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Lie Groups and Lie Algebras

The standard treatment of group theory for physicists usually begins with complicated definitions, lemmas, and existence proofs which are certainly necessary for a proper understanding of group theory. (See Appendix A at end of notes).

Most physicists have not yet learned to live with this abstract group theory in the same way that they have learned for other mathematical techniques such as differential equations, Fourier analysis, or complex integration.

This is very strange since many of the so-called group theoretical methods are, in principle, no different and no more complicated than some of the mathematical techniques that every physicist learns to use in a course on quantum mechanics, namely, the algebra of angular momentum operators.

It is possible for physicists to understand and to use many techniques which have a group theoretical basis without understanding all aspects of the abstract group theory. This can be done in the same way that she uses angular momentum algebra without spending much time studying the abstract properties of the 3-dimensional rotation group.

Review of Angular Momentum Algebra

The commutation relations that define angular momentum operators

Ji , J j⎡⎣ ⎤⎦ = iεijk Jk ≡ i εijk Jkk∑

where

εijk =+1 if ijk = even permutation of 123−1 if ijk = odd permutation of 123 0 if any two indices are identical

⎧⎨⎪

⎩⎪

and the Einstein summation convention over repeated indices is understood if the summation sign is left out (unless an explicit override is given).

In addition, these are all Hermitian operators, i.e., Ji( )+ = Ji+ = JiSince these three operators form a closed commutator algebra, we can solve for the eigenvectors and eigenvalues using only the commutators.

The three operators J1 , J2 and J3 do not commute with each other and

hence do not share a common set of eigenvectors (called a representation).

However, there exists another operator that commutes with each of the angular momentum components separately. If we define

J 2 = Ji

2

i=1

3

∑which is the square of the total angular momentum vector, we have

J 2 , Ji⎡⎣ ⎤⎦ = 0 for i = 1 , 2 , 3

In addition, we have J 2( )+ = J 2, so that J 2 is Hermitian also.The commutation relations and the Hermitian property say that J 2 and any one of the components share a complete set of common eigenvectors. By convention, we choose to use J 2 and J3 as the two operators, whose eigenvalues(good quantum numbers) will characterize the set of common eigenvectors.

As we shall see, J 2 is a so-called Casimir invariant operator that characterizes the representations(set of eigenvectors). In particular, the eigenvalue of J 2 characterizes the representation and the eigenvalues of one of the components of the angular momentum (usually J3 ) will characterize the

eigenvectors within a representation.

We define the eigenvector/eigenvalue relations by the equations

J 2 λm = λ2 λm

J3 λm = m λmwhere the appropriate factors of that have been explicitly put into the relations make m and λ dimensionless numbers.

We now define some other operators and their associated commutators so that we can use them in our derivations.

J± = J1 ± iJ2

J− = J+( )+ → they are not Hermitian operators

We then have

J 2 , J±⎡⎣ ⎤⎦ = J 2 , J1⎡⎣ ⎤⎦ ± i J2 , J2⎡⎣ ⎤⎦ = 0

J3, J±⎡⎣ ⎤⎦ = J3, J1⎡⎣ ⎤⎦ ± i J3, J2⎡⎣ ⎤⎦ = iJ2 i iJ1( ) = ±J±

and

J+ , J−⎡⎣ ⎤⎦ = J1, J1⎡⎣ ⎤⎦ + i J2 , J1⎡⎣ ⎤⎦ − i J1, J2⎡⎣ ⎤⎦ − J2 , J2⎡⎣ ⎤⎦ = −2i J1, J2⎡⎣ ⎤⎦ = 2J3

and

J+ J− = J1 + iJ2( ) J1 − iJ2( ) = J12 + J2

2 − i J1, J2⎡⎣ ⎤⎦ = J 2 − J3

2 + J3

and

J− J+ = J

2 − J32 − J3

Finally, we have

J 2 =

J+ J− + J− J+2

+ J32

Derivation of Eigenvalues

Now the definitions tell us that

λm J 2 λm = λ2 λm λm = λm Ji2 λm

i∑

λ2 λm λm = λm Ji Ji λmi∑ = λm Ji

+ Ji λmi∑

Let us define the new vector α i = Ji λm . Remember that the norm

of any vector is non-negative, i.e., a a ≥ 0 . Therefore

λm λm ≥ 0 and α i α i ≥ 0

Now since α i = λm Ji+ we have

λ2 λm λm = λm Ji+ Ji λm

i∑ = α i α i

i∑ ≥ 0

or we have

λ ≥ 0 or the eigenvalues of J 2 are greater than 0In fact, we can even say more than this using these equations. We have

λ2 λm λm = α i α ii∑

= α1 α1 + α2 α2 + α 3 α 3

= α1 α1 + α2 α2 + λm J32 λm

= α1 α1 + α2 α2 + m2

2 λm λm ≥ 0

which says that

λ2 λm λm ≥ m2

2 λm λm

λ ≥ m2

This says that for a fixed value of λ (the eigenvalue of J 2 ),

which characterizes the representation, there must be maximum and minimum values of m (the eigenvalue of J3 ), which characterizes the eigenvectors within a representation.

Now we have

J3J+ λm = J3(J+ λm ) = (J+ J3 + J3, J+⎡⎣ ⎤⎦) λm = (J+ J3 + J+ ) λm

= (m +1)J+ λm = (m +1)(J+ λm )

which says that J+ λm is an eigenvector of J3 with the raised

eigenvalue (m +1),i.e., J+ λm ∝ λ,m +1 (remember the harmonic

oscillator discussion).

Since we already showed that for fixed λ , there must be a maximum value of m , say mmax , then it must be the case that for

that particular m -value we have

J+ λmmax = 0If this were not true, then we would have

J+ λmmax ∝ λ,mmax +1

but this violates the statement that mmax was was the maximum m -value.

Using this result we find

J− J+ λmmax = 0 = J 2 − J32 − J3( ) λmmax

2 λ − mmax

2 − mmax( ) λmmax = 0

λ − mmax2 − mmax = 0→ λ = mmax

2 + mmax

λ = mmax (mmax +1)It is convention to define

mmax = j and hence λ=j(j+1)In the same way we can show

J3J− λm = J3(J− λm ) = (J− J3 + J3, J−⎡⎣ ⎤⎦) λm = (J− J3 − J+ ) λm

= (m −1)J− λm = (m −1)(J− λm )

which says that J− λm is an eigenvector of J3 with the lowered

eigenvalue (m −1),i.e., J− λm ∝ λ,m −1 .

If we let the minimum value of m be mmin , then as before we must

have

J− λmmin = 0or mmin is not the minimum value of m . This says that

J+ J− λmmin = 0 = J 2 − J32 + J3( ) λmmin

2 λ − mmin

2 + mmin( ) λmmin = 0

λ − mmin2 + mmin = 0→ λ = mmin

2 − mmin

λ = mmin (mmin −1) = j( j +1)which says that

mmin = − j

We have thus shown that the pair of operators J 2 and J3 have a

common set of eigenvectors jm (we now use the labels j and m ),

where we have found that

− j ≤ m ≤ j

and the allowed m -values change by steps of one, i.e., for a given j -value, the allowed m -values are

− j , − j +1 , − j + 2 , ............ , j − 2 , j −1 , jwhich implies that

2 j = integeror

j =

integer2

≥ 0 (the allowed values)

Thus, we have the allowed sets or representations of the angular momentum commutation relations given by

j = 0 , m = 0

j =12

, m =12

, − 12

j = 1 , m = 1 , 0 , -1

j =32

, m =32

, 12

, − 12

, − 32

and so on.

For each value of j , there are 2 j +1 allowed m -values in the

eigenvalue spectrum(representation) and

J 2 jm = 2 j( j +1) jm

J3 jm = m jmBefore proceeding, we need a few more relations. We found earlier that

J+ jm = C+ j,m +1 = α+

J− jm = C− j,m −1 = α−

and from these we have

α+ = C+* j,m +1

α− = C−* j,m −1

We can then say that

α+ α+ = C+2 j,m +1 j,m +1 = C+

2

= ( jm (J+ )+ )(J+ jm ) = jm J− J+ jm

= jm J 2 − J32 − J3( ) jm = jm 2 ( j( j +1) − m2 − m) jm

= 2 ( j( j +1) − m2 − m) jm jm = 2 ( j( j +1) − m2 − m)or

C+ = j( j +1) − m(m +1)

and similarly

α− α− = C−2 j,m −1 j,m −1 = C−

2

= ( jm (J− )+ )(J− jm ) = jm J+ J− jm

= jm J 2 − J32 + J3( ) jm = jm 2 ( j( j +1) − m2 + m) jm

= 2 ( j( j +1) − m2 + m) jm jm = 2 ( j( j +1) − m2 + m)or

C− = j( j +1) − m(m −1)

Therefore, we have the very important relations for the raising/lowering or ladder operators

J± jm = j( j +1) − m(m ±1) j,m ±1

= ( j ± m +1)( j m) j,m ±1

Some of these features can be demonstrated simply in diagrams of the type shown below.

These diagrams are 1-dimensional plots of the eigenvalues of Jz.

The top figure represents the operators J± as vectors which change the eigenvalue of Jz by ±1. The bottom figure illustrates

the structure of a typical multiplet, in this case one with J = 7 / 2, in which a point is plotted for each value of Jz where a

state exists in the multiplet. It is called a "state" diagram. The operation of any of the operators in the top figure on the states in the multiplet of the bottom figure is represented graphically by taking the top figure and placing it on the state diagram (bottom figure) and noting which states are connected by the vectors.

There are also well-known rules for combining multiplets(in group terminology these are called "representations"). A system

!

Jz

J+

J-

0-1 1

! ! !!!!!!

Jz

-7/2 -5/2 -3/2 -1/2 1/2 3/2 5/2 7/2

may consist of several parts, each of which is characterized by a multiplet having a particular value of J, i.e., orbital and spin angular momentum. These multiplets can be combined to form a multiplet describing the whole system. The rules for combining two multiplets are derived below. This process is easily extended to more than two mutiplets by taking them two at a time.

General Addition of Two Angular MomentaGiven two angular momenta

J1,op and

J2,op, we have the operators and

states for each separate system

J1,op →

J1,op2 , J1z , J1± → j1,m1

J2,op →

J2,op2 , J2z , J2± → j2 ,m2

with

J1,op

2 j1,m1 = 2 j1( j1 +1) j1,m1 , J1z j1,m1 = m1 j1,m1

J1± j1,m1 = j1( j1 +1) − m1(m1 ±1) j1 ±1,m1J2,op

2 j2 ,m2 = 2 j2 ( j2 +1) j2 ,m2 , J2z j2 ,m2 = m2 j2 ,m2

J2± j2 ,m2 = j2 ( j2 +1) − m2 (m2 ±1) j2 ±1,m2

Remember that there are 2 j1 +1 possible m1 values and 2 j2 +1 possible m2 values.

Since all of the "1" operators commute with all of the "2" operators, we can find a common eigenbasis for the four operators

J1,op2 , J1z ,

J2,op2 , J2z in terms of the direct-product states

j1, j2 ,m1,m2 = j1,m1 ⊗ j2 ,m2

For the combined system we define total operators as before

Jop =

J1,op +

J2,op= total angular momentum

Jz = J1z + J2z , Jop

2 , Jz⎡⎣ ⎤⎦ = 0Jop

2 , J1,op2⎡⎣ ⎤⎦ = 0 ,

Jop

2 , J2,op2⎡⎣ ⎤⎦ = 0 , J1,op

2 , Jz⎡⎣ ⎤⎦ = 0 , J2,op2 , Jz⎡⎣ ⎤⎦=0

These commutators imply that we can construct a common

eigenbasis of

J1,op2 ,J2,op2 , J 2 , Jz using the states j1, j2 , j,m where

Jop

2 j,m = 2 j( j +1) j,m and Jz j,m = m j,m

There are 2 j +1 possible m values for each allowed j value.

We cannot use the operators J1z , J2z to construct the eigenbasis

for the combined system because they do not commute with

Jop2 .

Remember that in order for a label to appear in a ket vector it must be one of the eigenvalues of a set of commuting observables since only such a set shares a common eigenbasis.

We now determine how to write the j1, j2 , j,m basis in terms of the

linear combinations of the j1, j2 ,m1,m2 = j1,m1 ⊗ j2 ,m2 basis. We have

j1, j2 , j,m = j '1, j '2 ,m1,m2

m2∑

m1∑

j '2∑

j '1∑ j '1, j '2 ,m1,m2 j1, j2 , j,m

where

j '1, j '2 ,m1,m2 j1, j2 , j,m = Clebsch-Gordon coefficients

This corresponds to inserting an identity operator of the form

I = j '1, j '2 ,m1,m2 j '1, j '2 ,m1,m2

m2∑

m1∑

j '2∑

j '1∑

Since

j '1, j '2 ,m1,m2

J1,op

2 j1, j2 , j,m = 2 j '1( j '1+1) j '1, j '2 ,m1,m2 j1, j2 , j,m

= 2 j1( j1 +1) j '1, j '2 ,m1,m2 j1, j2 , j,m

the Clebsch-Gordon(CG) coefficients must vanish unless j '1 = j1 and similarly unless j '2 = j2 . Thus we have

j1, j2 , j,m = j1, j2 ,m1,m2

m2∑

m1∑ j1, j2 ,m1,m2 j1, j2 , j,m

Also, as we saw earlier, since Jz = J1z + J2z we must have

j1, j2 ,m1,m2 Jz j1, j2 , j,m = (m1 + m2 ) j1, j2 ,m1,m2 j1, j2 , j,m

= m j1, j2 ,m1,m2 j1, j2 , j,m

which implies that the CG coefficients must vanish unless m1 + m2 = m. Thus the only non-vanishing coefficients are

j1, j2 ,m1,m2 j1, j2 , j,m = m1 + m2

and we can write

j1, j2 , j,m = j1, j2 ,m1,m2m2 =m−m1

∑m1

∑ j1, j2 ,m1,m2 j1, j2 , j,m

= j1, j2 ,m1,m2 = m − m1m1

∑ j1, j2 ,m1,m2 = m − m1 j1, j2 , j,m

For fixed j1 and j2 there are 2 j1 +1 possible m1 values and 2 j2 +1 possible m2 values. Thus, there are (2 j1 +1)(2 j2 +1) linearly independent states of the form

j1, j2 ,m1,m2 = j1,m1 ⊗ j2 ,m2

and hence the vector space describing the combined system is (2 j1 +1)(2 j2 +1)-dimensional.

This says that there must be (2 j1 +1)(2 j2 +1) states of the form

j1, j2 , j,m also.

We notice that there is only one state with m = m1 + m2 = j1 + j2, namely,

m1 = j1 and m2 = j2This state has the maximum possible m value.

There are two states with m = m1 + m2 = j1 + j2 −1, namely,

m1 = j1 −1 , m2 = j2 and m1 = j1 , m2 = j2 -1 and so on.

For example,

j1 = 2 , j2 = 1→ (2(2 +1))(1(1+1)) = 15 statesIf we label these states by the m-values only (since j1 and j2 do

not change) or m1,m2 we have(in this example)

m = 3→ 1 state → 2,1m = 2 → 2 states → 2,0 , 1,1m = 1→ 3 states → 1,0 , 0,1 , 2,−1m = 0 → 3 states → 0,0 , 1,−1 , −1,1m = −1→ 3 states → −1,0 , 0,−1 , −2,1m = −2 → 2 states → −2,0 , −1,−1m = −3→ 1 state → −2,−1

for a total of 15 states. The combined system, as we shall see by construction, has these states

j = 3→ m = 3,2,1,0,−1,−2,−3→ 7 statesj = 2 → m = 2,1,0,−1,−2 → 5 statesj = 1→ m = 1,0,−1→ 3 states

for a total of 15 states.

The general rules, which follows from group theory, are

(1) The combined system has allowed j values given by

j1 + j2 ≥ j ≥ j1 − j2 in integer steps (2) The total number of states is given by the total number of m-values for all the allowed j -values or

(2 j +1) =

j= j1 − j2

j1 + j2

∑ (2 j1 +1)(2 j2 +1)

We write the addition of two angular momenta symbolically as

j1⊗ j2 = j1 − j2 ⊕ j1 − j2 +1⊕ j1 − j2 + 2⊕ ......⊕ j1 + j2 −1⊕ j1 + j2Examples:

Our original special case of adding two spin = 1/2 systems gives

j1 = j2 =

12→ j = 0,1→ 1

2⊗

12= 0⊕1→ 4 states

which is the result we found earlier.

Other cases are:

j1 = j2 = 1→ j = 0,1,2 → 1⊗1 = 0⊕1⊕ 2 → 9 statesj1 = 2 , j2 = 1→ j = 1,2,3→ 2⊗1 = 1⊕ 2⊕ 3→ 15 statesj1 = 2 , j2 = 3→ j = 1,2,3,4,5 → 2⊗ 3 = 1⊕ 2⊕ 3⊕ 4 ⊕ 5 → 35 states

Actual Construction of States

Notation: (1) states labelled by numbers 7,6 are j,m states

(2) states labelled by symbols 3,2 ⊗ are m1,m2 states

(3) we suppress the j1, j2 labels everywhere

Procedure : (1) choose j1 and j2 (2) write down the direct-product m1,m2 basis

(3) determine the allowed j values (4) write down the maximum m state (m = j1+ j2 ); it is unique (5) the maximum m-value corresponds to the maximum j -value (6) use the lowering operator to generate all other m-states for this j -value; there are 2j+1, i.e.,

J− j,m = j( j +1) − m(m −1) j,m −1

(7) find the maximum m-state for the next lowest j-value; it is constructed from the same basis states as in the corresponding m-states for higher j-values; use orthonormality properties to figure out coefficients (8) repeat (6) and (7) until all j-values have been dealt with.

More Detailed Examples (we must learn this process by doing it)

#1 - Individual system values:

j1 = j2 =

12→ m1,m2 =

12

,− 12

(we already did this example)

The basis states are (we use the notation + − here instead of

1 / 2,−1 / 2 ⊗)

+ + + − − + − −m = m1 + m2 = 1 0 -1 0

Construction Algebra

Allowed j-values are j = 1, 0

j = 1 has 2j+1 = 3 m-values = 1, 0, -1 j = 0 has 2j+1 = 1 m value = 0 1,1 = + + maximum or topmost (j,m) state is always unique

J− 1,1 = 2 1,0 = (J1− + J2− ) + + = + − + − +

1,0 =12

+ − +12

− +

J− 1,0 = 2 1,−1 = (J1− + J2− )(12

+ − +12

− + ) = 2 − −

1,−1 = − −

We must now have 0,0 = a + − + b − + (rule 7) with

a 2 + b 2 = 1 and 1,0 0,0 =

12a +

12b = 0

which gives

a =

12= −b

or

0,0 =

12

+ − −12

− +

All the j-values are now done. We end up with the Clebsch-Gordon coefficients

#2 - Individual system values:

j1 = j2 = 1→ m1,m2 = 1,0,−1The basis states are

1,1 ⊗ 1,0 ⊗ 1,−1 ⊗ 0,1 ⊗ 0,0 ⊗ 0,−1 ⊗ −1,1 ⊗ −1,0 ⊗ −1,−1 ⊗

m = m1 + m2 = 2 1 0 1 0 -1 0 -1 -2 Construction Algebra

Allowed j-values are j = 2, 1, 0

j = 2 has 2j+1 = 5 m-values = 2, 1, 0, -1, -2

j = 1 has 2j+1 = 3 m-values = 1, 0, -1

j = 0 has 2j+1 = 1 m-value = 0 2,2 = 1,1 ⊗ maximum or topmost (j,m) state is always unique

J− 2,2 = 2 2,1 = (J1− + J2− ) 1,1 ⊗ = 2 1,0 ⊗ + 2 0,1 ⊗

2,1 =12

1,0 ⊗ +12

0,1 ⊗

J− 2,1 = 6 2,0 = (J1− + J2− )( 12

1,0 ⊗ +12

0,1 ⊗ )

= 1,−1 ⊗ + 2 0,0 ⊗ + −1,1 ⊗

2,0 =16

1,−1 ⊗ +26

0,0 ⊗ +16

−1,1 ⊗

Continuing we have

2,−1 =12

−1,0 ⊗ +120,−1 ⊗

2,−2 = −1,−1 ⊗

which completes the five j = 2 states. We must now have 1,1 = a 1,0 ⊗ + b 0,1 ⊗ with

a 2 + b 2 = 1 and 2,1 1,1 =

12a +

12b = 0

which gives

a =

12= −b

or

1,1 =

121,0 ⊗ −

120,1 ⊗

We now find all of the j = 1 states

J− 1,1 = 2 1,0 = (J1− + J2− )( 12

1,0 ⊗ −12

0,1 ⊗ )

= 1,−1 ⊗ − −1,1 ⊗

1,0 =12

1,−1 ⊗ −12

−1,1 ⊗

and continuing

1,−1 =

120,−1 ⊗ −

12

−1,0 ⊗

which completes the three j = 1 states.

We must now have 0,0 = a 1,−1 ⊗ + b 0,0 ⊗ + c −1,1 ⊗ with

a 2 + b 2 + c 2 = 1 and

2,0 0,0 =16a +

26b +

16c = 0 and 1,0 0,0 =

16a −

16c = 0

which gives

a = −b = c =

13

or

0,0 =

131,−1 ⊗ −

130,0 ⊗ +

13−1,1 ⊗

All the j-values are now done.

So far we have considered only the consequences of the angular momentum commutation rules and have made no mention of any Lie group. All the results obtained so far depend only on the existence of operators satisfying the angular momentum commutation rules and are not in any way dependent on the existence of a continuous group of transformations.

We learned in the quantum seminar that the angualr momentum operator are the generators of rotations about the corresponding axes. We, in fact, derived the commutation relations from this property. The statement that a Hamiltonian is invariant under rotations is equivalent to the statement that it commutes with the angular momentum operators. By studying the properties of these rotations and the way in which sate vectors and operators transform under them, many interesting and useful results can be obtained.

In our study of particle physics in this seminar, we will be able to proceed directly from the commutation rules and will not need to delve into the more complicated aspects of group theory.

First, we will extend the angular momentum ideas to isospin and develop a graphical method using "weight" diagrams to discuss the corresponding group properties of SU(2). We will then extend these graphical methods to the group SU(3) to introduce the mathematical properties associated with quarks and gluons.

Generalization by Analogy of the Angular Momentum Results

Starting from the commutation rules one finds a classification scheme for the states of a quantum mechanical system in which angular momentum operators have simple properties. The states are divided into sets or multiplets or representations such that the matrix elements of all angular momentum operators vanish between states belonging to different multiplets. Within each multiplet the action of the angular momentum operators is simple. Appropriate linear combinations of these operators so that they are either diagonal( Jz ) or step operators( J± ). The

latter simply change the eigenvalue of the state on which they are operating and thus jump from one state to another within the

multiplet. The matrix elements of all angular momentum operators vanish between states belonging to different multiplets.

We now assert without proof: Whenever one encounters a set of operators satisfying similar commutation rules, one can play the same game. One can define multiplets and suitable linear combinations of the operators such that these operators are either diagonal in the representation defining the multiplets or act like step operators within a multiplet.

Let us now assume that we have a finite number of operators Xρ

which satisfy commutation rules similar to those of the angular momentum operators, namely that the commutator of of any two of the operators is a linear combination of the operators of the set:

Xρ ,Xσ⎡⎣ ⎤⎦ = Cρσ

τ Xττ∑

where the coefficients Cρστ are constants. A set of operators

satisfying such commutation rules is called a Lie algebra. We now assert that from these operators we can construct operators like J 2 which commute with all the operators of the set. There may only be one such independent operator as in the case of ordinary angular momentum. We will call such operators Cµ. These

operators are sometimes called Casimir operators.

Cµ ,Xρ⎡⎣ ⎤⎦ = 0 for all µ,ρ

We now choose one of the operators of the set (like Jz earlier)

to be diagonal in the representation we will define. It may be possible to choose more than one such operator. If there are many operators in the set there may be operators within the set that commute with one another. We will choose as many commuting operators as we can find and denote them by the letters Hi . These

operators Hi also commute with the Casimir operators Cµ since the

latter commute with all operators of the set. We can therefore find a complete set of states in any problem which are simultaneous eigenvectors of all the operators Cµ and Hi with

eigenvalues as shown by the labels cµ ,hi . These are analogous to

the complete set of states J,M .

We now further assert that the remaining operators of the set can all be expressed in terms of a linearly independent set of

step operators. We call these operators Eα and state that they

satisfy the following simple commutation rules:

Cµ ,Eα⎡⎣ ⎤⎦ = 0

Hi ,Eα[ ] = α iEα

These are directly analogous to the commutation rules satisfied by the operators J± . The operators Eα are thus step operators

which shift the eigenvalue of the operators Hi by an amount α i .

In the same way as with angular momentum, one can generate sets of states or multiplets and determine their structure, i.e., numbers and spacings. State diagrams (and weight diagrams once we define them) can be constructed to represent any Lie algebra and the associated multiplets. However, if there are several operators Hi which are simultaneously diagonal, several quantum

numbers are then required to specify the position of a state in the multiplet. In such a case, the diagrams are not 1-dimensional as was the case for angular momentum, but r-dimensional where r is the number of simultaneously commuting operators Hi which exist in the set. The Lie algebra is then said

to be of rank r. The angular momentum algebra is thus of rank 1.

There are general rules for combining multiplets like there are rules for coupling angular momenta We will develop a graphical technique using the weight diagrams to combine multiplets.

SU(2), SU(3) and the Eightfold Way - A Graphical Procedure(This section follows the work of Gasiorowicz - ANL-6729)

The basic elements of a group are unitary operators (matrices of determinant = +1 or unimodular matrices) that transform the basis vectors qi (we switch to Dirac notation for convenience) among themselves as

q 'i = uji

i=1

n

∑ qi

We assume that the observed elementary particle states or vectors, as linear combinations (or composites) of these basis vectors also transform into one another under the unitary operations of the group.

It is most convenient to work with infinitesimal unitary transformations where

U = exp i

2ε jλ j

j∑

⎛

⎝⎜⎞

⎠⎟= I +

i2

ε jλ jj∑

where the ε j are infinitesimal. Now

detU = 1 = det I +

i2

ε jλ jj∑

⎛

⎝⎜⎞

⎠⎟= det I + i

2ε jTrλ j

j∑ +O(ε 2 )

Therefore we have, in the general case,

ε jTrλ j

j∑ = 0

We can satisfy this condition by choosing matrices λ j such that

Trλ j = 0

for all j. For n x n matrices, there are n2 −1 linearly independent traceless matrices. Since these matrices now have all of the properties of the generators of a group as discussed in the earlier parts of these notes, we will make that designation. The λ j , j = 1,2,3,....,n

2 −1 are the generators of the

group.

We need, therefore, only study the commutation relations among the generators in order to understand all of the group properties.

To see how this all works and to connect it to earlier ideas we will now use SU(2) as an illustration. As we stated earlier, the algebra is defined by the commutation relations among a certain group of operators(the generators). In order to find

(1) all the operators and (2) the defining commutation relations of the algebra we start with the simplest physical realization of this algebra.

We assume the existence some entity with 2 states (we end up with the 2-dimensional representation) and represent it by a 2-element column vector, i.e.,

ab

⎛⎝⎜

⎞⎠⎟=

amount of "upness"amount of "downness"

⎛⎝⎜

⎞⎠⎟

The example we will use is a quantity called isospin (I1, I2 , I3) and

the “upness” and “downness” will refer to the values of I3 = Iz , i.e.,

proton = p =

10

⎛⎝⎜

⎞⎠⎟, neutron = n =

01

⎛⎝⎜

⎞⎠⎟

The operator (2 x 2 matrix) which converts a neutron into a proton (a raising operator) is given by

τ + =

0 10 0

⎛⎝⎜

⎞⎠⎟

where

τ + n = p , τ + p = 0Similarly, its Hermitian adjoint (a lowering operator) is

τ− = τ +

+ =0 01 0

⎛⎝⎜

⎞⎠⎟

where

τ− p = n , τ− n = 0

Note that τ + "annihilates" protons and τ− "annihilates"

neutrons.

Forming their commutator we find that

τ + ,τ−[ ] = τ 3 =

1 00 −1

⎛⎝⎜

⎞⎠⎟

and

τ 3 p = p , τ 3 n = − n

We also find that

τ 3 ,τ ±[ ] = ±2τ ±

We then define I = 1

2 τ → (I1, I2 , I3) = 12 (τ1,τ 2 ,τ 3) (defined below)

so that the "upness" values are given by ± 12 . i.e.,

I3 p = 12 p , I3 n = − 1

2 n

That completes the discussion of the 2-dimensional representation.

The same commutation relations are also used to derive higher order representations.

Note that these three commutators only involve the same three operators. This means that these three commutators complete the commutation relations (all further commutators can be expressed in terms of these three operators or we have a closed commutator algebra). The three operators τ + ,τ− ,τ 3 are the basic set for SU(2). An equivalent set is

τ1 = i τ− − τ +( ),τ 2 = i τ− + τ +( ),τ 3This also correlates with the earlier statements that we would have n2 −1 = 3 generators for SU(2).

To obtain the other representations of the SU(2) algebra, we must only use these defining commutation relations.

It is clear from the commutators that only one of the Hermitian operators τ1 ,τ 2 ,τ 3 can be diagonalized in a given representation of the space(using a given set of basis vectors). We will choose a representation in which τ 3 is diagonal (this was the choice in the 2-dimensional representation above also).

Remember that these diagonal elements are the eigenvalues of τ 3 .We label the state by the eigenvalue m of τ 3 , i.e.,

τ 3 m = m m

Now let us use the commutation relations.

τ 3τ + m = τ +τ 3 + 2τ +( ) m = m + 2( )τ + m

This implies that

τ + m = c m + 2In a similar manner, we can show that

τ− m = c ' m − 2Now, if we start with any state in a given representation, we can generate states with higher eigenvalues, until we reach the state with maximum eigenvalue M (which we assume exists).

Since we cannot go any higher, this state must have the property that

τ + M = 0This was the case for the 2-dimensional representation where τ + p = 0. We now start with this highest weight state, which it turns out is unique for an irreducible representation. We then have

τ− M = λ1 M − 2

which implies that (assuming λ1 is real)

M τ + = λ1 M − 2 (think matrices)As usual, we choose to normalize our basis states to 1, i.e.,m m = 1 and we then obtain

M τ +τ− M = M − 2 λ1λ1 M − 2 = λ12 M − 2 M − 2 = λ1

2

= M ( τ + ,τ−[ ] + τ−τ + ) M = M τ + ,τ−[ ] M + M τ−τ + M

= M τ 3 M = M

where we have used

τ + M = 0 , τ + ,τ−[ ] = τ 3 , τ 3 M = M M

So we get

λ12 = M

We also note that

M − 2 τ− M = λ1 = M τ + M − 2 *

which implies that

τ + M − 2 = λ1 M

Now consider

τ− M − 2 = λ2 M − 4which gives (same algebra as before)

λ22 = M − 2 τ +τ− M − 2 = M − 2 ( τ + ,τ−[ ] + τ−τ + ) M − 2

= M − 2 τ + ,τ−[ ] M − 2 + M τ−τ + M

= M − 2 τ 3 M − 2 + M λ12 M = λ1

2 + M − 2In general, we can state..... if

τ− M − 2(p −1) = λp M − 2p

we get

λp2 = λp−1

2 + M − 2(p −1)

Now λ02 = 0 (definition of maximum eigenvalue and see below). Thus,

we have

λ12 = λ0

2 +1(M −1+1) = M as before!λ2

2 = λ12 + M − 2(2 −1) = λ1

2 + M − 2 = 2(M − 2 +1)λ3

2 = λ22 + M − 2(3−1) = 3(M − 3+1)

or generalizing

λp2 = p(M − p +1)

When λp2 = 0 (p ≠ 0) we reach the minimum eigenvalue. This gives

p = M +1

which implies the minimum eigenvalue is −M , that is,

τ− M − 2(p −1) = τ− M − 2(M +1−1) = 0 (λM +1 = 0)→τ− −M = 0

The total number of states contained in this set is

p = M +1 = multiplicityLet us represent the states by a linear array

The operators τ ± represent steps between equally spaced points.

There is only one state for a given value of m and only one set of states (an irreducible representation) is generated from a given maximum state M . This result implies that there exists

one independent operator that can be constructed from τ ± and τ 3 which commutes with τ ± and τ 3 (has simultaneous eigenstates with τ 3 and thus can simultaneously label the states) and which serves to distinguish representations.

We consider the operator (the square of the isospin)

Γ =

12τ +τ + + τ−τ +( ) + 1

4τ 32 =

14(τ1

2 + τ 22 + τ 3

2 ) = 14τ 2

We have then have a Casimir operator since

Γ,τ i[ ] = 0which implies that

Γ m = c m for all mor

m Γ m = M Γ M = c

Using the commutation relations we have

Γ =

14τ 32 +12τ 3 + τ−τ +

which gives

c = M Γ M =

14M 2 +

12M =

12M

12M +1⎛

⎝⎜⎞⎠⎟

If we let t = M / 2 then we have c = t(t +1). This means that

p = M = 1 = 2t +1 = multiplicitywhere 2t +1 is an integer.

We note the connection to ordinary Isospin I:

I3 =

12τ 3 , I 2 = Γ =

14τ 2

Now to delineate the representations. First, the 2-dimensional representation (so that we can see the general stuff give back the specific case that we started with). This is the smallest or lowest representation.

We choose M = 1 corresponding to

t =

12and p = multiplicity = 2

We are most interested in finding the basis states or the simultaneous eigenvectors of I 2 and I3 since these will turn out to represent elementary particles.

We have (as before)

proton = p =

10

⎛⎝⎜

⎞⎠⎟, neutron = n =

01

⎛⎝⎜

⎞⎠⎟

I =

12

I3 p =12p , I3 n = −

12n

I 2 p =1212+1⎛

⎝⎜⎞⎠⎟p =

34p , I 2 n =

1212+1⎛

⎝⎜⎞⎠⎟n =

34n

I+ n = p , I− p = n

where we have used

I3 =

1 / 2 00 −1 / 2

⎛⎝⎜

⎞⎠⎟, I+ =

0 10 0

⎛⎝⎜

⎞⎠⎟, I− =

0 01 0

⎛⎝⎜

⎞⎠⎟

Therefore the state diagram for the 12 -representation

(representations are labeled by I value) is given by

where

M = 2I3(max imum) = 1 , p = multiplicity = M +1 = 2We now define a weight diagram for SU(2) by the following symbol:

Generating the 1-representation is just as easy and leads to a general graphical method for generating higher representations via the weight diagrams.

Generalizing, the I-representation (M=2I) has a multiplicity 2I+1 or 2I+1 equally spaced points (ΔI3 = constant) representing states in the representation.

We now consider the higher representation generated by combining two 1

2 -representations. The formal name for this product is a

direct product and it is written

12⊗12

The rules for using the weight diagrams are as follows:

(1) draw the first weight diagram

(2) superpose a set of weight diagrams centered at the arrow tips of the first weight diagram

(3) allowed states are at the final arrow tips

For the case we are considering this looks like:

We observe that

In words we say, the direct product of two 12 -representations is

equal to the sum of a 0- and a 1-representation.

In a similar manner we can generate the representations for

12⊗12⊗12

The superposed weight diagram picture looks like:

Note how multiple arrow tips require multiple weight diagrams in the next stage and multiple arrow tips at the end imply multiple states with that value of I3. Rearranging we have

which gives the result

12⊗12⊗12=12⊕12⊕32

Another way to think about it is to write

12⊗12⊗12= 0⊕1( )⊗ 1

2= 0⊗ 1

2+1⊗ 1

2

0⊗ 12=12

, 1⊗ 12=12⊕32

12⊗12⊗12=12⊕12⊕32

where we have used the fact that the 0-representation represents no additional states

0⊗ 1

2=12

and

1⊗ 1

2=12⊕32

as can be seen from the weight diagrams below:

This is the same as we found for angular momentum because it is exactly the same as the angular momentum coupling rule (there is a homomorphism between SU(2) and the rotation group O(3)). The same weight diagrams and graphical methods would work for angular momentum.

We have the general rule:

I1⊗ I2 = I1 − I2 ⊕ I1 − v2 +1⊕ .........⊕ I1 + I2

Now if SU(2) were the correct group for describing the real world, then all possible states (all known particles) should be generated from the 1

2 -representation (most fundamental building

block), i.e., from p and n as the fundamental buiilding

blocks.

We know this is not possible, however, because:

(1) we have not included any anti-baryons, which implies that we cannot have any states (particles) with baryon number B=0.

(2) we have not introduced strangeness (or hypercharge) so our states cannot represent the real world

Let us correct (1) first, within the context of SU(2), and then attack (2), which will require that we look at SU(3). Let us now add particle labels to our weight diagrams. The weight diagram for the 1

2 -representation or as we now rename it, the pn

weight diagram is given by

where we assumed Baryon number = B = 1 for each particle. We also assume that Baryon number is additive. We then define the12*-representation by the weight diagram

where we assumed Baryon number = B = -1 for each antiparticle. Now let us see the effect of this assumption on higher representations.

Remember

12⊗12= 0⊕1

Now we would have

Now consider

The isospin structure is the same as

12⊗12

but B = 0 (these could be mesons!). The original identifications were a “triplet” of mesons (π + ,π 0 ,π − ) and a “singlet” meson η0 .

The mesons would somehow be bound states made from the baryon pairs! It was the correct idea but the wrong fundamental baryons!

SU(3) - The Eightfold Way

In order to include hypercharge Y or strangeness S into our structure, we must start with three basic states (no identification with any known particles will be attempted at this point).

We choose:

q1 =100

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , q2 =

010

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , q3 =

001

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Since we will be working with vectors containing three elements and our operators will be 3 x 3 matrices, there will be 32 −1 = 8 independent matrices representing the generators of the transformations (this is the group SU(3)). These 8 operators will : (1) represent I3 (2) represent Ywhere we will want I3,Y[ ] = 0 so that our states can be simultaneous eigenstates of both operators and hence our particles can be characterized by values of both quantum numbers.

(3) represent q1 → q2




(7) represent q3 → q1 (8) represent q3 → q2which correspond to raising and lowering operators.

I will now write down(just straightforward algebra) the shift operators, the isospin and hypercharge operators and a whole mess of commutators and state equations. After that we will discuss their meanings. Shift Operators

E1 =16

0 1 00 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−1

+

E2 =16

0 0 10 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−2

+

E3 =16

0 0 00 0 10 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−3

+

where

E1 q2 = q1 , E2 q3 = q1 , E3 q3 = q2E−1 q1 = q2 , E−2 q1 = q3 , E−3 q2 = q3

Commutation Relations

E1,E2[ ] = E1,E−3[ ] = E2 ,E3[ ] = E−1,E−2[ ] = E−1,E−3[ ] = E−2 ,E−3[ ] = 0E1,E−2[ ] = −

16E−3 , E1,E3[ ] = 1

6E2 , E2 ,E−3[ ] = 1

6E1

E−1,E2[ ] = 16E3 , E−1,E−3[ ] = −

16E−2 , E−2 ,E3[ ] = −

16E−1

and

E1,E−1[ ] = 13H1 , E2 ,E−2[ ] = 1

2 3H1 +

12H2

E3,E−3[ ] = −12 3

H1 +12H2

where

H1 =12 3

1 0 00 −1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , H2 =

16

1 0 00 1 00 0 −2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

and finally

H1,E1[ ] = 13E1 , H1,E2[ ] = 1

2 3E2 , H1,E3[ ] = −

12 3

E3

H2 ,E1[ ] = 0 , H2 ,E2[ ] = 12 E2 , H2 ,E3[ ] = 12 E3 , H1,H2[ ] = 0It is clear from the commutators that we have a closed algebra (a Lie algebra), i.e., all of the commutators among the 8 operators only involve the same 8 operators. If we write out the commutators in general, we can express the RHS in terms of

so-called "structure" constants, i.e., for angular momentum we have the Lie algebra

Ji , J j⎡⎣ ⎤⎦ = iεijk Jk

where the εijk are the structure constants and the angular

momentum operators Jk are the generators.

For SU(2) we have the same Lie algebra (remember the homeomorphism).

For SU(3) we have the Lie algebra

λi ,λ j⎡⎣ ⎤⎦ = 2ifijkλk

where the fijk are the structure constants and the operators λk

(the E's and H's above) are the generators. See Appendix B for a discussion of the structure constants.

We note that q1 , q2 , q3 are eigenstates of both H1 and H2

simultaneously, which suggests that we will be able to use H1

and H2 to represent I3 and Y .

Now we have

H1 q1 =12 3

q1 ,H1 q2 = −12 3

q2 , H1 q3 = 0

H2 q1 =16q1 ,H2 q2 =

16q2 , H2 q3 = −

13q3

Thus, if we define I3 = 3H1 , Y = 2H2 we get

I3 Y

q1

q2

q3

1/2 1/3

-1/2 1/3

0 -2/3

Construction of Representations

We define eigenstates m1,m2 where we assume

H1 m1,m2 = m1 m1,m2 , H2 m1,m2 = m2 m1,m2

Now 6E1 , 6E−1 , 2 3H1 obey the same commutation relations as

τ + ,τ− ,τ 3 -- in fact, they form a sub-algebra of the closed algebra of the 8 operators. Therefore, we construct the operator

I 2 = 3(E1E−1 + E−1E1) + 3H12

in analogy to

Γ =

12τ +τ + + τ−τ +( ) + 1

4τ 32

Since I 2 ,H1⎡⎣ ⎤⎦ = I 2 ,H2⎡⎣ ⎤⎦ = 0 the states m1,m2 are also eigenstates of

I 2 or

I 2 m1,m2 ,i = i(i +1) m1,m2 ,iWe note that

I 2 ,E±1⎡⎣ ⎤⎦ = 0

but

I 2 ,E±2⎡⎣ ⎤⎦ ≠ 0 , I 2 ,E±3⎡⎣ ⎤⎦ ≠ 0

Which implies that I 2 is not an invariant operator or that it can change within a representation.

This suggests the following possible picture:

where each line is the same as in our earlier discussion ( I 2 = constant), but there are many different lines (values of I 2 ) in the representation. It is not difficult to deduce the following table:

operator/effect Δm1 Δm2

E1 1/√3 0

operator/effect Δm1 Δm2

E-1

E2

E-2

E3

E-3

-1/√3 0

1/2√3 1/2

-1/2√3 -1/2

-1/2√3 1/2

1/2√3 -1/2

This table implies relation like:

H1E1 m1,m2 = E1 H1 +

13

⎛⎝⎜

⎞⎠⎟m1,m2 = m1 +

13

⎛⎝⎜

⎞⎠⎟E1 m1,m2

or

E1 m1,m2 = constant m1 +

13,m2

Graphically, this table is represented by the diagram:

Let us now represent our states on a Y − I3 diagram and indicate the effect of the operators

I± = E±1 ΔI3 = ±1,ΔY = 0( )V± = E±2 ΔI3 = ±1 / 2,ΔY = ±1( )U± = E±3 ΔI3 = ±1 / 2,ΔY = ±1( )

Remember that I3 = 3H1 and Y = 2H2.

This gives the simplest representation of SU(3):

and suggests a weight diagram as shown below:

We have other related quantum numbers:

Q(charge) = I3 +

Y2

, Y = B + S

This leads to the table below:

Quark I I3 Y Q B S

u

d

s

1/2 1/2 1/3 2/3 1/3 0

1/2 -1/2 1/3 -1/3 1/3 0

0 0 -2/3 -1/3 1/3 -1

The baryon number of each of these states or particles is B = 1/3 and they represent the original three quarks.....up, down and strange.

Another way of visualizing the shifts (shift operators) is the diagram below:

These shift operator transform quarks into each other. They will turn out to represent gluons later on.

It turns out that there exists an equivalent simple representation. It is shown below:

It corresponds to the weight diagram below:

It is the 3* representation and the particles or states are the

anti-quarks. The corresponding table looks like:

Quark I I3 Y Q B S

anti-u=ū

anti-d=ƌ

anti- s=ŝ

1/2 -1/2 -1/3 -2/3 -1/3 0

1/2 1/2 -1/3 1/3 -1/3 0

0 0 2/3 1/3 -1/3 1

We now use these simplest representations or simplest weight diagrams to build higher representations. We note that

3⊗ 3* will have B = 0 (quark + anti-quark) and should represent mesons and

3⊗ 3⊗ 3 will have B = 1 and should represent baryons.

So we now consider 3⊗ 3* using the weight diagrams as shown

below:

where the black dots represent the final states of the new (higher) representations. Looking only the states we have

The I-values associated with a line are determined by the multiplicity of a line (remember each line is a different I-value). Note that there were three(3) final states at the same point. Symbolically we have

3⊗ 3* = 8⊕ 1 = octet + singlet

We now look more carefully at the OCTET representation. If we label the states as

We get the table

# quarks I I3 Y Q B S meson

1 dŝ 1/2 -1/2 1 0 0 1 K0

2 uŝ 1/2 1/2 1 1 0 1 K+

3 dū 1 -1 0 -1 0 0 π-

4 dƌ-uū 1 0 0 0 0 0 π0

5 dƌ+uū-2sŝ 0 0 0 0 0 0 η0

6 -uƌ 1 1 0 1 0 0 π+

7 sū 1/2 -1/2 -1 -1 0 -1 K-

8 sƌ 1/2 1/2 -1 0 0 -1 K0-bar

Our SU(3) scheme has produced the known mesons (including the strange particles) as linear combination of quarks and anti-quarks.

Let us push it even further and consider the baryon states. This means we look at the

3⊗ 3⊗ 3 representation. Using weight diagrams we get

These states separate as follows:

or symbolically we have

3⊗ 3⊗ 3 = 10

~⊕ 8⊕ 8⊕ 1

that is, we end up with a decuplet, two octets(turns out to be two ways of representing the same one) and a singlet. The particle identifications are given in the table below:

# States I I3 Y Q B S Particle

1 ddd 3/2 -3/2 1 -1 1 0 Δ-

2 udd+dud+ddu 3/2 -1/2 1 0 1 0 Δ0

3 duu+udu+uud 3/2 1/2 1 1 1 0 Δ+

4 uuu 3/2 3/2 1 2 1 0 Δ++

5 sdd+dsd+dds 1 -1 0 -1 1 -1 Σ*-

6 sdu+dsu+dus +sud+usd+uds

1 0 0 0 1 -1 Σ*0

7 suu+usu+uus 1 1 0 1 1 -1 Σ*+

8 dss_sds+ssd 1/2 -1/2 -1 -1 1 -2 Ξ*-

9 uss+sus+ssu 1/2 1/2 -1 0 1 -2 Ξ*0

10 sss 0 0 -2 -1 1 -3 Ω-

11 d[ud] ---- ---- ---- ---- ---- ---- ----

12 u[ud] ---- ---- ---- ---- ---- ---- ----

13 d[ds] ---- ---- ---- ---- ---- ---- ----

14 u[ds]+d[su]-2s[ud] ---- ---- ---- ---- ---- ---- ----

15 u[ds]-d[su] ---- ---- ---- ---- ---- ---- ----

16 u[su] ---- ---- ---- ---- ---- ---- ----

17 s[ds] ---- ---- ---- ---- ---- ---- ----

18 s[su] ---- ---- ---- ---- ---- ---- ----

19 ---- ---- ---- ---- ---- ---- ---- ----

20 uud+udu-2duu ---- ---- ---- ---- ---- ---- ----

---- ---- ---- ---- ---- ---- ---- ---- ----

27 u[ds]+d[su]+s[ud] ---- ---- ---- ---- ---- ---- ----

where [ab] = ab=ba.

Consider the DECUPLET representation. All of the particles had been previously observed except for the Ω− . , which we will discuss further later. We end up with two OCTET representations. The first is antisymmetric in the second and third quarks and the second is symmetric in the second and third quarks. They are completely equivalent representations within SU(3) and represent the same physical particles.

The Mass Formula

We will simply state and discuss the mass formula here. The most general form for the mass formula is given by:

M = A + BY + C I(I +1) − 1

4Y 2⎛

⎝⎜⎞⎠⎟+ DI3

where the A, B, and C terms represent the average mass (within a row) and the D term represents electromagnetic splittings. Some OCTET experimental data implies that:

M 8 = 1109.80 −189.83Y + 41.49 I(I +1) − 1

4Y 2⎛

⎝⎜⎞⎠⎟− 2.45I3

which gives the following tables :

particle experiment formula

n 939.5 943.2

p 938.2 938.3

Λ0 1115.4 1109.8

Σ- 1189.4 1190.3

Σ0 1193.2 1192.8

Σ+ 1197.6 1195.2

Ξ- 1316 1317.9

Ξ0 1321 1322.8

This is an average error of only 0.19%. The experimental data in the table below was also available. As we said above, the Ω−

particle had not yet been discovered.

particle experiment

Δ 1238

Σ* 1385

Ξ* 1530

This data gives the DECUPLET mass formula below:

M10 = 1501.8 − 58.5Y − 58.4 I(I +1) − 1

4Y 2⎛

⎝⎜⎞⎠⎟

The mass formula then allows us to predict the mass of the Ω− particle (SU(3) also predicts all of the other quantum numbers).

The prediction is MΩ− = 1677 (1685 is observed or about a 0.5%

error).

Introduction of Color

A basic problem with the quark model of hadrons is with the wave function antisymmetry. Consider the Ω− state. It is strangeness S = -3 baryon with isospin I = 3 / 2 . In the quark model

Ω− = s s s

Each of these quarks has the following states

s = 1

2 ,12 spin

12 ,

12 isospin

B = 13 ,S = −1

and, thus, the final state is symmetric in all of its quantum numbers (the orbital angular momentum states are also symmetric when one solves the bound state problems), which violates the Pauli Principle for fermions.

The solution to the problem is to generalize the definition of the quarks such that each quarks can have one of three colors (r=red, b=blue, g=green) or

uds

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ =

ur ub ugdr db dgsr sb sg

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

The associated observable is called flavor. Quarks now are assumed to come in three flavors, namely, red, blue and green.

The problem with the Ω− state is fixed by assuming that it is

completely antisymmetric in the flavor space or we write it as

Ω− = sr sb sg − sr sg sb − sb sr sg

+ sb sg sr + sg sr sb − sg sb srIt is equal parts red, green and blue!

Another type of problem was that certain reactions were not occurring when SU(3) said they were allowed. This usually signals that an extra (new) quantum number is needed to prevent the reaction. The new quantum number would have a new conservation law that would be violated if the reaction occurred and hence it is disallowed.

The new quantum number introduced was charm. In the quark model this is treated as the appearance of a new quark. I will delineate the starting point of this discussion below. It leads to the formalism of SU(4) if we carried out the same steps as we did with SU(3).

Quark states:

u =

1000

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

, d =

0100

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

, s =

0010

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

, c =

0001

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

We now have

Q = I3 +

B + S + C2

, C = charm

The four quarks then have quantum numbers

Quark Y Q B S C

u

d

s

c

1/3 2/3 1/3 0 0

1/3 -1/3 1/3 0 0

-2/3 -1/3 1/3 -1 0

1/3 2/3 1/3 0 1

The charm quark is the same as the up quark except for the charm value.

The corresponding lowest representation weight diagram is

This is the 4 representation.

That concludes this discussion of the Eightfold Way .........

Appendix A - Abstract Group Theory

Introduction to Group Theory

The theory of finite groups and continuous groups is a very useful tool for studying symmetry and invariance. In order to use group theory we need to introduce some definitions and concepts.

A group G is defined as a set of objects or operations (called elements) that may be combined or multiplied to form a well-defined product and that satisfy the following four conditions. If we label the elements a, b, c, .......... , then the conditions are:

(1) If a and b are any two elements, then the product ab is an element.

(2) The defined multiplication is associative, (ab)c = (a(bc)

(3) There is a unit element I, with Ia = aI = a for all elements a.

(4) Each element has an inverse b = a-1, with aa-1 = a-1a = I for all elements a.

In physics, these abstract conditions will take on direct physical meaning in terms of transformations of vectors and tensors.

As a very simple, but not trivial, example of a group, consider the set {I,a,b,c} that combine according to the group multiplication table:

I a b c

I

a

b

c

I a b c

a b c I

b c I a

c I a b

Clearly, the four conditions of the definition of group are satisfied. The elements {I,a,b,c} are abstract mathematical entities, completely unrestricted except for the above multiplication table.

A representation of the group is a set of particular objects {I,a,b,c} that satisfy the multiplication table.

Some examples are: {I=1, a=i, b=-1, c=-i}

where the combination rule is ordinary multiplication. This group representation is labelled C4. Since multiplication of the group elements is “commutative”, i.e., ab = ba for any pair of elements, the group is labelled “commutative or abelian”. This group is also a “cyclic” group, since all of the elements can be written as successive powers of one element. In this case,

{I=i4, a=i, b=i2, c=i3}

Another representation is given by the successive 90° in a plane. Remember the matrix that represents a rotation through angle φ in 2-dimensions is given by

R =

cosφ − sinφsinφ cosφ

⎛⎝⎜

⎞⎠⎟

with φ = 0,π / 2,π , 3π / 2 , we have

I =

1 00 1

⎛⎝⎜

⎞⎠⎟, a =

0 −11 0

⎛⎝⎜

⎞⎠⎟, b =

−1 00 −1

⎛⎝⎜

⎞⎠⎟, c =

0 1−1 0

⎛⎝⎜

⎞⎠⎟

Thus, we have seen two explicit representations. One with numbers(complex) using ordinary multiplication and the other with matrices using matrix multiplication.

There maybe a correspondence between the elements of two groups. The correspondence can be one-to-one , two-to-one or, many-to-one. If the correspondence satisfies the same group multiplication table, then it is said to be homomorphic. If it is also one-to-one, then it is said to be “isomorphic”. The two representations of that we discussed above are one-to-one and preserve the multiplication table and hence, are isomorphic.

The representation of group elements by matrices is a very powerful technique and has been almost universally adopted among physicists. It can be proven that all the elements of finite groups and continuous groups of the type important in physics can be represented by unitary matrices. For unitary matrices we have

A+ = A−1

or, in words, the complex conjugate/transposed is equal to the inverse.

If there exists a transformation that will transform our original representation matrices into diagonal or block-diagonal form, for example,

A11 A12 A13 A14A21 A22 A23 A24A31 A23 A33 A34A41 A24 A34 A44

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

⇒

P11 P12 0 0P21 P22 0 00 0 Q11 Q120 0 Q21 Q22

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

such that the smaller portions or submatrices are no longer coupled together, then the original representation is said to be reducible.

Equivalently, we have

SAS−1 =

P 00 Q

⎛⎝⎜

⎞⎠⎟

which is a "similarity" transformation.

We write

R = P⊕Q and say that R has been decomposed into the representations P and Q. The irreducible representations play a role in group theory that is roughly analogous to the unit vectors of vector analysis. They are the simplest representations -- all others may be built up from them.

If a group element x is transformed into another group element yby means of a similarity transformation involving a third group element gi

gixgi−1 = y

then y is said to be conjugate to x. A class is a set of mutually conjugate group elements. A class is generally not a group.

The trace of each group element (each matrix of our representation) is invariant under unitary transformations. We define

tr(A) = Aii

i=1

n

∑ = trace(A)

Then we have

tr(SAS−1) = (SAS−1)iii=1

n

∑ = sikk=1

n

∑j=1

n

∑i=1

n

∑ akjs ji−1 = akj

k=1

n

∑j=1

n

∑ s ji−1sik

i=1

n

∑

= akjk=1

n

∑j=1

n

∑ (S−1S) jk = akjk=1

n

∑j=1

n

∑ (I ) jk = akjk=1

n

∑j=1

n

∑ δ jk

= ajjj=1

n

∑ = tr(A)

We now relabel the trace as the character. Then we have that all members of a given class(in a given representation) have the same character. This follows directly from the invariance of the trace under similarity or unitary transformations.

If a subset of the group elements(including the identity element I) satisfies the four group requirements (means that the subset is a group also), then we call the subset a subgroup. Every group has two trivial subgroups: the identity element alone and the group itself. The group C4 we discussed earlier has a nontrivial subgroup comprised of the elements {I, b}.

The order of a group is equal to the number of group elements.

Consider a subgroup H with elements hi and a group element x not

in H. Then xhi and hix are not in the subgroup H. The sets generated by

xhi , i = 1,2,.... and hix , i = 1,2,....are called the left and right cosets, respectively, of the subgroup H with respect to x. The coset of a subgroup has the same number of distinct elements as the subgroup. This means that the original group G can be written as the sum of H and its cosets:

G = H + x1H + x2H + .......Since H and all of its cosets have the same order(same number of elements), the order(number of elements) of any subgroup is a divisor of the order(number of elements) of the group.

The similarity transformation of a subgroup H by a fixed group element x not in H

xHx−1

yields a subgroup. If this subgroup is identical to H for all x, i.e., if

xHx−1 = Hthen H is called an invariant, normal or self-conjugate subgroup.

In physics, groups usually appear as a set of operations that leave a system unchanged or invariant. This is an expression of symmetry. A symmetry is usually defined as the invariance of the Hamiltonian of a system under a group of transformations. We now investigate the symmetry properties of sets of objects such as the atoms in a molecule or a crystal.

Two Objects - Twofold Symmetry Axis

Consider first the two-dimensional system of two identical atoms in the xy-plane at (1,0)and (-1,0) as shown below.

What rotations can be carried out(keeping both atoms in the xy-plane) that will leave the system invariant? The only ones are I (always works) and a rotation by π about the z-axis. So we have a rather uninteresting group of two elements. The z-axis is labelled a twofold symmetry axis, corresponding to the two rotation angles 0 and π that leave the system invariant.

Our system becomes more interesting in -dimensions. Now imagine a molecule (or part of a crystal) with atoms of element X at ±a on the x-axis, atoms of element Y at ±b on the y-axis, and atoms of element Z at ±c on the z-axis as shown below:

Clearly, each axis is now a twofold symmetry axis. The matrix representation of the corresponding rotations are:

Rx (π ) =1 0 00 −1 00 0 −1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , Ry (π ) =

−1 0 00 1 00 0 −1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , Rz (π ) =

−1 0 00 −1 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , I =

1 0 00 1 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

These four elements I ,Rx (π ),Ry (π ),Rz (π ){ } form an abelian group with a group multiplication table:

I Rx Ry Rz

I

Rx

Ry

Rz

I Rx Ry Rz

Rx I Rz Ry

Ry Rz I Rx

Rz Ry Rx I

The products in this table can be obtained in either of two distinct ways:

(1) We may analyze the operations themselves -- a rotation of π about the x-axis followed by a rotation of π about the y-axis is equivalent to a rotation of π about the z-axis

Ry (π )Rx (π ) = Rz (π )

and so on.

(2) Alternatively, once the matrix representation is established , we can obtain the products by direct matrix multiplication. This latter method is especially important when the system is too complex for direct physical interpretation. This symmetry group is often labelled D2, where the D signifies a dihedral group and the subscript 2 signifies a twofold symmetry axis (and that no higher symmetry axis exists).

Three Objects -- Threefold Symmetry Axis

Consider now three identical atoms at the vertices of an equilateral triangle as shown below:

Rotations (in this case we assume counterclockwise) of the triangle of 0, 2π / 3, and 4π / 3 leave the triangle invariant. In matrix form, we have

I =1 00 1

⎛⎝⎜

⎞⎠⎟

A = Rz (2π / 3) =cos2π / 3 − sin2π / 3sin2π / 3 cos2π / 3

⎛⎝⎜

⎞⎠⎟=

−1 / 2 − 3 / 2

3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟

B = Rz (4π / 3) =−1 / 2 3 / 2

− 3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟

The z-axis is a threefold symmetry axis. {I, A, B} form a cyclic group, which is a subgroup of the complete -element group that we now discuss.

In the x-y plane there are three additional axes of symmetry -- each atom (vertex) and the origin define the twofold symmetry axes C, D, and E. We note that twofold rotation axes can also be described via reflections through a plane containing the twofold axis.

In this case, we have for rotation of π about the C-axis.

C = RC (π ) =

−1 00 1

⎛⎝⎜

⎞⎠⎟

The rotation of π about the D-axis can be replaced by a rotation of 4π / 3 about the z-axis followed by a reflection in the y-z plane (x→ −x) or

D = RD (π ) = CB =−1 00 1

⎛⎝⎜

⎞⎠⎟

−1 / 2 3 / 2

− 3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟ =

1 / 2 − 3 / 2

− 3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟

Similarly the rotation of π about the E-axis can be replaced by a rotation of 2π / 3 about the z-axis followed by a reflection in the y-z plane (x→ −x) or

E = RE (π ) = CA =−1 00 1

⎛⎝⎜

⎞⎠⎟

−1 / 2 − 3 / 2

3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟ =

1 / 2 3 / 2

3 / 2 −1 / 2

⎛

⎝⎜

⎞

⎠⎟

The complete group multiplication table is :

I A B C D E

I I A B C D E

I A B C D E

A

B

C

D

E

A B I D E C

B I A E C D

C E D I B A

D C E A I B

E D C B A I

Notice that each element of the group appears only once in each row and in each column.

It is clear from the multiplication table that the group is not abelian. We have explicitly constructed a 6-element group and a 2 x 2 irreducible matrix representation of it. The only other distinct 6-element group is the cyclic group I ,R,R2 ,R3,R4 ,R5{ } with

R =cosπ / 3 − sinπ / 3sinπ / 3 cosπ / 3

⎛⎝⎜

⎞⎠⎟=

1 / 2 − 3 / 2

3 / 2 1 / 2

⎛

⎝⎜

⎞

⎠⎟

The group {I,A,B,C,D,E} is labeled D3 in crystallography, the dihedral group with a threefold axis of symmetry. The three axes (C,D,E) in the x-y plane are twofold symmetry axes and, as a result, (I,C), (I,D), and (I,E) all form two-element subgroups. None of these two-element subgroups are invariant.

There are two other irreducible representations of the symmetry group of the equilateral triangle. They are

(1) the trivial {1,1,1,1,1,1}

(2) the almost as trivial {1,1,1,-1,-1,1}

Both of these representations are homomorphic to D3.

A general and most important result for finite groups of h elements is that

ni2

i∑ = h

where ni is the dimension of the matrices of the ith irreducible representation. This equality is called the dimensionality

theorem and is very useful in establishing the irreducible representation of a group. Here for D3 we have

12 +12 + 22 = 6for our three representations. This means that no other representations of the symmetry group of three objects exists.

We note from the examples above that for the transformations involving rotations and reflections, the transformations involving only pure rotations have determinant = 1 and the those involving a rotation and a reflection have determinant = -1.

Continuous Groups

The groups that we have been discussing all have a finite number of elements. If we have a group element that contain one or more parameters that vary continuously over some range, then variation of the parameter will produce a continuum of group elements (an infinite number). These groups are called continuous groups.

If continuous group has a rule for determining combination of elements or the transformation of an element or elements into a different element which is an analytic function of the parameters, where analytic means having derivatives of all orders, then it is called a Lie group. For example, suppose we have the transformation rule

x 'i = fi (x1, x2 , x3,θ)then for this transformation group to be a Lie group the functions fi must be analytic functions of the parameter θ .

This analytic property will allow us to define infinitesimal transformations thereby reducing the study of the whole group to a study of the group elements when they are only slightly different than the identity element I.

Orthogonal Groups

We begin our study by looking at the orthogonal group O3. In particular, we consider the set of n x n real, orthogonal matrices with determinant = +1 (no reflections). The defining property of a real orthogonal matrix is

AT = A−1

where T represents the transpose operation. In terms of matrix elements this gives

(AT ) ji

i∑ Aik = (A−1) ji

i∑ Aik = I jk = δ jk = Aij

i∑ Aik

The ordinary rotation matrices are an example of real 3 x 3 orthogonal matrices:

Rx (φ) =1 0 00 cosφ sinφ0 − sinφ cosφ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , Ry (θ) =

cosθ 0 − sinθ0 1 0sinθ 0 cosθ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , Rz (γ ) =

cosγ sinγ 0− sinγ cosγ 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

These matrices follow the convention that the rotation is a counterclockwise rotation of the coordinate system to a new orientation.

Each n x n real orthogonal matrix with determinant = +1 has n(n −1) / 2 independent parameters. For example, 2-dimensional rotations are described by 2 x 2 real orthogonal matrices and they need only one independent parameter (2(2 −1) / 2 = 1), namely the rotation angle. On the other hand, 3-dimensional rotations require three independent angles (3(3−1) / 2 = 3) .

Let us see this explicitly for the 2 x 2 case: Suppose we have the general 2 x 2 matrix

a bc d

⎛⎝⎜

⎞⎠⎟

where all the elements are real. Imposing the condition determinant = +1 gives ad − bc = 1Imposing the condition that the transpose is the inverse gives

a bc d

⎛⎝⎜

⎞⎠⎟a cb d

⎛⎝⎜

⎞⎠⎟=

a2 + b2 ac + bdac + bd c2 + d 2

⎛⎝⎜

⎞⎠⎟= I =

1 00 1

⎛⎝⎜

⎞⎠⎟

These equations imply the solution

a 1− a2

− 1− a2 a

⎛

⎝⎜⎜

⎞

⎠⎟⎟

which shows that there is only one independent parameter.

Special Unitary Groups, SU(2)

The set of complex n x n unitary matrices also forms a group. This group is labeled U(n). If we impose an additional restriction that the determinant of the matrices be +1, then we

get the special unitary group, labeled SU(n). The defining relation for complex unitary matrices is

A+ = A−1

where + represents the complex conjugate operation.

In this case we have n2 −1 independent parameters. This can be seen explicitly below:

Suppose we have the general 2 x 2 matrix representing the transformation U

a bc d

⎛⎝⎜

⎞⎠⎟

where all the elements are complex. Imposing the condition determinant = +1 gives ad − bc = 1Imposing the condition that the transpose is the inverse gives

a bc d

⎛⎝⎜

⎞⎠⎟a* c*

b* d*⎛⎝⎜

⎞⎠⎟=

a 2 + b 2 ac* + bd*

a*c + b*d c 2 + d 2

⎛

⎝⎜

⎞

⎠⎟ = I =

1 00 1

⎛⎝⎜

⎞⎠⎟

These equations imply the solution

a b−b* a*

⎛⎝⎜

⎞⎠⎟

where

a 2 + b 2 = aa* + bb* = 1Since each complex number has two independent components, this shows that there are three (22 −1) independent parameters. For n = 3, there will be eight (32 −1) independent parameters. This will become the famous eightfold way of elementary particle physics.

An alternative general form is for the 2 x 2 case is

U(ξ,η,ζ ) =

eiξ cosη eiζ sinη−e− iζ sinη e− iξ cosη

⎛⎝⎜

⎞⎠⎟

Now we will determine the irreducible representations of SU(2). Since the transformations are 2 x 2 matrices, the objects being transformed (the vectors or states) will be a two-component complex column vector (called a spinor):

u 'v '

⎛⎝⎜

⎞⎠⎟=

a b−b* a*

⎛⎝⎜

⎞⎠⎟uv

⎛⎝⎜

⎞⎠⎟

or

u ' = au + bv , v ' = −b*u + a*vFrom the form of this result, if we were to start with a homogeneous polynomial of the nth degree in u and v and carry out the unitary transformation given by the last two equations, we would still have a homogeneous nth degree polynomial.

To illustrate what happens we use the function(due to Wigner):

fm (u,v) =

u j+mv j−m

( j + m)!( j − m)!

The index m will range from -j to j, covering all terms of the form u pvq with p + q = 2j. The denominator makes sure that our representation will be unitary. Now we can show that

Ufm (u,v) = fm (u ',v ')which becomes

Ufm (u,v) = fm (au + bv,−b

*u + a*v) = (au + bv)j+m (−b*u + a*v) j−m

( j + m)!( j − m)!

We now express this last equation as a linear combination of terms of the form fm (u,v). The coefficients in the linear combination will give the matrix representation in the standard way. We expand the two binomials using the binomial theorem:

(au + bv) j+m =( j + m)!

k!( j + m − k)!k=0

j+m

∑ a j+m− ku j+m− kbkvk

(−b*u + a*v) j−m = (−1) j−m−n

n=0

j−m

∑ ( j − m)!n!( j − m − n)!

b*( j−m−n)u j−m−na*nvn

Upon rearranging and changing some dummy indices we get

Ufm (u,v) = fm (u ',v ') = Umm '

m '=− j

j

∑ fm ' (u,v)

where the matrix element Umm ' is given by

Umm ' = (−1)m '−m+ k

k=0

j+m

∑ ( j + m)!( j − m)!( j + m)!( j − m)!k! ( j − m '− k)!(( j + m − k)!(m '− m + k)!

a j+m− ka*( j−m '− k )bkb*(m '−m+ k )

The index k starts with zero and runs up to j + m, but the factorials in the denominator guarantee that the coefficient will vanish if any exponent goes negative (since (−n)!= ±∞ for

n = 1,2,3,...).

Since m and m' each range from -j to j in unit steps, the matrices Umm ' representing SU(2) have dimensions (2j+1) x (2j+1).

If j = 1/2, then we get

U =

a b−b* a

⎛⎝⎜

⎞⎠⎟=

Um=1/2,m '=1/2 Um=1/2,m '=−1/2

Um=−1/2,m '=1/2 Um=−1/2,m '=−1/2

⎛⎝⎜

⎞⎠⎟

which is identical to the earlier equation.

SU(2) −O3 Homomorphism (for the mathematically inclined)

The operation of SU(2) on a matrix is given by a unitary transformation

M ' =UMU +

Now any 2 x 2 matrix M can be written in terms of I and the three Pauli matrices

I =

1 00 1

⎛⎝⎜

⎞⎠⎟, σ1 =

0 11 0

⎛⎝⎜

⎞⎠⎟, σ 3 =

0 −ii 0

⎛⎝⎜

⎞⎠⎟, σ 3 =

1 00 −1

⎛⎝⎜

⎞⎠⎟

that is, these four matrices span the 2 x 2 world.

Let M be a zero-trace matrix(then I does not contribute) so thatin general

M = xσ1 + yσ 2 + zσ 3 =

z x − iyx + iy −z

⎛⎝⎜

⎞⎠⎟

Since the trace is invariant under a unitary transformation, M' must have the same form

M ' = x 'σ1 + y 'σ 2 + z 'σ 3 =

z ' x '− iy 'x '+ iy ' −z '

⎛⎝⎜

⎞⎠⎟

The determinant is also invariant under a unitary transformation. Therefore

−(x2 + y2 + z2 ) = −(x '2+ y '2+ z '2 )

or x2 + y2 + z2 is invariant under this operation of SU(2) (same as

with O3, the rotation group). SU(2) must, therefore, describe

some kind of rotation. In fact, it is easy to check that the square of the length of a spinor(2-element column vector)

s+s = u* v*( ) u

v⎛⎝⎜

⎞⎠⎟= u*u + v*v

is invariant under the transformation U. This suggests that SU(2) and O3 may be isomorphic or homomorphic.

We can figure out what rotation SU(2) describes by considering some special cases. Consider

Uz (α / 2) =

eiα /2 00 e− iα /2

⎛⎝⎜

⎞⎠⎟

Now we carry out the transformation of each of the three Pauli matrices. For example,

Uzσ1Uz

+ =eiα /2 00 e− iα /2

⎛⎝⎜

⎞⎠⎟0 11 0

⎛⎝⎜

⎞⎠⎟e− iα /2 00 eiα /2

⎛⎝⎜

⎞⎠⎟=

0 eiα

e− iα 0⎛⎝⎜

⎞⎠⎟

or

Uzxσ1Uz+ = (x cosα )σ1 − (x sinα )σ 2

Similarly,

Uzyσ 2Uz+ = (ysinα )σ1 + (ycosα )σ 2

Uzzσ 3Uz+ = (z)σ 3

Our earlier expression for M' then gives

x ' = x cosα + ysinαy ' = −x sinα + ycosαz ' = z

Thus, the 2 x 2 unitary transformation using Uz (α / 2) is

equivalent to the rotation operator Rz (α ). The same correspondence can be shown for

Uy (β / 2) =

cos β / 2( ) sin β / 2( )− sin β / 2( ) cos β / 2( )

⎛⎝⎜

⎞⎠⎟

, Ux (ϕ / 2) =cos ϕ / 2( ) i sin ϕ / 2( )i sin ϕ / 2( ) cos ϕ / 2( )

⎛⎝⎜

⎞⎠⎟

In general, we have

Uk (γ / 2) = I cos(γ / 2) + iσ k sin(γ / 2)The correspondence

Uz (α / 2) =eiα /2 00 e− iα /2

⎛⎝⎜

⎞⎠⎟⇔ Rz (α ) =

cosα sinα 0− sinα cosα 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

is not a simple one-to-one correspondence. Specifically, as α in Rz (α ) ranges from 0 to 2π the parameter in Uz (α / 2) , α / 2 , goes

from 0 to π . We find

Rz (α + 2π ) = Rz (α )

Uz (α / 2 + π ) =−eiα /2 00 −e− iα /2

⎛⎝⎜

⎞⎠⎟= −Uz (α / 2)

Therefore both Uz (α / 2) and Uz (α / 2 + π ) = −Uz (α / 2) correspond to Rz (α ).

The correspondence is 2 to 1, or SU(2) and O3 are homomorphic.

The establishment of the correspondence between representations of SU(2) and those of O3 means that the known representations of

SU(2) automatically provide us with representations of O3.

Combining the various rotations, we find that a unitary transformation using

U(α,β,γ ) =Uz (γ / 2)Uy (β / 2)Uz (α / 2) corresponds to the general Euler rotation (you will study these in Physics 111 and Physics 113) Rz (α )Ry (β)Rz (γ ). By direct

multiplication we get

U(α,β,γ ) =Uz (γ / 2)Uy (β / 2)Uz (α / 2)

=eiγ /2 0

0 e− iγ /2

⎛⎝⎜

⎞⎠⎟

cos β / 2( ) sin β / 2( )− sin β / 2( ) cos β / 2( )

⎛⎝⎜

⎞⎠⎟eiα /2 0

0 e− iα /2

⎛⎝⎜

⎞⎠⎟

=ei(γ +α )/2 cos β / 2( ) ei(γ −α )/2 sin β / 2( )−e− i(γ −α )/2 sin β / 2( ) e− i(γ +α )/2 cos β / 2( )

⎛⎝⎜

⎞⎠⎟

From our alternative original definition of U, using

ξ = (γ +α ) / 2 ,η = β / 2 ,ψ = (γ −α ) / 2we identify the parameters a and b as

a = ei(γ +α )/2 cos β / 2( ) , b = ei(γ −α )/2 sin β / 2( )Finally the SU(2) representation of Umm ' becomes

Umm ' (α,β,γ )

= (−1)kk=0

j+m

∑ ( j + m)!( j − m)!( j + m)!( j − m)!k! ( j − m '− k)!(( j + m − k)!(m '− m + k)!

eimγ (cos(β / 2))2 j+m−m '−2k (− sin(β / 2))m−m '+2k )eim 'α

Here the irreducible representations are in terms of the Euler angles. This result allows us to calculate the (2j+1) x (2j+1) irreducible representations of SU(2) for all j (j = 0, 1/2, 1, 3/2, ...) and the irreducible representations of O3 for integral

orbital angular momentum j (j = 0,1,2,3,....) (Physics 113).Generators

In all cases, rotations about a common axis combine as

Rz (ϕ2 )Rz (ϕ1) = Rz (ϕ1 +ϕ2 )

that is, multiplication of these matrices is equivalent to addition of the arguments. This suggests that we look for an exponential representation of the rotations, i.e.,

exp(ϕ2 )exp(ϕ1) = exp(ϕ1 +ϕ2 )Suppose we define the exponential function of a matrix by the following relation:

U = eiaH = I + iaH +

(iaH )2

2!+ ........

where a is a real parameter independent of matrix(operator) H. It is easy to prove that if H is Hermitian, then U is unitary and if U is unitary, the H is Hermitian.

In group theory, H is labeled a Lie algebra generator, the generator of U.

The following matrix describes a finite rotation of the coordinates through an angle ϕ counterclockwise about the z-axis:

Rz (ϕ ) =cosϕ sinϕ 0− sinϕ cosϕ 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Now let Rz be an infinitesimal rotation through an angle δϕ . We can then write

Rz (δϕ ) = I + iδϕMz

where

Mz =

1i∂Rz (ϕ )∂ϕ ϕ =0

so that

Mz =0 −i 0i 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

Similarly, we find

Mx =0 0 00 0 −i0 i 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , My =

0 0 i0 0 0−i 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

A finite rotation ϕ can be compounded out of successive

infinitesimal rotations δϕ .

Rz (δϕ2 + δϕ1) = (I + iδϕ2Mz )(I + iδϕ1Mz )Letting δϕ = ϕ / N for N rotations, with N →∞ , we get

Rz (ϕ ) = limN→∞

I + iϕ / N )Mz( )( )N = exp(iϕMz )

This implies that Mz is the generator of the group Rz (ϕ ), a

subgroup of O3. Before proving this result, we note:

(a) Mz is Hermitian and thus, Rz (ϕ ) is unitary

(b) tr(Mz ) = 0 and det(Rz (ϕ )) = +1 In direct analogy with Mz , Mx may be identified as the

generator of Rx , the subgroup of rotations about the x-axis, and

My as the generator of Ry .

Proof for Mz :

Rz (ϕ ) = exp(iϕMz ) = I + iϕMz +(iϕMz )

2

2!+

(iϕMz )3

3!+ ......

=0 0 00 0 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

1 0 00 1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ + iϕMz +

(iϕ )2

2!

1 0 00 1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

(iϕ )3

3!Mz + ......

=0 0 00 0 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

1 0 00 1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ (1− ϕ 2

2!+ ...) + i(ϕ −

ϕ 3

3!+ ...)Mz + ......

=0 0 00 0 00 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ +

1 0 00 1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ cosϕ + i sinϕMz + ...... =

cosϕ sinϕ 0− sinϕ cosϕ 0

0 0 1

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

where we have used the relations

Mz2 =

1 0 00 1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , Mz

3 = Mz , etc

and we recognized the two series as cosϕ and sinϕ . Thus, we get the same matrix Rz (ϕ ) as earlier.

Other relations we can write are:

(1) all infinitesimal rotations commute

Ri (δϕ i ),Rj (δϕ j )⎡⎣ ⎤⎦ = 0

(2) an infinitesimal rotation about an axis defined by unit

vector n is

R(δϕ ) = I + iδϕn ⋅

M

(3) the generators satisfy the commutation rules

Mi ,M j⎡⎣ ⎤⎦ = iεijkMk

k∑

This will turn out to be the commutation relations for angular momentum operators in quantum mechanics.In a similar way the elements (U1,U2 ,U3) of the two-dimensional unitary group, SU(2), may be generated by

exp(aσ1 / 2) , exp(aσ 2 / 2) , exp(aσ 3 / 2)where σ1 , σ 2 , σ 3 are the Pauli spin matrices and a, b, and c are

real parameters. We note that the σ 's are Hermitian and have zero trace and thus, the elements of SU(2) are unitary and have determinant = +1.

We also note that the generators in diagonal form such as σ 3

will lead to conserved quantum numbers.

Finally, if we define si = 12σ i then we have

si , s j⎡⎣ ⎤⎦ = iεijk sk

k∑

which are the angular momentum commutation rules implying that the si are angular momentum operators.

Appendix B - SU(3) Structure Constants

We start first by repeating the matrix expressions for the eight generators of SU(3):

Shift Operators

E1 =16

0 1 00 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−1

+

E2 =16

0 0 10 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−2

+

E3 =16

0 0 00 0 10 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = E−3

+

where

E1 q2 = q1 , E2 q3 = q1 , E3 q3 = q2E−1 q1 = q2 , E−2 q1 = q3 , E−3 q2 = q3

Commutation Relations

E1,E2[ ] = E1,E−3[ ] = E2 ,E3[ ] = E−1,E−2[ ] = E−1,E−3[ ] = E−2 ,E−3[ ] = 0E1,E−2[ ] = −

16E−3 , E1,E3[ ] = 1

6E2 , E2 ,E−3[ ] = 1

6E1

E−1,E2[ ] = 16E3 , E−1,E−3[ ] = −

16E−2 , E−2 ,E3[ ] = −

16E−1

and

E1,E−1[ ] = 13H1 , E2 ,E−2[ ] = 1

2 3H1 +

12H2

E3,E−3[ ] = −12 3

H1 +12H2

where

H1 =12 3

1 0 00 −1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , H2 =

16

1 0 00 1 00 0 −2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

and finally

H1,E1[ ] = 13E1 , H1,E2[ ] = 1

2 3E2 , H1,E3[ ] = −

12 3

E3

H2 ,E1[ ] = 0 , H2 ,E2[ ] = 12 E2 , H2 ,E3[ ] = 12 E3 , H1,H2[ ] = 0It is clear from the commutators that we have a closed algebra (a Lie algebra), i.e., all of the commutators among the 8 operators only involve the same 8 operators. If we write out the commutators in general, we can express the RHS in terms of so-called "structure" constants. For SU(3) we have the Lie algebra

λi ,λ j⎡⎣ ⎤⎦ = 2ifijkλk

where the fijk are the structure constants and the operators λk

(the E's and H's above) are the generators. They also obey the anticommutation rules

λi ,λ j{ } = 43δ ij + 2 dijk

k∑ λk

One standard form for the λk (linear combinations of the E's and

H's)is:

λ1 =0 1 01 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6 E1 + E−1( ) , λ2 =

0 −i 0i 0 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6i E−1 − E1( )

λ3 =1 0 00 −1 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 2 3H1 , λ4 =

0 0 10 0 01 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6 E2 + E−2( )

λ5 =0 0 −i0 0 0i 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6i E−2 − E2( ) , λ6 =

0 0 00 0 10 1 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6 E3 + E−3( )

λ7 =0 0 00 0 −i0 i 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6i E−3 − E3( ) , λ8 =

1 0 00 1 00 0 −2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = 6H2

We note a few properties that these matrices possess: (1) All have Tr λi( ) = 0 (2) All obey Tr λiλ j( ) = δ ij (3) λ1,λ2 and λ3 are a reproduction of SU(2) and therefore obey

the SU(2) commutation and anticommutation rules: (4) λ4 ,λ5 and 1

2 λ3 +32 λ8 are also a reproduction of SU(2).

(5) λ6 ,λ7 and − 12 λ3 +

32 λ8 are also a reproduction of SU(2).

These facts will greatly reduce our calculations.

Note that i, j, and k can each run from 1 to 8. Thus, there are a total of 2x8x8x8 = 1024 total f 's and d 's that we need to solve for. After a groan, we think a little and eventually apply the basic principle of physics that says, "There must be an easier way". That "easier way" involves using the trace theorems. If we multiply the commutation relations by λm we can take the trace of both sides of the equation and apply some trace theorems we get

Tr λm λi ,λ j⎡⎣ ⎤⎦( ) = 2i fijkTr λmλk( )k∑ = 2i ⋅2δkm fijm

fijk =14iTr λk λi ,λ j⎡⎣ ⎤⎦( )

We can now exploit the cyclical property of a trace to see that:

fijk = f jki = fkij = − fikj = − fkji = − f jikThus, the fijk are totally antisymmetric. In particular, any fijk

value with a repeated index must equal zero. Thus, we have

succeeded in reducing the number of fijk values we have to solve

for from 8x8x8 = 512 to (1x2+3x4+4x5+5x6+6x7)/2 = 56. Those 56 coefficients are listed below:

f123 f124 f125 f126 f127 f128 f134f135 f136 f137 f138 f145 f146 f147f148 f156 f157 f158 f167 f168 f178f234 f235 f236 f237 f238 f245 f246f247 f248 f256 f257 f258 f267 f268f278 f345 f346 f347 f348 f356 f357f358 f367 f368 f378 f456 f457 f458f467 f468 f478 f567 f568 f578 f678

But most of these values equal zero! In order to see this, we exploit the commutation relations for the SU(2) subgroups within SU(3):

(1) Since λ1,λ2 and λ3 form an SU(2) subgroup and since Tr λiλ j( ) = δ ij , it

follows that fijk = 0 whenever two of the indices are (1,2),

(1,3) or (2,3) unless the third index is 3, 2, or 1 respectively.

(2) Since λ4 ,λ5 and 12 λ3 +

32 λ8 form an SU(2) subgroup and since

Tr λiλ j( ) = δ ij , it follows that fijk = 0 whenever two of the indices

are (3,4), (3,5), (4,5), (4,8) or (5,8) unless the third index is 5, 4, (3 or 8), 5 or 4 respectively.

(3) Since λ6 ,λ7 and − 12 λ3 +

32 λ8 form an SU(2) subgroup and since

Tr λiλ j( ) = δ ij , it follows that fijk = 0 whenever two of the indices

are (3,6), (3,7), (6,7), (6,8) or (7,8) unless the third index is 7, 6, (3 or 8), 7 or 6 respectively.

These facts reduce the possible nonzero fijk values to the

following 9 values:

f123 f147 f156 f246 f257 f345 f367 f458 f678Each of these values is straightforward to evaluate. For example:

f123 =14iTr λ3 λ1,λ2[ ]( ) = 1

4iTr

2i 0 00 2i 00 0 0

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= 1

We get

f123 = 1

f147 = f516 = f246 = f257 = f345 = f637 =12

f458 = f678 =32

In a similar manner we have

Tr λm λi ,λ j{ }( ) = 43δ ijTr λm( )2 dijkTr λmλk( )k∑ = 2 ⋅2δkmdijm

dijk =14Tr λk λi ,λ j{ }( )

We can use the same procedure used to find the f's to find the d's.

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