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is homogenous (the right hand side being zero). So, in order to get non-trivial
eigenvectors,we require
det(A λI)=0. (5.55)
−
Thisequation,calledthesecular equation,isann-thdegreeequationforthe
unknownλandthusithasncomplex,notnecessarilydistinct,roots. Theseare
the eigenvalues of the operator A. Once the eigenvalues are known, the corre-
sponding eigenvectorscanbe calculated. Severaldistinct, linearlyindependent,
eigenvectors might correspond to one and the same eigenvalue. In that case
the eigenvalue is said to be degenerate. The degree of degeneracy is equal to
the number of distinct linearly independent eigenvectors corresponding to the
degenerate eigenvector.
Diagonalization
Amatrixis saidto bediagonalifithasnon-zeroelements onlyonthe diagonal.
Using outer products of basis vectors i a diagonal operator can be written as
| i
A= λ i i. (5.56)
i
| ih |
Xi
That this actually representsa diagonalmatrix is clear from using equation
(5.34) to compute the matrix elements. We get
k Aj = λ k i ij =λ δ .
i k kj
h | | i h | ih | i
Xi
That is, only the diagonal elements are non-zero, and equal to the numbers
λ .
i
It would be natural to identify the λ with the eigenvalues of the operator.
i
Indeed,adiagonaloperatortriviallysatisfiestheeigenvalueequation(1.35)with
eigenvectors equal to the basis vectors i . So the question arises, when is an
| i
operator diagonalizable?
111
5.4.1 Spectral decomposition
Here we will just state an important theorem that allows us to use diagonal
representations for certain classes of operators.
An operator A on a vector space is said to be normal if A A = AA . It
† †
follows immediately that a Hermitean operator is normal. Also, any unitary
operator U is also normal. This follows from the simple calculation
U U =U 1U =I =UU 1 =UU .
† − − †
Spectral decomposition theorem
Let A be a normal operator on a vector space V. Then A can be diagonalized
with respect to some orthonormal basis for V. Conversely, any diagonalizable
operator is normal.
This means that the eigenvalue equation (5.53) can be solved and the oper-
ator can be represented explicitly as in equation (5.56). To be definite,
A= λ i i
i
| ih |
Xi
where λ are the eigenvalues of A, i is an orthonormal basis and each i is
i
| i | i
an eigenvector of A. This equation can also be trivially rewritten in terms of
projectors P
i
A= λ P . (5.57)
i i
Xi
In particular, hermitean and unitary operators can be diagonalized. Proofs
of the spectral decomposition theorem can be found in [38] and [39].
Diagonalization using unitary transformations
We saw in section 5.3 that symmetry transformations are effected by unitary
operators. Choosing a suitable unitary operator, a normal operator can be
transformed into a diagonal form. Let A be a normal operator. We want to
find a unitary operator D that transforms A into diagonal form. Explicitly as
in (5.52)
A =DAD 1. (5.58)
′ −
wherewe demandthatA isdiagonal. Inordertobe concrete,we representthe
′
operatorsbythecorrespondingmatrices,sothatA =A δ . Thenmultiplying
′kl ′k kl
the equation (5.58) by D from the left gives DA = AD. Writing this last
′
equationintermsofmatricesyieldsashortcalculation(notethesubtlechanges
of indices)
D A = A D = A δ D =A D = A δ D ,
km ml ′km ml ′k km ml ′k kl ′k ml km
Xm Xm Xm Xm
112
or
D (A A δ ).
km ml ′k ml
−
Xm
Now, fixing the index k, we have n homogeneousequations for the transfor-
mation matrix elements D . These equations have non-trivial solution if and
km
only if the determinant of the coefficients vanish, i.e. that the determinant of
the matrix A A δ is zero,
ml ′k ml
−
det(A A δ )=0.
ml ′k ml