Labelle P. Supersymmetry DeMYSTiFied
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426
Supersymmetry Demystified
Some identities involving σ
μ
and ¯σ
μ
:
¯χ ¯σ
μ
λ =−λσ
μ
¯χ (A.49)
χσ
μ
¯
λ =−
¯
λ ¯σ
μ
χ (A.50)
( ¯χ ¯σ
μ
λ)
†
=
¯
λ ¯σ
μ
χ (A.51)
(χσ
μ
¯
λ)
†
= λσ
μ
¯χ (A.52)
Identities Involving Three Left-Chiral Spinors
α
a
(β · γ)+ β
a
(γ · α) + γ
a
(α · β) = 0 (A.53)
Identities Involving Four Left-Chiral Spinors
Particularly useful are Weyl spinor Fierz identities:
χ ·αχ · β =−
1
2
χ ·χα · β (A.54)
¯χ · ¯α ¯χ ·
¯
β =−
1
2
¯χ · ¯χ ¯α ·
¯
β (A.55)
(λσ
μ
¯χ)(λσ
ν
¯χ) =
1
2
η
μν
λ · λ ¯χ · ¯χ (A.56)
( ¯χ ¯σ
μ
λ)( ¯χ ¯σ
ν
λ) =
1
2
η
μν
λ · λ ¯χ · ¯χ (A.57)
APPENDIX B
Solutions to Exercises
2.1 The case μ = ν = 0 is trivially true because σ
0
= ¯σ
0
= 1 (the 2 ×2 unit
matrix) and η
00
= 1.
Consider now μ = 0 and ν = i. We use ¯σ
μ=i
=−σ
i
. The left-hand side
then is
−σ
i
+ σ
i
= 0
On the other hand, the right-hand side also vanishes because η
μν
is diagonal,
so η
0i
= η
i0
= 0.
Obviously, if ν = 0 and μ = i , both sides of the equation are also equal to
zero.
The case of two spatial indices is more interesting. Setting μ = i and ν = j
(where i and j may or may not be equal), the left-hand side of Eq. (2.25) is
σ
i
¯σ
j
+ σ
j
¯σ
i
=−σ
i
σ
j
− σ
j
σ
i
428
Supersymmetry Demystified
where we have again used the definitions of σ
μ
and ¯σ
μ
. Using now (2.9), we
find
−σ
i
σ
j
− σ
j
σ
i
=−{σ
i
,σ
j
}
=−2δ
ij
1 (B.1)
As for the right-hand side of Eq. (2.25), it is equal to η
ij
times the unit matrix.
This is zero when i = j and −1 when i = j, which is nothing other than
−δ
ij
! Therefore, the right-hand side is
2η
ij
1 =−2δ
ij
1
which again agrees with the left-hand side. We thus have shown the validity
of Eq. (2.24) for all possible values of the indices.
2.2 We have
(η
R
)
†
χ
L
=
1 +
i
2
·σ −
1
2
β ·σ
η
R
†
1 +
i
2
·σ +
1
2
β ·σ
χ
L
= η
†
R
1 +
i
2
·σ −
1
2
β ·σ
†
1 +
i
2
·σ +
1
2
β ·σ
χ
L
= η
†
R
1 −
i
2
·σ −
1
2
β ·σ
1 +
i
2
·σ +
1
2
β ·σ
χ
L
= η
†
R
1 −
i
2
·σ −
1
2
β ·σ +
i
2
·σ +
1
2
β ·σ
χ
L
= η
†
R
χ
L
(B.2)
where in the third step we used the fact that the Pauli matrices are hermitian,
and in the fourth step we have only kept the terms linear in and
β. The proof
that χ
†
L
η
R
is also invariant is similar.
2.3 Let us first consider the transformation of η
†
σ
0
η, which is simply η
†
η because
σ
0
is the identity matrix (we will not show explicitly the label R on the right-
chiral spinor η in this solution). We get (following steps that are similar to the
previous exercise)
(η
)
†
η
= η
†
1 −
i
2
·σ −
1
2
β ·σ +
i
2
·σ −
1
2
β ·σ
η
= η
†
η − η
†
β ·ση
= η
†
η −
β · η
†
ση
= η
†
σ
0
η −
β · η
†
ση (B.3)
APPENDIX B Solutions to Exercises
429
This is of the same form as the transformation of the zeroth component of a
four-vector [see the transformation of the energy E in Eq. (2.33)] with η
†
ση
playing the role of the three-vector part.
Now consider the transformation of η
†
σ
j
η [we used an index j instead of i
only because it will make it less confusing to use Eq. (2.14)]. We get, to first
order in
β and ,
(η
)
†
σ
j
η
= η
†
1 −
i
2
·σ −
1
2
β ·σ
σ
j
1 +
i
2
·σ −
1
2
β ·σ
η
= η
†
σ
j
η + η
†
−
i
2
·σσ
j
−
1
2
β ·σσ
j
+
i
2
σ
j
·σ −
1
2
σ
j
β ·σ
η
(B.4)
Consider the sum of the second and fourth terms on the right-hand side.
Using Eq. (2.14), we obtain
η
†
−
i
2
·σσ
j
+
i
2
σ
j
·σ
η =−
× (η
†
ση)
j
For the third and fifth terms on the right-hand side of Eq. (B.4), we use instead
Eq. (2.9):
η
†
−
1
2
β ·σσ
j
−
1
2
σ
j
β ·σ
η = η
†
−
1
2
β
i
σ
i
σ
j
+ σ
j
σ
i
η
=−β
i
δ
ij
η
†
η
=−β
j
η
†
σ
0
η (B.5)
We therefore get for Eq. (B.4)
(η
)
†
ση
= η
†
ση− × (η
†
ση)−
βη
†
σ
0
η
This is of the same form as the transformation of the three-vector part of a
four-vector [see the transformation of p in Eq. (2.33)] with, again, η
†
σ
0
η
playing the role of the zeroth component and η
†
ση playing the role of the
three-vector.
We have therefore proved that η
†
σ
μ
η indeed transforms as a four-vector.
The proof that the quantity χ
†
¯σ
μ
χ is also a four-vector is similar.
430
Supersymmetry Demystified
2.4 Consider taking the hermitian conjugate of iσ
2
χ
†T
L
and contracting this with
−iσ
2
η
†T
R
.Weget
iσ
2
χ
†T
L
†
−iσ
2
η
†T
R
=−
σ
2
χ
†T
L
†
σ
2
η
†T
R
=−χ
T
L
(σ
2
)
†
σ
2
η
†T
R
=−χ
T
L
σ
2
σ
2
η
†
R
=−χ
T
L
η
†T
R
= η
†
R
χ
L
(B.6)
where we have used the fact that σ
2
is hermitian and that σ
2
σ
2
= 1. This
invariant is already listed in Eq. (2.42).
Is is now obvious that
−iσ
2
η
†T
R
†
iσ
2
χ
†T
L
=−η
T
R
χ
†T
L
= χ
†
L
η
R
which is also on the list of Eq. (2.42).
2.5 For a the vector quantity, we may simply take Eq. (2.45), which is a Lorentz
invariant, and drop the derivative. The result,
χ
†
L
¯σ
μ
iχ
L
is a contravariant vector.
To get a second rank tensor, we need a second sigma matrix. Let’s start
with
¯σ
μ
χ
L
From Eqs. (2.43) and (2.44), we see that if the index μ is contracted with
something, this behaves like a right-chiral spinor. Equation (2.43) tells us that
we must multiply a right-chiral spinor by the matrix σ
ν
to get something with
well-defined Lorentz properties, so let’s consider
σ
ν
¯σ
μ
χ
L
(B.7)
Imagining again for an instant that the Lorentz indices are contracted, we
know from Eq. (2.43) that this quantity transforms like a left-chiral spinor.
The first term in Eq. (2.42) shows that we must contract a left-chiral spinor
APPENDIX B Solutions to Exercises
431
with −χ
T
L
iσ
2
to get a Lorentz scalar. We therefore have that
−χ
T
L
iσ
2
σ
ν
¯σ
μ
χ
L
transforms like a second-rank contravariant tensor.
2.6 For example, we have
λ · χ = λ
T
(−iσ
2
)χ
=−χ
T
(−iσ
2
)
T
λ using Eq. (2.48)
= χ
T
(−iσ
2
)λ using the antisymmetry of −iσ
2
= χ ·λ (B.8)
What happens therefore is that we pick up two minus signs: one from the
anticommutativity of the spinors and one from the antisymmetry of the matrix
iσ
2
. Likewise, one can show that
¯
λ · ¯χ = ¯χ ·
¯
λ.
2.7 We will only prove the first one because the two proofs are almost identical.
In terms of explicit components, we have
θ ·χ = θ
T
(−iσ
2
)χ = θ
2
χ
1
− θ
1
χ
2
and a similar expression for θ ·λ. Therefore,
θ ·χθ · λ = (θ
2
χ
1
− θ
1
χ
2
)(θ
2
λ
1
− θ
1
λ
2
)
=−θ
2
χ
1
θ
1
λ
2
− θ
1
χ
2
θ
2
λ
1
(B.9)
(all the other terms are identically zero because all the parameters here are
Grassmann quantities).
On the other hand, we have
−
1
2
θ ·θχ · λ =−
1
2
(θ
2
θ
1
− θ
1
θ
2
)(χ
2
λ
1
− χ
1
λ
2
)
=−θ
2
θ
1
(χ
2
λ
1
− χ
1
λ
2
)
=−θ
1
χ
2
θ
2
λ
1
− θ
2
χ
1
θ
1
λ
2
432
Supersymmetry Demystified
This is identical to Eq. (B.9), which completes the proof. An alternative deriva-
tion using Eq. (2.54) is
θ ·χθ·λ = θ
a
(−iσ
2
)
ab
χ
b
θ
c
(−iσ
2
)
cd
λ
d
=−(−iσ
2
)
ab
(−iσ
2
)
cd
θ
a
θ
c
χ
b
λ
d
=−
1
4
(−iσ
2
)
ab
(−iσ
2
)
cd
(iσ
2
)
ac
(iσ
2
)
bd
θ ·θχ·λ
=−
1
4
(−iσ
2
)
ab
(iσ
2
)
bd
(−iσ
2
)
dc
(iσ
2
)
ca
θ ·θχ·λ
=−
1
4
Tr(σ
2
σ
2
σ
2
σ
2
)θ· θχ· λ
=−
1
4
Tr(1)θ· θχ· λ
=−
1
2
θ ·θχ·λ
3.1 By definition,
η · χ = η
T
(−iσ
2
)χ
where on the right-hand side it is implicit that the two spinors have lower
undotted components. To be more explicit, we have
η · χ =−η
1
χ
2
+ η
2
χ
1
(B.10)
On the other hand,
χ ·η =−χ
1
η
2
+ χ
2
η
1
(B.11)
The expressions (B.10) and (B.11) are equal because −η
1
χ
2
= χ
2
η
1
and
η
2
χ
1
=−χ
1
η
2
.
3.2 We must express ¯η
˙a
in terms of components with upper dotted indices because
this is the type of index that is implicitly assigned to all the right-chiral spinors
in Eq. (2.42). Using Eq. (3.17), we get
¯η
˙a
¯η
˙a
= (−iσ
2
)
˙a
˙
b
¯η
˙
b
¯η
˙a
= ¯η
˙
b
(−iσ
2
)
˙a
˙
b
¯η
˙a
= ¯η
˙
b
(iσ
2
)
˙
b ˙a
¯η
˙a
which corresponds to the expression η
T
iσ
2
η of Eq. (2.42).
APPENDIX B Solutions to Exercises
433
3.3 We have
χ ·χ = χ
2
χ
1
− χ
1
χ
2
= 2χ
2
χ
1
=−2χ
1
χ
2
(B.12)
Using Eq. (3.16), we can rewrite this as
χ ·χ =−2 χ
1
χ
2
= 2 χ
2
χ
1
(B.13)
We could, of course, have instead used
χ ·χ = χ
a
χ
a
= χ
1
χ
1
+ χ
2
χ
2
and then Eq. (3.16) to recover Eq. (B.13). Either way, the result is
χ
2
χ
1
=
1
2
χ ·χχ
1
χ
2
=−
1
2
χ ·χ
And, of course, χ
1
χ
1
= χ
2
χ
2
= 0. On the other hand,
(iσ
2
)
12
= 1 (iσ
2
)
21
=−1
whereas the diagonal elements are zero. We therefore find
χ
a
χ
b
=−
1
2
(iσ
2
)
ab
χ ·χ
as was to be shown.
On the other hand, using Eq. (3.13), we have
¯χ · ¯χ = ¯χ
˙
1
¯χ
˙
1
+ ¯χ
˙
2
¯χ
˙
2
=−¯χ
˙
2
¯χ
˙
1
+ ¯χ
˙
1
¯χ
˙
2
so that
¯χ
˙
2
¯χ
˙
1
=−
1
2
¯χ · ¯χ ¯χ
˙
1
¯χ
˙
2
=
1
2
¯χ · ¯χ
434
Supersymmetry Demystified
which gives us
¯χ
˙a
¯χ
˙
b
=
1
2
(iσ
2
)
ab
¯χ · ¯χ
3.4 This is actually a trick question. One can tell that this expression is zero without
doing any work. Indeed, χ ·χ contains one factor of χ
1
and one factor of χ
2
,
whereas χ ·η has two terms that contain either χ
1
or χ
2
. Therefore, all terms
contain the square of either χ
1
or χ
2
, and the whole expression is identically
zero.
3.5 Following the hint, we take the trace of both sides of Eq. (2.24). This gives
Tr(σ
μ
¯σ
ν
) + Tr(σ
ν
¯σ
μ
) = 2 η
μν
Tr(1)
From the properties of the Pauli matrices, each trace on the left is nonzero if
and only if μ = ν. Therefore the two traces are equal to one another and we
have
2Tr(σ
μ
¯σ
ν
) = 4η
μν
which, after dividing by 2, completes the proof.
4.1 We have
χ
†
¯σ
μ
i∂
μ
χ
†
= (i∂
μ
χ)
†
( ¯σ
μ
)
†
χ
Note that we did not pick up a minus sign when we switched the order of the
spinors, according to the rule (2.56). Applying the hermitian conjugate to the
parentheses and using the fact that the Pauli matrices are hermitian, we get
(i∂
μ
χ)
†
( ¯σ
μ
)
†
χ = (−i∂
μ
χ
†
) ¯σ
μ
χ
= χ
†
¯σ
μ
i∂
μ
χ
using an integration by parts in the last step. This is the original kinetic term,
as was to be shown.
4.2 We have
−m
χ
¯
p
· χ
p
+ ¯χ
p
· ¯χ
¯
p
†
=−m
χ
¯
p
· χ
p
)
†
− m
¯χ
p
· ¯χ
¯
p
†
=−m ¯χ
¯
p
· ¯χ
p
− mχ
p
· χ
¯
p
(B.14)
APPENDIX B Solutions to Exercises
435
using in the last step Eq. (A.44). This is indeed equal to Eq. (4.11) because
the order in spinor dot products does not matter.
4.3 The second term of Eq. (4.13) agrees with the second term of Eq. (4.12). Let’s
focus on the first term. First, use Eq. (A.5) to get rid of the σ
2
matrices:
iχ
T
¯
e
σ
2
σ
μ
∂
μ
σ
2
χ
†T
¯
e
= iχ
T
¯
e
¯σ
μT
∂
μ
χ
†T
¯
e
Next, we use Eq. (A.36) to get rid of all the annoying transpose operations:
iχ
T
¯
e
¯σ
μT
∂
μ
χ
†T
¯
e
=−i
∂
μ
χ
†
¯
e
¯σ
μ
χ
¯
e
Finally, we do an integration by parts to get the derivative to act on the spinor
on the right:
−i
∂
μ
χ
†
¯
e
¯σ
μ
χ
¯
e
= iχ
†
¯
e
¯σ
μ
∂
μ
χ
¯
e
which is the first term of Eq. (4.13).
4.4 In Eq. (4.14), an integration by parts has been performed on one of the deriva-
tives, which introduced a minus sign. We must replace ∂
μ
by ∂
μ
−ieA
μ
before
any integration by parts on ∂
μ
has been made. After an integration by parts,
we have to make the replacement ∂
μ
→ ∂
μ
+ieA
μ
, but this gets confusing
because we must keep track of which partial derivative has been integrated
by parts!
5.1 For the hermitian conjugate of the scalar field transformation, the proof is
trivial if we use Eq. (A.44) and the fact that the order does not matter in spinor
dot products:
(δφ)
†
= (ζ · χ)
†
=
¯
ζ · ¯χ = ¯χ ·
¯
ζ = χ
†
(iσ
2
)ζ
∗
where the last equality follows from the definition of the dot product between
barred spinors.
Now consider
(δχ )
†
=
−C
∗
(∂
μ
φ) σ
μ
iσ
2
ζ
∗
†
Using the fact that (AB)
†
= B
†
A
†
,weget
(δχ )
†
=−C(∂
μ
φ
†
)ζ
T
(iσ
2
)
†
(σ
μ
)
†