SU(2), the Special Unitary Group of Order 2

San José State University

applet-magic.com Thayer Watkins Silicon Valley & Tornado Alley USA

SU(2),
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
and
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₊

Likewise
[J₃, J₋] = −J₋

Furthermore
[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 J₋X is an eigenvector of J₃ and its eigenvalue is (λ−1).
Raising and Lowering Limits

Now consider the eigenvector X_max of J₃ which has the maximum eigenvalue λ_max; i.e.,
J₃X_max = λ_maxX_max

The ordering of the eigenvalues is meaningful by virtue of their being real numbers.
According to the previous section J₊X_max is an eigenvector of J₃ and its eigenvalue is (λ_max+1). But λ_max was chosen to be the maximum eigenvalue. The contradiction can only be avoided if J₊X_max is equal to the zero vector,
J₊X_max = 0

Derivation of an Iteration Scheme

Now consider the sequence of vectors { X_max, J₋X_max, … J₋ⁿX_max} up to some maximum n with eigenvalues of {λ_max, (λ_max−1), …, (λ_max−n}. There is an eigenvector X_min with a minimum eigenvalue and if J₋ is applied to it the result has to be the zero vector; i.e.,
J₋X_min = (λ_min−1)X_min=0

This could occur either because J₋X_min is equal to the zero vector or the eigenvalue (λ_min−1) is equal to zero. This latter would imply that λ_min=1 and hence that the eigenvalues are equal to consecutive integers. The issue of the relationship between λ_min and λ_max will be dealt with later.
At this point for typographic convenience let λ_max be denoted as k. The nature of k other than being a real number has not yet been determined. Likewise λ_min will be denoted as n.
Let p be an element of the set of numbers {k, k-1, k-2,… n}. Let {Y_p; p=k,k-1, …, } be the normalized versions of the above sequence of eigenvectors of J₃. Now define
J₋Y_p = α_pY_p-1
and
J₊Y_p = β_p+1Y_p+1

Then
J₊Y_k-1 = β_kY_k

However, since J₋Y_k=α_kY_k then Y_k=(1/α_k)J₋Y_k.
Thus
J₊Y_k-1 = = J₊(1/α_k)J₋Y_k) = (1/α_k)J₊J₋Y_k

Since
J₊J₋ = [J₊, J₋] + J₋J₊
and from previous analysis
[J₊, J₋] = J₃
this means that
J₊Y_k-1 = (1/α)(J₃Y_k + J₋J₊Y_k)

However J₊Y_k=0 and J₃Y_k=kY_k so
J₊Y_k-1 = (1/α_k)(kY_k) = (k/α_k)Y_k
but it is also true that
J₊Y_k-1 = β_kY_k

Therefore
β_k = k/α_k
or, equivalently
α_kβ_k = k

Now consider J₊Y_p-1. Since J₋Y_p=α_pY_p-1
J₊Y_p-1 = (1/α_p)J₊J₋Y_p
= (1/α_p){[J₊, J₋]Y_p + J₋J₋Y_p}
= (1/α_p){J₃Y_p + J₋β_p+1Y_p+1}
= (1/α_p){pY_p + β_p+1J₋Y_p+1}
= (1/α_p){pY_p + β_p+1α_p+1Y_p} = (1/α_p)(p + β_p+1α_p+1)Y_p

But J₊Y_p-1 is also equal to β_pY_p. Therefore
β_p = (1/α_p)(p + β_p+1α_p+1
or, equivalently
α_pβ_p = p + α_p+1β_p+1

Let N_p=α_pβ_p for all p. Then
N_p = p + N_p+1
for p=k, k-1,…, n.

Thus written out
N_k = k
N_k-1 = (k-1) + k
N_k-2 = (k-2) + (k-1) + k
…

The solution is
N_p = k + (k-1) + (k-2) + … + (p+1) + p
which evaluates to
N_p = (k-p+1)k − ½(k-p)(k-p+1)
= (k-p+1)(k − ½(k-p))
N_p = (k-p+1)(k+p)/2

The Equality of α_p and β_p

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₊Y_p-1, Y_p>. First of all J₊Y_p-1 is equal to β_pY_p. Therefore
<J₊Y_p-1, Y_p> = <β_pY_p, Y_p>
= β_p<Y_p, Y_p> = β_p
since all of the Y vectors
are normalized

On the other hand
<J₊Y_p-1, Y_p> = (J₊Y_p-1)^*TY_p
= Y_p-1^*TJ₊^*TY_p
= Y_p-1^*TJ₋Y_p = <Y_p-1, J₋Y_p>

But J₋Y_p is equal to α_pY_p-1 and therefore
<J₊Y_p-1, Y_p> = <Y_p-1, α_pY_p-1>
= α_p<Y_p-1, Y_p-1> = α_p

Since <J₊Y_p-1, Y_p> is equal to both α_p and β_p this means that
α_p = β_p

Therefore N_p=α_p² and hence
α_p = ((k+p)(k−p+1)/2)^½

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

Consider J₋J₊. It can be evaluated as
J₋J₊ = (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
J₋J₊ = (1/2)(J₁² + J₂² − J₃)
which is equivalent to
J₋J₊ = (1/2)(J₁² + J₂² + J₃² − J₃² − J₃)

Let (J₁² + J₂² + J₃²) be denoted as J². Then
J₋J₊ = (1/2)(J² − J₃² − J₃)

Again let X_max be the eigenvector of J₃ which has the maximum eigenvalue λ_max. Remember that J₊X_max is also an eigenvector of J₃ and likewise for J₋X_max. However J₊X_max is equal to a zero vector and likewise for J₋J₊X_max. Thus
(1/2)(J² − J₃² − J₃)X_max = 0
and hence
J²X_max = (J₃² − J₃)X_max = (λ_max² + λ_max)X_max

This implies that X_max is an eigenvector of J₋J₊ and that its eigenvalue is equal to (λ_max² + λ_max) which is λ_max(λ_max+1).
The same procedure can be carried out with J₊J₋ and X_min with the implication that the eigenvalue of J² is equal to λ_min(λ_min−1). Thus
λ_max(λ_max+1) = λ_min(λ_min−1)

Note that this equation is satisfied if λ_min is replaced on the RHS by −λ_max. Therefore
λ_min = −λ_max

The eigenvalue of λ_max can also be obtained by repeated applications of J₊ so
λ_max = λ_min + m
for some integer m.

But this implies
2λ_max = m
and hence
λ_max = m/2
and
λ_min = −m/2

Thus the eigenvalues of J₃ must be integers or half-integers.
(To be continued.)

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[A, B] = AB − BA

[B, A] = −[A, B]

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

J+ = (J1 + iJ2)/√2 and J− = (J1 − iJ2)/√2

[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

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

J3X = λX

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

Raising and Lowering Limits

J3Xmax = λmaxXmax

J+Xmax = 0

Derivation of an Iteration Scheme

J−Xmin = (λmin−1)Xmin=0

J−Yp = αpYp-1 and J+Yp = βp+1Yp+1

J+Yk-1 = βkYk

J+Yk-1 = = J+(1/αk)J−Yk) = (1/αk)J+J−Yk

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

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

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

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

Np = p + Np+1 for p=k, k-1,…, n.

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

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

<U, V> = U*TV

<Y, Y> = 1

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

J+*T = (1/√2)(J1 − iJ2) = J−

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

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

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

αp = βp

αp = ((k+p)(k−p+1)/2)½

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

J−J+ = (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 J−J+ = (1/2)(J1² + J2² − J3) which is equivalent to J−J+ = (1/2)(J1² + J2² + J3² − J3² − J3)

J−J+ = (1/2)(J² − J3² − J3)

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

λmax(λmax+1) = λmin(λmin−1)

λmin = −λmax

λmax = λmin + m for some integer m.

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

[J₁, J₂] = iJ₃

[J₂, J₃] = iJ₁

[J₃, J₁] = iJ₂

J₊ = (J₁ + iJ₂)/√2
and
J₋ = (J₁ − iJ₂)/√2

[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

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

J₃X = λX

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

J₃X_max = λ_maxX_max

J₊X_max = 0

J₋X_min = (λ_min−1)X_min=0

J₋Y_p = α_pY_p-1
and
J₊Y_p = β_p+1Y_p+1

J₊Y_k-1 = β_kY_k

J₊Y_k-1 = = J₊(1/α_k)J₋Y_k) = (1/α_k)J₊J₋Y_k

J₊J₋ = [J₊, J₋] + J₋J₊
and from previous analysis
[J₊, J₋] = J₃
this means that
J₊Y_k-1 = (1/α)(J₃Y_k + J₋J₊Y_k)

J₊Y_k-1 = (1/α_k)(kY_k) = (k/α_k)Y_k
but it is also true that
J₊Y_k-1 = β_kY_k

β_k = k/α_k
or, equivalently
α_kβ_k = k

J₊Y_p-1 = (1/α_p)J₊J₋Y_p
= (1/α_p){[J₊, J₋]Y_p + J₋J₋Y_p}
= (1/α_p){J₃Y_p + J₋β_p+1Y_p+1}
= (1/α_p){pY_p + β_p+1J₋Y_p+1}
= (1/α_p){pY_p + β_p+1α_p+1Y_p} = (1/α_p)(p + β_p+1α_p+1)Y_p

β_p = (1/α_p)(p + β_p+1α_p+1
or, equivalently
α_pβ_p = p + α_p+1β_p+1

N_p = p + N_p+1
for p=k, k-1,…, n.

N_k = k
N_k-1 = (k-1) + k
N_k-2 = (k-2) + (k-1) + k
…

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

The Equality of α_p and β_p

<U, V> = U^*TV

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

J₊^*T = (1/√2)(J₁ − iJ₂) = J₋

<J₊Y_p-1, Y_p> = <β_pY_p, Y_p>
= β_p<Y_p, Y_p> = β_p
since all of the Y vectors
are normalized

<J₊Y_p-1, Y_p> = (J₊Y_p-1)^TY_p
= Y_p-1^TJ₊^TY_p
= Y_p-1^TJ₋Y_p = <Y_p-1, J₋Y_p>

<J₊Y_p-1, Y_p> = <Y_p-1, α_pY_p-1>
= α_p<Y_p-1, Y_p-1> = α_p

α_p = β_p

α_p = ((k+p)(k−p+1)/2)^½

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

J₋J₊ = (1/2)(J² − J₃² − J₃)

(1/2)(J² − J₃² − J₃)X_max = 0
and hence
J²X_max = (J₃² − J₃)X_max = (λ_max² + λ_max)X_max

λ_max(λ_max+1) = λ_min(λ_min−1)

λ_min = −λ_max

λ_max = λ_min + m
for some integer m.

2λ_max = m
and hence
λ_max = m/2
and
λ_min = −m/2