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Appendix
A.3 Proof of Wick’s Theorem
The key step in the proof of Wick’s theorem [6] is the following
Lemma 1: Let ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t be a product of time-dependent fermion operators and
ˆ
b an operator with a time argument smaller than those of the factors in the product.
Then the following identity holds:
ˆ
N N N
ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t
ˆ
b = ˆ
N N N
ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t ˆ
b
+ ˆ
N N N
ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t ˆ
b
+ · · · + ˆ
N N N
ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t ˆ
b
+ ˆ
N N N
ˆ
a i ˆ
a j . . . ˆ
a s ˆ
a t ˆ
b
(A.3.1)
Proof: The operator ˆ
b can either be a physical operator, ˆ
b = ˆ
v, or an unphysical
operator, ˆ
b = ˆ
u. In the latter case, the lemma is essentially trivial. All contractions
on the right-hand side vanish,
ˆ
a
l ˆ
u
= ˆ
T T T
ˆ
a l ˆ
u
− ˆ
N N N
ˆ
a l ˆ
u
= 0
since each of the products ˆ
a l ˆ
u is both time- and normal-ordered. Moreover,
ˆ
N N N
ˆ
a i . . . ˆ
a t ˆ
u
= ˆ
N N N
ˆ
a i . . . ˆ
a t
ˆ
u
as ˆ
u is unphysical, so that Eq. (A.3.1) is fulfilled.
Now let us consider the case, where ˆ
b = ˆ
v is a physical operator. Without loss of
generality, we can assume that the factors in the operator product are already normalordered (otherwise a corresponding rearrangement could be performed on both sides
of Eq. A.3.1). In particular, we may assume that there are μ physical and ν unphysical
factors in the original product, so that the left side of Eq. (A.3.1) can be written as
ˆ
N N N
ˆ
v 1 . . . ˆ
v μ ˆ
u 1 . . . ˆ
u ν
ˆ
v = ˆ
v 1 . . . ˆ
v μ ˆ
u 1 . . . ˆ
u ν ˆ
v
(A.3.2)
To proceed, we commute ˆ
v successively to the left. The first step gives
ˆ
v 1 . . . ˆ
v μ ˆ
u 1 . . . ˆ
u ν ˆ
v = − ˆ
v 1 . . . ˆ
v μ ˆ
u 1 . . . ˆ
u ν−1 ˆ
v ˆ
u ν
+ ˆ
v 1 . . . ˆ
v μ ˆ
u 1 . . . ˆ
u ν−1
ˆ
u ν , ˆ
v
(A.3.3)
Now the anticommutator
ˆ
u ν , ˆ
v
on the right-hand side can be replaced by the
contraction ˆ
u
ν ˆ
v
,
ˆ
u ν , ˆ
v
= ˆ
u ν ˆ
v + ˆ
v ˆ
u ν = ˆ
T T T
ˆ
u ν ˆ
v
− ˆ
N N N
ˆ
u ν ˆ
v
= ˆ
u
ν ˆ
v
(A.3.4)