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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 J_{1}, J_{2} and J_{3}. 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
The commutation operation is not only non-commutative; it is anti-commutaive; i.e.,
Thus [A, A] is equal to a square matrix of zeroes for any A.
Let J_{1}, J_{2} and J_{3} 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 J_{1}, J_{2} and J_{3} are
where i is the imaginary unit of complex numbers.
Two new operators (matrices) are defined as
Now consider the commutation of J_{3} and J_{+}. Since the commutation relation is linear in its components
Likewise
Furthermore
Written out the commutation relation [J_{3}, J_{+}] = J_{+} means
Suppose X is an eigenvector of J_{3} and λ is its eigenvalue; i.e.,
At this point the character of λ is not known other than it is a real number by virtue of J_{3} being Hermitian.
Now consider the vector J_{+}X and the product of J_{3} with it.
Thus J_{+}X is an eigenvector of J_{3} and its eigenvalue is (λ+1).
Likewise J_{−}X is an eigenvector of J_{3} and its eigenvalue is (λ−1).
Now consider the eigenvector X_{max} of J_{3} which has the maximum eigenvalue λ_{max}; i.e.,
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 J_{3} 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,
Now consider the sequence of vectors { X_{max}, J_{−}X_{max}, … J_{−}^{n}X_{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.,
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 J_{3}. Now define
Then
However, since J_{−}Y_{k}=α_{k}Y_{k} then Y_{k}=(1/α_{k})J_{−}Y_{k}.
Thus
Since
However J_{+}Y_{k}=0 and J_{3}Y_{k}=kY_{k} so
Therefore
Now consider J_{+}Y_{p-1}. Since J_{−}Y_{p}=α_{p}Y_{p-1}
But J_{+}Y_{p-1} is also equal to β_{p}Y_{p}. Therefore
Let N_{p}=α_{p}β_{p} for all p. Then
Thus written out
The solution is
The inner product of two complex valued column vectors U and V is denoted as <U, V> and defined in matric form as
where U^{*T} is the transpose of the complex conjugate of U.
For any normalized vector Y,
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}.
But J_{1} and J_{2} are Hermitian so J_{1}^{*T}=J_{1} and J_{2}^{*T}=J_{2} and therefore
Consider now <J_{+}Y_{p-1}, Y_{p}>. First of all J_{+}Y_{p-1} is equal to β_{p}Y_{p}. Therefore
On the other hand
But J_{−}Y_{p} is equal to α_{p}Y_{p-1} and therefore
Since <J_{+}Y_{p-1}, Y_{p}> is equal to both α_{p} and β_{p} this means that
Therefore N_{p}=α_{p}^{2} and hence
Consider J_{−}J_{+}. It can be evaluated as
Let (J_{1}² + J_{2}² + J_{3}²) be denoted as J². Then
Again let X_{max} be the eigenvector of J_{3} which has the maximum eigenvalue λ_{max}. Remember that J_{+}X_{max} is also an eigenvector of J_{3} and likewise for J_{−}X_{max}. However J_{+}X_{max} is equal to a zero vector and likewise for J_{−}J_{+}X_{max}. Thus
This implies that X_{max} is an eigenvector of J_{−}J_{+} and that its eigenvalue is equal to (λ_{max}² + 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
Note that this equation is satisfied if λ_{min} is replaced on the RHS by −λ_{max}. Therefore
The eigenvalue of λ_{max} can also be obtained by repeated applications of J_{+} so
But this implies
Thus the eigenvalues of J_{3} must be integers or half-integers.
(To be continued.)
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