Labelle P. Supersymmetry DeMYSTiFied
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436
Supersymmetry Demystified
Now we use the fact that Pauli matrices are hermitian to obtain
(δχ )
†
= C(∂
μ
φ
†
)ζ
T
iσ
2
σ
μ
where the change of sign comes uniquely from the complex conjugation of
the i factor.
5.2 We have
σ
μ
¯σ
ν
∂
μ
∂
ν
φ = (σ
0
∂
t
+ σ
i
∂
i
)(σ
0
∂
t
− σ
j
∂
j
)φ
= ∂
2
t
φ −σ
i
σ
j
∂
i
∂
j
φ
Now we use the fact that σ
i
σ
j
= δ
ij
+i
ijk
σ
k
. The second term is antisym-
metric in the indices i and j, so it won’t contribute when contracted with the
symmetric expression ∂
i
∂
j
φ. We therefore end up with
∂
2
t
φ −∂
i
∂
i
φ = ∂
μ
∂
μ
φ
6.1 Using the antisymmetry of the coefficients ω
μν
, we write
φ(x
μ
+ ω
μν
x
ν
) ≈ ϕ(x) + ω
μν
x
ν
∂
μ
φ(x)
= φ(x) +
1
2
(ω
μν
− ω
νμ
)x
ν
∂
μ
φ(x)
= φ(x) +
1
2
ω
μν
x
ν
∂
μ
φ(x) −
1
2
ω
νμ
x
ν
∂
μ
φ(x)
= φ(x) +
1
2
ω
μν
(x
ν
∂
μ
− x
μ
∂
ν
)φ(x)
where in the last step we simply relabeled μ ↔ ν in the last term.
Using Eq. (6.12), we have
−
1
2
ω
μν
M
μν
,φ
=−
i
2
ω
μν
(x
ν
∂
μ
− x
μ
∂
ν
)φ (B.15)
where the minus sign on the left is due to the fact that the argument of the
exponential in our definition of U is negative. We immediately get the result
we were seeking, i.e.,
[M
μ,ν
,φ] = i(x
ν
∂
μ
− x
μ
∂
ν
)φ (B.16)
APPENDIX B Solutions to Exercises
437
6.2 Using the derivation made in the solution of Exercise 6.1, we write
φ
x
μ
+ ω
μν
x
ν
≈ ϕ +
1
2
ω
μν
(x
ν
∂
μ
− x
μ
∂
ν
)φ
which we set equal to
e
i
2
ω
μν
ˆ
M
μν
ϕ ≈ φ +
i
2
ω
μν
ˆ
M
μν
φ
which immediately leads to
ˆ
M
μν
= i(x
μ
∂
ν
− x
ν
∂
μ
)
6.3 Consider, for example,
[
ˆ
M
μν
,
ˆ
P
λ
] = i
2
[x
μ
∂
ν
− x
ν
∂
μ
,∂
λ
]
The trick to handle this is familiar from quantum mechanics: We simply
imagine that the whole expression is acting on a test function. With this in
mind, we get
ˆ
M
μν
,
ˆ
P
λ
=−x
μ
∂
ν
∂
λ
+ (∂
λ
x
μ
)∂
ν
+ x
μ
∂
ν
∂
λ
+ x
ν
∂
μ
∂
λ
− (∂
λ
x
ν
)∂
μ
− x
ν
∂
λ
∂
μ
= (∂
λ
x
μ
)∂
ν
− (∂
λ
x
ν
)∂
μ
= η
λμ
∂
ν
− η
λν
∂
μ
(B.17)
You might have expected the derivative of the coordinates to give a Kronecker
delta instead of a metric. The reason we get the metric is clear if we explicitly
write the variable with respect to which the derivative is taken. For example,
∂
λ
x
μ
=
∂x
μ
∂x
λ
If both indices are equal to zero, we get
∂x
0
∂x
0
= 1
438
Supersymmetry Demystified
because x
0
= x
0
(with our choice of metric). On the other hand, when the
indices are spatial, we have
∂x
i
∂x
i
=−1
because x
i
=−x
i
,so∂
λ
x
μ
= η
λμ
.
Coming back to Eq. (B.17), we may use ∂
ν
=−i
ˆ
P
ν
and ∂
μ
=−i
ˆ
P
μ
to
finally obtain
[
ˆ
M
μν
,
ˆ
P
λ
] = i (η
λν
ˆ
P
μ
− η
λμ
ˆ
P
ν
)
We thus have shown that the representation as differential operators obeys
the desired algebra. But the algebra is a fundamental property of the group,
which is independent of the representation used, so our derivation gives the
general result.
It is straightforward to verify the commutator of two Lorentz generators
using the same approach.
6.4 We consider for the first transformation a translation with parameter a
λ
and
for second transformation a Lorentz transformations with parameters ω
μν
:
δ
ω
δ
a
φ −δ
a
δ
ω
φ = δ
ω
(a
λ
∂
λ
φ) −
1
2
ω
μν
δ
a
(x
ν
∂
μ
φ − x
μ
∂
ν
φ)
=
1
2
a
λ
ω
μν
∂
λ
(x
ν
∂
μ
φ − x
μ
∂
ν
φ) −
1
2
a
λ
ω
μν
(x
ν
∂
μ
∂
λ
φ − x
μ
∂
ν
∂
λ
φ)
=
1
2
a
λ
ω
μν
(η
νλ
∂
μ
φ −η
λμ
∂
ν
φ)
=
i
2
a
λ
ω
μν
(η
νλ
[P
μ
,φ] −η
λμ
[P
ν
,φ])
From Eq. (6.40), this is equal to
a
λ
P
λ
, −
1
2
ω
μν
M
μν
,φ
where we used exp(ia
λ
P
λ
) for the operator generating translations and
exp(−iω
μν
M
μν
/2) for the operator generating Lorentz transformations. We
APPENDIX B Solutions to Exercises
439
thus find
iη
νλ
[P
μ
,φ] −i η
λμ
[P
ν
,φ] =
[P
λ
, −M
μν
],φ
or
iη
νλ
P
μ
−i η
λμ
P
ν
= [M
μν
, P
λ
]
as was to be shown.
6.5 a. Let’s work out the right-hand side of Eq. (6.82). Consider the first term:
δ
ω
δ
ζ
φ = δ
ω
(−ζ
T
iσ
2
χ) (B.18)
From Eq. (6.81), we have
δ
ω
δ
ζ
φ =−ζ
T
iσ
2
δ
ω
χ
=−
i
2
ω
μν
iζ
T
σ
2
σ
μν
χ + x
ν
ζ
T
σ
2
∂
μ
χ − x
μ
ζ
T
σ
2
∂
ν
χ
(B.19)
On the other hand,
δ
ζ
δ
ω
φ ≈ δ
ζ
1
2
ω
μν
x
ν
∂
μ
φ − x
μ
∂
ν
φ
=−
i
2
ω
μν
x
ν
ζ
T
σ
2
∂
μ
χ − x
μ
ζ
T
σ
2
∂
ν
χ
(B.20)
We see the reason the order of the two transformations matters is entirely
due to the spin generators in the Lorentz transformations, which appear
in Eq. (B.19) but not in Eq. (B.20) because in the first case the Lorentz
transformation acts on a spinor, whereas in the second case it acts on a
scalar field.
Substituting Eqs. (B.19) and (B.20) into Eq. (6.82), we get
−
1
2
[ζ · Q,ω
μν
M
μν
],φ
=
1
2
ω
μν
ζ
T
σ
2
σ
μν
χ (B.21)
Using Eq. (6.85), we have
−
1
2
[ζ · Q,ω
μν
M
μν
],φ
=
i
2
ω
μν
ζ
T
σ
2
σ
μν
[Q,φ] (B.22)
440
Supersymmetry Demystified
from which we conclude
[ζ · Q, M
μν
] =−i ζ
T
σ
2
σ
μν
Q (B.23)
Let’s write the indices explicitly. As usual, we assign lower indices to both
ζ and Q and make sure that repeated indices appear once in an upper
position and once in a lower position, which uniquely fixes the positions
of all the indices. This gives us
ζ
a
(−iσ
2
)
ab
Q
b
, M
μν
=−iζ
a
(σ
2
)
ab
(σ
μν
)
c
b
Q
c
(B.24)
Note that the matrices σ
μν
are antisymmetric, so we need to keep track
of which index comes first, which is why the c is shifted a bit, to clearly
indicate that it is the second index. Since ζ
a
is arbitrary, we finally obtain
[Q
b
, M
μν
] = (σ
μν
)
c
b
Q
c
as was to be shown.
b. Let’s first evaluate
δ
ω
δ
ζ
φ
†
= δ
ω
( ¯χ ·
¯
ζ)
= (δ
ω
¯χ
˙a
)
¯
ζ
˙a
(B.25)
We need the Lorentz transformation of ¯χ, which we can obtain simply by
taking the hermitian conjugate of Eq. (6.81):
δ
ω
¯χ =
1
2
ω
μν
−i ¯χ ¯σ
μν
+ x
ν
∂
μ
¯χ − x
μ
∂
ν
¯χ
(B.26)
where we have used (σ
μν
)
†
= ¯σ
μν
, which can be confirmed easily using
the definitions in Eqs. (6.78) and (6.79).
Now let’s write Eq. (B.26) with the indices explicitly shown. In Eq.
(6.81), the spinors have lower undotted indices. This means that in Eq.
(B.26), the spinors also have lower indices, except that they are now dotted
because we have taken the hermitian conjugate. Therefore, the transforma-
tion is
δ
ω
¯χ
˙a
=
1
2
ω
μν
−i ¯χ
˙
b
( ¯σ
μν
)
˙
b
˙a
+ x
ν
∂
μ
¯χ
˙a
− x
μ
∂
ν
¯χ
˙a
(B.27)
APPENDIX B Solutions to Exercises
441
The indices of ¯σ
μν
have been chosen to match the rest of the expression,
but we can check that they are indeed correct by using Eqs. (3.33) and
(3.34) in the definition of ¯σ
μν
[Eq. (6.79)]. Indeed, the right-hand side of
Eq. (6.79) is
i
4
( ¯σ
μ
)
˙ab
(σ
ν
)
b˙c
− ( ¯σ
ν
)
˙ab
(σ
μ
)
b˙c
which carries indeed indices ()
˙a
˙c
.
Substituting Eq. (B.27) into Eq. (B.25), we finally obtain
δ
ω
δ
ζ
φ
†
=
1
2
ω
μν
−i ¯χ
˙
b
( ¯σ
μν
)
˙
b
˙a
¯
ζ
˙a
+ x
ν
(∂
μ
¯χ
˙a
)
¯
ζ
˙a
− x
μ
(∂
ν
¯χ
˙a
)
¯
ζ
˙a
=
1
2
ω
μν
−i ¯χ(¯σ
μν
¯
ζ)+ x
ν
(∂
μ
¯χ)·
¯
ζ − x
μ
(∂
ν
¯χ)·
¯
ζ
(B.28)
where we are using the shorthand notation discussed in Chapter 3 to avoid
having to show the indices explicitly.
On the other hand, we find [using Eq. (6.23)]
δ
ζ
δ
ω
φ
†
=
1
2
ω
μν
δ
ζ
x
ν
∂
μ
φ
†
− x
μ
∂
ν
φ
†
=
1
2
ω
μν
x
ν
(∂
μ
¯χ)·
¯
ζ − x
μ
(∂
ν
¯χ)·
¯
ζ
so that, finally, Eq. (6.87) gives
−
1
2
¯
Q
˙a
¯
ζ
˙a
,ω
μν
M
μν
,φ
†
=−
i
2
ω
μν
¯χ
˙
b
( ¯σ
μν
)
˙
b
˙a
¯
ζ
˙a
(B.29)
which can be simplified to
¯
Q
˙a
, M
μν
,φ
†
= i ¯χ
˙
b
( ¯σ
μν
)
˙
b
˙a
(B.30)
The only missing ingredient is the commutator of
¯
Q
˙a
with φ
†
. Instead of
Eq. (6.84), we now have
¯
Q ·
¯
ζ,φ
†
=−i ¯χ ·
¯
ζ (B.31)
442
Supersymmetry Demystified
which implies
¯
Q
˙
b
,φ
†
=−i ¯χ
˙
b
(B.32)
Using this to replace ¯χ
˙
b
on the right-hand side of Eq. (B.30), we obtain our
final result, i.e.,
¯
Q
˙a
, M
μν
=−
¯
Q
˙
b
( ¯σ
μν
)
˙
b
˙a
which confirms the first relation in Eq. (6.86). To obtain the second relation,
we express the supercharges in terms of their components with upper dotted
indices:
˙a ˙c
¯
Q
˙c
, M
μν
=−
˙
b
˙
d
¯
Q
˙
d
( ¯σ
μν
)
˙
b
˙a
To get rid of the on the left side, let’s contract both sides with
˙e ˙a
.Onthe
left, we may use Eq. (A.25), namely
˙e ˙a
˙a ˙c
= δ
˙e
˙c
, so the equation becomes
¯
Q
˙e
, M
μν
=−
˙e ˙a
˙
b
˙
d
¯
Q
˙
d
( ¯σ
μν
)
˙
b
˙a
=
˙e ˙a
˙
d
˙
b
¯
Q
˙
d
( ¯σ
μν
)
˙
b
˙a
where we have used the antisymmetry of the symbol. Using Eq. (6.88)
(which we will prove in a second), we finally obtain the second commutator
of Eq. (6.86), after relabeling some of the indices.
That leaves us the proof of Eq. (6.88). Let’s do it! Using Eq. (6.79) with
all the indices raised, we have
˙c
˙
b
˙
d ˙a
( ¯σ
μν
)
˙a
˙
b
=
i
4
˙c
˙
b
˙
d ˙a
( ¯σ
μ
)
˙ae
(σ
ν
)
e
˙
b
− μ ↔ ν
=
i
4
˙c
˙
b
˙
d ˙a
δ
e
f
( ¯σ
μ
)
˙af
(σ
ν
)
e
˙
b
− μ ↔ ν
=
i
4
˙c
˙
b
˙
d ˙a
fg
ge
( ¯σ
μ
)
˙af
(σ
ν
)
e
˙
b
− μ ↔ ν [using Eq. (A.24)]
=−
i
4
(σ
μ
)
g
˙
d
( ¯σ
ν
)
˙cg
− μ ↔ ν [using Eqs. (A.26) and (A.27)]
= ( ¯σ
μν
)
˙c
˙
d
(B.33)
as was to be shown.
APPENDIX B Solutions to Exercises
443
6.6 When the spinor dot products are written fully, Eq. (6.92) is
α
a
[β
T
(−iσ
2
)γ ] + β
a
[γ
T
(−iσ
2
)α] +γ
a
[α
T
(−iσ
2
)β] = 0
with
β
T
(−iσ
2
)γ = β
2
γ
1
− β
1
γ
2
and so on. The identity then is
α
a
β
2
γ
1
− α
a
β
1
γ
2
+ β
a
γ
2
α
1
− β
a
γ
1
α
2
+ γ
a
α
2
β
1
− γ
a
α
1
β
2
= 0
If we set a = 1, the left side is
α
1
β
2
γ
1
− α
1
β
1
γ
2
+ β
1
γ
2
α
1
− β
1
γ
1
α
2
+ γ
1
α
2
β
1
− γ
1
α
1
β
2
To see the cancellation explicitly, let’s move all the components in the order
αβγ . Including a minus sign every time we move a component passed another
one, we get
α
1
β
2
γ
1
− α
1
β
1
γ
2
+ α
1
β
1
γ
2
− α
2
β
1
γ
1
+ α
2
β
1
γ
1
− α
1
β
2
γ
1
which is clearly zero because all terms cancel pairwise. The proof with a = 2
is similar.
6.7 In the case of the scalar field, we want to show that the new term in the
transformation of the spinor field does not ruin Eq. (6.61), which was of the
desired form. We have
δ
ζ
φ = ζ · χ
so
δ
β
δ
ζ
φ = δ
β
(ζ · χ)
= previous expression + Fζ · β (B.34)
On the other hand,
δ
ζ
δ
β
φ = previous expression + Fβ · ζ (B.35)
444
Supersymmetry Demystified
so
δ
β
δ
ζ
φ −δ
ζ
δ
β
φ = previous expression + Fζ · β − Fβ · ζ
=−i
ζ
†
¯σ
μ
β − β
†
¯σ
μ
ζ
∂
μ
φ + Fζ · β − Fβ · ζ (B.36)
using our result (6.61). But clearly, the last two terms cancel out because
ζ · β = β · ζ ! So our result of the commutator of two SUSY transformations
on the scalar field is, thankfully, unaffected.
Consider now the transformation of the spinor field. We have
δ
β
δ
ζ
χ
a
= δ
β
(−iσ
μ
iσ
2
ζ
∗
)
a
∂
μ
φ + Fζ
a
= previous expression − i(β
†
¯σ
μ
∂
μ
χ)ζ
a
(B.37)
whereas
δ
ζ
δ
β
χ
a
= previous expression − i(ζ
†
¯σ
μ
∂
μ
χ)β
a
(B.38)
Therefore,
δ
β
δ
ζ
χ
a
− δ
ζ
δ
β
χ
a
= previous expression − i(β
†
¯σ
μ
∂
μ
χ)ζ
a
+i (ζ
†
¯σ
μ
∂
μ
χ)β
a
(B.39)
with the previous expression being given in Eq. (6.93). Substituting Eq. (6.93)
into Eq. (B.39), we get that the two offending terms in Eq. (6.93) that did not
vanish off-shell are exactly canceled by the two extra terms in Eq. (B.39),
leaving us with the result we were wishing for:
δ
β
δ
ζ
χ
a
− δ
ζ
δ
β
χ
a
=−i
ζ
†
¯σ
μ
β − β
†
¯σ
μ
ζ
∂
μ
χ
a
It remains to check the auxiliary field. This is done easily:
δ
β
δ
ζ
F = δ
β
(−iζ
†
¯σ
μ
∂
μ
χ)
=−iζ
†
¯σ
μ
σ
ν
σ
2
β
∗
∂
μ
∂
ν
φ −i(ζ
†
¯σ
μ
β)∂
μ
F (B.40)
The first term on the right-hand side can be simplified using Eq. (5.13):
−iζ
†
¯σ
μ
σ
ν
σ
2
β
∗
∂
μ
∂
ν
φ =−i ζ
†
σ
2
β
∗
φ
APPENDIX B Solutions to Exercises
445
which is simply
−
¯
ζ ·
¯
β
φ
by definition. Using this in Eq. (B.40), we have
δ
β
δ
ζ
F =−
¯
ζ ·
¯
βφ − i(ζ
†
¯σ
μ
β)∂
μ
F (B.41)
From this we readily obtain
δ
ζ
δ
β
F =−
¯
β ·
¯
ζ φ − i(β
†
¯σ
μ
ζ)∂
μ
F
so that
δ
β
δ
ζ
F −δ
ζ
δ
β
F =−
¯
ζ ·
¯
βφ +
¯
β ·
¯
ζ φ − i(ζ
†
¯σ
μ
β − β
†
¯σ
μ
ζ)∂
μ
F
=−i(ζ
†
¯σ
μ
β − β
†
¯σ
μ
ζ)∂
μ
F (B.42)
where we have used the fact that
¯
ζ ·
¯
β =
¯
β ·
¯
ζ to cancel the two terms con-
taining the scalar field. We finally see that, indeed, the auxiliary field also
obeys Eq. (6.100).
7.1 We write
P
μ
W
μ
= W
μ
P
μ
+ [P
μ
, W
μ
]
The first term on the right-hand side is zero, from Eq. (7.3). The commutator
is equal to
[P
μ
, W
μ
] =
1
2
μνρσ
[P
μ
, M
ρσ
]P
ν
=−
i
2
μνρσ
(η
μσ
P
ρ
− η
μρ
P
σ
) (B.43)
using Eq. (6.34). This is equal to zero because the flat metric is symmetric,
whereas the Levi-Civita symbol is totally antisymmetric.
7.2 We have to simplify
1
2
ijk
M
jk
=
1
2
ijk
(σ
jk
+ x
j
p
k
− x
k
p
j
)