Labelle P. Supersymmetry DeMYSTiFied

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456

Supersymmetry Demystified

In terms of explicit components, this gives

=−

(B.59)

Now consider Eq. (12.47):

(θ · ∂

χ)(θ σ

λ)

The parentheses were added only to show the grouping more clearly. Writing

the indices explicitly on all quantities, this is

∂

(σ

)

b˙c

˙c

This is of the form of the left side of Eq. (B.59) with the identiﬁcation

γ = θ and

= ∂

= (σ

)

b˙c

˙c

so that Eq. (B.58) gives

θσ

λθ ·∂

χ =−

θ ·θ∂

(σ

)

d ˙c

˙c

=−

θ ·θ(∂

χ)σ

as was to be shown.

Now consider Eq. (12.48), which may be written as

θ ·

λ(∂

χ)(σ

)

θ =

˙a

(∂

)(σ

)

d ˙c

˙c

(B.60)

Showing all the indices, the identity (A.55) corresponds to

¯α

˙a

¯γ

˙a

¯γ

=−

¯γ

˙c

¯γ

˙c

¯α

(B.61)

The left-hand side is exactly the same as Eq. (B.60) if we identify ¯γ =

θ and

¯α

˙a

= (∂

)(σ

)

APPENDIX B Solutions to Exercises

457

The identity (B.61) then gives us

θ ·

λ(∂

)(σ

)

d ˙c

˙c

=−

θ ·

θ(∂

)(σ

)

=−

θ ·

θ(∂

χ)σ

which is what we needed to prove.

13.1 One has to be careful with this calculation because the components with

upper indices are not independent of the components with lower indices. If

one does not keep this in mind, it is tempting to conclude that

˙a

θ ·

θ =

∂

˙a

∂

˙a

will simply give 1. But this is incorrect. The simplest way to do the calculation

is to expand

θ ·

θ out in terms of lower components:

∂

˙a

∂

˙a

= 2

∂

− 2

∂

Now we use

and

=−

, we ﬁnally get

∂

− 2

∂

= 2

∂

+ 2

∂

= 4

as was to be shown.

14.1 Since we are only interested in reading off the masses, not in the interactions,

we will consider only the terms quadratic or bilinear in the ﬁelds:

V =

− g

+ g

+ 2g

φ



+2mgφ

R

+ 2mgφ

I

+ 2g

φ



··· (B.62)

where the dots indicate terms that don’t contribute to the mass matrix.

We see that since there are no terms containing R

or I

, these ﬁelds are

massless. The ﬁelds R

and R

get mixed, and their mass matrix is (in the

458

Supersymmetry Demystified

basis R

, R

)

⎛

⎜

⎝

2mgφ



2mgφ

 m

− 2g

+ 4g

φ



⎞

⎟

⎠

whose eigenvalues are

− g

+ 2g

φ



± g

)



2gφ



− gM



+ 4m

φ



(B.63)

This is not a very pleasant expression. It simpliﬁes greatly if we use the

freedom in the choice of φ

 to set it equal to zero, in which case the matrix

is already diagonal, and we get

= m

− g

= m

Now consider the terms in I

and I

in Eq. (B.62). The corresponding

squared-mass matrix is (in the basis I

, I

)

⎛

⎜

⎝

2mgφ



2mgφ

 m

+ 2g

+ 4g

φ



⎞

⎟

⎠

whose eigenvalues are

+ g

+ 2g

φ



± g

)



2gφ



+ gM



+ 4m

φ



(B.64)

Setting again φ

=0, we get once more a diagonal matrix and

= m

+ g

= m

14.2 All we need are Eqs. (B.63) and (14.16). If we ﬁrst take the limit g → 0,

then we simply get

= m

with the other ﬁelds being massless, as previously.

APPENDIX B Solutions to Exercises

459

On the other hand, if we take the limit M → 0, the masses of the four real

scalars split exactly the same way as the fermion masses; i.e., we have

= m



φ



+ m

± gφ



For each fermion, there is a boson of equal mass, and supersymmetry is

unbroken.

14.3 Two of the bosons are massless, whereas the other four have masses squared

given by Eqs. (B.63) and (B.64) so that



scalars

= 4m

+ 8g

φ



On the other hand, one fermion is massless, with the other two having masses

squared given by Eq. (14.16) so that



spinors

= 4m

+ 8g

φ



and we see that the supertrace still vanishes.

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APPENDIX C

Solutions to Quizzes

Chapter 2

1. We have

¯χ ·

λ = χ

†

iσ

†

=−χ

†

+ χ

†

2. A Weyl spinor has a deﬁnite helicity only when it is massless. However, by

construction, a Weyl spinor has a deﬁnite chirality, no matter what its mass

is. So Weyl spinors are always states of deﬁnite chirality.

3. They transform the same way under rotations but differently under boosts (the

boost parameter has an opposite sign).

4. A Dirac mass term is

−m

 =−m

†

 =−m(η

†

χ +χ

†

η)

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Supersymmetry Demystified

5. The operation of hermitian conjugate on a product of spinor ﬁelds is deﬁned

to switch their order without introducing a minus sign, i.e.,

(χ

)

†

= χ

†

Chapter 3

1. We ﬁrst need to move the component with the upper index to the right of

the one with the lower index (recall that dotted indices must go diagonally

upward), so we get

¯η

˙a

¯χ

˙a

=−¯χ

˙a

¯η

˙a

=−¯χ · ¯η

This also may be written as −¯η · ¯χ.

2. 

= 0 and 

=−1.

3. Recall that we ﬁrst must get the last indices of the  to match the indices of

the σ

. We therefore must switch the order of the indices of the ﬁrst epsilon,

which introduces a sign:



(σ

)

=−



(σ

)

and now we can contract the indices to get

−( ¯σ

)

Note the order of the indices!

4. Since the common index appears in the same position in both , this is equal

to minus one times the Kronecker delta, i.e.,



˙a



˙a ˙c

=−δ

˙c

5. Applying a hermitian conjugation changes a dotted index into an undotted

index, and vice versa, without changing the position of the index. So (χ

)

†

¯χ

˙a

, and so on.

6. The matrix −iσ

lowers indices.

7. We can do it in two different ways. We can choose to contract the  together

ﬁrst:



=−δ

=−χ

APPENDIX C Solutions to Quizzes

463

where there is a minus sign in the ﬁrst step because the index common to the

two  is in the same position. Or we can contract one at a time the two  with

the spinor:



=−



=−

=−χ

where the minus sign comes from switching the order of the indices of 

which is necessary before contracting with the spinor.

Chapter 4

1. It is the corresponding left-chiral antiparticle state. Of course, this state also

will be the left-chiral particle state if we are dealing with a Majorana spinor.

2. The Dirac mass term contains particle and antiparticle Weyl spinors dotted

together, i.e.,

· χ

+ ¯χ

· ¯χ

whereas the Majorana mass term contains the same expressions but with all

spinors being particle spinors, i.e., χ

· χ

+ ¯χ

· ¯χ

3. There is no difference besides the fact that a Majorana spinor is written in four-

component language, whereas a lagrangian of a Weyl fermion is expressed

in terms of two-component spinors. Physically, the two are equivalent. This

is the convention followed in this book. However, some references deﬁne a

Weyl spinor to be necessarily massless and use the term Majorana spinor to

denote the massive case, no matter if it is expressed in two- or four-component

language.

4. The answer is yes to both questions. We saw explicitly how a Dirac spinor

can be written in terms of left-chiral spinors only in Section 4.6 (i.e., in terms

of the particle and antiparticle left-chiral Weyl spinors). It is also possible to

use right-chiral spinors only.

5. The mass term is

−



· χ

+ ¯χ

· ¯χ



It is clear that this is real because (χ

· χ

)

†

= ¯χ

· ¯χ

and ( ¯χ

· ¯χ

)

†

= χ

, so taking the hermitian conjugate of the mass term we just wrote gives

back the same expression. Note that we absolutely need both terms to make

it work.

464

Supersymmetry Demystified

6. Parity. This is why in parity-invariant theories, such as QED or QCD, Dirac

spinors are necessary.

Chapter 5

1. Rigid supersymmetry is another name for global supersymmetry, i.e., super-

symmetry with a spacetime-independent inﬁnitesimal parameter ζ .

2. It has dimension −1/2.

3. Since [φ] = 1, [χ ] = 3/2, and [ζ ] =−1/2, we need to combine ζφ with

something with a dimension equal to one. The only available quantity that is

not a ﬁeld is a derivative ∂

.Sowehaveδχ  ζ∂

χ. Of course, this does not

have the right Lorentz transformation, so it needs to be modiﬁed as shown in

the text.

4. It is the abbreviation of supergravity, which is the theory obtained when

supersymmetry is made local (which forces the introduction of a graviton).

5. The right-hand side transforms like a right-chiral spinor, whereas χ is a left-

handed spinor.

Chapter 6

1. The supercharges commute with the four-momentum operator but not with

the Lorentz generators. This is why the supercharges do not change the four-

momentum of a state but change the helicity.

2. It is a ﬁeld that has no kinetic energy term and is therefore nondynamical. The

equation of motion for an auxiliary ﬁeld is algebraic, and the ﬁeld therefore

can always be eliminated trivially.

3. By counting the number of degrees of freedom off-shell. A Weyl spinor has

four degrees of freedom off-shell (two complex components), whereas a com-

plex scalar ﬁeld has only two of them. Therefore, we needed two more scalar

degrees of freedom.

4. The transformation must contain the spinor χ to contract with the ζ . Thus we

could try δF  ζ · χ , but this is of dimension 1. To increase the dimension,

we cannot use φ because we want the transformation to be linear in the ﬁelds.

We need to use a derivative, so we may try δ F  ζ · ∂

χ, but now we need

to contract the Lorentz index with some Pauli matrices. We need to use ¯σ

when we act on a left-chiral spinor in order to have a quantity with well-

deﬁned Lorentz transformation properties, so let’s try δF  ζ ¯σ

∂

χ. This

is not quite right because ¯σ

χ transforms like a right-chiral spinor. In order

APPENDIX C Solutions to Quizzes

465

to get a Lorentz scalar for δ F, we need to take the hermitian conjugate of ζ .

Thus the ﬁnal answer is δF = ζ

†

¯σ

∂

χ.

5. We needed a total of four charges, Q

, Q

†

, and Q

†

. We needed that many

because the SUSY parameter ζ is a Weyl spinor, so we needed enough charges

to construct Lorentz invariants containing ζ and

ζ .

Chapter 7

1. It does not change the momentum but increase the helicity by 1/2.

2. Only two on-shell states, differing in helicity by one-half. If we impose CPT

invariance in addition, we need four states.

3. It is supersymmetry with more than one set of supercharges (where one set is

deﬁned as Q

, Q

and their hermitian conjugate).

4. It is necessarily larger or equal to zero (and strictly larger than zero if SUSY

is broken spontaneously).

5. By considering charges that anticommute.

Chapter 8

1. W =

y φ

2. It depends only on the ﬁelds 

and not on their hermitian conjugate φ

†

3. W

∂W

∂φ

, W

∂

∂φ

4. Since F

has dimension 2, W

can be at most of dimension 2 as well (if we

consider only renormalizable theories). Therefore, the superpotential is at

most of dimension 3 and can contain at most three scalar ﬁelds.

5. It is the same as in Question 1 except that we may add a liner term cφ.

Chapter 9

1. A Dirac propagator has only one contribution, from the contraction of  and

. A Majorana propagator also contains the contractions of two 

and two



, so there are three distinct contributions to a Majorana propagator.

2. Only two: m and g. The parameter m is the mass of all the particles but also

appears as a coupling constant in some interactions.

3. No, logarithmic divergences are present, but these can be cured by wavefunc-

tion renormalization.