symmetrieslinkinginternaland spacetimeproperties....

Paradise is exactly like where you are right now.

. . . only much, much better.

Laurie Anderson (1947)

25Symmetries Linking Internal andSpacetime Properties. Supersymmetry

Besides space-time symmetries under Lorentz- and Poincaré-group, and under ap-proximate internal symmetries groups SU(2), SU(3), etc., there exists a further classof approximate symmetries which combine external and internal indices. Histori-cally, the first example for this class of symmetries arose in nuclear physics whereisospin was combined with spin to form the larger group SU(4).

25.1 Approximate SU(4)-Symmetry of Nuclear Forces

Soon after the discovery of the approximate isospin symmetry in nuclear physics, thesearch went on for higher approximate symmetries. It turned out that the nuclearforces exhibited also an approximate independence of spin. Moreover, this could becombined with isospin to form an approximate invariance under the larger groupSU(4). The different terms in the potential between nucleons are conventionallyclassified by their behavior under certain interchange operators acting upon two-nucleon wave functions. These are defined as follows:

PMψ(x1a1, x2a2) ≡ ψ(x2a1, x1a2) Majorana operator,PBψ(x1a1, x2a2) ≡ ψ(x1a2, x2a1) Bartlett operator, (25.1)PHψ(x1a1, x2a2) ≡ ψ(x2a2, x1a1) Heisenberg operator,

where x denotes the positions and a the third components of the spins of the par-ticles. The effect of the interchange operators on two-nucleon wave functions ofdefinite orbital angular momentum l and spin (singlet or triplet) is shown in Table25.1. Using these operators, the most general two-particle potential can be writtenas follows:

V = VW (r) + VM(r)PM + VB(r)P

B + VH(r)PH

+ VTW (r)S12 + VTM(r)S12PM , (25.2)

where S12 is the spin-orbit coupling operator

S12 ≡ 3(�1r̂)(�2r̂)− (�1�2). (25.3)

1381

1382 25 Symmetries Linking Internal and Spacetime Properties

Table 25.1 Action of the different interchange operators.

even l odd ltriplet singlet triplet singlet

PM 1 1 −1 −1PB 1 −1 1 −1PH 1 −1 −1 1

Here r̂ ≡ r/|r| is the direction of the distance vector r between the two particles.The �-matrices �1, �2 act on the spin indices of the particles 1, 2, respectively. Thisis, in fact, the most general way of describing noncentral forces which can be derivedfrom a potential. The potential VW (r) is called Wigner potential, which is purelycentral. The potential VTW that is proportional to the operator S12 parametrizes theso-called tensor force that is responsible for mixing the 31D wave into the

31S ground

state wave function1 of the deuteron (≈ 2− 6%). Note that for a spin singlet state,the eigenvalue of S12 is zero. The potential VM is called Majorana potential. TheBartlett operator PB can be expressed in terms of the Pauli spin matrices as follows:

PB =1

2(1 + �1�2). (25.4)

In this parametrization, each of the six V -terms in (25.8) is proportional to the unitmatrix in the space of charge indices.

Let us now define, in this space, the operator

P τ ≡ 12(1 + �1�2), (25.5)

where � denotes a three-component vector of Pauli matrices � = (τ 1, τ 2, τ 3), which,instead of acting on spin indices, acts on the isospin indices of the nucleons. It iseasy to verify that the operators PH and P τ have opposite signs

PH = −P τ , (25.6)

when acting on wave functions satisfying the Pauli principle. We can also verifythat

PM = PBPH . (25.7)

The action of the spin-isospin operators is shown in Table 25.2.From this we see that it is possible to rewrite the potential (25.1) in terms of the

�-matrices as follows:

V = Va(r) + Vσ(r)(�1�

2) + Vτ (r)(�1�

2) + Vστ (�1�

2)(�1�2)

+ VTd(r)S12 + VTτ (�1�

2)S12. (25.8)

1Notation: JSL where L is the orbital angular momentum, S the combined spin, and J the total

angular momentum.

25.1 Approximate SU(4)-Symmetry of Nuclear Forces 1383

Table 25.2 Action of spin and isospin operators in the expansion (25.8).

even l odd ltriplet singlet triplet singlet

�

1�

2 −3 1 −3 1�

1�

2 1 −3 −3 1(�1�2)(�1�2) −3 −3 9 1

The coefficient functions are related to the earlier ones by

Va = VW +12VB − 12VH − 14VM , Vσ = 12VB − 14VM ,

Vτ =−12VH − 14VM , Vστ = −14VM ,VTd = VTW − 12VTM , VTτ = −12VTM .

(25.9)

In 1937, Wigner observed [1] that if the spin-orbit forces are small and Wignerand Majorana forces are dominant, the potential can be written as

V = V1(r) + V2(r)(1 + �1�

2)(1 + �1�2). (25.10)

This would be invariant under spin rotations, isospin rotations, and the extensionof these two SU(2)-groups to SU(4). Experimentally, however,

VHVM

≈ 0.4− 0.5, (25.11)

so that there exists really quite a large symmetry breaking. Nevertheless, the grouphas turned out to be useful in understanding nuclear spectra.

The group SU(4) is generated by the fifteen 4 × 4 matrices acting on a nucleonfield Nαα′ with isospin index α (α = 1, 2) and spin index α

′ (α′ = 1, 2):

�

2× 1, 1× �

2,�

2× �

2. (25.12)

The commutation rules are given by the known ones for �,�, an additional set ofobvious commutation rules between the �’s and �’s, and the combined generators(�/2)× (�/2). They express their vector nature with respect to each subgroup, plusthe new ones

[

τa × σi2

,τ b × σj

2

]

= iδabǫijk1×σk

2+ iǫabcδij

τ c

2× 1. (25.13)

If SU(4) is a good approximate symmetry of nuclear forces, then nuclei shouldoccur in irreducible multiplets of SU(4). Moreover, since both isospin and spin areinvariant under space reflections, each multiplet should have the same parity.

The fundamental representation of SU(4) is four-dimensional and denoted by 4.Its four components are identified with the four states of the nucleon as follows:

N =

p↑

p↓

n↑

n↓

. (25.14)


The higher representations are obtained as usual. One forms an appropriate multi-nucleon state of highest spin and isospin and applies the lowering operators

τ−

2× 1, 1× σ

−

2,

τ− × σ34

,τ 3 × σ−

4,

τ− × σ−4

(25.15)

to generate irreducible multiplets.Consider the symmetric Young tableau and take the state of highest weightin isospin and spin

|1, 1; 1, 1〉 ≡∣

∣

∣p↑p↑〉

, (25.16)

where the notation on the left-hand side specifies the isospins and spins T, S, andtheir third components as |T, T3;S, S3〉.2 From the state (25.16) we run through theentire isovector triplet of spin 1 by repeated application of the lowering operator τ−.For these the other spin components are obtained by applying σ−.

The application of τ−/2 to (25.16) gives the state

|1, 0; 1, 1〉 = 1√2

∣

∣

∣n↑p↑ + p↑n↑〉

, (25.17)

whereas σ−/2 leads to

|1, 0; 1, 0〉 = 12

∣

∣

∣n↓p↑ + n↑p↓ + p↓n↑ + p↑n↓〉

. (25.18)

Since the two operators τ−, σ− are taken from the SU(2)T×SU(2)S subgroup, theydo not change T, S. This is only possible by applying a proper SU(4) generator suchas τ− × σ−/4. It creates the mixed state

1√2

∣

∣

∣n↓p↑ + p↑n↓〉

. (25.19)

Combining this with (25.18), we now form the state

|0, 0; 0, 0〉 = 1√4

∣

∣

∣n↓p↑ − p↓n↑ − n↑p↓ + p↑n↓〉

, (25.20)

which is orthogonal to (25.18) and has the properties:

τ−

2|0, 0; 0, 0〉 = 1

2

∣

∣

∣n↓n↑ − n↓n↑ − n↑n↓ + n↑n↓〉

= 0, (25.21)

σ−

2|0, 0; 0, 0〉 = 1

2

∣

∣

∣n↓n↓ − p↓n↓ − n↓p↓ + p↓n↓〉

= 0, (25.22)

showing that it is a singlet under isospin and spin operations, as indicated by thezeros in the ket on the left-hand side of (25.20). Thus we have found an irreducibledecuplet of SU(4) with 10 states, and the [T, S] decomposition

10 = [1, 1] + [0, 0]. (25.23)

2In accordance with the standard names for isospin and spin matrices �,� we shall denote halfthe isospin generators τ/2 by T rather than I, as is customary in nuclear physics, and �/2 by S.


Figure 25.1 Pole position of the would-be SU(4) -partner of the deuteron with spin-1

and isospin-0. It is close to the D-pole but not a particle since it hides in the complex

energy plane just below the two-particle cut.

In addition, there is a six-dimensional representation consisting of the states

|0, 0; 1, 1〉 = 1√2

∣

∣

∣p↑n↑ − n↑p↑〉

,

σ−

2|0, 0; 1, 1〉 = 1√

2

∣

∣

∣p↓n↑ + p↑n↓ − n↓p↑ − n↑p↓〉

=√2 |0, 0; 1, 0〉 ,

σ−

2|0, 0; 1, 0〉 =

∣

∣

∣p↓n↓ − n↓p↓〉

=√2 |0, 0; 1,−1〉

σ− × τ−4

|0, 0; 1, 1〉 = 1√2

∣

∣

∣−n↓n↑ + n↑n↓〉

= |1,−1; 0, 0〉

τ+

2|1,−1; 0, 0〉 = 1√

2

∣

∣

∣−p↓n↑ − n↓p↑ + p↑n↓ + n↑p↓〉

=√2 |1, 0; 0, 0〉

τ+

2|1, 0; 0, 0〉 =

∣

∣

∣−p↓p↑ + p↑p↓〉

=√2 |1, 1; 0, 0〉 . (25.24)

These have the [T, S]-content denoted by

6 = [1, 0] + [0, 1]. (25.25)

The representation 6 is an irreducible content of the tensor product, i.e.,

4× 4 = 10 + 6. (25.26)

The deuteron is a state in the irreducible representation 6. It has [T, S] = [0, 1],i.e., it is an isosinglet vector particle transforming like the SU(4) generator �/2× 1.If SU(4) is supposed to be a good approximate symmetry, there should also bea partner [1, 0], i.e., a spin-zero isotriplet transforming like the SU(4) generator1 × �/2. This does not really exist. However, the singlet NN scattering length isvery large in the T = 1 state, indicating a bound state just around the cut of thetwo-particle continuum in the energy plane. The symmetry breaking prevents theseSU(4) partners from appearing as physical particles [2].


In elementary-particle physics, the group SU(4) combining isospin and spin hasplayed quite an important historical role. There, up and down quarks with spin upand down are identified with the representation 4 of SU(4):

4 =

u↑

u↓

d↑

d↓

. (25.27)

For mesons, the representation 15 in 4 × 4̄ = 15 + 1 extends the isospin tripletpseudoscalars π+, π0, π−, which transform in the same way as the generators of theisospin subgroup τa. The vector mesons ρ+, ρ0, ρ− are assigned to the generatorsτa × σi, plus an isosinglet σi which is identified with the ω meson. The symmetrybreaking, however, is extremely large since

mπ ≈ 135 MeV,mρ ≈ 770 MeV,mω ≈ 783 MeV. (25.28)

For the nucleons p, n, the SU(4) symmetry is much better. Taking three up anddown quarks with spin up and down, the three Young tableaux

, , (25.29)

possess, according to the general dimension formulas (24.156)–(24.158), the dimen-sions

n(n+ 1)(n+ 2)

6,

(n− 1)n(n+ 1)3

,n(n− 1)(n− 2)

6, (25.30)

the second appearing twice. These tableaux represent a symmetric representation20, two mixed representations 20, and one antisymmetric representation 4. Actually,the latter 4 is equivalent to the complex conjugate 4 of the fundamental represen-tation. To see this we note that the completely antisymmetric representation

(25.31)

of rank 4 is one-dimensional which means that the antisymmetric 4 must have metwith a 4 to form an invariant.

In order to find the isospin and spin contents of the representations (25.29), wewrite the SU(4) index as α, α′, where α are the two isospin indices and α′ the spinindices. The representation 4 is a doublet with respect to each subgroup

aα = 2× 2. (25.32)


The Young tableau = 10 is symmetric under the simultaneous exchange ofboth indices. Hence, with respect to the two subgroups it can become a product ofsymmetric tensors or antisymmetric tensors

aα bβ = a b × α β + ab

× αβ.

10 3 3 1 1 (25.33)

Similarly

aα

bβ=

a

b× α β + a b × α

β.

6 1 3 3 1 (25.34)

To obtain all irreducible tensors of rank 3 we multiply and by another

box, and find within SU(4) the decomposition

× = +10 4 20 20

, (25.35)

and

× = +

6 4 20 4̄ .

(25.36)

As far as the isospin and spin contents are concerned, we multiply the decomposition(25.33) of the 10 by

= ×4 2 2

. (25.37)

Thus the product 10× 4 contains the isospin and spin products

× = +3 2 4 2

(25.38)

× = .1 2 2

(25.39)

In the second case there is only one product since is one-dimensional. Hence

we can decompose 10× 4 as follows

× =(

+)

×(

+)

10 4 4 2 4 2(25.40)


+ ×2 2

. (25.41)

On the left-hand side, the product reduces to

+ .

20 20(25.42)

It is easy to read off which products on the right-hand side belong to these tworepresentations:

= × + ×20 4 4 2 2

(25.43)

= × + ×20 4 2 2 4

+× .

2 2(25.44)

Thus the symmetric representation 20 consists of a spin-12isospin-1

2, and a spin-3

2

isospin-32multiplet. They are associated with the following particles: The spin-1

2

isospin-12states are the two isospin states of the nucleon p and n. The spin-3

2isospin-

32states contain the first pion-nucleon resonance ∆(1230), plus its isospin partners

with the charge states −, 0,+,++. All particles have intrinsic parity +.The mixed 20 consists of a spin-3

2isospin-1

2, a spin-1

2isospin-3

2, and spin-1

2isospin-1

2

multiplet. These states are also found experimentally in pion-nucleon scattering asresonances of negative parity.

The content of the antisymmetric SU(4) representation

= 4̄

is obviously 2 × 2, i.e., an isospin-12spin-1

2multiplet. It is associated with the

spin-parity pion-nucleon resonance N(1440) of spin-parity SP = 12

+.

25.2 Approximate SU(6)-Symmetry in Strong Interactions

With the success of the approximate symmetry SU(4) in nuclear and particle physics,the question arose whether there exists an extension of the symmetry group which

25.2 Approximate SU(6)-Symmetry in Strong Interactions 1389

includes also the internal symmetry SU(3). The three quarks with spin-up and-down form a sextet of particles

q =

u↑

u↓

d↑

d↓

s↑

s↓

. (25.45)

Hence the most straightforward extension would be SU(6). This group is generatedby the 35 matrices

λa

2× 1, 1× σ

i

2,

λa

2× σ

i

2. (25.46)

The only non-trivial commutation rules are[

λa × σi2

,λb × σj

2

]

= idabcǫijk1

2λc × σk + ifabcδij

1

2λc × 1 + i2

3δabǫijk1×

1

2σk. (25.47)

Since in SU(4) the representation 4×4 contains the pseudoscalar π and vector mesonsρ, ω, we expect the representation 6× 6̄ to extend these multiplets by their SU(3)-partners. The representation 6× 6̄ decomposes into 35+1, with the particles in the35-representation having the transformation properties of the 35 group generators(25.46). Thus there is now an SU(3)-octet of pseudoscalar mesons which transformlike the generators λa × 1/2. In addition, there is an SU(3)-octet of vector mesonsρ, ω,K∗. These are 24 states which transform like the generators λa×σi/2, and thereis an SU(3)-singlet vector meson ϕ(1020) which transforms like the three generators1× σi/2. More accurately the states ω and φ will be mixed with each other.

These meson multiplets are quite widely separated in mass, implying that thesymmetry SU(6) is strongly broken. Still, this would-be symmetry group organizesnicely the most important low-lying meson states, as illustrated in Fig. 25.2.

As in the case of SU(4)-symmetry, the situation is much better for baryons. Thenucleons and their strange partners as well as the first nucleon resonances and theirstrange partners all lie in a single irreducible multiplet of three-quark states. Theydecompose as follows:

× ×6 6 6

=56

+ 2 ×70

+

20

.(25.48)

In order to find the SU(3)- and spin-contents of these multiplets we decompose, asin the case of SU(4),

aα

6=

a × α3 2

, (25.49)


K∗

ρ

ω

K̄∗

K

π

η

K̄

φ

JP = 0− JP = 1− JP = 1−

Figure 25.2 Pseudoscalar and vector mesons of the 35-representation of SU(6).

and form the symmetric and antisymmetric products of two representations 6. Theyare 21- and 15-dimensional and their SU(3) and spin contents are given by

aα bβ = a b × α β + ab

× αβ,

21 6 3 3̄ 1

aα

bβ= a b × α

β+

a

b × α β .

15 6 1 3̄ 3

Multiplying the 21 by a further 6, we find

×21 6

=56

+70

=

10+

8

×

4+

2

+

1

×8

×2

. (25.50)

It is then easy to read off the SU(3) and spin contents on the right-hand side

56=

10×

4+

8×

2,


70=

10×

2+

8× 4

+8

×2

+

1

×2

. (25.51)

Thus the 56 consists of a decuplet of spin 32and an octet of spin 1

2. These can

be associated with the decuplet of nucleon resonances and the nucleon octet, all ofpositive parity as shown in Fig. 25.3.

Figure 25.3 The SU(3)-content of particles in the 56-representation of SU(6) containing

the JP = 1/2+ nucleons and the JP = 3/2+ resonances.

∆∗

Σ∗

Ξ∗

Ω

N

Σ

Ξ

Λ

JP = 1/2+ JP = 3/2+

The 70 consists of a decuplet of spin 12, an octet of spin 3

2, an octet of spin 1

2, and

a singlet of spin-12states, thus extending the isospin SU(4)-multiplets of negative

parity in Eq. (25.44) by their SU(3)-partners as shown in Fig. 25.4.

The SU(3)× SU(2)-contents of the antisymmetric SU(6) representation

20

are easily written down as

20

=8

×2

+

1

×4

. (25.52)

The octet states could possibly be the spin 12-particles of positive parity shown in

Fig. 25.5. The SU(3)-singlet 32

+ partner of them has not yet been found.


∆(1670)

Σ(1750)

Ξ(?)

Ω(?)

N(1520)

Σ(1670)

Ξ(1820)

Λ(1520)

10× 2JP = 1/2−

8× 4JP = 3/2−

N(1650)

Σ(1750)

Ξ(1950)?

Λ(1405)

8× 2JP = 1/2−

1× 2

JP = 1/2− Λ(1800)

Figure 25.4 Nucleon resonances of negative parity in the 70-representation of SU(6).

Let us now study the states of the 56-representation in more detail. The stateof highest weight is the ∆++(1235) resonance of spin 3

2

∆++(1235)↑↑ =∣

∣

∣u↑u↑u↑〉

. (25.53)

By lowering the third components of spin and isospin once, we obtain the normalizedstate

∆+(1235)↑ =1

3

∣

∣

∣d↓u↑u↑ + u↓d↑u↑ + u↓u↑d↑

+d↑u↓u↑ + u↑d↓u↑ + u↑u↓d↑

+d↑u↑u↓ + u↑d↑u↓ + u↑u↑d↓〉

. (25.54)

Applying to this the operator λ3σ3, we find

λ3σ3∆+(1235)↑=1

3

∣

∣

∣3(d↓u↑u↑ + u↑d↓u↑ + u↑u↑d↓)


N(1440)

Σ(1660)

Ξ(?)

Λ(1660)

Figure 25.5 Octet of spin-parity 12+-baryons.

−(u↓d↑u↑ + u↓u↑d↑ + d↑u↓u↑ + u↑u↓d↑ + d↑u↑u↓ + u↑d↑u↓)〉

.(25.55)

This should be a mixture of ∆+(1235)↑ and a proton p↑.Since the proton is orthogonal to ∆+(1235)↑, we can immediately guess a nor-

malized proton state:

p↑ =

√2

3

∣

∣

∣

∣

∣

d↓u↑u↑ + u↑d↓u↑ + u↑u↑d↓ (25.56)

−12(u↓d↑u↑ + u↓u↑d↑ + d↑u↑u↓ + u↑u↓d↑ + d↑u↑u↓ + u↑d↑u↓)

〉

.

To ensure that it is a pure proton state we simply apply the isospin raising operatorand find that the state is annihilated, thus guaranteeing the total isospin to be 1/2.Applying the isospin lowering operator, we obtain the neutron state

n↑ = −√2

3

∣

∣

∣

∣

∣

d↑d↑u↓ + d↑u↓d↑ + u↓d↑d↑ (25.57)

− 12(d↓u↑d↑ + d↓d↑u↑ + u↑d↑d↓ + d↑d↓u↑ + u↑d↑d↓ + d↑u↑d↓)

〉

.

As an interesting consequence of SU(6), we calculate the ratio between the mag-netic moment of neutron and proton. For this we assume that the magnetic momentoperator transforms under SU(3) like the charge operator which is given by the gen-erators

Q = λQ =1

2

(

λ3 +1√3λ8)

. (25.58)

Under rotations, it must transform like a spin operator �. Thus the magneticmoment operator should be proportional to the following generator in SU(6):

� =1

2(λ3 +

1√3λ8)× �. (25.59)

We now apply the third component of this

µ3 =1

2(λ3 +

1√3λ8)× σ3, (25.60)


to the quark states and use the eigenvalues listed in Table 25.3. In front of the states|p↑〉 and |n↑〉, this yields

µ3|p↑〉 =√2

3

∣

∣

∣

∣

∣

5

3(d↓u↑u↑ + u↑d↓u↑ + u↑u↑d↓) (25.61)

− 12

(

−13

)

(u↓d↑u↑ + u↓u↑d↑ + d↑u↑u↓ + u↑u↓d↑ + d↑u↑u↓ + u↑d↑u↓)

〉

,

µ3|n↑〉 =√2

3

∣

∣

∣

∣

∣

− 43(d↑d↑u↓ + d↑u↓d↑ + u↓d↑d↑) (25.62)

− 12

(

2

3

)

(d↓u↑d↑ + d↓d↑u↑ + u↑d↑d↓ + d↑d↓u↑ + u↑d↑d↓ + d↑u↑d↓)

〉

.

Hence we find the matrix elements of the magnetic moments as being proportionalto

〈

p↑∣

∣

∣µ3∣

∣

∣p↑〉

=2

9

(

5

3· 3− 1

3· 64

)

= 1, (25.63)

〈

n↑∣

∣

∣µ3∣

∣

∣n↑〉

=2

9

(

−43· 3 + 2

3· 64

)

= −23. (25.64)

These imply a ratio of magnetic moments

µnµp

= −23. (25.65)

Experimentally, the magnetic moments are

µp = 2.7928444 ± 0.0000011,µn = −1.91304308± 0.00000054, (25.66)

in units of ch̄/2mpc = 505 × 10−24 erg/Gauss. Thus the ratio is ≈ −2.055/3, inexcellent agreement with the SU(6)-prediction (25.65).

By the same method we can calculate all rates of the radiative decays of thedecuplet resonances 10 → 8+γ in terms of only one unknown overall size parameter.

The prediction of the ratios of the magnetic moments is possible due to the factthat in the irreducible multiplet 56 of the symmetry group SU(6), the SU(3)-octetoperator λa × σ3/2 has, between the octet states of nucleons, a specific ratio of

Table 25.3 Eigenvalues of charge and other operators on quark states.

q u↑ u↓ d↑ d↓ s↑ s↓

λ3 × σ3 1 −1 1 −1 0 0(λ8/

√3)× σ3 1

3−1

313

−13

−23

23

Q× σ3 23 −23 −13 13 −13 13


symmetric and antisymmetric couplings (which within SU(3) would be arbitrary).Let us express the two couplings in terms of the irreducible matrix elements F andD as follows

〈

b

∣

∣

∣

∣

∣

λa

2× σ3

∣

∣

∣

∣

∣

c

〉

= F · (−ifab∗c) +D · (dab∗c)

≡ F · (Fa)b∗c +D · (Da)b∗c. (25.67)

Then we find between protons with the octet components (4 + i√5)/

√2

(F3)pp = f345 =1

2, (D3)pp = d344 =

1

2, (25.68)

and between neutrons with the components (6 + i7)/√2

(F3)nn = −f367 = −1

2, (D3)nn = d366 = −

1

2. (25.69)

The a = 8 components of Fa, Da have the matrix elements

(F8)pp = f845 =√

32, (D8)pp = d866 = − 12√3

(F8)nn = f867 =√

32, (D8)nn = d866 = − 12√3 .

(25.70)

The matrix elements of the magnetic moment operator

µ3 = λQ × σ3 ≡ 1

2(λ3 +

1√3λ8)σ3 (25.71)

are therefore given in terms of the irreducible matrix elements F and D by

〈p|µ3|p〉 = F ·(

1

2+

1

2

)

+D ·(

1

2− 1

6

)

=(

F +1

3D)

,

〈n|µ3|n〉 = F ·(

−12+

1

2

)

+D ·(

−12− 1

6

)

= −23D. (25.72)

The ratio is〈p|µ3|p〉〈n|µ3|n〉

=F/D + 1

3

−23

. (25.73)

This holds on the basis of only SU(3) symmetry. Inserting the SU(6) result (25.65),we see that SU(6) fixes the F/D ratio

F

D=

2

3. (25.74)

The absolute values of the irreducible matrix elements of the operators

λ3

2× σ3, λ

8

2× σ3 (25.75)


are also known from the above results(25.63) and (25.64):

F +D1

3= 1, − 2

3D = −2

3, (25.76)

so that

F =2

3, D = 1. (25.77)

It is straight-forward to calculate the magnetic moment of any other membersof the octet, for instance of Σ+ =̂ (1 + i2)/

√2:

〈

Σ+|µ3|Σ+〉

= F · (−ifQ12) +D · dQ11

= F ·[

1

2

(

f312 +1√3f812

)]

+D ·[

1

2

(

d311 +1√3d811

)]

=1

2F +

1

6D =

1

2. (25.78)

This implies the ratio〈

p∣

∣

∣µ3∣

∣

∣p〉

〈

Σ+∣

∣

∣µ3∣

∣

∣Σ+〉 = 2. (25.79)

In fact, the fastest way to find the F/D-ratio is by calculating the matrix elementsof λ3 × σ3/2 from the SU(6) wave functions. Between protons we find immediately

〈

p

∣

∣

∣

∣

∣

λ3

2× σ3

∣

∣

∣

∣

∣

p

〉

=5

6. (25.80)

Using the wave function

∣

∣

∣Σ+↑〉

=

√2

3

∣

∣

∣

∣

∣

(

u↑u↑s↓ + u↑s↓u↑ + s↓u↑u↑)

(25.81)

− 12(u↑u↓s↑ + u↓s↑u↑ + s↑u↑u↓ + u↓u↑s↑ + u↑s↑u↓ + s↑u↓u↑)

〉

,

we obtain〈

Σ+∣

∣

∣

∣

∣

λ3

2× σ3

∣

∣

∣

∣

∣

Σ+〉

=2

3. (25.82)

In terms of the irreducible matrix elements F,D, the corresponding matrix elementswould be in general

〈

p

∣

∣

∣

∣

∣

λ3

2× σ3

∣

∣

∣

∣

∣

p

〉

=F

2+D

2,

〈

Σ+∣

∣

∣

∣

∣

λ3

2× σ3

∣

∣

∣

∣

∣

p

〉

= F. (25.83)

Hence we identify, once more,

F =2

3, D = 1. (25.84)


The magnetic moment µ∗ for the transition p → ∆ is found by multiplyingµ3∣

∣

∣p↑〉

from the left by 〈∆+↑|. This gives

µ∗ ≡〈

∆+↑∣

∣

∣µ3∣

∣

∣p↑〉

=

√2

9

[

5

3· 3 + 1

6· 6]

=2

3

√2. (25.85)

A comparison with the proton matrix element (25.63) yields the SU(6) prediction

µ∗ =2

3

√2µp. (25.86)

Experimentally one finds, from the ∆(1236) resonance in the cross section of thephotoproduction process

γ + p→ π0 + p, (25.87)the value

µ∗exp ≈ (1.28− 0.01) · µ∗. (25.88)This is not in quite such a good agreement as the proton-to-neutron ratio (25.65).We must, however, keep in mind that the symmetry SU(6) is significantly broken,and that the comparison with the data requires matrix elements of µ between stateswhose normalization factors can depend on the masses of the particles. Thus oneis bound to commit errors of the order of the relative mass differences within theexperimental multiplets, herethere between the ∆(1232) and the proton p(938).These can easily account for the 28 % discrepancy in the magnetic moment prediction(25.88).

It should be noted that the ratio g∗/g ≈ 1.07, found in Eq. (24.212) from theexperimental decay rates of the ∆-resonance, is a manifestation of SU(4)-symmetry.Since the ∆-resonance and the nucleon are in the same SU(4)-multiplet 20 (corre-sponding to the SU(6)-multiplet 56), all pion couplings are controlled by only oneirreducible matrix element. In calculating them we have to remember that, in or-der to preserve parity, the pseudoscalar pion must emerge in a p-wave. The matrixelement between a ∆+-resonance and a proton

1

MNūµ(0, 12)u(pN , s

′3)iq

µπ (25.89)

has, for small momentum pN in the z-direction, the limiting form

−i

√

2

3

pNMN

ū(0, 12)u(0, 1

2) +

1√6MN

(p1N − ip2N )u(0, 12)

. (25.90)

Thus it can be written in terms of two component spinors as

−2√

2

3

pN

MNχ( 1

2)�χ( 1

2). (25.91)


The p-wave emission leads to the appearance of a spin matrix σi acting upon thehelicity index of the nucleon. The same thing is observed for the matrix elementsin the pseudoscalar pion nucleon coupling

ū(0, 12)iγ5u(pN , 12). (25.92)

For small momentum, this becomes

−i (χ( 12), χ( 1

2))

(

1 00 −1

)

γ0

(

e−��/2χ( 12)

e��/2 χ( 12)

)

≈ −i pNMN

χ( 12)�χ( 1

2). (25.93)

We now add the internal SU(2)-structure and the SU(4)-matrix associated withthe emission of the pion in the z-direction of ∆+. The pion transforms like thetensor operator

πa=̂τaσ3, (25.94)

so that the ratio of the couplings pπ0p and ∆+π0p can be found quite simply byapplying τ 3σ3 to the proton state

p =

√2

3

[

u↑u↑d↓ + u↑d↓u↑ + d↓u↑u↑ (25.95)

−12(u↑u↓d↑ + u↑d↑u↓ + u↓d↑u↑ + u↓u↑d↑ + d↑u↑u↓ + d↑u↓u↑)

]

,

which gives

τ 3σ3p =

√2

3

[

3(u↑u↑d↓ + u↑u↓u↑ + d↓u↑u↑) (25.96)

+1

2(u↑u↓d↑ + u↑d↑u↓ + u↓d↑u↑ + u↓u↑d↑ + d↑u↑u↓ + d↑u↓u↑)

]

.

From these equations and (25.54) we extract the matrix elements

p τ 3σ3p =2

9

(

3 · 3− 14· 6)

=5

3,

∆+τ 3σ3p =

√2

9

(

3 · 3 + 126)

=4

3

√2. (25.97)

For the ratio between g and g∗, we have to go from ∆+τ 3σ3p to the matrix element∆++τ+σ3p which involves a Clebsch-Gordan factor

∆++τ+σ3p =1√2

4

3

√2 =

4

3. (25.98)

Thus we obtain

g∗

g=

4

3

5

3=

4

5= 0.8, (25.99)

25.3 From SU(6) to Current Algebra 1399

in reasonable agreement with the experimental ratio 0.92 from the analysis of pion-nucleon scattering amplitudes.

By a similar quark calculation, we can predict from SU(6)-symmetry the F/D-ratio of the coupling between nucleon octets and pseudoscalar mesons fitted to ex-perimental data in Eq. (24.188). According to the interaction Lagrangian (24.181),the coupling Σ+Σ+π0 is given by

gΣ+Σ+π0 = g(1− α) = g2F

F +D. (25.100)

But this matrix element is easily calculated. The SU(6) state is obtained from Σ+↑

by applying V− to it which changes the d-quark into an s-quark. Hence

Σ+↑ = −√2

3(u↑u↑s↓ + u↑s↓u↑ + s↓u↑u↑) (25.101)

−12

(

u↑u↓s↑ + u↑s↑u↓ + s↑u↑u↓ + u↓u↑s↑ + u↓s↑u↑ + s↑u↓u↑)

.

Applying τ3σ3 gives

τ3σ3Σ+↑ = −

√2

32(u↑u↑s↓ + u↑s↓u↑ + s↓u↑u↑), (25.102)

leading to a scalar product with Σ+↑:

Σ+↑τ3σ3Σ+↑ =

4

3. (25.103)

In comparison with the pion-nucleon matrix element gπNN = g, this amounts to therelation

gπ0Σ+Σ+ =4

5g. (25.104)

We therefore find, for the F/D-ratio, the SU(6)-value

F/D = 2/3, (25.105)

in good agreement with the experimental determination.

25.3 From SU(6) to Current Algebra

Historically, the SU(6) symmetry had an important impact upon the developmentof the quark model. Since the symmetry is only approximate there was need to finda way of controlling the symmetry-breaking corrections. For the group SU(6) thishas remained impossible. For an important subgroup of SU(6), however, this can bedone. It is the subgroup generated by λa/2, λaσ3/2. It is called SU(3)R × SU(3)L,and its generators are taken to be the combinations of

Qa± =λa

2× (1± σ

3)

2. (25.106)


These generators are intimately related to the SU(3)×SU(3) -current algebra whichwas discussed before in the context of chiral symmetries. The charges and axialcharges observed in electromagnetic and weak interactions are related to Qa± asfollows:

Qa = Qa+ +Qa−, Q

a5 = Q

a+ −Qa−. (25.107)

This relation makes it possible to calculate the symmetry-breaking corrections interms of experimentally observable quantities.

The good SU(6)-results for the magnetic moments were based on the predictionof the F/D ratio 2/3 and the ratio 〈p|µ3|p〉 /〈∆+|µ3|p〉 = 2

√2/3. It is gratifying to

observe that these predictions are also contained in the SU(3)R × SU(3)L subgroupof SU(6). Within the framework of current algebra it will, moreover, be possible tocalculate symmetry-breaking corrections to these ratios.

The representations 56 and 70 of SU(6) have the following irreducible contentswith respect to this subgroup

56 =

32

12

−12

−32

(10, 1) (6, 3)10 −(3, 6)10 (1, 10)(6, 3)8 (3, 6)8

, (25.108)

70 =

32

12

−12

−32

(8, 1) 1√2[(6, 3)− (3̄, 3)]8 1√2 [(3, 6) + (3, 3̄)]8 −(1, 8)

1√2[(6, 3) + (3̄, 3)]8

1√2[(3, 6)− (3, 3̄)]8

(6, 3)10 (3, 6)10(3̄, 3)1 − (3, 3̄)1

. (25.109)

The columns in these matrices are ordered according to the third component of thespin, s3, pointing along the direction of the momentum of the particle, which isindicated in the upper row, and the subscripts denote the coupling of the two partsof SU(3)R × SU(3)L with respect to the ordinary SU(3) symmetry group generatedby Qa+ + Q

a−. Within this subgroup, the F/D ratio 2/3 is the F/D ratio found for

the operator Qa+ − Qa− between two (6, 3)8 representations. The transition ∆ → pcomes from the matrix element of Qa+ − Qa− = λaσ3/2. The representation (6, 3)contains an SU(3)-octet and a decuplet which can be identified with the nucleonoctet and the first resonance decuplet at helicity s3 = 1/2.

At this point it is useful to realize that the commutation rules of SU(3)L×SU(3)Rcan be rephrased as sum rules for the irreducible matrix elements of Qa, Qa5. Take,for example, the commutation rules between the axial charges

[Qa5, Qb5] = ifabcQ

c, (25.110)

and choose the commutator with the indices a = 1 + i2, b = 1− i2, c = 3:

[Q+5 , Q−5 ] = 2Q

3. (25.111)

Sandwiching this between proton states gives〈

p∣

∣

∣Q+5 Q−5

∣

∣

∣p〉

−〈

p∣

∣

∣Q−5 Q+5

∣

∣

∣p〉

= 2〈

p∣

∣

∣Q3∣

∣

∣p〉

. (25.112)


Inserting a complete set of intermediate states between the axial charges, we obtain

∣

∣

∣

〈

n∣

∣

∣Q−5∣

∣

∣p〉∣

∣

∣

2+∣

∣

∣

〈

∆0∣

∣

∣Q−5∣

∣

∣p〉∣

∣

∣

2 −∣

∣

∣

〈

∆++∣

∣

∣Q+5∣

∣

∣p〉∣

∣

∣

2= 2

〈

p∣

∣

∣Q3∣

∣

∣p〉

. (25.113)

We now introduce irreducible matrix elements of the axial charge Qa5 with respectto the SU(2) subgroup by defining between protons:

〈

p∣

∣

∣Q35∣

∣

∣p〉

≡ 12GA. (25.114)

Then〈

n∣

∣

∣Q−5

∣

∣

∣p〉

≡ GA. (25.115)This can either be derived directly by using the generator λ−σ3/2, or via the Wigner-Eckart-theorem for SU(2)-vector operators. We only have to keep in mind that Q35creates the quantum numbers of π0, while Q±5 creates those of

√2π+,−

√2π− if we

assume that π+, π0, π− satisfy the Condon-Shortley phase convention3. We definefurther between a proton and a ∆-resonance

〈

∆+∣

∣

∣Q35∣

∣

∣p〉

≡ 1√2G∗ (25.116)

and find, from the Wigner-Eckart theorem,

〈

∆++∣

∣

∣Q+5∣

∣

∣p〉

≡ −√

3

2G∗,

〈

∆0∣

∣

∣Q−5∣

∣

∣p〉

=1√2G∗. (25.117)

With the right-hand side of (25.113) being the isospin value of the proton 1/2 weobtain the sum rule

G2A −G∗2 = 1. (25.118)Similarly, we can take the commutator (25.111) between two ∆++ states and havethe sum rule

∣

∣

∣

〈

p∣

∣

∣Q−5∣

∣

∣∆++〉∣

∣

∣

2+∣

∣

∣

〈

∆+∣

∣

∣Q−5∣

∣

∣∆++〉∣

∣

∣ = 3. (25.119)

Let us now introduce the irreducible matrix element G∗A of the axial charge betweentwo ∆-states via

〈

∆+∣

∣

∣Q05

∣

∣

∣∆+〉

=1

2G∗A. (25.120)

Then, by the Wigner-Eckart theorem we have〈

∆+∣

∣

∣Q−5∣

∣

∣∆++〉

= −√3G∗A, (25.121)

and Eq. (25.119) becomes1

2G∗2 +G∗A

2 = 1. (25.122)

3One may verify this by applying Q± to Q3 giving ±Q∓.


Between p and ∆+, finally, the commutator (25.111) yields

〈

∆+∣

∣

∣Q+5∣

∣

∣n〉 〈

n∣

∣

∣Q−5∣

∣

∣p〉

+〈

∆+∣

∣

∣Q+5∣

∣

∣∆0〉 〈

∆0∣

∣

∣Q−5∣

∣

∣p〉

−〈

∆+∣

∣

∣Q−5∣

∣

∣∆++〉 〈

∆++∣

∣

∣Q+5∣

∣

∣p〉

= 0, (25.123)

anf this amounts to the sum rule

G∗GA − 5G∗AG∗ = 0. (25.124)

The unique solution to these SU(2)× SU(2) sum rules is

GA =5

3, G∗ =

4

3, G∗A =

1

3. (25.125)

A more covariant looking derivation of these sum rules is possible by rewritingthe matrix elements of Qa5 for a = 1, 2, 3 between isospin-

12and -3

2spinors as follows:

NN : (Qa5)11mm′ = GA χ

†(m)τa

2χ(m′) = G

√3

2〈 12, m |1, a; 1

2, m′〉

∆N : (Qa5)31mm′ = G

∗√3

2χ†a(m)χ(m

′) = G∗√3

2

〈

3

2, m∣

∣

∣1, a; 12, m′

〉

, (25.126)

∆∆ : (Qa5)33mm′ = G

∗A

3

2iǫabcχ

+b (m)χc(m

′) = G∗A

√15

2

〈

3

2, m∣

∣

∣1, a; 32, m′

〉

.

Here〈

SM∣

∣

∣s1m1s2m2〉

are the usual SU(2) Clebsch-Gordan coefficients and χa areRarita-Schwinger isospinors.They are obtained by coupling an ordinary isospinor χwith an isovector va by a Clebsch-Gordan coefficient

χa(m) =〈

3

2, m∣

∣

∣1, m1; 12 , m2〉

va(m1)χ(m2). (25.127)

They satisfy the constraintτaχa(m) = 0 (25.128)

and the identitiesiǫabcτ

bχc = χa, τaχb = −iǫabcχc. (25.129)

Using these we can calculate the matrix elements by sandwiching the commutationrelations (25.110) between nucleon and ∆-states, and summing over a complete setof intermediate states.

When employing the matrix elements (25.126), we can do the summation easilythe help of the completeness relations

∑

m

χα(m)χ∗β(m) = δαβ ,

∑

m

χa(m)χ∗b(m) =

2

3δab −

1

6[τa, τa]. (25.130)

In this way we obtainr once more the sum rules (25.118), (25.122), and (25.124).


So far, these relations hold for the SU(2)L×SU(2)R subgroup of SU(3)L×SU(3)R.Extending the procedure from SU(2) to SU(3), we decompose the irreducible matrixof Q35 between protons and remember (25.68) to see that

GA = F +D. (25.131)

Considering once more the commutator (25.111) between Σ+ states gives

∣

∣

∣

〈

Σ0∣

∣

∣Q−5

∣

∣

∣Σ+〉∣

∣

∣

2+∣

∣

∣〈Λ|Q−5∣

∣

∣Σ+〉∣

∣

∣

2+∣

∣

∣

〈

Σ∗0∣

∣

∣Q−5

∣

∣

∣Σ+〉∣

∣

∣

2= 2, (25.132)

where Σ∗0 is the strange partner of the ∆0 resonance. Using the same SU(3)-decomposition as in Eq. (25.67), the matrix element Q35 between two Σ

+ statesis

〈

Σ+∣

∣

∣Q05

∣

∣

∣Σ+〉

= F. (25.133)

Hence, by the Wigner-Eckart theorem for the SU(2) subgroup we have〈

Σ0∣

∣

∣Q−5

∣

∣

∣Σ+〉

=√2F. (25.134)

Using once more the SU(3)-decomposition (25.67), we calculate the matrix element〈

Λ0∣

∣

∣Q−5∣

∣

∣Σ+〉

= F ·(

−if1−i2,8,(1+i2)/√2)

+D · d1−i2,8,(1+i2)/√2

= D1√2(d118 + d228) =

√

2

3D. (25.135)

Similarly, the matrix element

〈

∆0∣

∣

∣Q−5

∣

∣

∣p〉

=1√2G∗ (25.136)

is related to 〈Σ∗0|Q−5 |Σ+〉 by SU(3)-symmetry. Using the Clebsch-Gordan coeffi-cients of Table 4.2 on p. 381 and the isoscalar factors of Table 24.5 on page 1378we have

〈

∆0∣

∣

∣Q−5

∣

∣

∣p〉

= −〈

3

2,− 1

2

∣

∣

∣1,−1; 12, 1

2

〉

−√

1

2

a,

=1√6a

〈

Σ∗0∣

∣

∣Q−5∣

∣

∣Σ+〉

= −〈1, 0|1,−1; 1, 1〉 1√6a

=1√12a, (25.137)

where a is some number. Using these equations we find from (25.136)

〈

Σ∗0∣

∣

∣Q−5∣

∣

∣Σ+〉

=1

2G∗. (25.138)


The sum rule (25.132) becomes therefore

F 2 +1

3D2 +

1

8G∗2 = 1. (25.139)

Given the solution of the SU(2) × SU(2)-algebra (25.125), we find from this oncemore the F/D ratio 2

3.

For completeness, let us also look at the commutation rule (25.111) between Ξ+

particles, which gives the sum rule

∣

∣

∣

〈

Ξ0∣

∣

∣Q−5∣

∣

∣Ξ+〉 ∣

∣

∣

2+∣

∣

∣

〈

Ξ∗0∣

∣

∣Q−5∣

∣

∣Ξ+〉 ∣

∣

∣

2= 1. (25.140)

Within SU(3), the particles Ξ+ and Ξ0 have the vector indices (6− i7)/√2 and

(4− i5)/√2, so that the decomposition (25.67) reads

〈

Ξ−∣

∣

∣Q−5∣

∣

∣Ξ+〉

= F · (−if1−i2,(4+i5)/√2,−(6−i7)/√2) +D · d1−i2,(4+i5)/√2,−(6−i7)/√2= F · (f146 + f157) +D · (−d146 − d157) = 2F · f146 +D · (−2d146)= F −D. (25.141)

The matrix element between Ξ∗0(1530) and Ξ+, on the other hand, comes fromthe SU(3)-matrix elements (which is found again by combining the Clebsch-Gordancoefficients of Table 4.2 on p. 381 and the isoscalar factors of Table 24.5 on page1378):

〈

Ξ∗0∣

∣

∣Q−∣

∣

∣Ξ+〉

= 〈 12,− 1

2|1,−1; 1

2, 1

2〉 12a

=

−√

2

3

1

2a. (25.142)

This has to be compared with

〈

∆0∣

∣

∣Q−5∣

∣

∣p〉

= 〈 32,− 1

2|1,−1; 1

2, 1

2〉(

− 1√2

)

a

=1√3

(

− 1√2

)

a. (25.143)

Hence〈

Ξ∗0∣

∣

∣Q−5

∣

∣

∣Ξ+〉

=1√2G∗, (25.144)

and the sum rule reads

(F −D)2 + 12G∗2 = 1. (25.145)

This is satisfied by

F =2

3, D =

1

3, G∗ =

4

3. (25.146)

25.4 Supersymmetry 1405

25.4 Supersymmetry

Until 1967, a search was going on to combine the approximate internal symmetriesSU(3) or SU(6) with the external symmetry of spacetime. That search went on until1967 when Coleman and Mandula proved a no-go theorem [4]. This stated that theonly combination between the two groups is the trivial direct product.

In 1974, a new type of symmetry was discovered [5], which extended the Poincarégroup in a completely novel way. Its representations possess irreducible multipletswhich contain at the same time both types of particles, fermions and bosons. Thebasic example is given by the free action

A =∫

d4x[

1

2(∂A)2 +

1

2(∂B)2 +

1

2F 2 +

1

2G2 +M(FA +GB) +

1

2χ̄ (/∂ −M) χ

]

,

(25.147)

where A,B,C,D are four real scalar fields, and χ is a four-component real Majoranaspinor. The four scalar fields can be transformed into the four components of theMajorana field by the following transformations

δχ = [F + iγ5G− i/∂ (A+ iγ5B)] ǫ,δχ̄ = ǭ[F + iγ5G+ i(A+ iγ5B)

←/∂ ], (25.148)

δA = ǭχ, δF = ǭ(−i/∂ χ),δB = ǭiγ5χ, δG = ǭ[−i/∂ (−iγ5)]χ. (25.149)

Here ǫ is a four-component real parameter with spinor indices. The spinor charactermakes it necessary to require ǫ and χ to be Grassmann variables which anticommutewith each other

ǫαχβ = −χβǫα. (25.150)It is easy to verify that these transformations leave the action invariant. For the

mass terms we calculate:

δ(FA) = ǭ(−i/∂ χ)A + F ǭχ, (25.151)δ(CB) = ǭ[−i/∂ (−iγ5)χ]B +Gǭiγ5χ, (25.152)δχ̄χ = χ̄[F + iγ5G− i/∂ (A+ iγ5B)]ǫ+ ǭ[F + iγ5G + i(A+ iγ5B)

←/∂ ]χ. (25.153)

The real Majorana spinor satisfies (recall Section 4.13)

χ̄T = γ0χ = Cχ =

(

0 σ2

σ2 0

)

χ, (25.154)

so that

χ̄ǫ = χTγT0 ǫ = −ǫTγ0χ = ǫTγT0 χ = ǭχ,χ̄γ5ǫ = χ

TγT0 γ5γ0ǭT = χTγ5ǫ

T = −ǭγT5 χ = ǭγ5χ. (25.155)


Recall that in the Majorana representation

γ5 =

(

σ2 00 −σ2

)

, (25.156)

so that γT5 = −γ5. Hence the F− and G-terms in the transformation (25.153) canbe combined to

δ(χ̄χ) = 2ǭ(F + iγ5G)χ, (25.157)

thus canceling in the transformation of the total mass term

FA+GB − 12χ̄χ. (25.158)

What about the A− and B-terms? We write these in (25.153) as

δ(χ̄χ) = ∂µ {χ̄[−iγµ(A+ iγ5B)]ǫ+ ǭi(A+ iγ5B)γµχ}−{

−χ̄←/∂ i(A+ iγ5B)ǫ− ǭi(A + iγ5B)

→/∂ χ

}

. (25.159)

The first term is a pure surface term and does not contribute to the total action. Inthe third term we use

χ̄γµǫ = −ǭγµχ, χ̄γµγ5ǫ = −ǭγ5γµχ, (25.160)

and see that it is equal to the fourth term. Both together cancel the AB-terms inδ(FA + GB). Thus the total action is invariant. The transformations are calledsupersymmetry transformation and the action supersymmetric.

Note that the fields F,G carry no gradients in the action and are therefore notdynamical. In principle they could be eliminated from the action using the equationsof motion, but then the verification of the supersymmetry would be much moreinvolved. Only if they are explicitly present, the four components of χ have fourreal scalar counterparts A,B,G, F . In a free field theory, the equations of motionmake F,G vanish. When interactions are included, they are still non-dynamical,but no longer zero.

In particle physics, no indication of some broken supersymmetry has so far beendiscovered. There exists, however, a supersymmetry between the unphysical contentin gauge fields and the Faddeev-Popov ghost fields in QED. This is essential forremoving the vacuum energy of the two unphysical modes. Recall the Lagrangian(7.525):

Ltot = L+LGF+Lghost = −1

4Fµν

2−D(x)∂µAµ(x)+αD2(x)/2−i∂µC̄∂µC. (25.161)

It is invariant under the following two sets of supersymmetry transformations:

δAµ = i∂µC,

δC̄ = D, (25.162)

δC = δD = 0,

Notes and References 1407

and

δ̄Aµ = i∂µC̄,

δ̄C = −D, (25.163)δ̄C̄ = δ̄D = 0.

These are known as the BRS transformations (see Footnote 14 in Chapter 7).The BRS supersymmetry is generated by charges

Q =∑

k

ωk(ak,lck + h.c.),

Q̄ =∑

k

ωk(ak,lc̄k + h.c.). (25.164)

Together with the subsidiary conditions (7.523), the conditions

Q|0〉 = 0, Q̄|0〉 = 0 (25.165)

are seen to enforce precisely the Gupta-Bleuler subsidiary condition (7.502) (see Ref.[3]). The full power of the BRS supersymmetry is needed in the renormalization ofnonabelian gauge theories. This is necessary for extracting finite results from higher-order perturbative calculations of weak interaction processes within the standardmodel of electroweak interactions.

Note that by postulating the absence of ghosts from the Hilbert space at a specifictime, they are excluded for all times since the Lagrangian conserves ghost charge.This is a consequence of its invariance under

C → eαC, C̄ → e−αC̄. (25.166)

The ghost charge conservation follows from Noether’s theorem (recall Chapter 8).

Notes and References

[1] E.P. Wigner and E. Feenberg, Phys. Rev. 51, 106 (1937) and Rep. Progr. Phys. 8, 274(1941).

[2] For more detail see the paper by A. Franzini, A. Radicati, Phys. Lett 6, 322 (1963).

[3] For details see T. Kugo and I. Ojima, Suppl. Prog. Theor. Phys. 66, 1 (1979) and J. Ambjornand R.J. Hughes, Nucl. Phys. B 217, 336 (1983).

[4] S. Coleman and J. Mandula, Phys. Rev. 139. 1251 (1967).

[5] D.V. Volkov, V.P. Akulov, Zh. Eksp. Teor. Fiz. (Pisma) 16, 621 (1972); Phys. Lett. B 46,109 (1973).V.P. Akulov, D.V. Volkov, Teor. Mat. Fiz. 18, 39 (1974).J. Wess and B. Zumino, Phys. Lett. B 49, 521 (1974), Nucl. Phys. B 70, 39 (1974), B 76,310, B 78, 1 (1974), Phys. Lett. B 59, 163 (1975).

[6] F. Buccella, P. Sorba, Mod. Phys. Lett. A 19, 1547 (2004);F. Buccella, H. Hogaasen, J.M. Richard, and P. Sorba, Eur. Phys. J. C 49, 743 (2007).

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