Labelle P. Supersymmetry DeMYSTiFied

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126

Supersymmetry Demystified

To simplify the notation, let’s deﬁne

α ≡ σ

∗

(6.90)

so that we may write Eq. (6.89) as

= α

β · ∂

χ (6.91)

Our ultimate goal is to check if the commutator of two SUSY transformations on

the spinor is of the same form as what we obtained for the scalar ﬁeld in Eq. (6.61).

To compare, we need to rewrite Eq. (6.91) in the form of something times ∂

i.e., we want the index a to be on χ. We can move the index using the following

identity, valid for any three spinors α, β and γ :

(β · γ)+ β

(γ · α) + γ

(α · β) = 0 (6.92)

Note that the spinor dot products are commuting quantities, so we are free to move

them around. For example, α

(β · γ) = (β ·γ)α

EXERCISE 6.6

Prove Eq. (6.92). Hint: Expand out all the terms in terms of the components of the

spinors, and check explicitly that the identity is satisﬁed for both a = 1 and a = 2.

Obviously, to apply Eq. (6.92) to Eq. (6.91), we simply have to deﬁne ∂

χ ≡ γ .

We then obtain [using Eq. (6.90)]

=−(∂

χ) · αβ

− α · β∂

=−(∂

χ) · αβ

− β · α∂

=−



(∂

χ)

(−iσ

)σ

∗



−



(−iσ

)σ

∗



∂

= i



(∂

χ)

∗





∗



∂

where in the second line we have rewritten α · β as β ·α only so that the transpose

would be applied to β, not to the complicated α [deﬁned in Eq. (6.90)]. Note that

the expressions inside the two parentheses in the last line are commuting quantities.

We can freely move β

or ∂

through these parentheses without picking up any

minus sign.

CHAPTER 6 The Supersymmetric Charges

127

Using Eq. (A.5) and then Eq. (A.36), we can simplify the result greatly:

= i



(∂

χ)

∗





∗



∂

= i



(∂

χ)

¯σ

μT

∗





¯σ

μT

∗



∂

=−i



†

¯σ

∂



−i



†

¯σ



∂

The second term on the right-hand side is what we wanted, something proportional

to ∂

, but the ﬁrst term will cause us some grief.

Obviously, by switching ζ and β,weget

=−i



†

¯σ

∂



−i



†

¯σ



∂

so that



− δ



=−i



†

¯σ

∂





†

¯σ

∂



−i



†

¯σ

β − β

†

¯σ



∂

(6.93)

Note that the last term is identical to the transformation we had for the scalar ﬁeld

(6.61) with φ replaced by χ! Unfortunately, the ﬁrst two terms of Eq. (6.93) spoil

the fun.

Note that if i ¯σ

∂

χ = 0, the two unwanted terms drop out. But this is nothing

other than the Weyl equation! In other words, if the spinor satisﬁes its equation of

motion (i.e., if it is on-shell), then we recover the algebra we had obtained previously

using the transformations of the scalar ﬁeld. However, for an off-shell spinor, one

cannot write the algebra of the SUSY and Poincar´e charges in terms of those same

charges, as we did in Section 6.5. The conclusion therefore is that the algebra on

the spinor ﬁeld closes only on-shell.

What we would like to do now is to modify the theory so that the algebra on the

spinor ﬁeld will close off-shell, i.e., even if the spinor does not obey its equations of

motion. Why do we want that? This is a subtle issue. There are SUSY theories that

have no off-shell formulations, i.e., whose algebra closes only on-shell, so it would

not be the end of the world if we would not modify our toy theory to close off-shell.

However, when the algebra does not close offshell, SUSY transformations become

nonlinear (they contain products of the ﬁelds) and depend on the coupling constants

present in the theory. This makes it much more difﬁcult to design lagrangians that

are invariant under SUSY. We will therefore modify the theory to enforce that the

algebra closes off-shell.

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Supersymmetry Demystified

6.7 Introduction of an Auxiliary Field

There is a simple argument requiring no calculations that we could have used to

predict that the SUSY algebra would not close off-shell. SUSY requires the number

of bosonic and fermionic degrees of freedom to be equal (this will be demonstrated

in the next chapter). A Weyl spinor has two complex components and so four

degrees of freedom when it is off-shell. On-shell, the equation of motion imposes

two constraints, reducing the number of degrees of freedom to two. A complex

scalar ﬁeld has two degrees of freedom, so we see that on-shell, the number of

bosonic and fermionic degrees of freedom match. However, off-shell, the number

of bosonic and fermionic degrees of freedom do not match anymore. This explains

why the SUSY algebra only closes on-shell if we have a single complex scalar ﬁeld

and a Weyl spinor.

But the argument also suggests a solution. We need to add a bosonic ﬁeld that

will provide two degrees of freedom off-shell but no on-shell degrees of freedom!

A ﬁeld that has no on-shell degrees of freedom is called an auxiliary ﬁeld, so what

we need is to add to the theory a complex scalar auxiliary ﬁeld, commonly called

F. How do we get the ﬁeld F to have no on-shell degrees of freedom? The trick

is to make sure that the equation of motion for this ﬁeld is F

†

= F = 0! This will

clearly be the case if we add a term in the lagrangian that contains a product of F

and F

†

; i.e., we do not include a kinetic term for F.

The simplest real term depending on F and F

†

is FF

†

. So our lagrangian now

takes the form

L = ∂

φ∂

†

+ χ

†

i ¯σ

∂

χ +FF

†

(6.94)

Note that despite being a scalar ﬁeld, the ﬁeld F is of dimension 2! This is something

important to remember, so let’s emphasize it:

[F] = 2

What should we take for the transformation of F under SUSY? Since it has

a different dimension than φ, we cannot use the same transformation. We want

a transformation that is linear in the SUSY parameter ζ and linear in one of the

ﬁelds. Recall that ζ has dimension −1/2, so it must be multiplied by something

of dimension 5/2 (because F has a dimension equal to 2). This means that the

transformation must contain the spinor ﬁeld χ, so our starting point is

δF  ζχ

CHAPTER 6 The Supersymmetric Charges

129

Since the dimensions of χ and ζ add up to 1, we need to include something with

a dimension equal to 1. We don’t want to use the scalar ﬁeld to avoid a nonlinear

transformation. What else do we have with dimension equal to 1? A derivative, of

course! So let’s try

δF  ζ∂

which has the right dimension. But, of course, it does not have the right Lorentz

transformation because F is Lorentz-invariant. We need to include something that

is dimensionless and yet carries a Lorentz index. Obviously, we want either σ

¯σ

. We know, from the Dirac equation [see Eq. (2.26)] that it is the matrices ¯σ

that must be applied to a left-chiral spinor to yield an expression with well-deﬁned

Lorentz properties. So let’s consider

 ζ ¯σ

∂

Unfortunately, this still does not have the correct behavior under Lorentz transfor-

mations. From Eq. (2.26) again, we know that ¯σ

∂

χ transforms like a right-chiral

spinor. On the other hand, ζ is a left-chiral spinor. So we must recall how to com-

bine a left- and right-chiral spinor to form a Lorentz invariant. The answer is that

we must contract the hermitian conjugate of either spinor with the other spinor [see

Eq. (2.42)]. It will be easier here to take the hermitian conjugate of ζ and contract

it with ¯σ

∂

χ, so we ﬁnally take

δF = K ζ

†

¯σ

∂

where K is some constant that we will determine shortly. This has the correct

dimensions and Lorentz properties.

By taking the dagger, we get

δF

†

= K

∗

(∂

†

) ¯σ

What we must now ﬁnd out is whether it is possible to add the term FF

†

without

spoiling the invariance of the lagrangian under SUSY. The variation of FF

†

δ(FF

†

) = K ζ

†

¯σ

(∂

χ)F

†

+ K

∗

(∂

†

) ¯σ

ζ F (6.95)

which is not a total derivative, so we are in trouble. What we need to do is to

modify the SUSY transformation of either φ or χ in such a way that the extra

terms generated will cancel Eq. (6.95). Note that the variation of the spinor kinetic

130

Supersymmetry Demystified

energy is

δ(χ

†

i ¯σ

∂

χ) = (δχ

†

)i ¯σ

∂

χ +χ

†

i ¯σ

∂

(δχ ) (6.96)

which looks a lot like Eq. (6.95)! To make the similarity more apparent, let’s rewrite

Eq. (6.95) in the form (after doing an integration by parts on the second term)

δ(FF

†

) = (K

∗

ζ F)

†

¯σ

∂

χ −χ

†

¯σ

∂

∗

ζ F) (6.97)

It is now obvious by comparing Eqs. (6.96) and (6.97) that if we take

δχ =−iK

∗

ζ F + previous transformation

the variation of the kinetic term of the spinor ﬁeld will cancel out the variation of

†

(modulo a total derivative).

We see that the value of the constant K is left completely arbitrary. The con-

vention is to make the new term in the transformation of χ as simple as possible,

namely,

−iK

∗

ζ F ≡ ζ F

which ﬁxes K =−i.

Our ﬁnal result is that the lagrangian

L = ∂

φ∂

†

+ χ

†

i ¯σ

∂

χ +FF

†

(6.98)

is invariant (up to total derivatives, as always) under the following SUSY transfor-

mations:

δφ = ζ · χ

δχ =−iσ

(iσ

∗

)∂

φ + Fζ

δF =−iζ

†

¯σ

∂

(6.99)

CHAPTER 6 The Supersymmetric Charges

131

6.8 Closure of the Algebra

Let’s not lose track of the original motivation for introducing an auxiliary ﬁeld,

which was to ensure that the algebra closes off-shell for all the ﬁelds! What we now

need to show is that for all three ﬁelds φ,χ, and F, the commutator of two SUSY

transformations gives the same result, without recourse to any equation of motion.

We should start by repeating the calculation that led to Eq. (6.61) for the scalar

ﬁeld. Since we have modiﬁed the transformation of the spinor, there is no guarantee,

at ﬁrst sight, that we will recover the same result. However, it’s easy to check that the

new term present in the transformation of χ does not change the result of Eq. (6.61).

Not only that, but now we do obtain the same result for all three ﬁelds without using

the equations of motion; i.e., we get (see Exercise 6.7)

X − δ

X =−i



†

¯σ

β − β

†

¯σ



∂

X (6.100)

where X stands for any of the three ﬁelds φ, χ,orF.

This ﬁnally proves that the introduction of the auxiliary ﬁeld allows closure of

the algebra off-shell on all the ﬁelds of the theory. And the consequence of this is

that the algebra we found for the SUSY charges holds in general.

EXERCISE 6.7

Prove that Eq. (6.100) is satisﬁed by all three ﬁelds φ, χ , and F.

6.9 Quiz

1. Do the supercharges commute with the momentum operator? Do they com-

mute with the Lorentz generators?

2. What is the deﬁnition of an auxiliary ﬁeld?

3. Explain how we could have predicted that the SUSY algebra would not close

off-shell before the introduction of an auxiliary ﬁeld, even before doing any

calculation.

4. Without looking at the text, write down the most general transformation law

of the auxiliary ﬁeld F that is linear in the ﬁelds, of the right dimension,

and with the correct Lorentz property. Of course, it must contain the SUSY

parameter ζ . Recall that F is of dimension 2.

5. How many supercharges did we need to introduce (count two charges related

by hermitian conjugation as independent). Why did we need to introduce that

many?

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CHAPTER 7

Applications of the SUSY

Algebra

We have spent a considerable amount of energy working out the supersymmetry

(SUSY) algebra, so it better be useful to something! Indeed, it proves to be a

powerful tool for establishing general properties of supersymmetric theories without

the need to ever write down any lagrangian! We will work out several such general

properties in this chapter, but it seems like a good idea to ﬁrst review how some of

this works in the familiar context of the Poincar´e group.

134

Supersymmetry Demystified

7.1 Classification of States Using the Algebra:

Review of the Poincar´e Group

Let’s brieﬂy review some basic facts about the representations of the Poincar´e

group (no SUSY here) on one-particle states. The details were worked out by

Wigner in 1939

in what has become a classic paper. Here, we simply highlight

some key results.

Casimir operators are operators that commute with all the generators of the group.

These operators are of key importance in building representations because their

eigenvalues can be used to classify the representations of the group. One Casimir

operator of the Poincar´e group is the squared momentum operator P

. This

obviously commutes with the momentum generators, and the proof that it commutes

with the Lorentz generators as well is straightforward. The eigenvalue of P

, which is therefore one of the parameters used to classify one-particle states.

The second Casimir operator is not as obvious. It is built out of the so-called

Pauli-Lubanski operator W

, deﬁned as

≡



μνρσ

ρσ

(7.1)

where 

μνρσ

is the totally antisymmetric Levi-Civita symbol in four dimensions.

We will use the normalization



0123

= 1

which implies



0123

=−1

With this normalization, it is clear that



0ijk

=−

ijk

(7.2)

where 

ijk

is the usual three-dimensional Levi-Civita symbol. Note that we auto-

matically have

= 

μνρσ

ρσ

= 0 (7.3)

which is obvious because P

is symmetric under the exchange of those two

indices, whereas the Levi-Civita symbol is antisymmetric. It is not as obvious that

CHAPTER 7 Applications of the SUSY Algebra

135

is also equal to zero (recall that P

and W

are operators), but it turns out

to be true, as you will show in Exercise 7.1. Therefore,

, W

] = 0 (7.4)

EXERCISE 7.1

Prove that P

= 0.

With some effort, one can show that W

also commutes with all the generators

of the Poincar´e group, so that’s our second Casimir operator.

To uncover the physical signiﬁcation of the Pauli-Lubanski operator, consider

ﬁrst massive particles. In that case, we can choose to work in the rest frame of the

particle, where its four-momentum is simply

= p

= (m,



0) (7.5)

We will be careful to distinguish the operator P

from its eigenvalues, which we

will write as p

For the state with four-momentum given by Eq. (7.5), we then have

|p=P

|p=m|p (7.6)

Let us now apply the Pauli-Lubanski operator to this state:

|p=



μνρσ

ρσ

|p



μ0ρσ

ρσ

|p



μ0ρσ

ρσ

|p

Because of the antisymmetry of the Levi-Civita symbol, all three indices μ, ρ,

and σ must be spacelike, so we conclude that

|p=0 (7.7)