1QFT in zero dimensions

1.5 Fermions
So far, we have been dealing with Bosons. These particles obey Bose statistics,
so different fields
φ
commute with each other. However, we want to study
Fermions as well. This would involve variables that anti-commute with each
other. When we did canonical quantization, this didn’t pose any problem to our
theory, because fields were represented by operators, and operators can obey
any commutation or anti-commutation relations we wanted them to. However,
in the path integral approach, our fields are actual numbers, and they are forced
to commute.
So how can we take care of fermions in our theory? We cannot actually use
path integrals, as these involves things that commute. Instead, we will now treat
the fields as formal symbols.
The theory is rather trivial if there is only one field. So let’s say we have many
fields
θ
1
, ··· , θ
n
. Then a “function” of these fields will be a formal polynomial
expression in the symbols
θ
1
, ··· , θ
n
subject to the relations
θ
i
θ
j
=
θ
j
θ
i
. So,
for example, the action might be
S[θ
1
, θ
2
] =
1
2
m
2
θ
1
θ
2
=
1
2
m
2
θ
2
θ
1
.
Now the path integral is defined as
Z
dθ
1
···dθ
n
e
S[θ
1
,...,θ
n
]
.
There are two things we need to make sense of the exponential, and the
integral.
The exponential is easy. In general, for any analytic function
f
:
R R
(or
f : C C), we can write it as a power series
f(x) =
X
i=0
a
i
x
i
.
Then for any polynomial p in the fields, we can similarly define
f(p) =
X
i=0
a
i
p
i
.
Crucially, this expression is a finite sum, since we have only finitely many fields,
and any monomial in the fields of degree greater than the number of fields must
vanish by anti-commutativity.
Example.
e
θ
1
+θ
2
= 1 + θ
1
+ θ
2
+
1
2
(θ
1
+ θ
2
)
2
+ ···
= 1 + θ
1
+ θ
2
+
1
2
(θ
1
θ
2
+ θ
2
θ
1
)
= 1 + θ
1
+ θ
2
How about integration? Since our fields are just formal symbols, and do not
represent real/complex numbers, it doesn’t make sense to actually integrate it.
However, we can still define a linear functional from the space of all polynomials
in the fields to the reals (or complexes) that is going to act as integration.
If we have a single field
θ
, then the most general polynomial expression in
θ
is a + . It turns out the correct definition for the “integral” is
Z
dθ (a + ) = b.
This is known as the Berezin integral .
How can we generalize this to more fields? Heuristically, the rule is that
R
d
θ
is an expression that should anti-commute with all fields. Thus, for example,
Z
dθ
1
dθ
2
(3θ
2
+ 2θ
1
θ
2
) =
Z
dθ
1
3
Z
dθ
2
θ
2
+ 2
Z
dθ
2
θ
1
θ
2
=
Z
dθ
1
3 2θ
1
Z
dθ
2
θ
2
=
Z
dθ
1
(3 2θ
1
)
= 2.
On the other hand, we have
Z
dθ
1
dθ
2
(3θ
2
+ 2θ
2
θ
1
) =
Z
dθ
1
3
Z
dθ
2
θ
2
+ 2
Z
dθ
2
θ
2
θ
1
=
Z
dθ
1
3 + 2
Z
dθ
2
θ
2
θ
1
=
Z
dθ
1
(3 2θ
1
)
= 2.
Formally speaking, we can define the integral by
Z
dθ
1
···dθ
n
θ
n
···θ
1
= 1,
and then sending other polynomials to 0 and then extending linearly.
When actually constructing a fermion field, we want an action that is “sensi-
ble”. So we would want a mass term
1
2
m
2
θ
2
in the action. But for anti-commuting
variables, this must vanish!
The solution is to imagine that our fields are complex, and we have two
fields
θ
and
¯
θ
. Of course, formally, these are just two formal symbols, and bear
no relation to each other. However, we will think of them as being complex
conjugates to each other. We can then define the action to be
S[
¯
θ, θ] =
1
2
m
2
¯
θθ.
Then the partition function is
Z =
Z
dθ d
¯
θ e
S(
¯
θ,θ)
.
Similar to the previous computations, we can evaluate this to be
Z =
Z
dθ d
¯
θ e
S(
¯
θ,θ)
=
Z
dθ
Z
d
¯
θ
1
1
2
m
2
¯
θθ

=
Z
dθ
Z
d
¯
θ
1
2
m
2
Z
d
¯
θ
¯
θ
θ
=
Z
dθ
1
2
m
2
θ
=
1
2
m
2
.
We will need the following formula:
Proposition.
For an invertible
n × n
matrix
B
and
η
i
, ¯η
i
, θ
i
,
¯
θ
i
independent
fermionic variables for i = 1, . . . , n, we have
Z(η, ¯η) =
Z
d
n
θ d
n
¯
θ exp
¯
θ
i
B
ij
θ
j
+ ¯η
i
θ
i
+
¯
θ
i
η
i
= det B exp
¯η
i
(B
1
)
ij
η
j
.
In particular, we have
Z = Z(0, 0) = det B.
As before, for any function f, we can again define the correlation function
hf(
¯
θ, θ)i =
1
Z(0, 0)
Z
d
n
θ d
n
¯
θ e
S(
¯
θ,θ)
f(
¯
θ, θ).
Note that usually,
S
contains even-degree terms only. Then it doesn’t matter
if we place f on the left or the right of the exponential.
It is an exercise on the first example sheet to prove these, derive Feynman
rules, and investigate further examples.