Labelle P. Supersymmetry DeMYSTiFied

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Supersymmetry Demystified

density Eq. (5.3) is

δL = ∂

(δφ) ∂

†

+ ∂

φ∂

δ(φ

†

) + δ(χ

†

) i ¯σ

∂

χ +χ

†

i ¯σ

∂

δχ

= ∂

(δφ) ∂

†

+ ∂

φ∂

(δφ)

†

+ (δχ )

†

i ¯σ

∂

χ +χ

†

i ¯σ

∂

δχ (5.9)

where we have used δ(φ

†

) = (δφ)

†

and δ(χ)

†

= (δχ)

†

. To see that this is true, let’s

take the dagger of Eq. (5.4):

†

→ φ

†

+ (δφ)

†

On the other hand, by deﬁnition, we have

†

→ φ

†

+ δ(φ

†

)

so, clearly, δ(φ

†

) = (δφ)

†

, and we may simply write δφ

†

without any ambiguity.

So the variation of φ

†

and χ

†

are simply the hermitian conjugates of Eqs. (5.5)

and (5.8). One ﬁnds (see Exercise 5.1):

(δφ)

†

= ¯χ ·

ζ = χ

†

(iσ

)ζ

∗

(5.10)

and

(δχ )

†

= C (∂

†

)ζ

iσ

(5.11)

EXERCISE 5.1

Prove Eqs. (5.10) and (5.11).

Using Eqs. (5.5), (5.8), (5.10), and (5.11) into Eq. (5.9), we get

δL = (∂

†

)iσ

∗

∂



 

−(∂

†

)ζ

(iσ

)∂



 

+C(∂

†

)ζ

(iσ

)σ

i ¯σ

∂

  

III

−C

∗

†

i ¯σ

(∂

∂

φ)(i σ

∗

)

  

(5.12)

Note that in all the expressions we will write down, derivatives always will be

acting only on the ﬁeld that is immediately to their right. To make this explicit, we

will always group together a derivative and the ﬁeld it is acting on inside parentheses.

For example, instead of writing ∂

φχ, which could cause confusion, we will write

(∂

φ)χ.

CHAPTER 5 Building the Langrangian

Keep in mind that the scalar ﬁeld (or the scalar ﬁeld with any number of deriva-

tives acting on it) can be moved freely through spinors and matrices. Also recall that

two lagrangian terms differing by a total derivative are to be considered equivalent.

These points are important to remember if you do some calculations and obtain a

result that seems at ﬁrst sight to be different from the one given in this book or

in any other SUSY reference. In addition, several spinor identities we have seen

in previous chapters, the most important of which are listed in Appendix A, may

be used to rearrange expressions. It takes some practice to be able to tell quickly

whether or not two apparently different expressions are actually equivalent.

Coming back to Eq. (5.12), we observe that the ﬁrst and last term contain ζ

∗

whereas the second and third term contain ζ . Thus each pair of terms must cancel

separately (up to total derivatives).

Consider the ﬁrst and last terms. At ﬁrst sight, these two terms look very different:

The last term contains ¯σ

and σ

, whereas these matrices do not appear at all in

the ﬁrst term. However, we have already shown how the two are related! Indeed, in

Eq. (4.34) we proved that

¯σ

∂

χ = ∂

∂

The proof makes it evident that the identity also holds if the derivatives are acting

on a scalar ﬁeld, so we have

¯σ

∂

φ = ∂

∂

φ (5.13)

In Section 4.9 we proved this using the identity (A.9). It also can be proven by

directly contracting the Lorentz indices, as you are invited to do in Exercise 5.2.

EXERCISE 5.2

Prove Eq. (5.13) by expanding explicitly the Lorentz indices (i.e., by writing σ

∂

+ σ

∂

and so on).

Therefore, the sum of the ﬁrst and last terms of Eq. (5.12) is

I + IV = (∂

†

)iσ

∗

∂

φ −iC

∗

†

φi σ

∗

Of course, we can move φ around because it is not a matrix quantity and write

I + IV = (∂

†

)iσ

∗

∂

φ −iC

∗

†

iσ

∗

φ

Supersymmetry Demystified

We cannot make those two terms cancel because the derivatives act on different

ﬁelds. But we have a trick to move derivatives around: integrations by parts! If we

integrate by parts the derivative acting on χ

†

and use the fact that ∂

∗

= 0, we get

I + IV =−χ

†

iσ

∗

φ −iC

∗

†

iσ

∗

φ (5.14)

where, as usual, we have dropped the surface term. We see that this vanishes if we

take C =−i .

Now consider the sum of the second and third terms of Eq. (5.12):

II + III =−(∂

†

)ζ

iσ

∂

χ +C(∂

†

)ζ

iσ

i ¯σ

∂

which, after integrating the derivatives acting on the spinor ﬁeld by parts and using

Eq. (5.13), may be written as

II + III = (

φ

†

)ζ

iσ

χ −iC (φ

†

)ζ

iσ

χ (5.15)

which again vanishes if we take C =−i !

We have therefore achieved our goal: We have introduced a SUSY transformation

that leaves our action invariant! We have shown that the lagrangian

free

= ∂

φ∂

†

+ χ

†

i ¯σ

∂

χ (5.16)

is invariant under the SUSY transformations

δφ = ζ · χ

δφ

†

ζ · ¯χ

δχ =−i(∂

φ) σ

iσ

∗

δχ

†

=−i(∂

†

)ζ

iσ

(5.17)

Building our ﬁrst supersymmetric theory was not so hard after all! Unfortunately,

there is one subtlety that will require us to modify these results signiﬁcantly in two

ways: we will need to include a new ﬁeld in the lagrangian (with its own SUSY

transformation) and modify the transformation of the spinor. To understand why

anyone would want to do that, we will ﬁrst need to review the concept of symmetry

charges and their algebra.

CHAPTER 5 Building the Langrangian

5.4 Quiz

1. What is rigid supersymmetry?

2. What is the dimension of the inﬁnitesimal SUSY parameter ζ ?

3. We want the transformation of the spinor ﬁeld to be linear in the scalar ﬁeld φ

and in the SUSY parameter ζ . Using dimensional analysis alone, what form

must the variation of the spinor ﬁeld δχ take (the answer does not have to

have the correct Lorentz transformation property)?

4. What is SUGRA?

5. What would be wrong with the transformation δχ = C(∂

φ) ¯σ

ζ ?

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CHAPTER 6

The Supersymmetric

Charges and

Their Algebra

When a lagrangian is observed to possess some continuous symmetry, the natural

thing to do is to ﬁnd the charges generating the symmetry, also called the generators

of the symmetry. Actually, as we will see, what really matters is the algebra obeyed

by the charges, i.e., the commutation or anticommutation rules, not the charges

themselves (as an aside, when the algebra involves anticommutators, it is often

referred to as a graded Lie algebra). We will start by reviewing some elementary

results concerning charges and their algebra before tackling the supersymmetry

(SUSY) algebra in Section 6.5. We will be led to modify our supersymmetric

theory of the preceding chapter substantially in order for the SUSY algebra to be

well-deﬁned for off-shell ﬁelds (see Sections 6.6 to 6.8).

102

Supersymmetry Demystified

6.1 Charges: General Discussion

To explain what charges are, let’s go back to the ﬁeld transformations δφ and δχ,

which leave the action invariant. We have written these variations explicitly in terms

of the ﬁelds and of the inﬁnitesimal spinor ζ . Now, if we think in terms of canonical

quantization, φ and χ are operators, and the calculation of any physical process

always boils down to the evaluation of expectation values of some combination of

the ﬁelds.

To simplify the discussion, consider for now a scalar ﬁeld operator φ (unrelated

to the scalar ﬁeld φ that appeared in our supersymmetric lagrangian). If a certain

transformation is a symmetry of the theory, it must leave unchanged the expectation

value

a|φ|b (6.1)

We will be mostly interested in spacetime transformations and, later, the implemen-

tation of the supersymmetric transformations as transformations in superspace. Let

us then focus for now on spacetime transformations. Transformations that leave a

classic scalar ﬁeld unchanged should leave the expectation value (6.1) unaffected

as well. To be speciﬁc, we should have

a



|φ(x



)|b



=a|φ(x)|b (6.2)

Note that there are two differences between the two expressions: The ﬁeld operator

is evaluated at different spacetime points, and it is sandwiched between different

states. We now introduce a new ﬁeld operator φ



that we deﬁne by

a



|φ(x



)|b



≡a|φ



)|b (6.3)

In other words, the effects of the change of states is contained in the deﬁnition of

the new ﬁeld operator. If we substitute the right-hand side of Eq. (6.3) back into

Eq. (6.2), we get

a|φ



)|b=a|φ(x)|b

Since the states are arbitrary, we must conclude that



) = φ(x)

This is not surprising, this is exactly how a classic scalar ﬁeld behaves under a

Lorentz transformation!

CHAPTER 6 The Supersymmetric Charges

103

Let’s write x



as x −δx (the reason for the choice of sign of δx will become

clear shortly). This gives us



(x −δx) = φ(x)



(x) = φ(x + δx) (6.4)

We see that our choice of a negative sign of the variation in x



was so that we would

get a positive sign on the right-hand side of Eq. (6.4).

We will use this in a moment, but ﬁrst, let’s talk about what happens if the ﬁeld

is not a scalar. Let’s consider instead a ﬁeld operator F that transforms under some

representation of the Lorentz group (a vector, spinor, etc.). Then, Eq. (6.2) will be

replaced by

a



|F(x



)|b



=S() a|F(x)|b (6.5)

where S() is the Lorentz transformation matrix corresponding to the representa-

tion of the ﬁeld operator. Correspondingly, Eq. (6.4) is replaced by



(x) = S() F(x + δx) (6.6)

We will need to apply this result to the case of a spinor ﬁeld later in this chapter.

Now we will work out a different expression for the left-hand side of Eq. (6.2),

an expression that will allow us to establish a connection with the charges.

Physically meaningful transformations preserve the norm of states and therefore

can be written in the form of unitary operators (or antiunitary operators in the

case of time reversal), namely, operators satisfying U

†

= U

−1

. We will take the

transformation of the states to be

|a→U

†

|a|b→U

†

|b (6.7)

so the left-hand side of Eq. (6.2) takes the form

a



|φ(x



)|b



=a|U φ(x



†

|b

It is, of course, arbitrary whether we deﬁne the states to transform with the matrix

U or with the matrix U

†

, and you will encounter both conventions in the literature.

Now, using Eq. (6.3), we can relate this expression to the primed ﬁeld φ



a|φ



)|b=a|U φ(x



†

|b

104

Supersymmetry Demystified

which implies



) = U φ(x



†



(x) = U φ(x)U

†

(6.8)

Now we are ready to introduce the charges generating the transformations.

Assume that the transformations we are interested in are parametrized by a set

of n inﬁnitesimal parameters 

,

,...,

, in which case there are necessarily n

charges Q

, Q

,...,Q

. The charges are deﬁned through

U ≡ exp(±i

) = exp(±i · Q) (6.9)

where we introduced a shorthand dot notation for the sum 

. We allowed for

two possible signs in the argument of the exponential because we will encounter

both cases in our calculations.

It is important to keep in mind that here the  could either be ordinary (commut-

ing) constants or Grassmann quantities, such as the Weyl spinor ζ that we intro-

duced to parametrize supersymmetric transformations. The corresponding charges

are then, respectively, bosonic or fermionic.

Substituting Eq. (6.9) in Eq. (6.8) and expanding to ﬁrst order in ,weget



(x) ≈ (1 ± i  · Q)φ(x)(1 ∓i  · Q)

= φ(x) ± i · Q φ(x ) ∓ iφ(x) · Q

= φ(x) ± i[ · Q,φ(x)] (6.10)

Let’s deﬁne



(x) = φ(x) + δφ(x) (6.11)

By comparing Eqs. (6.10) and (6.11), we obtain a very useful result:

±[ · Q,φ] =−i δφ (6.12)

where the sign on the left-hand side is the same sign as the argument of the expo-

nential in the operator U.

In the case of spacetime symmetries, Eqs. (6.4) and (6.11) imply the obvious

relation

δφ = φ(x + δx) − φ(x) (6.13)

CHAPTER 6 The Supersymmetric Charges

105

For a general ﬁeld F, we have instead

δF(x) = S() F(x + δx) − F (x) (6.14)

Equations (6.12) and (6.4) [and its generalization, Eq. (6.6)] are the key results of

this section. They show us how to relate the transformation of a ﬁeld to a commutator

with a charge and to the transformation of the coordinates, respectively.

As a very simple example, consider a lagrangian density that is a function of a

single real scalar ﬁeld L(φ). The action S = i



xL is obviously invariant under

a translation of the ﬁeld

φ(x

) → φ



) = φ(x

+ a

) (6.15)

where a

is a constant-displacement four-vector. Therefore, the variation of the

ﬁeld δφ under this symmetry transformation is

δφ(x

) = φ(x

+ a

) − φ(x

)

≈ a

∂

φ (6.16)

This is an obvious invariance because the lagrangian density is integrated over all

spacetime, so we can always do a shift of the integration variable x

→ x

+ a

which implies that



x L(φ) =



x L(φ



). (6.17)

Since there are four parameters, the four a

, we introduce four charges that we

suggestively call P

, and write the unitary operator as

U ≡ exp(ia

) (6.18)

Substituting Eqs. (6.16) and (6.18) into Eq. (6.12), we get

,φ] =−ia

∂

φ (6.19)

Since this must be true for an arbitrary inﬁnitesimal four-vector a

, we conclude

that

,φ] =−i∂

φ (6.20)

a relation that will soon prove useful.