
Because ((-1)^{n-1}= -\epsilon) (recall (\epsilon = -1) for fermions), the overall factor produced by the two steps—cyclic permutation and re‑ordering inside (T_\tau)—is exactly (-\epsilon).

Putting everything together:

[ \begin{aligned} G(\tau_1,\dots ,0,\dots ,\tau_n) &= \frac{1}{Z}\, \operatorname{Tr}!\Big[ e^{-\beta H}\, T_\tau{ \dots A_k(0) \dots}\Big] \[4pt] &= \frac{1}{Z}\, \epsilon\, \operatorname{Tr}!\Big[ e^{-\beta H}\, T_\tau{ \dots A_k(\beta) \dots}\Big]\;(-\epsilon) \[4pt] &= -\epsilon\; \frac{1}{Z}\, \operatorname{Tr}!\Big[ e^{-\beta H}\, T_\tau{ \dots A_k(\beta) \dots}\Big] \[4pt] &= -\epsilon\; G(\tau_1,\dots ,\beta ,\dots ,\tau_n). \end{aligned} \tag{11} ]

Equation (11) is exactly the statement (1) that Rickayzen wrote.

2.6 Why the “counter‑example’’ with three operators fails

In the attempted counter‑example the author compared

[ \langle A(\beta) B(\tau_2) C(\tau_3) \rangle \quad\text{with}\quad \langle B(\tau_2) C(\tau_3) A(0) \rangle . ]

Both expressions miss the crucial ingredients:

The time‑ordering operator (T_\tau) (which would reorder the three operators according to their times).
The cyclic permutation sign that appears when moving the operator at (\tau=0) through the trace.

If the operators are fermionic, moving (A(0)) from the rightmost position to the leftmost position of the trace indeed produces a minus sign. Together with the extra sign that comes from putting (A(\beta)) at the leftmost place in the time‑ordered product, the overall factor is (-\epsilon), exactly as derived above.
Therefore the apparent discrepancy disappears once the full definition (2) is used.

3. Final answer

For a thermal (imaginary‑time) Green function defined by

[ G(\tau_1,\dots ,\tau_n)=\frac{1}{Z} \operatorname{Tr}!\big[e^{-\beta H}\, T_\tau{A_1(\tau_1)\cdots A_n(\tau_n)}\big], ]

the boundary condition

[ \boxed{\,G(\tau_1,\dots ,0,\dots ,\tau_n)= -\;\epsilon\;G(\tau

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