Nakahara M. Geometry, Topology and Physics
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1.5.4 Integration
Supprisingly enough, integration with respect to a Grassmann variable is
equivalent to differentiation. Let D denote differentiation with respect to a
Grassmann variable and let I denote integration, where integration is understood
as a definite integral. Suppose they satisfy the relations
(1) ID = 0,
(2) DI = 0,
(3) D(A) = 0 ⇒ I (BA) = I (B)A,
where A and B are arbitrary functions of Grassmann variables. The first relation
states that the integration of a derivative of any function yields the surface term
and it is set to zero. The second relation states that a derivative of a definite
integral vanishes. The third relation implies that A is a constant if D( A) = 0and
hence it can be taken out of the integral. These relations are satified if we take
I ∝ D. Here we adopt the normalization I = D and put
dθ f (θ) =
∂ f (θ )
∂θ
. (1.179)
We find from the previous definition that
dθ =
∂1
∂θ
= 0
dθθ =
∂θ
∂θ
= 1.
If there are n generators {θ
k
}, equation (1.179) is generalized as
dθ
1
dθ
2
...dθ
n
f (θ
1
,θ
2
,...,θ
n
) =
∂
∂θ
1
∂
∂θ
2
...
∂
∂θ
n
f (θ
1
,θ
2
,...,θ
n
).
(1.180)
Note the order of dθ
k
and ∂/∂θ
k
.
The equivalence of differentiation and integration leads to an odd behaviour
of integration under the change of integration variables. Let us consider the case
n = 1 first. Under the change of variable θ
= aθ (a ∈ ), we obtain
dθ f (θ) =
∂ f (θ )
∂θ
=
∂ f (θ
/a)
∂θ
/a
= a
dθ
f (θ
/a)
which leads to dθ
= (1/a)dθ. This is readily extended to the case of n variables.
Let θ
i
→ θ
i
= a
ij
θ
j
.Then
dθ
1
...θ
n
f (θ) =
∂
∂θ
1
...
∂
∂θ
n
f (θ )
=
n
k
i
=1
∂θ
k
1
∂θ
1
...
∂θ
k
n
∂θ
n
∂
∂θ
k
1
...
∂
∂θ
k
n
f (a
−1
θ
)
=
n
k
i
=1
ε
k
1
...k
n
a
k
1
1
...a
k
n
n
∂
∂θ
k
1
...
∂
∂θ
k
n
f (a
−1
θ
)
= det a
dθ
1
...θ
n
f (a
−1
θ
).
Accordingly, the integral measure transforms as
dθ
1
dθ
2
...θ
n
= det a dθ
1
dθ
2
...dθ
n
. (1.181)
1.5.5 Delta-function
The δ-function of a Grassmann variable is introduced as
dθδ(θ− α) f (θ) = f (α) (1.182)
for a single variable. If we substitute the expansion f (θ ) = a + bθ into this
definition, we obtain
dθδ(θ− α)(a + bθ) = a + bα
from which we find that the δ-function is explicitly given by
δ(θ −α) = θ − α. (1.183)
Extension of this result to n variables is easily verified to be (note the order of
variables)
δ
n
(θ − α) = (θ
n
− α
n
)...(θ
2
− α
2
)(θ
1
− α
1
). (1.184)
The integral form of the δ-function is obtained from
dξ e
iξθ
=
dξ(1 +iξθ) = iθ
as
δ(θ) = θ =−i
dξ e
iξθ
. (1.185)
1.5.6 Gaussian integral
Let us consider the integral
I =
dθ
∗
1
dθ
1
...dθ
∗
n
dθ
n
e
−
ij
θ
∗
i
M
ij
θ
j
(1.186)
where {θ
i
} and {θ
∗
i
} are two sets of independent Grassmann variables. The n × n
c-number matrix M is taken to be anti-symmetric since θ
i
and θ
∗
i
anti-commute.
The integral is evaluated with the help of the change of variables θ
i
=
j
M
ij
θ
j
as
I = det M
dθ
∗
1
dθ
1
...dθ
∗
n
dθ
n
e
−
i
θ
∗
i
θ
i
= det M
dθ
∗
dθ(1 + θ
θ
∗
)
n
= det M. (1.187)
We prove an interesting formula as an application of the Gaussian integral.
Proposition 1.3. Let a be an anti-symmetric matrix of order 2n and define the
Pfaffian of a by
Pf(a) =
1
2
n
n!
Permutations of
{i
1
,...,i
2n
}
sgn(P)a
i
1
i
2
...a
i
2n−1
i
2n
. (1.188)
Then
det a = Pf(a)
2
. (1.189)
Proof. Observe that
I =
dθ
2n
...dθ
1
exp
1
2
ij
θ
i
a
ij
θ
j
=
1
2
n
n!
dθ
2n
...dθ
1
ij
θ
i
a
ij
θ
j
n
= Pf(a).
Note also that
I
2
=
dθ
2n
...dθ
1
dθ
2n
...dθ
1
exp
1
2
ij
(θ
i
a
ij
θ
j
+ θ
i
a
ij
θ
j
)
.
Under the change of variables
η
k
=
1
√
2
(θ
k
+ θ
k
), η
∗
k
=
1
√
2i
(θ
k
− θ
k
),
we obtain the Jacobian = (−1)
n
and
θ
i
θ
j
+ θ
i
θ
j
= η
i
η
∗
j
− η
∗
j
η
i
dη
2n
...dη
i
dη
∗
2n
...dη
∗
1
= (−1)
n
2
dη
1
dη
∗
1
...dη
2n
dη
∗
2n
,
from which we verify that
Pf(a)
2
=
dη
1
dη
∗
1
...dη
2n
dη
∗
2n
exp
ij
η
∗
i
a
ij
η
j
= det a.
Exercise 1.9. (1) Let M be a skewsymmetric matrix and K
i
be Grassmann
numbers. Show that
dθ
1
...dθ
n
e
−
1
2
t
θ·M·θ +
t
K ·θ
= 2
n/2
√
det M e
−
t
K ·M
−1
·K /4
. (1.190)
(2) Let M be a skew-Hermitian matrix and K
i
and K
∗
i
be Grassmann numbers.
Show that
dθ
∗
1
dθ
1
...dθ
∗
n
dθ
n
e
−θ
†
·M·θ+K
†
·θ+θ
†
·K
= det M e
K
†
·M
−1
·K
. (1.191)
1.5.7 Functional derivative
The functional derivative with respect to a Grassmann variable can be defined
similarly to that for a commuting variable. Let ψ(t) be a Grassmann variable
depending on a c-number parameter t and F[ψ(t)] be a functional of ψ.Thenwe
define
δ F[ψ(t)]
δψ(s)
=
1
ε
{F[ψ(t) + εδ(t − s)]−F[ψ(t)]}, (1.192)
where ε is a Grassmann parameter. The Taylor expansion of F[ψ(t) −εδ(t −s)]
with respect to ε is linear in ε since ε
2
= 0. Accordingly, the limit ε → 0is
not necessary. A word of caution: division by a Grassmann number is not well
defined in general. Here, however, the numerator is proportional to ε and division
by ε simply means picking up the coefficient of ε in the numerator.
1.5.8 Complex conjugation
Let {θ
i
} and {θ
∗
i
} be two sets of the generators of Grassmann numbers. Define the
complex conjugation of θ
i
by (θ
i
)
∗
= θ
∗
i
and (θ
∗
i
)
∗
= θ
i
.Wedefine
(θ
i
θ
j
)
∗
= θ
∗
j
θ
∗
i
. (1.193)
Otherwise, the real c-number θ
i
θ
∗
i
does not satisify the reality condition
(θ
i
θ
∗
i
)
∗
= θ
i
θ
∗
i
.
1.5.9 Coherent states and completeness relation
The fermion annihilation and creation operators c and c
†
satisfy the anti-
commutation relations {c, c}={c
†
, c
†
}=0and{c, c
†
}=1 and the number
operator N = c
†
c has the eigenvectors |0 and |1. Let us consider the Hilbert
space spanned by these vectors
= Span{|0, |1}.
An arbitrary vector | f in
may be written in the form
| f =|0 f
0
+|1 f
1
,
where f
0
, f
1
∈ .
Now we consider the states
|θ=|0+|1θ (1.194)
θ|=0|+θ
∗
1| (1.195)
where θ and θ
∗
are Grassmann numbers. These states are called the coherent
states and are eigenstates of c and c
†
respectively,
c|θ=|0θ =|θ θ, θ|c
†
= θ
∗
0|=θ
∗
θ|.
Exercise 1.10. Verify the following identities;
θ
|θ=1 + θ
∗
θ = e
θ
∗
θ
,
θ| f = f
0
+ θ
∗
f
1
,
θ|c
†
| f =θ|1 f
0
= θ
∗
f
0
= θ
∗
θ| f ,
θ|c| f =θ |0 f
1
=
∂
∂θ
∗
θ| f .
Let
h(c, c
†
) = h
00
+ h
10
c
†
+ h
01
c + h
11
c
†
ch
ij
∈
be an arbitrary function of c and c
†
. Then the matrix elements of h are
0|h|0=h
00
0|h|1=h
01
1|h|0=h
10
1|h|1=h
00
+ h
11
.
It is easily found from these matrix elements that
θ|h|θ
=(h
00
+ θ
∗
h
10
+ h
01
θ
+ θ
∗
θ
h
11
)e
θ
∗
θ
. (1.196)
Lemma 1.3. Let |θ and θ | be defined as before. Then the completeness relation
takes the form
dθ
∗
dθ |θθ|e
−θ
∗
θ
= I. (1.197)
Proof. Straightforward calculation yields
dθ
∗
dθ|θ θ|e
−θ
∗
θ
=
dθ
∗
dθ(|0+|1θ)(0|+θ
∗
1|)(1 −θ
∗
θ)
=
dθ
∗
dθ
|00|+|1θ0|+|0θ
∗
1|+|1θθ
∗
1|
(1 −θ
∗
θ)
=|00|+|11|=I.
1.5.10 Partition function of a fermionic oscillator
We obtain here the partition fuction of a fermionic harmonic oscillator as an
application of the path integral formalism of fermions. The Hamiltonian is
H = (c
†
c − 1/2)ω, which has eigenvalues ±ω/2. The partition function is then
Z(β) = Tr e
−β H
=
1
n=0
n|e
−β H
|n=e
βω/2
+ e
−βω/2
= 2cosh(βω/2).
(1.198)
Now we evaluate Z (β) in two different ways using a path integral. We start our
exposition with the following lemma.
Lemma 1.4. Let H be the Hamiltonian of a fermionic harmonic oscillator. Then
the partition function is written as
Tr e
−β H
=
dθ
∗
dθ−θ |e
−β H
|θe
−θ
∗
θ
. (1.199)
Proof. Let us insert the completeness relation (1.197) into the definition of a
partition function to obtain
Z(β) =
n=0,1
n|e
−β H
|n
=
n
dθ
∗
dθe
−θ
∗
θ
n|θ θ |e
−β H
|n
=
n
dθ
∗
dθ(1 −θ
∗
θ)(n|0+n|1θ)(0|e
−β H
|n+θ
∗
1|e
−β H
|n)
=
n
dθ
∗
dθ(1 − θ
∗
θ)[0|e
−β H
|nn|0
− θ
∗
θ1|e
−β H
|nn|1+θ0|e
−β H
|nn|1+θ
∗
1|e
−β H
|nn|0].
The last term of the last line does not contribute to the integral and hence we may
change θ
∗
to −θ
∗
.Then
Z(β) =
n
dθ
∗
dθ(1 − θ
∗
θ)[0|e
−β H
|nn|0
− θ
∗
θ1|e
−β H
|nn|1+θ0|e
−β H
|nn|1−θ
∗
1|e
−β H
|nn|0]
=
dθ
∗
dθ e
−θ
∗
θ
−θ|e
−β H
|θ.
Accordingly, the coordinate in the trace is over anti-periodic orbits. The
Grassmann variable is θ at τ = 0 while −θ at τ = β andwehavetoimposean
anti-periodic boundary condition over [0,β] in the trace.
Use the expression
e
−β H
= lim
N→∞
(1 −β H /N)
N
and insert the completeness relation at each time step to find
Z(β) = lim
N→∞
dθ
∗
dθ e
−θ
∗
θ
−θ|(1 − β H /N)
N
|θ
= lim
N→∞
dθ
∗
dθ
N−1
k=1
dθ
∗
k
dθ
k
e
−
N−1
n=1
θ
∗
n
θ
n
×−θ|(1 − ε H )|θ
N−1
θ
N−1
|...|θ
1
θ
1
|(1 − εH )|θ
= lim
N→∞
N
k=1
dθ
∗
k
dθ
k
e
−
N
n=1
θ
∗
n
θ
n
×θ
N
|(1 − εH )|θ
N−1
θ
N−1
|...|θ
1
θ
1
|(1 −εH )|−θ
N
wherewehaveputε = β/N and θ =−θ
N
= θ
0
,θ
∗
=−θ
∗
N
= θ
∗
0
.
Each matrix element is evaluated as
θ
k
|(1 − ε H )|θ
k−1
=θ
k
|θ
k−1
1 −ε
θ
k
|H |θ
k−1
θ
k
|θ
k−1
θ
k
|θ
k−1
e
−εθ
k
|H|θ
k−1
/θ
k
|θ
k−1
= e
θ
∗
k
θ
k−1
e
−εω(θ
∗
k
θ
k−1
−1/2)
= e
εω/2
e
(1−εω)θ
∗
k
θ
k−1
.
The partition function is now expressed in terms of the path integral as
Z(β) = lim
N→∞
e
βω/2
N
k=1
dθ
∗
k
dθ
k
e
−
N
n=1
θ
∗
n
θ
n
e
(1−εω)
N
n=1
θ
∗
n
θ
n−1
= e
βω/2
lim
N→∞
N
k=1
dθ
∗
k
dθ
k
e
−
N
n=1
[θ
∗
n
(θ
n
−θ
n−1
)+εωθ
∗
n
θ
n−1
]
= e
βω/2
lim
N→∞
N
k=1
dθ
∗
k
dθ
k
e
−θ
†
·B·θ
, (1.200)
where
θ =
θ
1
θ
2
.
.
.
θ
N
θ
†
=
θ
∗
1
,θ
∗
2
,...,θ
∗
N
B
N
=
10... 0 −y
y 10...0
0 y 1 ...0
.
.
.
.
.
.
.
.
.
00... y 1
with y =−1 + εω in the last line. We finally find from the definition of the
Gaussian integral of Grassmann numbers that
Z(β) = e
βω/2
lim
N→∞
det B
N
= e
βω/2
lim
N→∞
[1 + (1 −βω/N)
N
]
= e
βω/2
(1 +e
−βω
) = 2cosh
1
2
βω. (1.201)
This should be compared with the partition function (1.151) of the bosonic
harmonic oscillator.
This partition function is also obtained by making use of the ζ -function
regularization. It follows from the second line of equation (1.200) that
Z(β) = e
βω/2
lim
N→∞
N
k=1
dθ
∗
k
dθ
k
e
−
n
[(1−εω)θ
∗
n
(θ
n
−θ
n−1
)/ε+ωθ
∗
n
θ
n
]
= e
βω/2
θ
∗
θ exp
−
β
0
dτθ
∗
(1 − εω)
d
dτ
+ ω
θ
= e
βω/2
Det
APBC
(1 −εω)
d
dτ
+ ω
.
Here the subscript APBC implies that the eigenvalue should be evaluated for the
solutions that satisfy the anti-periodic boundary condition θ(β) =−θ(0).It
might seem odd that the differential operator contains ε. We find later that this
gives a finite contribution to the infinite product of eigenvalues. Let us expand
the orbit θ(τ) in the Fourier modes. The eigenmodes and the corresponding
eigenvalues are
exp
πi(2n + 1)τ
β
,(1 −εω)
πi(2n + 1)
β
+ ω,
where n = 0, ±1, ±2,.... It should be noted that the coherent states are
overcomplete and that the actual number of degrees of freedom is N,whichis
related to ε as ε = β/N . Then we have to truncate the product at −N/4 ≤ k ≤
N/4 since one complex variable has two real degrees of freedom. Accordingly,
the partition function takes the form
Z(β) = e
βω/2
lim
N→∞
N/4
k=−N/4
i(1 − εω)
π(2n − 1)
β
+ ω
= e
βω/2
e
−βω/2
∞
k=1
2π(n − 1/2)
β
2
+ ω
2
=
∞
k=1
π(2k − 1)
β
2
∞
n=1
1 +
βω
π(2n − 1)
2
.
The first infinite product, which we call P, is divergent and requires
regularization. Note, first, that
log P =
∞
k=1
2log
2π(k − 1/2)
β
.
Define the corresponding ζ -function by
˜
ζ(s) =
∞
k=1
2π(k − 1/2)
β
−s
=
β
2π
s
ζ(s, 1/2)
with which we obtain P = e
−2
˜
ζ
(0)
.Here
ζ(s, a) =
∞
k=0
1
(k + a)
s
(0 < a < 1) (1.202)
is the generalized ζ -function (the Hurwitz ζ -function). The derivative of
˜
ζ(s)
at s = 0 yields
˜
ζ
(0) = log
β
2π
ζ(0, 1/2) + ζ
(0, 1/2) =−
1
2
log 2,
where use has been made of the values
17
ζ(0, 1/2) = 0 ζ
(0, 1/2) =−
1
2
log 2.
Finally we obtain
P = e
−2
˜
ζ
(0)
= e
log 2
= 2. (1.203)
Note that P is independent of β after regularization.
Putting them all together, we arrive at the partition function
Z(β) = 2
∞
n=1
1 +
βω
π(2n − 1)
2
. (1.204)
By making use of the well-known formula
cosh
x
2
=
∞
n=1
1 +
x
2
π
2
(2n − 1)
2
(1.205)
we obtain
Z(β) = 2cosh
βω
2
. (1.206)
Suppose, alternatively, we are ignorant about the formula (1.205). Then,
by equating equation (1.201) with equation (1.204), we have proved the formula
(1.205) with the help of path integrals. This is a typical application of physics
to mathematics: evaluate some physical quantity by two different methods
and equate the results. Then we often obtain a non-trivial relation which is
mathematically useful.
1.6 Quantization of a scalar field
1.6.1 Free scalar field
The analysis made in the previous sections may be easily generalized to a case
with many degrees of freedom. We are interested, in particular, in a system with
infinitely many degrees of freedom; the quantum field theory (QFT). Let us
start our exposition with the simplest case, that is, the scalar field theory. Let
φ(x) be a real scalar field at the spacetime coordinates x = (x, x
0
) where x is the
space coordinate while x
0
is the time coordinate. The action depends on φ and its
derivatives ∂
µ
φ(x) = ∂φ(x)/∂x
µ
:
S =
dx (φ, ∂
µ
φ). (1.207)
17
The first formula follows from the relation ζ(s, 1/2) = (2
s
− 1)ζ(s), which is derived from
the identity ζ(s, 1/2) + ζ(s) = 2
s
∞
n=1
[1/(2n − 1)
s
+ 1/(2n)
s
]=2
s
ζ(s). The second formula
is obtained by differentiating ζ(s, 1/2) = (2
s
− 1)ζ(s) with respect to s and using the formula
ζ(0) =−1/2.