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1 Spinors

Let us first discuss objects with upper and lower indices in a general finite dimensionalspace, without assuming a metric. We can get the required equations from many places,

for example Weyl, Space-Time-Matter pp. 33-35. Suppose we have contravariant vectorswith components i. The transformation law for such vectors under a transformation Tis

i = Ti j j (1)

where the transformation T is required to have an inverse. Covariant vectors i have thetransformation law

j = Ti

j i (2)

In terms of the inverse of T, the transformation laws read

j = (T1)

ij j

j

= (T1

)

j

i i

Once we have upper and lower indexed objects we can define an invariant product:

jj = j Ti j

i = T

ij

ji = ii (3)

1.1 Two Component Spinors

We now specialize to the case of two-component complex objects, called spinors. Greekletters with Greek indices usually denote these, e.g. , = 1, 2, with the transformationdenoted by M, so the transformation equation for a spinor with an upper index is

= M

(4)The general discussion given in the previous section applies here. However, in addition tohaving an inverse, for relativistic field theory the transformation matrix M is required tobe unimodular, i.e. det(M) = 1. The set of complex 22 matrices with unit determinantforms a group which is the covering group of the group of (d = 4) Lorentz transformations.This will become clear later in this section. For unimodular transformations, we canintroduce a metric in the spinor space which allows us to raise and lower indices. Let and be two spinors and consider (1221). Using our transformation equation, itis easy to show that

(12

21) = det(M)(12

21) = (12

21) (5)

Exercise 1 Write out the matrix M as

M =

a bc d

, (6)

and verify Eq.(5) explicitly.

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Antisymmetric Metric Using Eq.(5), we may define objects with lower indices by

(12 21) (11 + 22), (7)where the minus sign on the right hand side is conventional. Comparing the left and

right sides of Eq.(7), we have 1

= 2 and 2

= 1. Denoting the metric by the matrixC, this is equivalent to

= C, where C =

0 11 0

. (8)

We also have 1 = 2 and 2 = 1. This is equivalent to

= C, where C =

0 11 0

. (9)

Thus C is antisymmetric and CC = . The totally antisymmetric symbol

isoften written instead of C. As we shall see later, the C matrix is intimately tied to

the operation of charge conjugation. With it, we may define an invariant scalar productbetween spinors:

= C. (10)Eq.(10) is analagous to the scalar product between two 4-vectors x and y

x y xy = xgy (11)Under a Lorentz transformation , the 4-vector x transforms according to

x = x, (12)

and the requirement x y = x y gives the following condition on

g

= g. (13)Eq.(13) can be written as a pseudo-orthogonality condition,

= , (14)

where = g

g

. Likewise, the requirement = gives the followingcondition on M,

M C M

= C . (15)

Exercise 2 Show that Eq.(15) is equivalent to the statement that det(M) = 1.

Eq.(15) can also be written as a pseudo-orthogonality condition,M M

= , (16)

where 1

M = CM

C

. (17)

1For raising/lowering indices on matrices with spinor indices, our convention is that the factors of Care applied so that the indices read in the order of matrix multiplication, as in Eq.(17).

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Symmetric Scalar Product and Grassmann Variables In the scalar product oftwo spinors, , we make the convention that the first spinor has an upperindex and the second has a lower index. Since and are on the same footing, wewant to equal . Due to the anti-symmetry of C, this is possible only if and

are treated as anti-commuting or Grassmann variables, satisfying

=

. So-called anticommuting c-numbers are familiar in path integrals involving Fermi fields. Forexample the path integral description of QED treats the electron-positron Dirac spinor as four Grassmann variables or anti-commuting c-numbers. We will assume from nowon that all spinor objects are anti-commuting.

In the context of supersymmetry, spinor Grassmann variables will occur in two mainways. Those which are independent of the spacetime coordinates x will be coordinatesin superspace. Spinors which depend on x will correspond to Fermi fields which appearin the action and are variables of integration in the path integral.

Assuming that and are anti-commuting, we have

= = C = C = C = = , (18)so the anti-symmetry of C and the anti-commuting property of , indeed lead to asymmetric scalar product.

Dotted and Undotted Indices and Conjugation The set of unimodular complex2 2 matrices forms a group under matrix multiplication, e.g.

(MN) = M

N

(19)

The condition det(M) = 1 imposes two conditions on the eight complex elements of M,

so the group has six parameters, appropriate for the Lorentz group, so a more completenotation for the unimodular matrix M would be M(), analogous to U(R) for rotations.Unity for the determinant and the group multiplication property are preserved undercomplex conjugation. We may then ask whether the complex conjugate representation isequivalent to the original. While this is well-known to be true for unitarytwo dimensionalmatrices, it is false for the broader class of unimodular matrices. A simple exampleillustrates this. Consider the unimodular matrix

A =

exp(z) 0

0 exp(z)

, (20)

for z complex. A matrix S which purports to transform A to A must here satisfyS1AS = A, or AS = SA. But writing this out we find that for aribtrary z, it can onlybe satisified by S = 0. We conclude that no matrix S exists which can transform the fullset of unimodular matrices M into their complex conjugates. A representation and itscomplex conjugate are in fact two dimensional representations of the unimodular group.

It is conventional to distinguish the two representations by the use of dotted andundotted indices. An overbar is often used as well. We define (), and M

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(M ), and likewise for other matrices.. With these conventions, the analog of Eq.(4)

is = M

(21)

So far, it would appear that the use of dotted indices could be omitted, the required

information being in the overbar on M or . While this is true for spinors and matriceslike M which have indices of the same type, there are matrices one of whose indicesrefers to one representation and the other to the complex conjugate representation. Theoverbar on such a matrix is then insufficient, and distinguishing dotted and undottedindices is in fact necessary.

Conjugation It is useful to define the notion of conjugation so that it works in asimilar way to taking the adjoint in operator field theory. We write

() . (22)

Eq.(22) says two things. First, in taking the adjoint, we move the spinors past each otherwith no Grassmann minus sign. Second, in the scalar product of barred spinors, the firstfactor has a lower index, while the second has an upper index. Taking C = C

and C = C, it is easy to show that = , similar to Eq.(18). Referring back to

Eq.(22), it is implicit that (z) = z, where z is either an ordinary complex number,or matrix therof sandwiched between and .

Lorentz Transformations and Pauli Matrix Four-vectors

Pauli Matrix Three-vectors It is familiar in the non-relativistic quantum me-

chanics of a spin 12

particle that the Pauli matrices k transform as a three-vector underrotations. Consider the expression k, where is an ordinary two-component spinor.Under an active rotation, = U(R), where U(R) is a 2 2 unitary matrix. Thenwe have

()k = (UkU = Rkj j, (23)

which impliesUkU = Rkj j, (24)

where Rkj is the 3 3 matrix describing the rotation. For example if the spinor rotationis

U(R) = exp(i3

2 ), (25)

then Rij is

Rij =

cos sin 0sin cos 0

0 0 1

(26)

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Exercise 3

1. Write down the matrices Rij for the cases

U(R) = exp(i1

2 )andU(R) = exp(i2

2 ).

2. Check that the 2-dimensional matrices U(R) perform the correct active rotationson the Pauli matrices, as in Eq.(23).

A useful property of the 3-vector Pauli matrices is the anti-commutator

ij + ji = ij (27)

Pauli Matrix Four-vectors Under the full Lorentz group, the three Pauli ma-trices plus 2-dimensional identity form a 4-vector matrix, denoted either as or ,depending on the spinor index structure. Both forms have one dotted and one un-dottedindex. The group theoretic explanation for this will be given in a later section. It isconventional to give the index structure

and define it by

() (I, ) (28)

The second fvec matrix has the index structure (), and is defined by

() (I,) (29)

The connection between and is established as follows. We first raise indices on .

() C()C. (30)

Using Eq.(8) and C = C gives

() =I,

. (31)

Taking the complex conjugate and using Eq.(29), we finally have

() = () = (), (32)

where the last equality is just the statement that are hermitian matrices.Placing between spinors and we obtain a current

=

. (33)

An expression of the form Eq.(33) forms part of the electromagnetic current in QED.

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Placing between the spinors and gives another current,

= () (34)

The sum of expressions of the form of Eq.(34) and Eq.(33) gives the total electromag-

netic current in QED. Supersymmetric QED has an additional contribution from chargedbosons.

Under an active Lorentz transformation, () = M , and () =M

. This in turn induces a Lorentz transformation on :

= , (35)where is the usual 44 matrix describing a Lorentz transformation (combination ofrotations and boosts) acting on a 4-vector. Substituting for and and then strippingoff the spinor factors, we find the equation describing the transformation of ;

M ()M

=

(

). (36)

M () M

= (). (37)

Despite its formidable index structure, Eq.(36) is the analog for Lorentz transforma-tions of Eq.(24) for rotations. In fact by comparing Eq.(36) and Eq.(24) we can deducewhat the matrices M and M should be for a rotation. Suppose R is an arbitrary rota-tion involving three angles and specified by the two dimensional unitary matrix U(R).Looking at the right factor of U(R) in Eq.(24), we can see that to make Eq.(36) work forrotations we must take

M = U(R)

(38)

i.e. we simply equate the matrix elements of M with the ordinary matrix elements of

U(R). Once we know what M is, we know M, i.e. M = (M ), so we have

M = (U(R)) (39)

i.e. we set M equal to the matrix elements of fU(R). From Eq.(24) Finally, noting thatin Eq.(36) it is the left index of M which is summed over, and using the fact that thePauli matrices are hermitian, we see that Eq.(36) is equivalent to Eq.(24) for rotations,i.e. k transforms as a three-vector, and likewise, 0 transforms as a rotational scalar.

Velocity Transformations Now we turn to velocity transformations or boosts.Consider first a Lorentz transformation which is an active boost along the 3-axis with

rapidity . Without displaying all indices explicitly, Eq.(36) becomes the following setof four equations:

MT0M = cosh 0 + sinh 3

MT3M = cosh 3 + sinh 0

MT1M = 1

MT2M = 2,

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which implies after a trivial phase choice that

M = (cosh

20 + sinh

23) (40)

Generalizing to an arbitrary boost, the matrix B is given by

B = exp( 2

). (41)

Thus for an arbitrary Lorentz transformation, we have

M = (M

) = (BR) , (42)

with B and R given by Eqs.(41) and (??), respectively. It is clear that the most generalunimodular two-dimensional matrix can be written as the exponential of the most generaltraceless matrix. Eq.(42) displays this explicitly, the real part of the matrix in theexponential representing velocity transformations, while the imaginary part representsrotations.

Raising and Lowering Indices, Barring and Unbarring Let us return toEq.(4) and express it in terms of spinors with lowered indices. We obtain

C11 = M

C11 , (43)

or = M

(44)

whereM

= C1M1

1 C1, (45)

and we used CC = . Similarly, returning to Eq.(36) and expressing it in terms of

M and M

we have

M ()M

= (), (46)

where() = C1()11C

1 (47)

(Note the different ordering of indices in Eqs.(45) and (47).) A more useful object than

()

is its complex conjugate, defined by

()

()

(48)

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Exercise 4

1. Show using Eqs.(47) and (48), that

= I, = 0

k, = k = 1, 2, 3 (49)

2. Show using Eqs.(47) and (48), that

+ = 2g (50)

+ = 2g (51)

Just as we did previously for , we may sandwich between spinors,

= () (52)

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2 Transformations and the Lorentz Group

In this section we review space-time transformations, first in quantum mechanics, andthen in quantum field theory. The goal here is to recall how operators like the total mo-

mentum or total angular momentum act as differential operators on fields, in preparationfor a treatment of how the supertranslation operators act as differential operators on su-perfields. In the latter part of this section, the focus of attention is the Lorentz group.Here a group theoretic discussion of relativistic spin will be given. A particular applica-tion will be to derive the group theoretic labels of the two dimensional representationsdiscussed in the previous section.

We start by considering an ordinary quantum mechanical wave function (t, x) fora state . Suppose we want to perform a spacial translation of by an amount a. Wedenote the wave function of the translated state by

(t, x). Its relation to that for theoriginal state is given by

(t, x) =

(t, x a) = exp(a )

(t, x) (53)

For example, suppose (t, x) is large near x = 0. Then Eq.(53) says that the wavefunction of the translated state will be equally large near x = a.

In quantum mechanics, the momentum operator is given by p = i, so Eq.(53) canalso be written

(t, x) = exp(ia p)(t, x), (54)which is the familiar statement that the momentum operator is the generator of spacialtranslations.

In a similar manner, suppose we wish to transform the state to one rotated by

an angle around the 3-axis. Representing x by its spherical coordinates r,, andsuppressing all coordinates except for the azimuthal angle , we may write this out as

() = ( ) = exp(

)() (55)

In quantum mechanics, the orbital angular momentum operator is

L3 = (x1p2 x2p2) = 1i

, (56)

so we can write Eq.(55) as

() = ( ) = exp(iL3)(). (57)

If spin is present, the wave function carries spin indices, e.g. s , and there is now atotal angular momentum operator

Jk = Lk + Sk, (58)

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where Sk is a finite dimensional matrix which acts on the spin indices of the wave function.Eq.(55) becomes

s () = exp(iJ3)

s() =

s exp(iS3)

s ss( ). (59)

Eq.(59) exemplifies for the 3-axis the statement that the total angular momentum oper-ator components are the generators of rotations about their respective axes.

Field Transformations Let us move from non-relativistic quantum mechanics to rel-ativistic quantum field theory. We consider a quantum field operator (x), restrictingthe discussion temporarily to the case of a scalar field. A space-time transformation Twill be represented in the Hilbert space of states by a unitary transformation U(T). Theaction of the transformation on is

U(T)(x)U(T) = (T1x) (60)

For example, suppose T is a translation by a 4-vector a, with the corresponding unitarytransformation

U(a) = exp(iaP) = exp(ia0P0 ia P), (61)

where P is the 4-momentum operator in the Hilbert space of states.2

Eq.(60) becomes

exp(ia P)(x) exp(ia P) = (x a) = exp(a). (62)

If we take the limit of an infinitesimal translation, and match linear terms in a on both

sides of Eq.(62), we get the commutator

[(x), P] = i(x) P(x). (63)

Taking the matrix element of Eq.(62) between the vacuum |0 >, and an arbitrary state| >, we get

(x) = (x a) = exp(iaP)(x), (64)where < 0|| > is now a wave function-like object. Eq.(64) is in close analogy toEq.(53) and Eq.(54).

Turning to Lorentz transformations, a transformation which takes the 4-vector x tox, will be represented on the full Hilbert space of states by a unitary operator U().

The action of this on the scalar field (x) is given by

U()(x)U() = (1x). (65)

2Our notation is that when there is a need for a distinction, boldface capital letters, e.g. P areused for operators in the Hilbert space of states, while ordinary upper case letters denote correspondingdifferential operators, e.g. P, or matrices which act on fields or their indices. If the context is clear,ordinary capital letters are used.

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The full set of Lorentz transformations (rotations and velocity transformations) is gen-erated by an antisymmetric second rank tensor J, whose spacial components are theangular momentum operators,

J

23

= J23 = J

1

, J

31

= J31 = J

2

, J

31

= J31 = J

2

. (66)

In terms ofJ, and an anti-symmetric tensor of parameters , the operator U() can

be expressed as

U() = exp( i2

J). (67)

The 0k are rapidity parameters, and the ij are rotation angles, e.g. for the previouslyconsidered rotation by angle around the 3-axis, we would have 12 = 12 = . Takingthe limit of small and matching the linear term in on both sides of Eq.(67) givesthe commutator equation,

[(x), J] = i(x x)(x) L(x) (68)As it must, for both indices spacial, the differential operator on the right side of Eq.(68)agrees with the orbital angular momentum operator in quantum mechanics. For example,

L12 = L12 = i(x12 x21) = i(x1

x2 x2

x1) = (x1p2 x2p1) = L3 (69)

The commutation relations of the J with themselves (Lorentz group) and the addi-tional ones with the P (Poincare group) can easily be determined by making use of thedifferential operators L and P defined in Eqs.(63) and (68). We have

[J, J] = i g J + gJ gJ g J , (70)and

[J, P ] = i (g P g P ) . (71)

Velocity Transformations While the operators which generate rotations are veryfamiliar, those that generate velocity transformations or boosts are less so. In this sec-tion, we study the generators of velocity transformations in more detail. There are threeof these, commonly denoted as Kl, related to the three generators with space-time gen-erators, Jl0 . We start by setting Kl = zJl0 , where z = 1. Using Eq.(70), or the factthat the set of K

i

form a 3-vector under rotations, we have

[Jl , Km ] = ilmn Kn . (72)

From Eq.(70), the commutation relations of the Kl with themselves are

[Kl , Km ] = ilmn Jn . (73)

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It is worth noting that the minus sign on the right side of Eq.(73) is essential, and comesfrom the fact that the Lorentz group, as opposed to the rotation group, is non-compact.The result of Eqs.(72) and (73) is that these commutation relations are consistent witheither z = 1 or z = 1. To proceed further, we bring in commutation relations with the4-momentum operators. From Eq.(71) we have

[Jk0 , Pl ] = ikl P0 (74)

While this equation does not actually dictate the choice of z, it seems simplest to choosez = 1, or

Kl Jk0 , (75)and therefore

[Kl , Pm ] = ilm P0 ilm H, (76)where H = P0 is the Hamiltonian of the system. The only other non-trivial commutation

relation involving K

l

is, from Eq.(71),[Jk0 , P0 ] = [Kk , H] = iPk . (77)

Having chosen z, the next question is how to represent a velocity tranformation, i.e.for a positive boost with rapidity do we write exp(i K), or exp(i K) ? Wecan determine that the first case is correct by a simple arguement in the non-relativisticquantum mechanics of a particle with mass m. We return to the quantum mechanicalwave function for a state , but this time we consider the momentum space wave function(t, p). Suppose we want to perform a velocity transformation with velocity v on thisstate. We denote the boosted state by . The relation between boosted and unboosted

momentum space wave functions is

(t, p) = (t, pmv) = exp(mv p)(t, p). (78)Eq.(78) says that the boosted wave function has the same value at p+mv as the unboosted

wave function at p. Now in momentum space, x is an operator, given by x = ip. Wecan then write

(t, p) = exp(iv

cmcx)(t, p), (79)

where we have temporarily restored the velocity of light, c. For |v|

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where H = mc2 + . . .. Comparing to Eq.(76), we see that for small velocities we haveKl mcxl. Since the correct formula for a boost in terms of l and Kl will be of thesame structure for any value of the rapidity parameters, we conclude that the choiceEq.(75) implies that the form of a positive boost transformation is

exp(i K). (82)

We may now write a Lorentz transformation in terms of either the , or the rapidities and rotation angles ,

U() = exp( i2

J) = exp(i K i J). (83)

Many other six-parameter forms can be written down, but for Eq.(83), we can easilyrelate the to the rapidities and rotation angles;

k0

= k (84)lm = lmn n (85)

Matrix Reprentations of the Lorentz Group For a quantum field with spin, thefield carries spin indices, s, and Eq.(65) is modified to

U()s(x)U() =s

Dss()s(1x), (86)

where Dss() is a finite dimensional matrix representation of the Lorentz group. Thetask of finding all possible such representations is equivalent to finding all possible rep-

resentations of commutation relations of the Lorentz group given in Eq.(70). This job ismade much simpler by making use of the commutation relations expressed in terms ofthe Jl and Kl, as in Eq.(72) and Eq.(73) along with the well-known commutator of theJl with themselves. We define

Jl1 =1

2(Jl + iKl) (87)

Jl1 =1

2(Jl iKl). (88)

It follows directly from the commutation relation satisfied by the Jl and Kl, that the Jl1

and J

l

2 satisfy the commutation relation of two independent angular momenta. Writtenout, Jl1, J

m1

= ilmn Jn1 (89)

Jl2, Jm2

= ilmn Jn2 (90)

Jl1, Jm2

= 0. (91)

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Finding matrix representations of the Jl1 , Jl2 is straightforward using known results from

angular momentum theory. Each such representation defines a finite dimensional repre-sentation of the Lorentz group. These will be labelled by two quantum numbers j1, j2with values 0, 1/2, 1, 3/2, . . ., where

J1 J1 = j1(j1 + 1) (92)J2 J2 = j2(j2 + 1) (93)

Once a set of J1 , J2 matrices is specified, the J , K matrices are given by

J = J1 + J2 (94)

K = i( J1 + J2). (95)The corresponding finite dimensional representations of the Lorentz group generators J

are then obtained by using Eq.(66) and Eq.(75). It should be noted that the matrices sofound are notthe operators in the full Hilbert space, but appear only in the (non-unitary)

matrix Dss

of Eq.(86). The non-unitarity ofDss

has its origin in the non-compactness ofthe Lorentz group. This is associated with velocity transformations, and their generatorsK. From Eq.(95), we see that since angular momentum matrices are hermitian, the

matrices K must necessarily be anti-hermitian, and the resulting matrix representationsof velocity transformations, exp(i K), are non-unitary.

An irreducible finite dimensional representation is labelled by the two quantum num-bers j1, j2 , and we denote it by Dj1j2(). Two important results follow immediatelyfrom angular momentum theory.

1. The dimensionality of a representation Dj1j2 is (2j1 + 1)(2j2 + 1).

2. The product of two irreducible representions Dj1j2 and Dj

1j2 is obtained by the

rules for adding angular momenta.

From item 1, it follows that there are two distinct two-dimensional representations,D

1

20() and D0

1

2 (). These will be studied in more detail in the next section. Thefollowing table shows low-lying representations and the corresponding Lorentz group ob-

jects:

D00() Scalar

D1

2

1

2 () 4-vectorD11() Symmetric traceless second rank tensorD10()D01() Antisymmetric second rank tensor

As an example of reducing the product of two irreducible representations, considerthe product of two 4-vectors, D

1

2

1

2 D 12 12 (). From item 2 above, we getD

1

2

1

2 D 12 12 = D00 D10 D01 D11, (96)so the 16 = 4 4 components of the product comprise a scalar (1), antisymmetrictensor (6), and symmetric traceless tensor (9), where the numbers in parentheses denotenumbers of components.

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Spinor Representations We now return to the two dimensional spinor representa-tions first encountered in Sec.(1) and investigate their relation to the representations

D1

20() and D0

1

2 (). Consider first M which transforms a spinor according to

= M

. (97)

As for all representations of the Lorentz group, M can be written

M =

exp( i

2S

)

, (98)

where the matrices S must have the spinor index structure (S). The construction

of S uses products of the defined in Eq.(??) and the defined in Eq.(29). Weintroduce

() =

( )() ()( ), (99)

which is antisymmetric under and has the correct spinor index structure. FromEq.(99), ij = 4iijk k . Now for either 1

2, 0 or 0, 1

2representations, the Jk result from

adding spin 12

to spin 0 and thus are spin 12

. So

Jk =1

2k (100)

and

Sij = Sij = ijk Jk =

1

2ijk k . (101)

Comparing Sij and ij , we conclude

S =i

4. (102)

Exercise 5

Verify that the M defined in Eq.(102) satisfy the commutation relations Eq.(70).

From Eq.(75) and Eq.(99), the boost generators are

Kl =i

2l. (103)

Using J and K, we have for the Lorentz transformation Eq.(98),

M =

exp(

2 i

2)

. (104)

In Eq.(104), an arbitrary unimodular matrix is represented as the exponential of anarbitrary traceless matrix, with the hermitian part of the exponent representing velocity

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transformations, and the anti-hermitian part representing rotations. From J and K, wehave for J1 and J2,

J1 =1

2( J + i K) =

1

4( ) = 0, (105)

J2 = 12( J i K) = 14( + ) = 12. (106)These results state that M

is the representation with j1 = 0, and j2 =1

2, or D0

1

2 ().

For connecting to D1

20 (), we introduce a second matrix antisymmetric under

defined by() =

( )() ()( ) (107)

Exercise 6

Show using Eq.(??) that () = (() )

Defining

S = i4

, (108)

it is straightforward to verify that the S satisfy the commutation relations Eq.(70). Weuse S to generate a two dimensional representation of the Lorentz group by analogyto Eq.(98),

M =

exp( i2

S)

, (109)

Exercise 7

Consistency with our previous definitions demands that we have

M

= (C

M

C)

= (M

)

(110)

Verify the first equality. The second follows by definition.

From Eq.(99) and Eq.(107) it follows that Sij = Sij , so Jk is still given by Eq.(100).However, Sk0 = Sk0 , so now Kk is given by

Kk = i2

k. (111)

Returning to the equations for Jk1 and Jk2 , we now have

J1 =1

2( J + i K) =

1

4( + ) =

1

2, (112)

J2 =1

2( J i K) = 1

4( ) = 0. (113)

Together these equations state that M is D1

20 (). In this representation, an arbitrary

Lorentz transformation is

M =

exp(+

2 i

2)

. (114)

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As expected,Eq.(114) and Eq.(104) reduce to the same matrix for a pure rotation, sincethere is only one two dimensional representation of the rotation group.

With the group-theoretic classification of M and M in hand, we can understand

the index structure of

and (). Let us first note that

D01

2 ()D 120 () = D 12 12 (). (115)In words, the direct product of the two spinor representations is the vector representation.Returning to the equations that describe the action of a Lorentz tranformation on

and (), Eq.(36) and Eq.(37), we see that both and (essentially the complexconjugate of ), should be regarded as Clebsch-Gordon coefficients which accomplishthe reduction expressed in Eq.(115).

We can apply similar reasoning to () () . Taking (

) first, its two

undotted indices imply that it transforms according to D01

2 D0 12 . The reduction of thisproduct is

D

01

2

D01

2

= D

00

D01

(116)Only the D01 representation is relevant for ()

. This follows from the fact that()

is clearly not a Lorentz scalar, or more subtly, using the same arguement asExercise 1.

Another question arises when we note that is antisymmetric in . In Table2, a general antisymmetric tensor is listed as transforming as D01 D10 , which has 6independent components. What restriction on a general antisymmetric tensor leads tothe three component parts D01 or D10? Or, how does one project the D01 and D10 partsfrom a general antisymmetric tensor? The answers can be deduced from looking at theLorentz transformations properties of the most familiar such tensor, the field-strengthtensor in electrodynamics. This is a digression which will be useful in other ways later.

We start with the formulae for the transformations of electric and magnetic fields underan active velocity transformation of velocity v = c.

( E) = ( E B) (117)( B) = ( B + E) (118)( E) = E

( B) = B,

where =

1 . From these equations, it is easy to see that the combinationsE+ i B and E i B transform into each other under velocity transformations, and theyclearly also do so under rotations. Since each of these combinations has three independentcomponents, they must be the desired D01 and D10 parts of the field strength tensor.To find out which is which, we can write out the equations above for a boost along the+z axis. Denoting E+ i B as G, we have from Eq.(117) and Eq.(118)

Gx

Gy

Gz

= exp(iK3)

G

x

Gy

Gz

=

cosh() i sinh() 0i sinh() cosh() 0

0 0 1

.

G

x

Gy

Gz

(119)

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This gives for K3,

K3 =

0 1 01 0 00 0 0

. (120)

Now the ordinary cartesian 3 3 matrix J3 is given by (J3)ij = iij3, or

J3 = i 0 1 01 0 0

0 0 0

. (121)

Comparing Eq.(120) and Eq.(121), we have

K3 = iJ3. (122)The general result is Kl = iJl, for the combination E+i B. Returning to the quantitiesJl1 and J

l2 defined in Eq.(88), we obtain

Jl1 =1

2(Jl + iKl) = Jl (123)

Jl1 =1

2(Jl iKl) = 0. (124)

where here Jl are cartesian 3 3 matrices like J3 above. From Eq.(123) and Eq.(124)we conclude that for E+ i B, j1 = 1 and j2 = 0, so E+ i B transforms as D

10 . Entirelyanalogous steps lead to the conclusion that E i B transforms as D01 .

Dirac Spinors Having specified the two-dimensional representations of the Lorentz

group, we can reconstruct the Dirac 4-component representation, which has the groupstructure D0

1

2 D 120 . The Dirac matrices in our Weyl representation are given by

=

0

() 0

(125)

The standard anti-commutation relations of the Dirac -matrices follow from the prop-erties of and given in Eqs.(??). The index placement on the and matricesdefined in Eq.(??) and Eq.(29) dictates that the 4-component Dirac spinor is of theform

=

(126)In Dirac theory, Lorentz transformations involve the commutator of Dirac matrices. Ifwe define

= [, ], (127)

then we have

=

0

0

. (128)

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It follows that the Lorentz transformation matrices for Dirac spinors are given by SD =i4

. The fact that the SD are block diagonal makes explicit the D01

2 D 120 structureof a Dirac spinor.

The Dirac conjugate spinor c is defined in the usual way by c = CD0()T, where

CD

(CD)1

= (

)T

. In the representation Eq.(125), choosing a phase, we may takeCD = i20. Then

c =

0 C

C 0

=

, (129)

so that charge conjugation is equivalent to . A self-conjugate or Majorana spinorsatisfies c = , and is of the form

M =

(130)

The supersymmetry generators form a Majorana 4-component spinor, composed of Qand Q.

From a Dirac spinor, the two component parts are obtained by means of a chiralprojection. Defining

5 = i0123 =

I 0

0 I

, (131)

the left and right-handed parts of are

R =1

2(I+ 5) =

0

(132)

L =1

2(I 5) =

0

(133)

The handedness designations follow from the massless Dirac equation. In momentumspace for a particle travelling along the +3 axis, we have

p = 0, (134)

or

p0

0 00 2

1

2

= 0, (135)

which implies 1 = 0, 2 = 0, so the spinor has its spin aligned with the momentum.Similar steps show that the spinor has its spin aligned opposite to the momentum. Notethat since the spacial parts of and are the same, the meaning of spin projection

is the same for both and .Finally, it is useful to write the familiar Dirac bilinears in terms of their two-componentparts. We have

= + 5 = + = + 5 =

(136)

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3 The Supersymmetry Algebra

The supersymmetry algebra is an extension of the Poincare algebra 3. The latter con-sists of the operators M and P, generators of Lorentz transformations and space-time

translations respectively, satisfying the commutation relations Eq.(70) and Eq.(71). Thesimplest form of the supersymmetry algebra extends the Poincare algebra by introducingtwo spinor operators Q and Q

transforming according to the D01

2 and D1

20 represen-

tations of the Lorentz group. Both Q and Q are, like M and P, independent of x.

In addition, they are adjoints, Q (Q), where here the adjoint is taken in the senseof operator field theory.

3.1 The Poincare Group Commutation Relations

Before writing down the supersymmetry algebra, we study more closely the commutationrelations of fields with the Poincare group generators. Let us return to Eq.(86), take the

parameters in to zero, and match terms linear in on both sides. This gives

[, M(x)] = s(x) + l(x), (137)

where the spin indices are not displayed explicitly, s is the matrix representation of theLorentz generators appropriate for the field , and l is the same differential operatordefined in Eq.(68). In Eq.(137), can be any sort of bosonic or fermionic field. In aquantum field theory, operators like M , P and conserved charges arise as integralsover local densities constructed from such fields. For example, the electric charge operatorQ can be written as

Q=

J(x)d, (138)

where J is the conserved (J = 0) electromagnetic current, and is a spacelike

3-surface, i.e. one with a timelike normal. The realization of Eq.(137) for the 4-vector current is

[J (x), M] = (s) J(x) + lJ (x), (139)

where (s) is the spin matrix for the D1

2

1

2 (vector) representation of the Lorentzgroup. We expect Q to be a Lorentz scalar satisfying

[Q, M] = 0. (140)

Eq.(140) can be derived from Eq.(139) and current conservation, plus some extra rea-sonable physical requirements explored further in Exercise 8.

3In fact supersymmetry is the only possible extension of the Poincare algebra , although we will notpursue the proof here.

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Exercise 8

The simplest and most useful choice for the spacelike surface is the case where thenormal is always along the time axis. Then

d = d0 =1

3! 0ijkdxi

dxj

dxk

= d3

x, (141)

so the integration over a spacelike surface is just an ordinary 3-volume integration and

Q =

J0d3x. (142)

For this case, deduce the physical requirements that must be satisfied in order to deriveEq.(140) from Eq.(139) and current conservation. You will need to consider behaviorat spacial infinity and an integral over d3x of a certain quantity.

For a spacially uniform system, the total momentum 4-vector is a conserved quantitygiven by

P =

T(x)d, (143)

where T = T is the stress-energy tensor, which satisfies

[T(x), M] = (s) T(x) + (s) T

(x) + lT(x). (144)

As long as the stress tensor is conserved, T = 0, P is independent of x, and it

vector character implies[P , M] = (s) P

. (145)

Eq.(145) can be derived from Eq.(144) under the physical requirements of Exercise8. We note in passing that the vector representation spin matrix can be obtained bycomparing Eq.(145) and Eq.(71);

[P , M] = (s) P = i(gP gP). (146)

This gives for s in the vector representation,

(s) = i(g g), (147)

or

(s

)

= i(g

g

g

g

) (148)The discussion for commutators with the total momentum 4-vector runs parallel to

that just given for commutator with the Lorentz generators. For generators which areexpressed as integrals on a spacelike surface of a density, Eq.(63) plus integration byparts implies that the generator commutes with P. So

[Q, P] = 0 (149)

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[P, P] = 0. (150)

The exception is M which has non-zero commutators with the momentum 4-vector ,Eq.(71). This arises from the fact that the integrand of the expression for M is weightedwith powers of the coordinate x;

M =

(xT xT )d. (151)

In this case, starting from Eq.(63) applied to the stress tensor, integration by parts gen-erates non-zero terms which lead to Eq.(63) for the commutator between the momentum4-vector and M.

HERE SAT DEC 13, 2003 10:13 AM HERE TUE DEC 16, 2003 2:50 PM

Exercise 9

The commutation relations among P and M given in Eq.(70) and Eq.(71) imply

non-trivial commutators between the components of the stress tensor T

. These arevery general and must be satisfied in any relativistically invariant field theory. Sincethey are so general they can be found using free field theory. Consider the free masslessscalar field where

T = gL, (152)and the Lagrangian is L = 1

2

. In this theory, derive the non-vanishing elementsof

[T(t, x), T(t, x)]. (153)

The non-vanishing terms will again involve elements of the stress tensor, times spacialgradients of3(x

x). You can check your answers by looking at the paper by Boulware

and Deser, Journal of Mathematical Physics 8, p1470, Equations 7a-7e.

3.2 The Supersymmetry Algebra

We now take up the study of the operators Q and Q which extend the Poincare algebra.

By analogy with the previous cases, we expect these to be expressible as integrals ofdensities over a spacelike surface:

Q

J d (154)

and

Q

J d (155)

The supercurrents J and J are both conserved, J

= 0 and J

and are adjoints of

each other, J = (J )

. With one spinor and one vector index, the supercurrents are newobjects, not encountered in conventional quantum field theory. However, their commu-tation relations with the Poincare generators M and P are completely determined by

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the fact that they are fields with well-defined Lorentz group indices. So from translationinvariance we have

[J , P] = iJ , (156)

while from Lorentz invariance we have

[J , M] = (s)

J + (s) J

+ l

J . (157)

Although they are both denoted by s, the two occurences of s in Eq.(157) refer todifferent representations. The first, which acts on the spinor index of J is in the two

dimensional D01

2 representation, while the second, which acts on the vector index of J ,

is of course in the four dimensional D1

2

1

2 representation. The corresponding equationsfor J are very similar.

HERE AT 2:30 PM ON THURSDAY DECEMBER 18,2003By the same arguements and physical requirements we used previously for P and

M, it follows that

[Q, P

] = [QP

] = 0 (158)while for commutators with M we have

[Q, M] = (s)

Q (159)

and[Q, M] = (s) Q

. (160)

In Eq.(159), the D01

2 representation of s occurs, while in Eq.(160) it is the D01

2 repre-sentation.

The commutators of the Q and Q with the Poincare group generators which have

just been written down follow from general arguements and do not involve anything really

new. This changes when we turn to the algebra of the Qs and Qs with themselves.Returning to the supercurrents J and J

, it is clear that these are Fermi fields. This

follows from the fact that they have one spinor index and one vector index, and so describeeither spin 1

2or 3

2. The spin-statistics theorem is a general result in local quantum field

theory that states that an integer spin field is a Bose field, while a half-integer spin fieldis a Fermi field. Fermi fields naturally involve anti-commutators between themselves, sosince the Qs and Qs are integrals of Fermi fields, we are led to consider anti-commutatorsamongst the Qs and Qs.

The standard supersymmetry algebra is obtained by first setting

{Q, Q

}= 0 (161)

and{QQ} = 0 (162)

We delay for the moment a discussion of whether there is any possibility of having non-vanishing right hand sides to Eq.(161) and Eq.(162). Instead, we turn to considerationof

{Q, Q}. (163)

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We note that this anticommutator involves the product of an operator which transformsaccording to D0

1

2 and the other according to D01

2 . Thus the right hand side must involvean operator which transforms according to D0

1

2 . The only such operator in the Poincarealgebra is the momentum 4-vector , P. As discussed earlier, the Pauli matrix 4-vector

is a set of generalized Clebsch-Gordon coefficients which accomplish needed reduction,(See Eq.(115)). Thus we can write

{Q, Q} = c()P (164)where c is a constant.

3.3 The Supersymmetric Spectrum

Let us start with N = 1 supersymmetry, where we have one Q and its adjoint, Q =(Q)

. The only non-vanishing anti-commutator is

{Q, Q} = 2P, (165)

where recall

= (I, ). Writing out Eq.(165), we have

{Q1, Q1} = 2(011P0 + 311P3) = 2(P0 + P3) = 2(P0 P3) = 2P, (166)where we used

0 =

1 00 1

, 3 =

1 00 1

, (167)

Likewise

{Q2, Q2} = 2(022P0 + 322P3) = 2(P0 P3) = 2(P0 + P3) = 2P+, (168)The equations Eq.(166) and Eq.(168) are very similar to the corresponding equations intwo dimensional supersymmetry. Turning to components of momentum not along the3-axis,

{Q1, Q2} = 2(112P1 + 212P2) = 2(P1 iP2) = 2(P1 iP2) (169)and

{Q2, Q1} = 2(121P1 + 221P2) = 2(P1 + iP2) = 2(P1 + iP2) (170)Equations like Eq.(169) and Eq.(170) would have vanishing right sides in two dimensional

supersymmetry, where P

= (P

1

, P

2

) = 0.If we sum the equations in P, we get

{Q1, (Q1)}+ {Q1, (Q1)} = 4P0 = 4H. (171)If the ground or vacuum state |0 > is supersymmetric (annihilated by all Qs and Qs,then < 0|H|0 >= 0, or in words, the vacuum energy is automatically zero and does notneed infinite subtractions.

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Massless Particles Let us now investigate the particle spectrum and show that ithas bose-fermi symmetry. We will investigate single particle states, first taking the caseof massless particles. For massless particles, there is no rest state, but we can alwaysrotate so the linear momentum is along the +3 axis, so P+ > 0, and P = P = 0.

Let us denote a state of this type with helicity by |P+

>. We will act on thisstate with the Qs and Qs. This will not change the 4-momentum of the state since[Q, P

] = [Q, P] = 0. In our present sector of states with P = 0, Eq.(166) implies

that< P+, |

Q1Q

1 + Q

1Q1

|P+ >= 0. (172)

Now each term in Eq.(172) is in the form of < | >, i.e. the overlap of a state withitself. The result is necessarily positive, unless the state itself vanishes. Since the resultis zero, we conclude that Q1|P+ >= Q1|P+ >= 0. Thus on massless states with onlyP+ = 0, we can concentrate on Q2 and Q2 = Q2. From Eq.(159) and Eq.(102), we have

[Q2, J12

] = [Q2, J3

] =1

2(12

)22

Q2 = 1

2Q2. (173)

Applying Eq.(173) to |P+ >, and using the definition of helicity (J3|P+ >= |P+ >),we have

J3Q2|P+ >= ( + 12

)Q2|P+ > (174)which states that Q2 is a helicity raising operator. Taking the adjoint of Eq.(173) gives

[Q2, J3] ==

1

2Q2. (175)

A similar arguement to that just given implies that Q2 = Q

2 is a helicity loweringoperator.Now in a massless multiplet with our given value of P+, there will be a maximum

helicity state, |P+, m > with m as the maximum helicity. We will regard this state asthe vacuum with respect to Q2 and Q

2,

|0 > |P+, m > . (176)

Rescaling, we define creation and destruction operators

a =Q2

2P+

, a =(Q2)

2P+

. (177)

In terms of a and a, Eq.(168) becomes

{a, a} = 1. (178)

and we also have {a, a} = 0, and {a, a} = 0. Since a|0 >= 0 and aa = 0, ourspace space consists of just two states, |0 > and a|0 >. They both have the same

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value of 4-momentum and every charge-like quantum number. But if |0 > has helicitym, then a

|0 > has helicity m 12 . Thus the simplest supersymmetry algebra requiresthat massless particles occur in boson-fermion pairs, separated by 1

2in helicity. For

example in supersymmetric quantum electrodynamics, a photon of helicity 1 must be

accompanied in the spectrum by a photino of helicity1

2 . Although electrodynamics isparity invariant, and therefore supersymmetric electrodynamics will have photons andphotinos of helicity -1 and 1

2, respectively, the N =1 supersymmetry algebra alone

does not require a negative helicity for every positive helicity. Thus for N = 1, chiraltheories are possible.

It is simple to extend the treatment of massless particles to the case of N > 1 super-symmetry. Treating multiplets where only P+ is non-vanishing, we are again concernedonly with Q2 and its adjoint, but there is now an additional label, Q

I2, I = 1, 2, . . . N .

We can define destruction and creation operators as before,

aI =QI2

2P+, (aI) =

(QI2)

2P+, (179)

and we have{aI, (aJ)} = IJ. (180)

We again assume our massless multiplet has a maximum helicity state, |P+, m >. Thisis the vacuum as before, destroyed by all the aJ. But now we can create N states ofhelicity m 12 of the form (aI)|0 >. None of them can vanish, since aI(aI)+(aI)aI = 1.They are orthogonal, since the (aI) anti-commute. So there are N distinct states ofhelicity m 12 . States of helicity m 1 also exist and are of the form (aI)(aJ)|0 >,I = J. There are N(N 1)/2 m 1 states. The state of lowest helicity is unique;(a

1

)

(a

2

)

. . . (a

N

)

|0 >. This state has helicity m N

2 . To illustrate for N = 2, if m = 1, we have a multiplet consisting of = 1, two states with =1

2and one with

= 0. Also for N = 2, if m =1

2then we have a multiplet consisting of = 1

2, two

states with = 0 and a state with = 12

. Since both positive and negative helicitiesfor spin 1

2particles occur, an N = 2 theory cannot be chiral. The case of supersymmetry

with N = 4 and m = 1, naturally includes = 1, and has many properties whichcan be understood analytically. For theories containing a graviton with = 2, naturalinclusion of both of these helicities requires supersymmetry with N = 8. Values ofN > 8 necessarily involve massless particles with helicity greater than 2, for which nosatisfactory field-theoretic description exists.

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4 The Supersymmetry Generators as Differential Op-

erators

HERE AT 2PM ON DEC 31,2003

In both ordinary quantum mechanics as well as quantum field theory we are ac-customed to representing space-time generators as differential operators. This was dis-cussed in Section 2. Supersymmetric field theory is most efficiently expressed in su-perspace. In superspace, in addition to the usual space-time coordinate x, we haveGrassmann coordinates and . Just as the total 4-momentum operator P

moves ortranslates x, so the operators Q and Q will move and . Of course, the actionof Q and Q is not purely a translation in and because of the non-vanishinganti-commutatator {Q, Q}, written down in Eq.(164).

4.1 Preliminary

To learn how to represent Q and Q as differential operators, we will proceed heuris-tically and temporarily ignore the non-zero anti-commutator of Eq.(164). In quantummechanics, for a wave function , we write

(x + a) = exp(ad

dx)(x) exp(iap)(x), (181)

which defines the momentum operator p by

p =1

iddx (182)

Suppose now we have a wave function which depends only on a single Grassmann variable. By analogy with an ordinary translation, we write

( + ) = exp(d

d)() exp(iQ)(s), (183)

where is a Grassmann variable,

iQ =d

d. (184)

Derivatives of Grassmann variables are also anti-commuting. We take them to act fromthe left, so

dd

= 1, (185)

andd

d = . (186)

In words, we move the derivative operator from the left keeping track of signs fromGrassmann interchanges. Action of the operator on the variable takes place when the

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operator is just to the left of the variable. Right derivatives can also be defined, and willbe used later. For the moment, all Grassmann derivatives are left derivatives.

HERE AT AROUND 11AM JAN 1,2004In d = 4 supersymmetric field theory, we have either dotted or undotted indices on

the Q

s and Q

s. The corresponding superspace coordinates also carry these indices, and . Considering for the moment a wave function that depends only on

, we expandEq.(183) to

exp(iQ)() = ( + ), (187)

or explicitlyexp(iQ)(

) = ( + ), (188)

which requires

iQ =

, (189)

where

.Having defined , we raise the index and define

CThis relates and

in the same manner as the upper and lower components of anordinary spinor. Now consider

()() = =

=

= = (190)

Since the scalar product of spinors is symmetric we can also evaluate this as

()() = =

= () = = = . (191)

The correct term on the far right of the equation is obtained only if

= . (192)

Then defining

,

we have =

,

where the minus sign has its origin in the Grassmann nature of our variables and theirderivatives.

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Exercise 9

We have (i) = and (ii)

= . Derive (ii) from (i) by raising and lowerindices. Use Eq.(192) and the properties of the metric matrix C.

Summarizing, we have

iQ = =

, (193)

iQ = =

. (194)

We may carry out a similar set of steps to deduce the form of Q as a differentialoperator. We start from a wave function that depends only on and write

exp(iQ)() = ( + ), (195)

or explicitly

exp(iQ)() = ( + ), (196)

which requires

iQ =

. (197)

We now lower the index and define

= C , (198)

so and are related in the same manner as upper and lower components of a dotted

index or barred spinor. Making the natural definitions

=

, (199)

= , (200)

and repeating the steps of Eq.(190) and Eq.(191) using barred spinors we can deducethat

=

. (201)

Summarizing for the Q

s, we have

iQ = =

, (202)

iQ = =

. (203)

HERE ON FRIDAY JANUARY 2, 2004

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Right Derivatives So far the discussion has been given exclusively in terms of leftderivatives. This gives formulae like Eq.(188) and Eq.(196) a natural appearance, witha structure reminiscent of quantum mechanics. However, we can equally well write

()exp(iQ) = ( + ), (204)

and()exp(iQ

) = ( + ), (205)

In Eq.(204) and Eq.(205) the derivatives will be right derivatives. The basic right deriva-tives are defined by

= ,

= ,

and completely analogous equations for right derivatives of barred variables. In thisnotation, in our previous formulae left derivatives would have been denoted

, etc.

After the next few equations, we will eliminate the over arrows on both left and rightderivatives. The formulae analogous to those previously given in terms of left derivativesare easily obtained. We have

iQ =

=

, (206)

iQ =

=

, (207)

iQ = =

, (208)

iQ =

=

. (209)

HERE AT 928AM ON MONDAY JANUARY 5, 2004We now drop the over arrows on all derivatives, relying on the relative placement

of the and symbols to determine the direction in which the acts. 4 With thisunderstanding we have

= = , (210)

=

=

, (211)

and for variables with dotted indices,

= = , (212) = = . (213)

4In cases where confusion could arise, the over arrows will be restored.

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Both of these sets of equations are consistent with the Grassmann nature of the s ands. As an example, consider the evaluation of (). We have

() = = =

, (214)

but equally well() = = () = . (215)

HERE AT 12:40PM ON SATURDAY JANUARY 3, 2004

Exercise 10

There are two ways to evaluate ()() = ()() using left derivatives and two withright derivatives. Show that all four methods lead to ()() = () =

Finally, we note that under the bar or conjugation operation, a left derivative becomes

a (barred) right derivative and vice versa. Consider() = . (216)

Now take the comjugate of both sides, giving

() = . (217)

If we follow the rule that conjugation reorders factors with no minus signs (see thediscussion near Eq.(22)), we get

() = () = () = =

= , (218)

consistent with Eq.(217).HERE AT 250PM ON TUESDAY JANUARY 6, 2004

4.2 The Full Operators

The forms for the Qs and Qs of the previous subsection need to be augmented so thatthey satisfy the correct anticommutation equation Eq.(164). We start with the Hilbertspace operator form of the anticommutator and attach constant spinors and . Wehave

(QQ + QQ = 2

P)

, (219)

which gives the commutator equation,

[(Q), Q] = 2 P (220)

As a differential operator, the four-momentum is P = i, so we seek differential Q andQ satisfying

[i(Q), i(Q)] = 2i (221)

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Notation

To streamline the notation, we make the following definitions:

() , (222)

() = , (223)

A (A ) A , (224)where in the last equation, A could be a matrix like

. The same symbol can

denote a derivative with respect to the coordinate x, or the spinor variable . Noconfusion should arise, since when indices are displayed, indices from the beginning ofthe Greek alphabet like , imply a derivative with respect to , while those from themiddle or the end of the Greek alphabet like , , , . . . imply a derivative with respectto x. If indices are not displayed, spacial derivatives are preceded by a central dot asin Eq.(224), whereas no dot as in Eq.(222) implies a derivative with respect to . The

parentheses in the above equations have no particular significance, and are only usedto improve the readability of certain expressions.

We now introduce augmented differential operators Q and Q which satisfy Eq.(221).For iQ we write

iQ = + i( ), (225)and for iQ,

iQ = i( ). (226)

Exercise 12Using Eq.(225) and Eq.(226), check that Eq.(221) is satisfied.

The forms for Q and Q given in Eq.(225) and Eq.(226) are not unique; the partsinvolving can have different coefficients and still satisfy Eq.(221). We will not explorethis ambiguity here.

4.3 Supertranslations

We are now in a position to work out the effect of superymmetry transformations whichare often called supertranslations. The most general such transformation would be

exp(ia P + iQ + iQ)(x,, ). (227)

The function (x,, ) is closely connected to a superfield. These will be defined in moredetail in a later section. For the moment, it will suffice to regard as an object similar toa wave function, or quantum field matrix element, or variable in a path integral, except

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that it depends on Grassmann variables as well as x. We proceed in stages, and firstcompute

exp(iQ)(x,, ) = exp( i( ))(x,, ) (228)This is simple to evaluate. The right hand side of Eq.(228) is of the form

exp(+ b )(x,, ) (229)where b = i. Since and b commute, the result is simply to shift x by b and by . We have

exp(iQ)(x,, ) = (x + b,, + ). (230)

The new feature compared to an ordinary translation is that the coordinate x gets movedby an amount which depends on Grassmann variables. Now let us apply a factor ofexp(iQ) and compute

exp(iQ)exp(iQ)(x,, ) = exp(iQ)(x i,, + ). (231)We may write exp(iQ) as

exp(iQ) = exp( + i ) = exp(i )exp(). (232)The last equality of Eq.(232) is in the form of a translation of by , followed by atranslation of x by i. We finally obtain

exp(iQ)exp(iQ)(x,, ) = (x + i i( + ), + , + ) (233)The operation in Eq.(233) is as much of a supertranslation as that in Eq.(227). However,the final result takes a simpler form if we have both Q and Q in the same exponent asin Eq.(227). To accomplish this recall the Baker-Haussdorf theorem which states that

exp(A)exp(B) = exp(A + B +1

2[A, B] + . . .). (234)

In the general case, the terms not written out in Eq.(234) contain an infinite series ofmultiple commutators involving the operators A and B. However this series is absentif the operator [A, B] also commutes with both A and B. For the present case, whereA = iQ and B = iQ, we have

[A, B] = [iQ,iQ] = 2 P. (235)Since [P, Q] = [P, Q] = 0, there is no infinite series in Eq.(234) and we have

exp(iQ)exp(iQ) = exp(iQ + iQ 2 P), (236)or equivalently,

exp(iQ + iQ) = exp(2i )exp(iQ) exp(iQ) (237)Using Eq.(237) and Eq.(233), we finally have

exp(iQ + iQ)(x,, ) = (x + i i, + , + ). (238)The additional ordinary translation present in Eq.(227) is trivial, since it merely shiftsthe coordinate x by the ordinary 4-vector a.