Labelle P. Supersymmetry DeMYSTiFied
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Supersymmetry Demystified
be deceiving. Indeed, using Eq. (A.50) followed by an integration by parts, we have
λσ
μ
∂
μ
¯
λ =−(∂
μ
¯
λ) ¯σ
μ
λ
=
¯
λ ¯σ
μ
∂
μ
λ (13.47)
which is of the same form as the the spinor kinetic term appearing in Eq. (10.8).
Thus we may write
F
a
F
a
F
= 4 i
¯
λ(x) ¯σ
μ
∂
μ
λ(x) + 2 D
2
(x) − F
μν
(x) F
μν
(x)
−
i
2
μναβ
F
μν
(x) F
αβ
(x) (13.48)
We see that the first three terms are exactly four times the free abelian gauge
theory shown in Eq. (10.8) (where we used the notation λ
†
instead of
¯
λ). However,
we have an extra term, which is usually written as
iF
μν
F
∗μν
where
F
∗μν
≡
1
2
μναβ
F
αβ
is called the dual field strength. This term is actually a total derivative and is not
considered unless the gauge fields have a nontrivial topology, which is why we did
not include it in Chapter 10. We will ignore it from now on.
Our final expression therefore is
1
4
(F
a
F
a
)
F
=
1
2
D
2
+i
¯
λ ¯σ
μ
∂
μ
λ −
1
4
F
μν
F
μν
(13.49)
which, apart from the Fayet-Iliopoulos term, reproduces exactly our supersymmet-
ric lagrangian [Eq. (10.8)]! To include the Fayet-Iliopoulos term, we simply have
to add the D term of −2ξV (see the discussion after Eq. (13.12)). Of course, we
could simply write F
a
F
a
as F · F, using our notation for the spin dot products.
The obvious next step would be to apply a SUSY transformation to F
a
using the
supercharges in order to confirm the transformations of the component fields we
established in Chapter 10, but we won’t pursue this rather lengthy calculation here.
Rest assured that it does work!
CHAPTER 13 Gauge Field Theories
317
Our final SUSY and gauge-invariant lagrangian that couples a set of i left-chiral
superfields to a U(1) gauge field is the sum of Eqs. (13.11) and (13.49):
L =
W(
i
)
F
+ h.c.
+
i
†
i
e
2q
i
V
i
D
+
1
4
F
a
F
a
F
(13.50)
As noted at the end of Section 10.6, the superpotential is absent if there is only one
left-chiral superfield that is charged under the gauge group because it is impossible
to write down a gauge-invariant holomorphic function of .
13.6 Supersymmetric QED
We will now apply the results of this chapter to build a supersymmetric version of
quantum electrodynamics (QED). This will be useful as a warm-up exercise before
we construct the supersymmetric extension of the standard model, which we will
pursue in Chapter 15.
The first step is to determine what superfields we need. QED involves a photon
field and the electron field, which comes in right-chiral and left-chiral states, to-
gether with the corresponding positron states. Clearly, we will introduce an abelian
gauge superfield V that will contain the photon, its spin 1/2 left-chiral partner, the
photino λ, and a corresponding auxiliary field D. In the Wess-Zumino gauge, the
vector superfield is just what we wrote in Eq. (13.10). Of course, we also will use
the field-strength superfield F given in Eq. (13.40).
The more tricky issue is the electron. There are four physical degrees of freedom
associated with the electron field: the left-chiral and right-chiral electron states and
the left-chiral and right-chiral positron states. However, it is conventional to build
supersymmetric theories out of left-chiral superfields only. This can be done using
a trick that we will describe below.
But first, let’s introduce a left-chiral superfield E
e
containing the left-chiral elec-
tron state:
E
e
=
˜
φ
e
+ θ ·χ
e
+
1
2
θ ·θ F
e
It’s important to emphasize that E
e
is a scalar under Lorentz transformations. As
we have seen before, it is called a left-chiral superfield only because one of its
components is the left-chiral Weyl spinor χ
e
.
Here we have started using the notation that we will follow in Chapter 15
when we cover the minimal supersymmetric standard model (MSSM). First, we
will denote all superfields by calligraphic capital letters. Also, it will always be
318
Supersymmetry Demystified
understood that a spinor χ represents a left-chiral Weyl spinor, so we won’t include
a label L to avoid cluttering the notation. Finally, we will include a tilde (˜) over
the fields that are superpartners of the known particles (the Higgs field will require
a separate discussion).
To respect CPT invariance, our lagrangian must involve both E
e
and E
†
e
,so
the physical spectrum contains both the left-chiral electron state and its hermitian
conjugate, the right-chiral positron state. But there are two states still missing: the
right-chiral electron state and its hermitian conjugate, the left-chiral positron state.
To account for those, we have to include a second chiral superfield. We could make
it a right-chiral superfield containing for spinor component the right-chiral electron
state, but, as mentioned earlier, the convention is to express everything in terms of
left-chiral spinors.
The trick is to introduce a left-chiral superfield whose spinor component is the
left-chiral positron state! As we saw in Chapter 4, one always may trade off a right-
chiral particle Weyl spinor for the corresponding left-chiral antiparticle spinor
using [see Eq. (4.5)]
η
e
= i σ
2
χ
†T
¯
e
where we used the notation of Chapter 2: η is a right-chiral spinor, χ is a left-chiral
spinor, and
¯
e represents the positron. We use
¯
e to represent the positron instead
of e
+
because we want a notation that can be extended to denote all antiparticles,
including those with no electric charge. Obviously, this bar has nothing to do with
the bar notation introduced for dot products between Weyl spinors or the bar used
to denote Dirac conjugation!
We will write the left-chiral superfield containing the left-chiral positron state as
E
¯
e
=
˜
φ
¯
e
+ θ ·χ
¯
e
+
1
2
θ ·θ F
¯
e
We take, of course, E
e
and E
¯
e
to have U (1) charges q =−e and q = e, respectively.
In other words, under a gauge transformation, we have
E
e
→ e
−2ie
E
e
E
¯
e
→ e
2ie
E
¯
e
(13.51)
We are now ready to build our supersymmetric extension of QED. The gauge-
and Lorentz-invariant quantities that we have at our disposal are
E
e
E
¯
e
F
a
F
a
E
†
e
e
−2eV
E
e
E
†
¯
e
e
2eV
E
¯
e
(13.52)
as well as products of any combination of these terms and their hermitian conjugates
(more about this on next page).
CHAPTER 13 Gauge Field Theories
319
The first two combinations of Eq. (13.52) are left-chiral superfields because
E
e
, E
¯
e
, and F are all left-chiral superfields (recall that a product of left-chiral su-
perfields is also a left-chiral superfield). To obtain a supersymmetric lagrangian,
we therefore must take the F term of these expressions. The last two are not chiral
superfields, so we must take their D term to obtain supersymmetric lagrangians.
Now, let’s look at the dimensions. The field-strength superfield F
a
has dimension
3/2, whereas V is dimensionless, and both E
e
and E
¯
e
have dimension 1. Recall that
projecting out the F term increases the dimension by 1 because d
2
θ has dimension 1,
whereas projecting out the D term increases the dimension by 2 because d
2
θ d
2
¯
θ
has dimension 2. Thus we must consider combinations of superfields of at most
dimension 3 if we use the F term and at most dimension 2 if we use the D term.
Therefore, the following terms are all dimension 4 supersymmetric lagrangians:
F
a
F
a
F
, E
†
e
e
−2eV
E
e
D
, E
†
¯
e
e
2eV
E
¯
e
D
(13.53)
These are, of course, the kinetic terms we obtained previously for these superfields,
so no surprise here. We also have an interaction term of dimension 3:
E
e
E
¯
e
F
which must be multiplied by a constant with dimensions of mass. It turns out that
this coefficient corresponds precisely to the mass of a Dirac particle, as we will
demonstrate below, so let’s write this term as
mE
e
E
¯
e
F
(13.54)
It’e easy to see that we have exhausted all the possible terms that can contribute to
a renormalizable, supersymmetric, and gauge-invariant lagrangian. For example,
the F term of E
e
E
¯
e
E
e
E
¯
e
has dimension 5.
We must worry about one last thing before writing down the supersymmetric
version of QED: We must make sure that the lagrangian is real. All the terms in
Eq. (13.53) are real, but Eq. (13.54) is not, so we must add to it its hermitian
conjugate. Our supersymmetric QED lagrangian therefore is
L
super QED
=
1
4
F
a
F
a
F
+ E
†
e
e
−2eV
E
e
D
+ E
†
¯
e
e
2eV
E
¯
e
D
+ m E
e
E
¯
e
F
+ m E
†
e
E
†
¯
e
F
(13.55)
where we have taken m to be a real quantity.
320
Supersymmetry Demystified
Note that the last two terms of Eq. (13.55) represent the superpotential of the
theory (plus its hermitian conjugate), so we could write the lagrangian as
L
super QED
=
1
4
F
a
F
a
F
+ E
†
e
e
−2eV
E
e
D
+ E
†
¯
e
e
2eV
E
¯
e
D
+
W
F
+ h.c.
(13.56)
with
W ≡ m E
e
E
¯
e
Let’s write this out in terms of the component fields. Actually, let’s write the
lagrangian with the auxiliary fields eliminated, as an illustration of how to use
Eq. (10.56), which we repeat here for convenience (we will not include a Fayet-
Iliopoulos term):
L = L
F
i
=D=0
−
∂W
∂φ
i
2
−
1
2
q
i
φ
†
i
φ
i
2
(13.57)
To see how this is used, we first write the superpotential mE
e
E
¯
e
as a function of
the corresponding scalar fields
W = m
˜
φ
e
˜
φ
¯
e
This superpotential depends on two scalar fields, so we may take the φ
i
appearing
in Eq. (13.57) to be given by, say, φ
1
=
˜
φ
e
and φ
2
=
˜
φ
¯
e
. We then find
−
∂W
∂φ
i
2
=−
∂W
∂φ
i
&
∂W
∂φ
i
'
†
=−
∂W
∂
˜
φ
e
&
∂W
∂
˜
φ
e
'
†
−
∂W
∂
˜
φ
¯
e
&
∂W
∂
˜
φ
¯
e
'
†
=−m
2
˜
φ
¯
e
˜
φ
†
¯
e
− m
2
˜
φ
e
˜
φ
†
e
=−m
2
|
˜
φ
¯
e
|
2
− m
2
|
˜
φ
e
|
2
(13.58)
On the other hand, the last term of Eq. (13.57) is simply
−
1
2
q
i
φ
†
i
φ
i
2
=−
1
2
− e |
˜
φ
e
|
2
+ e |
˜
φ
¯
e
|
2
2
=−
e
2
2
|
˜
φ
e
|
2
−|
˜
φ
¯
e
|
2
2
(13.59)
CHAPTER 13 Gauge Field Theories
321
All we have to do now is to write down Eq. (13.56) in terms of component fields
with the auxiliary fields set to zero and add Eqs. (13.58) and (13.59).
We have actually worked out all the expressions appearing in Eq. (13.56) in
terms of component fields before. The first term is simply Eq. (13.49), the second
and third terms can be obtained from Eq. (13.23), and the last two terms can be
read off from Eq. (12.28). We finally get
L
super QED
= i
¯
λ ¯σ
μ
∂
μ
λ −
1
4
F
μν
F
μν
+ (D
μ
˜
φ
e
)
†
(D
μ
˜
φ
e
) +i ¯χ
e
¯σ
μ
D
μ
χ
e
+
√
2 e χ
e
· λ
˜
φ
†
e
+ h.c.
+ (D
μ
˜
φ
¯
e
)
†
(D
μ
˜
φ
¯
e
) +i ¯χ
¯
e
¯σ
μ
D
μ
χ
¯
e
−
√
2 e χ
¯
e
· λ
˜
φ
†
¯
e
+ h.c.
− m χ
e
· χ
¯
e
− m ¯χ
e
· ¯χ
¯
e
−m
2
|
˜
φ
¯
e
|
2
− m
2
|
˜
φ
e
|
2
−
e
2
2
|
˜
φ
e
|
2
−|
˜
φ
¯
e
|
2
2
(13.60)
Here, it is understood that when applied to fields belong to the electron multiplet
(
˜
φ
e
and χ
e
), the covariant derivative is (with the convention e > 0)
D
μ
= ∂
μ
−ieA
μ
whereas, when acting on the positron multiplet, the covariant derivative is
D
μ
= ∂
μ
+ieA
μ
Using Eq. (4.19), we see that the QED lagrangian is indeed contained in
Eq. (13.60):
L
QED
=
D
γ
μ
∂
μ
−ieA
μ
D
−
1
4
F
μν
F
μν
= i ¯χ
e
¯σ
μ
D
μ
χ
e
+i ¯χ
¯
e
¯σ
μ
D
μ
χ
¯
e
− m χ
e
· χ
¯
e
− m ¯χ
e
· ¯χ
¯
e
−
1
4
F
μν
F
μν
(13.61)
A few points deserve comments. First, let’s emphasize once more that we needed
to introduce two complex scalar fields:
˜
φ
e
and
˜
φ
¯
e
. The reason is clear: We have to
introduce a superpartner for each chiral state of the electron, which we have taken
to be the left-chiral particle and antiparticle states. So there are two supersymmetric
partners to a massive Dirac fermion. The next key observation is that those super-
partners have the same mass as the electron, a consequence of unbroken SUSY. This
immediately rules out this lagrangian as a viable model for the real world because
322
Supersymmetry Demystified
no scalar particle with the mass of an electron has ever been observed. What we
will need to do is to find ways to break SUSY spontaneously, but we will keep this
topic for next chapter.
13.7 Supersymmetric Nonabelian
Gauge Theories
We have seen how superfields may be used to obtain supersymmetric lagrangians
for a free abelian gauge field theory and for a chiral multiplet coupled to an abelian
gauge field. Here, we will generalize these results to nonabelian gauge theories.
The calculations are almost identical to what we did in previous sections, so we
will simply quote the main results without showing any explicit derivation.
The gauge superfield V is now taken to be a matrix in the fundamental represen-
tation of the Lie algebra [which we take to be su(n)], i.e.,
V ≡ V
i
T
i
F
where the sum is over the N
2
− 1 generators. This means that each component field
of V appears in N
2
− 1 copies, given by [see Eq. (13.10)]
V
i
=
1
2
θσ
μ
¯
θ A
i
μ
+
1
2
√
2
θ ·θ
¯
θ ·
¯
λ
i
+
1
2
√
2
¯
θ ·
¯
θθ· λ
i
−
1
8
θ ·θ
¯
θ ·
¯
θ D
i
Consider a left-chiral superfield belonging to the fundamental representation.
This means that is actually an N-component column vector that transforms as
→
= exp(2ig)
with being the matrix-valued quantity
≡
i
T
i
F
We take the vector superfield to transform as
exp(2gV) → exp(2gV
) = exp(2ig
†
) exp(2gV) exp(−2ig)
The quantity
†
exp(2gV) (13.62)
CHAPTER 13 Gauge Field Theories
323
is clearly gauge-invariant. This is, of course, exactly what we had for the abelian
case except that here the superfields do not commute because they are matrix valued.
The proper generalization of the abelian field-strength superfield F
a
given in Eq.
(13.24) is
F
a
=
¯
D ·
¯
D exp(−2gV) D
a
exp(2gV) (13.63)
where, as before, the label a is a spinor index, not a gauge group index. The field-
strength superfield is a matrix-valued quantity, like the other superfields. Showing
explicitly the matrix dependence, Eq. (13.63) reads
F
i
a
T
i
F
=
¯
D ·
¯
D exp(−2gV
i
T
i
F
)D
a
exp
2gV
i
T
i
F
The choice of Eq. (13.63) is motivated by the fact that the field-strength superfield
can be shown to transform in the usual way, i.e.,
F
a
→ exp(2ig) F
a
exp(−2ig)
Therefore, the trace
Tr
F
a
F
a
is gauge-invariant.
We now have all the pieces needed to write down a supersymmetric lagrangian
containing a nonabelian gauge superfield coupled to a left-chiral superfield. If there
is only one left-chiral superfield (which we assume is not invariant under the gauge
group), there is no superpotential because we can’t write down any gauge-invariant
holomorphic function of that superfield. There are then only two terms in the
lagrangian: the kinetic energy terms for the gauge superfield and for the left-chiral
superfield:
L =
1
2
Tr(F
a
F
a
)
F
+
†
e
2gV
D
If there are several left-chiral superfields, we obviously include a kinetic energy
term for each superfield, but in addition, we can now include a superpotential, i.e.,
W(
i
) + h.c.
assuming that the quantum numbers of the left-chiral superfields
i
make it possible
to write some gauge-invariant combination. Here the subscript i labels the different
324
Supersymmetry Demystified
left-chiral superfields, each of which is a gauge multiplet transforming in some
representation of the group.
13.8 Quiz
1. How does the field-strength superfield transform under Lorentz transforma-
tions?
2. What is the dimension of the gauge superfield V?
3. In the Wess-Zumino gauge, what are the component fields of the gauge su-
perfield?
4. In a nonabelian supersymmetric SU(N ) gauge theory, how many gauginos
are there?
5. How many complex scalar fields are needed to build a supersymmetric version
of QED? Why is that number of scalar fields required?
CHAPTER 14
SUSY Breaking
A key aspect of the theories we have considered so far is that all the particles of
a supersymmetric multiplet necessarily have the same mass. This follows from
the fact that the supercharges commute with the momentum operator, so applying
supercharges on a state does not change the eigenvalue of P
μ
P
μ
= m
2
. This is
obviously a disaster from the point of view of phenomenology because this degen-
eracy is clearly not observed in nature (e.g., there is no scalar particle of the same
mass and charge as the electron).
One can break supersymmetry (SUSY) explicitly by adding terms in the la-
grangian that are not invariant under SUSY. Of course, if we allow for any type of
SUSY-breaking interaction, we will loose all the benefits of SUSY, in particular the
cancellation of power-law divergences. The tricky question then becomes how to
identify the various SUSY-breaking interactions that do not spoil this cancellation—
in which case SUSY is said to be it softly broken—and how to justify why only
those should be included. We will come back to the issue of soft SUSY breaking
later in this chapter.
A second, and more desirable, method to break a symmetry is through spon-
taneous symmetry breaking (SSB). In this case, the lagrangian possesses the
symmetry, but the ground state does not, which then is reflected through the absence
of symmetry in the spectrum of the states.