1
The story of fermions
Consider a 1-d chain of lattice sites. At each site there is a fermion, represented by a Grassman
number
ψk , ψk ψl + ψl ψk = 0
(1)
so that these ψk are anticommuting objects. The path integral is performed with an action
Z
X
X
S = iα ψ∂ψ → iα
ψk (ψk+1 − ψk ) = iα
ψk ψk+1
k
(2)
k
where we have used the anticommuting nature of the ψk . Note that we get the same action if we
define the derivative differently
Z
X
X
X
X
S = iα ψ∂ψ → iα
ψk (ψk − ψk−1 ) = −i
αψk ψk−1 = iα
ψk−1 ψk = iα
ψk ψk+1 (3)
k
k
k
k
where again we had to have the anticommuting nature of the ψk .
To perform the path integral, we note some mathematical identities. Suppose that Ai , Bi are vectors
of Grassman numbers, and Mij is a matrix of commuting numbers. Consider
Z
d[Ai ]d[Bi ]e−Ai Mij Bj
(4)
For a single variable we have
Z
Z
dA = 0,
dAA = 1
(5)
Thus
Z
−AM B
dAdBe
=
Z
dAdB[1 − AM B] = −M
Z
dAdBAB = M
If M was a diagonal matrix we would get
Z
Y
Mii = det M
d[Ai ]d[Bi ]e−Ai Mii Bi =
Z
dAA
Z
dBB = M
(6)
(7)
i
More generally, we will get
Z
d[Ai ]d[Bi ]e−Ai Mij Bj = det M
Now suppose that we have just one kind of anticommuting vector
Z
d[Ci ]e−Ci Mij Cj
(8)
(9)
Now we must have
Mij = −Mji
(10)
M12 = −M21 = 1
(11)
As an example, let us take a 2 × 2 matrix
1
Then
Z
dC1 dC2 e−Ci Mij Cj =
More generally we get
Z
Z
dC1 dC2 [1 − 2C1 C2 ] = 2
(12)
d[Ci ]e−Ci Mij Cj = P haf f [M ]
(13)
where P haf f can be defined for any antisymmetric matrix of even dimension N , and is given by
summing terms like
1
ǫi ...i Mi i Mi i . . . MiN−1 iN
(14)
N! 1 N 1 2 3 4
We have
(P haf f [M ])2 = det M
(15)
1
but we lose the information of the sign if we write the result as (det M ) 2 .
Now let us compute the path integral with sources. For the case of two fields, we write
Z
d[Ai ]d[Bi ]e−Ai Mij Bj +Ai Ji +Ki Bi
We solve this by shifting the fields. We write
Z
Z
˜
−Ai Mij Bj +Ai Ji +Ki Bi
d[Ai ]d[Bi ]e
= d[Ai ]d[Bi ]e−(Ai +K̃i )Mij (Bj +Ji )+C
Then
˜
J = −M J,
(16)
(17)
J˜ = −M −1 J
(18)
K = −K̃M, K̃ = −KM −1
C = K̃M J˜ = KM −1 M M −1 J = KM −1 J
(19)
(20)
We then get
Z
d[Ai ]d[Bi ]e−Ai Mij Bj +Ai Ji +Ki Bi =
Z
−1
d[Ai ]d[Bi ]e−Ai Mij Bj eKi Mij
Jj
(21)
Thus the 2-point function will be given by
< Ak Bl >=
1 δ δ
−1
Z = Mlk
Z δJk δKl
(22)
For the case of a single field we have
Z
Z
J˜j
J˜i
1 ˜
˜
−Ci Mij Cj +Ci Ji
d[Ci ]e
= d[Ci ]e−(Ci + 2 )Mij (Cj + 2 )+ 4 Ji Mij Jj
(23)
where we have used the antisymmetry of M . We have
˜ J˜ = −M −1 J
J = −M J,
(24)
Thus
Z
−Ci Mij Cj +Ci Ji
d[Ci ]e
=
Z
−Ci Mij Cj
d[Ci ]e
e
1 ˜
J M J˜
4 i ij j
=
Z
1
−1
d[Ci ]e−Ci Mij Cj e− 4 Ji Mij
Jj
(25)
where we have used that
M −1T M M −1 = −M −1 M M −1 = −M −1
We have
< Ck Cl >=
1 δ δ
1 −1
Z = Mkl
Z δJk δJl
2
2
(26)
(27)
2
Fermions in 2-d
In 2-d the fermions will be 2-component spinors. The Γ matrices will be
Γ τ = σ1 ,
Γ σ = σ2
We write
(28)
z
τ
σ
+
0
0
1
0
(29)
z̄
τ
σ
−
0
1
0
0
(30)
1 0
0 −1
(31)
1
d2 z iαψ † (Γz ∂z + Γz̄ ∂z̄ )ψ
2
(32)
Γ = Γ + iΓ = 2σ = 2
Γ = Γ − iΓ = 2σ = 2
The chirality operator is
5
τ
3
σ
Γ = Γ Γ = iσ = i
The action is
S=
1
d z iαψ † Γa ∂a ψ =
2
Z
2
Z
where we will choose the constant α later. Let us write
+
ψ
ψ=
ψ−
Then we have
S=
Z
d2 ziα(ψ + )∗ ∂z ψ − +
Z
(33)
d2 ziα(ψ − )∗ ∂z̄ ψ +
(34)
Thus the right and left moving parts of S split up, and we can consider them one at a time.
3
2-point function of fermions
Consider the action
S = iα
Z
d2 zψ(z)∂z̄ ψ(z) +
Z
d2 zψ(z)J(z)
(35)
The correlation function will be
< ψ(z1 )ψ(z2 ) >=
1
δ
δ
1
< e−S ψ(z1 )ψ(z2 ) >=
Z
Z
Z δJ(z1 ) δJ(z2 )
(36)
The matrix M in this case is
M = iα∂z̄
Thus the inverse will satisfy
Z
(37)
d2 z ′ [iα∂z̄ ](z, z ′ )M −1 (z ′ , z ′′ ) = δ2 (z − z ′′ )
(38)
1
= πδ2 (z)
z
(39)
But we know that
∂z̄
3
Thus
1
1
′
iπα (z − z ′′ )
We then find, using the expression for the correlator in the case of a single field
M −1 (z ′ − z ′′ ) =
< ψ(z1 )ψ(z2 ) >= −
We would finally like to use a normalization where
< ψ(z1 )ψ(z2 ) >= −
Thus we choose
(40)
1
1
2πiα (z1 − z2 )
(41)
1
1
2πiα (z1 − z2 )
(42)
i
1
=
(43)
2πi
2π
The factor of i that arises in this normalization reflects the fact that we are working in Euclidean
signature, so that t → −iτ . Thus the action, which has a factor i in the Lorentzian signature, does
not have such a factor in Euclidean signature.
α=−
4
Currents
Let us take a set of fermions
ψk ,
k = 1, . . . N
(44)
These are anticommuting objects, and the 2-point functions are
< ψ k (z1 )ψ l (z2 ) >=
δkl
(z1 − z2 )
(45)
Now consider the matrices T a , a = 1, . . . r forming a Lie algebra
[T a , T b ] = fcab T c
(46)
We assume that these have been brought to an antisymmetric form
Tija = −Tjia
(47)
We also assume that they are normalized by
tr(T a T b ) = δab
(48)
Make the following bilinears in the fermions
1 a i
T ψ (z)ψ j (z)
2 ij
These are called currents. Note that the scaling dimension is
J a (z) =
¯ = (1, 0)
(∆, ∆)
since the fermions had holomorphic dimension
1
2
Qa =
Z
(49)
(50)
each. Thus we can define charges
dzJ a (z)
C
where C is a contour that encircles the region to which we wish to apply the charge operator.
4
(51)
5
OPE of currents
Consider the OPE of two currents
J a (z)J b (z ′ ) =
1 b k ′ l ′
1 a i
ψ (z )ψ (z )
T ψ (z)ψ j (z) Tkl
2 ij
2
(52)
The most singular term arises from contracting all fermions. This gives
1 ab
1 a b jk kl
1
1
1
a b
2δ
Tij Tkl [δ δ − δik δjl ]
=
tr(T
T
)
=
′
2
′
2
4
(z − z )
2
(z − z )
(z − z ′ )2
(53)
The next term comes when one pair of fermions is contracted. This is
1
1 a b jk i
T T [δ ψ (z)ψ l (z ′ ) − δjl ψ i (z)ψ k (z ′ ) − δik ψ j (z)ψ l (z ′ ) + δil ψ j (z)ψ k (z ′ )]
4 ij kl
(z − z ′ )
(54)
In this term let us put
ψ(z) ≈ ψ(z ′ )
(55)
since the corrections will be terms with no singularity. Then we get from the first part of the above
expression
1 a b jk i
1
1
1
Tij Tkl δ ψ (z)ψ l (z ′ )
= (T a T b )il ψ i (z ′ )ψ l (z ′ )
(56)
′
4
(z − z )
4
(z − z ′ )
Doing this with all four terms, and using the antisymmetry of the T a we find
1
1
1 ab c i ′ l ′
fcab J c (z ′ )
1 a b
ψ
(z
)ψ
(z
)
(T T − T b T a )il ψ i (z ′ )ψ l (z ′ )
=
f
T
=
c
il
2
(z − z ′ )
2
(z − z ′ )
(z − z ′ )
(57)
Thus overall we get the OPE
J a (z)J b (z ′ ) =
6
1 ab
2δ
(z − z ′ )2
+
fcab J c (z ′ )
+ ...
(z − z ′ )
(58)
Z
(59)
The current algebra
Define the operators
Jna =
Z
′
dzJ a (z)z n =
C
1
2πi
dzJ a (z)z n
C
We wish to compute the commutator
b
[Jna , Jm
]
We have
b
=
Jna Jm
Z
′
C2
dz ′
Z
′
dz ′ J b (z ′ )J a (z)z ′n z m
(61)
C1
where C2 is outside C2 . In the other order we will have
Z ′
Z ′
b a
′
Jm Jn =
dz
dz ′ J b (z ′ )J a (z)z ′n z m
C2
(60)
C1
5
(62)
with C2 inside C1 . Thus in the commutator we will get
Z ′
Z ′
b
dz ′
[Jna , Jm
]=
dz ′ J b (z ′ )J a (z)z ′n z m
C
(63)
C1
where C is a circle the encircles z counterclockwise. Let us first do this z ′ integral. The leading
term in the OPE gives
Z ′
δab
δab n−1
′
nz
(64)
dz ′ 2 2 z ′n =
(z − z)
2
C
The dz integral then is
δab
n
2
′
Z
dzz n+m−1 =
C1
n ab
δ δn+m,0
2
Now let us look at the term with the single pole. The dz ′ integral gives
Z ′
J c (z) ′n
dz ′ fcab ′
z = fcab J c (z)z n
z
−
z
C
The dz integral then gives
fcab
Z
′
C1
c
dzJ c (z)z n+m = fcab Jn+m
(65)
(66)
(67)
Thus we find the algebra
1
b
c
[Jna , Jm
] = fcab Jn+m
+ δab nδn+m,0
2
This is called a current algebra of level 1. More generally we have
k
b
c
[Jna , Jm
] = fcab Jn+m
+ δab nδn+m,0
2
(68)
(69)
which is called the current algebra of level k.
Consider the limit k → ∞. Then we can ignore the fosrt term on the RHS, and we get
k ab
δ nδn+m,0
2
(70)
[αan , αbm ] = nη ab δn+m,0
(71)
b
[Jna , Jm
]≈
This is just like the algebra of free bosons
Thus we describe a flat Euclidean space of dimension r, where r is the dimension of the group.
The physics here is that of strings propagating on the group manifold
√ which corresponds to the
Lie algebra that we have taken. The string has a string length ls = α′ , and we can ask how this
compares to the curvature length scale of the group manifold. In the limit k → ∞ the string is
very small compared to the size of the group manifold, so we do not see the curvature of the group
manifold and it just looks like flat space. Thus we get the oscillator algebra noted above. In the
opposite limit k = 1, the string length is comparable to the size of the group manifold, and the
entire motion is very quantum; we cannot ignore the curvature of the target space. It is remarkable
that we can solve the motion of the string exactly in this situation. We will see later that the
central charge contributed by such a target space is
c=
kD
cv + k
6
(72)
where D = r is the dimension of the group manifold, and cv is the second quadratic Casimir
′
′
fcab fba c = −cv δaa
(73)
f312 f213 + f213 f312 = −2
(74)
cv = 2
(75)
c→D
(76)
Thus for SU (2) we will have
so that
We see that for k → ∞
which agreed with the central charge of D free bosons, representing a target space that is D flat
dimensions.
7