San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Special Unitary
Group of Order 2

The abstract group SU(2) is represented by a set of 2×2 matrices of complex elements which have determinant of unity. The group can be generated by three matrices J1, J2 and J3. In the following, standard matrix notation will be used. Physics has developed a brilliant esoteric notation for this topic but it obscures the analysis for the those unaccustomed to it.

It is convenient to develop the properties of the group by working with the commutation relationships among the generating matrices. The commutation of two square matrices A and B of the same order is given by

[A, B] = AB − BA

The commutation operation is not only non-commutative; it is anti-commutaive; i.e.,

[B, A] = −[A, B]

Thus [A, A] is equal to a square matrix of zeroes for any A.

Let J1, J2 and J3 be the Hermitian matrices which generate a group. Hermitian means that a matrix is equal to the transpose of its complex conjugate. This guarantees that a matrix's eigenvalues are real numbers.

The commutation relations for the generating matrices J1, J2 and J3 are

[J1, J2] = iJ3
[J2, J3] = iJ1
[J3, J1] = iJ2

where i is the imaginary unit of complex numbers.

Two new operators (matrices) are defined as

J+ = (J1 + iJ2)/√2
J = (J1 − iJ2)/√2

Now consider the commutation of J3 and J+. Since the commutation relation is linear in its components

[J3, J+] = [J3, J1/√2] + i[J3, J2/√2]
= (1/√2)([J3, J1] + (i/√2)[J3, J2])
= (1/√2)iJ2 + (i/√2)(−iJ1) = (1/√2)(iJ2 + J1) = J+


[J3, J] = −J


[J+, J] = (1/2)[(J1+iJ2), (J1−iJ2]1, J1] −i[J1, J2] + i[J2, J1] + [J2. J2])
which reduces to
(1/2)(−2i[J1, J2] = −i[J1, J2] = −i[iJ3] = J3

J+ and J+ as the Eigenvalue Raising
and Lowering Operators, Respectively

Written out the commutation relation [J3, J+] = J+ means

J3J+ − J+J3 = J+
and hence
J3J+ = J+J3 + J+

Suppose X is an eigenvector of J3 and λ is its eigenvalue; i.e.,

J3X = λX

At this point the character of λ is not known other than it is a real number by virtue of J3 being Hermitian.

Now consider the vector J+X and the product of J3 with it.

J3J+X = J+J3X + J+X
which reduces to
J3J+X = λJ+X + J+X
J3J+X = (λ+1)J+X

Thus J+X is an eigenvector of J3 and its eigenvalue is (λ+1).

Likewise JX is an eigenvector of J3 and its eigenvalue is (λ−1).

Raising and Lowering Limits

Now consider the eigenvector Xmax of J3 which has the maximum eigenvalue λmax; i.e.,

J3Xmax = λmaxXmax

The ordering of the eigenvalues is meaningful by virtue of their being real numbers.

According to the previous section J+Xmax is an eigenvector of J3 and its eigenvalue is (λmax+1). But λmax was chosen to be the maximum eigenvalue. The contradiction can only be avoided if J+Xmax is equal to the zero vector,

J+Xmax = 0

Derivation of an Iteration Scheme

Now consider the sequence of vectors { Xmax, JXmax, … JnXmax} up to some maximum n with eigenvalues of {λmax, (λmax−1), …, (λmax−n}. There is an eigenvector Xmin with a minimum eigenvalue and if J is applied to it the result has to be the zero vector; i.e.,

JXmin = (λmin−1)Xmin=0

This could occur either because JXmin is equal to the zero vector or the eigenvalue (λmin−1) is equal to zero. This latter would imply that λmin=1 and hence that the eigenvalues are equal to consecutive integers. The issue of the relationship between λmin and λmax will be dealt with later.

At this point for typographic convenience let λmax be denoted as k. The nature of k other than being a real number has not yet been determined. Likewise λmin will be denoted as n.

Let p be an element of the set of numbers {k, k-1, k-2,… n}. Let {Yp; p=k,k-1, …, } be the normalized versions of the above sequence of eigenvectors of J3. Now define

JYp = αpYp-1
J+Yp = βp+1Yp+1


J+Yk-1 = βkYk

However, since JYkkYk then Yk=(1/αk)JYk.


J+Yk-1 = = J+(1/αk)JYk) = (1/αk)J+JYk


J+J = [J+, J] + JJ+
and from previous analysis
[J+, J] = J3
this means that
J+Yk-1 = (1/α)(J3Yk + JJ+Yk)

However J+Yk=0 and J3Yk=kYk so

J+Yk-1 = (1/αk)(kYk) = (k/αk)Yk
but it is also true that
J+Yk-1 = βkYk


βk = k/αk
or, equivalently
αkβk = k

Now consider J+Yp-1. Since JYppYp-1

J+Yp-1 = (1/αp)J+JYp
= (1/αp){[J+, J]Yp + JJYp}
= (1/αp){J3Yp + Jβp+1Yp+1}
= (1/αp){pYp + βp+1JYp+1}
= (1/αp){pYp + βp+1αp+1Yp} = (1/αp)(p + βp+1αp+1)Yp

But J+Yp-1 is also equal to βpYp. Therefore

βp = (1/αp)(p + βp+1αp+1
or, equivalently
αpβp = p + αp+1βp+1

Let Nppβp for all p. Then

Np = p + Np+1
for p=k, k-1,…, n.

Thus written out

Nk = k
Nk-1 = (k-1) + k
Nk-2 = (k-2) + (k-1) + k

The solution is

Np = k + (k-1) + (k-2) + … + (p+1) + p
which evaluates to
Np = (k-p+1)k − ½(k-p)(k-p+1)
= (k-p+1)(k − ½(k-p))
Np = (k-p+1)(k+p)/2

The Equality of αp and βp

The inner product of two complex valued column vectors U and V is denoted as <U, V> and defined in matric form as

<U, V> = U*TV

where U*T is the transpose of the complex conjugate of U.

For any normalized vector Y,

<Y, Y> = 1

A Hermitian matrix is one such that it is equal to the transpose of its complex conjugate; i.e., J*T = J.

Consider J+*T.

J+*T = (1/√2)(J1 + iJ2)*T = (1/√2)(J1*T −iJ2*T)

But J1 and J2 are Hermitian so J1*T=J1 and J2*T=J2 and therefore

J+*T = (1/√2)(J1 − iJ2) = J

Consider now <J+Yp-1, Yp>. First of all J+Yp-1 is equal to βpYp. Therefore

<J+Yp-1, Yp> = <βpYp, Yp>
= βp<Yp, Yp> = βp
since all of the Y vectors
are normalized

On the other hand

<J+Yp-1, Yp> = (J+Yp-1)*TYp
= Yp-1*TJ+*TYp
= Yp-1*TJYp = <Yp-1, JYp>

But JYp is equal to αpYp-1 and therefore

<J+Yp-1, Yp> = <Yp-1, αpYp-1>
= αp<Yp-1, Yp-1> = αp

Since <J+Yp-1, Yp> is equal to both αp and βp this means that

αp = βp

Therefore Npp2 and hence

αp = ((k+p)(k−p+1)/2)½

The Minimum Eigenvalue of J3 is the Negative of the Maximum Eigenvalue

Consider JJ+. It can be evaluated as

JJ+ = (1/2)(J1 − iJ2)(J1 + iJ2)
= (1/2)(J1² + iJ1J2 − iJ2J1 + J2²)
= (1/2)(J1² + J2² + i[J1, J2])
but [J1, J2] is equal to iJ3, hence
JJ+ = (1/2)(J1² + J2² − J3)
which is equivalent to
JJ+ = (1/2)(J1² + J2² + J3² − J3² − J3)

Let (J1² + J2² + J3²) be denoted as J². Then

JJ+ = (1/2)(J² − J3² − J3)

Again let Xmax be the eigenvector of J3 which has the maximum eigenvalue λmax. Remember that J+Xmax is also an eigenvector of J3 and likewise for JXmax. However J+Xmax is equal to a zero vector and likewise for JJ+Xmax. Thus

(1/2)(J² − J3² − J3)Xmax = 0
and hence
J²Xmax = (J3² − J3)Xmax = (λmax² + λmax)Xmax

This implies that Xmax is an eigenvector of JJ+ and that its eigenvalue is equal to (λmax² + λmax) which is λmaxmax+1).

The same procedure can be carried out with J+J and Xmin with the implication that the eigenvalue of J² is equal to λminmin−1). Thus

λmaxmax+1) = λminmin−1)

Note that this equation is satisfied if λmin is replaced on the RHS by −λmax. Therefore

λmin = −λmax

The eigenvalue of λmax can also be obtained by repeated applications of J+ so

λmax = λmin + m
for some integer m.

But this implies

max = m
and hence
λmax = m/2
λmin = −m/2

Thus the eigenvalues of J3 must be integers or half-integers.

(To be continued.)

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