Labelle P. Supersymmetry DeMYSTiFied

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106

Supersymmetry Demystified

As an aside, the charges of a symmetry sometimes are also called the genera-

tors of the transformation. For example, the P

are often called the generators of

translation in spacetime.

As a second simple example, consider a theory containing a single complex

scalar ﬁeld and which is described by a lagrangian density that is invariant under

a U (1) transformation, φ → e

iα

φ, with α being a constant parameter. This is an

example of an internal symmetry, i.e., a symmetry that is not related to a change

of coordinates but to an equivalence between different ﬁelds at the same spacetime

point. In this case, the variation of the ﬁeld is δφ = iαφ. Writing U = exp(iα Q),

Eq. (6.12) simply gives

[Q,φ] = φ (6.21)

A less trivial example is the object of Exercise 6.1.

EXERCISE 6.1

Consider the Lorentz transformation

→ 

For an inﬁnitesimal transformation, we may write



≈ x

+ ω

= x

+ ω

μν

(6.22)

where ω

μν

is antisymmetric, ω

μν

=−ω

νμ

. Because of this antisymmetry, it is

important to keep track of which index is the ﬁrst one and which is the second one

when we write ω

, which is why the lower index is shifted to the right.

We write the unitary operator implementing Lorentz transformation on a scalar

ﬁeld as

U = e

−

μν

where the charges M

μν

are also antisymmetric. The factor of 1/2 is included to allow

the sum to be over all values of the indices without double counting (since ω

, for example). Note the sign in the exponent! It is natural to take the Lorentz

transformation to have the opposite sign to the translation transformation by analogy

with classic physics (see Section 2.5 of Ref. 20 for more details).

The transformation of the ﬁeld is given by

δφ = φ(x

+ ω

μν

) − φ(x

)

CHAPTER 6 The Supersymmetric Charges

107

First, prove that we may write

δφ =

μν

∂

− x

∂

)φ (6.23)

Then prove that the commutation relations between the charges M

μν

and a scalar

ﬁeld are

μν

,φ] = i(x

∂

− x

∂

)φ

The combination of the Lorentz transformations introduced in Exercise 6.1 and

the translations generated by the P

form the Poincar

e group. You will work out

the Poincar´e algebra, the set of commutation relations obeyed by the Lorentz and

translation generators, in Exercise 6.3.

There is unfortunately no standard in the literature for the signs used in the expo-

nents of the operators U , i.e., for the choice between U and U

†

in the transformations

of the states (6.7) or for the choice between active and passive transformations in

the change of coordinates (which affects the sign of δx). A difference of conven-

tion in any of these aspects affects the signs of some of the equations containing

charges. Because of this, it is typical to see equations differing in signs from one

book to the next. I have chosen a set of conventions that maximizes agreement with

most introductory references on SUSY, but it’s impossible to agree with all of them

because they don’t even agree with one another!

Note that Eqs. (6.20) and (6.21) are formal expressions in the sense that they tell

us what the commutations of the charges with the scalar ﬁeld are without giving us

explicit expressions for those charges. It is possible to obtain explicit representations

of the charges, as we will discuss in the next section, but what is even more useful is

to work out the algebra of the charges, a topic to which we now turn our attention.

6.2 Explicit Representations of the Charges

and the Charge Algebra

Given that we already have ﬁgured out the SUSY transformations of the ﬁelds,

it might seem like there is no point in ﬁnding the charges. However, the charges

associated with a symmetry provide a powerful tool to learn a great deal about

theories having this symmetry without even writing down any lagrangian! Actually,

we don’t even need to work out the explicit charges; all we really need is the

algebra of the charges, i.e., the commutation rules (or anticommutation rules) they

obey.

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Supersymmetry Demystified

As an example of how the knowledge of mere commutation rules may be so

useful, think about the harmonic oscillator in quantum mechanics. Given only the

commutation rules of the raising and lowering operators, it is possible to ﬁnd the

allowed energies of the system without even working out any explicit wavefunc-

tion! Or consider the spin operators S

and S

. Using only the algebra between

those operators, it is possible to show that particles must have either integer or

half-integer spins and that for a given spin s there are 2s + 1 states. In a similar

vein, the algebra of the SUSY charges will reveal many general properties of su-

persymmetric theories, such as the types of particles that must be grouped together

in order to form SUSY-invariant theories without us having to write down any

lagrangian.

Now that we have motivated the usefulness of knowing the algebra of the charges

generating a symmetry, let us turn to the question of how to actually determine it.

One approach is to build explicit representations for the charges and then to

compute by brute force their algebra. It is difﬁcult to use Eq. (6.12) to guess the

correct expressions for the charges because of the commutator on the left-hand

side. Fortunately, a systematic way to obtain the charges is available. It consists

of building a set of currents J

associated with the symmetry from which the

corresponding charges are found by integrating over space (not spacetime) the

zeroth component of the currents:



(x, t)

It is important to note that in this approach, the charges (and the currents) are

themselves quantum ﬁeld operators. The algebra of the charges then can found

using the equal time commutation relations of the ﬁelds present in the theory.

We will review the use of currents in detail in Section 7.7. For now, let us simply

summarize the strategy underlying this approach. The starting point is a set of ﬁeld

transformations leaving the action invariant (i.e., leaving the lagrangian density

invariant up to total derivatives). From that, one can work out the algebra of the

charges through the following steps:

Field transformations ⇒ currents ⇒ charges as quantum ﬁelds ⇒ algebra

(6.24)

Of course, as a double-check that the charges are correct, one then can use Eq. (6.12)

to recover the transformations of the ﬁelds.

In the case of spacetime symmetries, there is a second explicit representation of

the charges available: as differential operators acting on the ﬁelds. In this chap-

ter we will use carets over the differential operator representation of the charges

CHAPTER 6 The Supersymmetric Charges

109

to distinguish them from their representation as quantum ﬁeld operators. Let us

now see how these differential operators can be found. The differential operator

representation of the charges is deﬁned via

φ(x



) ≡ exp(±i

)φ(x)

≈ φ(x) ± i

φ(x) (6.25)

where now we think of the ﬁeld φ as an ordinary function of spacetime. If we

know explicitly the transformation of the coordinates, we simply write x



in terms

of x and Taylor expand φ(x



) to ﬁrst order in the change of coordinates, which

allows us to read off the charges

Q directly. Once we have differential operators

representations for the charges, it is a simple matter to compute their algebra.

As an example, consider again spacetime translations, Eq. (6.15). Using

represent the differential operator representation of the charges associated with this

symmetry, we write

U = exp(−ia

) (6.26)

Note the sign difference with respect to Eq. (6.18). There is much freedom in the

choices of signs in most formula we deal with in this chapter, which makes it quite

frustrating to compare equations from different references that typically do not use

the same conventions. The absence of standard notation means that the signs of the

commutation relations and the signs in the deﬁnitions of the charges will vary from

one source to another. This must be kept in mind when consulting the literature.

With our convention, we ﬁnd

φ(x) +a

∂

φ(x) = φ(x) − ia

φ(x) (6.27)

which leads to

= i∂

(6.28)

Note the difference between Eqs. (6.28) and (6.20). It goes beyond the difference

of sign on the right-hand side! In Eq. (6.20), the charge P

is a quantum ﬁeld

operator, not a differential operator.

EXERCISE 6.2

Deﬁne the action of the differential operators

μν

on a scalar ﬁeld through

φ(x



) ≡ exp



μν



φ (6.29)

110

Supersymmetry Demystified

with the change of coordinates given again by Eq. (6.22). Show that

μν

= i(x

∂

− x

∂

) (6.30)

Be warned that most books do not use a different notation for the charges rep-

resented by quantum ﬁeld operators and the charges represented by differential

operators, which can lead to some confusion. Even worse, the same symbol is often

used to represent the eigenvalues of the charges. For example, consider again P

Depending on the context, this may represent a differential operator, a quantum

ﬁeld operator, or the actual four-momentum of a state! Always make sure that you

know which of these meaning is implied when looking at an equation that contains

a charge. In this chapter we use a caret, as in

, for the differential operator repre-

sentation and a simple capital letter, as in P

, for the quantum ﬁeld operator. Later,

when we will need to represent the eigenvalues of the momentum operator, we will

use a lowercase letter, as in p

Once we know the representation of the charges as differential operators, it is a

trivial matter to work out the algebra they obey (see Exercise 6.3 for the example

of the Poincar´e algebra).

To summarize, the strategy is to start from transformations of the coordinates

that leave the lagrangian density invariant (up to total derivatives) and from there

to go through the following steps:

Transformation of coordinates ⇒ charges as differential operators ⇒ algebra (6.31)

At ﬁrst, it may seem as if this approach is irrelevant to SUSY, a symmetry relating

boson and fermion ﬁelds. However, as already mentioned in Chapter 1, and as we

will show explicitly shortly, there is a deep connection between SUSY and Lorentz

transformations, a connection that can be formalized by deﬁning an extension of

spacetime called superspace, which contains Grassmann coordinates in addition

to the usual spatial and time coordinates. One then can write the SUSY charges as

differential operators acting in that extended space. This approach will be pursued

in Chapter 10.

If the algebra of the charges is known, one actually can reverse the steps illustrated

in Eq. (6.31). One may use the algebra to work out the transformation of the

coordinates. This is how we will use Eq. (6.31) in Chapter 10, where we will start

from the algebra of the SUSY charges to work out the SUSY transformations of

the superspace coordinates.

We have so far discussed two ways to obtain the algebra of the charges by building

explicit representations for them. It turns out that it is also possible to obtain the

algebra without ever writing down any explicit representation of the charges. In

CHAPTER 6 The Supersymmetric Charges

111

other words, one may start from the transformations of the ﬁelds leaving the action

invariant and obtain the algebra directly:

Field transformations ⇒ algebra (6.32)

There is also a shortcut in the case of spacetime symmetries: One can start from

the transformation of the coordinates and obtain the algebra of the charges directly:

Coordinate transformations ⇒ algebra (6.33)

In the case of SUSY, the only information we have so far is the ﬁeld transforma-

tions that leave the action invariant, so the only two routes available to us to ﬁnd

the algebra are Eqs. (6.24) and (6.32) (as noted earlier, the representation of SUSY

transformations as changes of coordinates will have to wait until Chapter 10, where

we introduce superspace). In the next section we will work out the algebra using

the shortcut of Eq. (6.32).

EXERCISE 6.3

Using the representations of the operators P

and M

μν

as differential operators

[as given in Eqs. (6.28) and (6.30)], show that they obey the so-called Poincar´e

algebra, i.e.,

[

] = 0

[

μν

] = i(η

νλ

− η

λμ

)

[

μν

ρσ

] = i(η

νρ

μσ

− η

νσ

μρ

− η

μρ

νσ

+ η

μσ

νρ

) (6.34)

Warning: Again, several different conventions are used in the literature, which leads

to commutation relations differing by an overall sign.

6.3 Finding the Algebra Without

the Explicit Charges

We will now show how the knowledge of the ﬁeld transformations can be used to

obtained the algebra of the charges directly, following Eq. (6.32). In this section,

all the charges have to be thought of as quantum ﬁeld operators, not differential

operators.

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Supersymmetry Demystified

To ﬁnd the algebra of the charges generating a symmetry, we clearly need to

consider two transformations in order to obtain an expression with a product of two

charges. Let us then consider two successive transformations applied to a ﬁeld φ:a

ﬁrst transformation with parameters α

, followed by a second transformation with

parameter β

. We will use a dot-product notation for the combinations α

and

We have

φ U

†

= exp(iβ · Q) exp(iα · Q)φ exp(−iα · Q) exp(−iβ · Q)

≈ φ + i[β · Q,φ] +i [ α · Q ,φ]

−



β · Q , [ α · Q ,φ]



+··· (6.35)

where we neglected all terms of order α

or β

but kept the term bilinear in α and β.

By deﬁnition, Eq. (6.35) is simply a variation with parameter α followed by a

variation with parameter β so that we can, of course, also write

φ U

†

= δ

φ (6.36)

We now consider the commutator of two transformations, more precisely

φ −δ

φ (6.37)

This is equal to

φ U

†

−U

φU

†

≈−



β · Q , [ α · Q ,φ]





α · Q , [ β · Q ,φ]



(6.38)

Setting Eq. (6.38) equal to Eq. (6.37), we get

φ −δ

φ =−



β · Q , [ α · Q ,φ]





α · Q , [ β · Q ,φ]



(6.39)

CHAPTER 6 The Supersymmetric Charges

113

If we expand this out, being careful about the order of all the quantities involved,

we ﬁnd that half the terms cancel out, leaving

φ −δ

φ = α · Q β · Q φ − β · Q α · Q φ −φα· Q β · Q

+φβ· Q α · Q

= (α· Q β · Q − β · Q α · Q )φ

−φ(α · Q β · Q − β · Q α · Q )

= [ α · Q ,β· Q ] φ − φ[ α · Q ,β· Q ]



[ α · Q ,β· Q ],φ



(6.40)

The same result, of course, can be obtained using the Jacobi identity, which may

be written in the form



A, [B

, C]



−



B, [A, C]





[A, B], C



(6.41)

Applying this to Eq. (6.39) gives right away the last line of Eq. (6.40).

The next step is to factor out the inﬁnitesimal parameters, but the details depend

on whether the charges are bosonic, which is the case for all continuous symmetries

encountered in introductory quantum ﬁeld theory classes, or fermionic, as is the

case in SUSY. When the charges are bosonic, Eq. (6.40) leads to the commutation

relations between the charges, whereas for fermionic charges, the equation yields

anticommutation rules, as we will see explicitly soon.

6.4 Example

Before tackling the more complex SUSY case, it might be useful to pause and do

a couple of examples with more familiar symmetries.

We will illustrate the use of Eq. (6.40) in the case of the algebra between the

charges P

. First, consider the left side of the equation. If we apply a second

translation to δ

φ(x

) with an inﬁnitesimal displacement b

this time, we get

φ(x

) = δ



φ(x

+ a

) − φ(x

)



= δ

φ(x

+ a

) − δ

φ(x

)

= φ(x

+ a

+ b

) − φ(x

+ a

) − φ(x

+ b

) + φ(x

) (6.42)

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Supersymmetry Demystified

This result is symmetric under the exchange a

↔ b

, which is not surprising

because the order of the two translations does not matter. This implies that

φ(x

) = δ

φ(x

) (6.43)

so the left-hand side of Eq. (6.40) is zero.

Now let us work on the right-hand side of Eq. (6.40). We take the ﬁrst transfor-

mation to be

= exp(ia

) (6.44)

and the second one to be

= exp(ib

) (6.45)

In Eq. (6.40) we therefore replace α · Q by a

and β · Q by b

, so Eq. (6.40)

is ﬁnally given by



, b

],φ



= a



, P

],φ



= 0 (6.46)

which implies



, P

],φ



= 0 (6.47)

Now we want to obtain from this the commutator [P

, P

]. Obviously,

, P

] = 0 is a possible solution, but is it the most general one?

This is one difﬁculty in using Eq. (6.40): The ﬁnal step involves extracting from

an expression of the form [[Q

, Q

],φ] the commutator between the charges (or

the anticommutator for fermionic charges). This is not always trivial. One must

make sure that one writes the most general commutator satisfying Eq. (6.40), and

there is, at ﬁrst sight, always more than one possible answer. However, symmetry

arguments are usually sufﬁcient to pin down the solution.

For example, one could argue that taking the commutator [P

, P

] to be a constant

still would satisfy Eq. (6.47). But we need to have something with two Lorentz

indices, so we could try

, P

] = cη

μν

(6.48)

This obviously satisﬁes Eq. (6.47), but there is a problem: The left-hand side of Eq.

(6.48) is antisymmetric under the exchange of the two Lorentz indices, whereas the

CHAPTER 6 The Supersymmetric Charges

115

right-hand side is symmetric. This forces us to set c = 0, giving

, P

] = 0 (6.49)

which is indeed the correct answer.

The key point of this example is that we obtained the commutation relations of

the charges P

without ever writing them down explicitly! As a less trivial example,

you are invited to prove again the second equation of Eq. (6.34), i.e.,

μν

, P

] = i (η

νλ

− η

λμ

). (6.50)

EXERCISE 6.4

Prove the commutation relations (6.50) using Eqs. (6.40) and (6.20).

6.5 The SUSY Algebra

We will now use Eq. (6.40) to ﬁnd the algebra of the SUSY charges!

But ﬁrst let us ask: how many charges do we have? Well, how many inﬁnitesimal

parameters are involved in SUSY tranformations? There is the two-component

spinor ζ and its complex conjugate ζ

∗

for a total of four parameters. We therefore

need four charges Q

, Q

†

, and Q

†

, which we obviously (see below) can group

into a Weyl spinor, that we will simply call Q, and its hermitian conjugate Q

†

These SUSY charges are also often referred to as supercharges (what a surprise).

Until now, we have used Q to represent charges in general, but for the rest of this

book it will be used to speciﬁcally represent the SUSY charges.

The argument of the exponential in the unitary operator U must be Lorentz-

invariant, so we must combine Q and Q

†

with ζ and ζ

∗

in a Lorentz-invariant way

(which is why it was obvious that the supercharges had to be combined into a Weyl

spinor). We choose to make Q a left-chiral spinor. We know from Chapter 2 that

the two possible invariant combinations are then

Q ·ζ = Q (−iσ

)ζ

and

Q ·

ζ = Q

†

iσ

∗

(recall that ζ is not a quantum ﬁeld operator, so we may write ζ

†T

simply as ζ

∗

Written in terms of the supercharges, the unitary operator U generating SUSY