Medidas Represivas:

2.4.3.1 Important theorems

Theorem I: If two operators A and B commute, and | i is an eigenvector of A, then B_{| i is also an eigenvector of A, with the same eigenvalue.}

Proof: Acting with B on the eigenvalue equation for A,

A| i = a| i, (2.4.17) leads to

BA| i = a[B| i] = A[B| i], (2.4.18) which proves the theorem.

Let us explore some consequences of this theorem, by noting that two possibilities arise:

(i) a is non-degenerate. Then, necessarily B_{| i / | i, which means that | i is also} an eigenvector of B.

(ii) a is degenerate. Then, all one can state is that B| i belongs to the eigensubspace Ea, corresponding to the eigenvalue a of A. One therefore says that Ea is globally

2.4. EIGENVALUE EQUATIONS. OBSERVABLES 41

With the definition of globally invariant subspaces, one can rephrase Theorem I as

Theorem I’: If two operators A and B commute, every eigensubspace of A is globally invariant under the action of B.

We now state theorems about observables.

Theorem II: If two observables A and B commute, and _| 1i and | 2i are two

eigenvectors of A, with di↵erent eigenvalues, then

h 1|B| 2i = 0. (2.4.19)

Proofs:

The first proof is extremely simple: from Theorem I, _| 2i is globally invariant un-

der B, and since A is Hermitian, eigenvectors belonging to di↵erent eigenvalues are necessarily orthogonal.

We can provide another proof without resorting to Theorem I: with a1 and a2 being

the corresponding eigenvalues of A, the Hermiticity of A implies

h 1|AB| 2i = a1h 1|B| 2i (2.4.20)

h 1|BA| 2i = a2h 1|B| 2i, (2.4.21)

in terms of which we can write

h 1|[A, B]| 2i = (a1 a2)h 1|B| 2i = 0. (2.4.22)

Since, by hypothesis, a16= a2, Eq. (2.4.19) follows.

The following comments are in order. First, as we will see later, if A represents a symmetry operation on a system described by the Hamiltonian _{H, one has [A, H] = 0.} Theorem II [expressed by Eq. (2.4.19)] then guarantees that _{H does not connect eigen-} states of A with distinct eigenvalues. For instance, if A probes the parity (even or odd) of a state, A2 ⌘ , so that its eigenvalues must be ±1; then, matrix elements of H between states with di↵erent parities are necessarily zero. The second comment is to point out that Theorem II is actually less stringent than stated above, since only A being Hermitian suffices for it to hold.

Theorem III: If two observables A and B commute, one can construct an orthonor- mal basis of the state space with eigenvectors common to A and B.

Let A be an observable, with spectrum

A_|ui_{ni = a}n|uini, (2.4.23)

where i runs over the gndegenerate eigenstates corresponding to the eigenvalue an. Since

A is an observable, its eigenvectors form an orthonormal basis, and we want to express B in this basis.

According to Theorem II, B does not connect eigenvectors of A corresponding to distinct eigenvalues, i.e.,

huin|B|ui

0

n0i = 0, when n 6= n0, (2.4.24)

but nothing can be said when n = n0 and i6= i0_.

In view of Theorem I’, it is convenient to group the eigenstates of A into subspaces E1, E2, . . ., with dimensions g1, g2, . . ., corresponding to the eigenvalues a1, a2, . . .. In so

doing, when B is expressed in this grouped basis it becomes block-diagonal, as illustrated in Fig. 2.1: one says that the eigensubspaces En are globally invariant under the action

of B.

Figure 2.1: The observable B expressed in terms of degenerate subspaces of eigenvectors of A: only the shaded parts contain non-zero matrix elements. (Extracted from CT, Page 141).

Two possibilities arise for each En:

(1) an is non-degenerate. The corresponding Enis necessarily one-dimensional, meaning

that the corresponding eigenvector of A is also an eigenvector of B.

(2) an is degenerate. In the basis {|uini}, for a given n and i = 1, 2, . . . , gn, the sub-

matrix representing A is diagonal, A = an ; it is also diagonal on any basis formed

by linear combinations of _{|ui

2.4. EIGENVALUE EQUATIONS. OBSERVABLES 43

one that diagonalises B; call this the {|vi

ni} basis, which is also gn-dimensional.

Then, in the En eigensubspace, A and B can be simultaneously diagonalised. It

should be stressed that if one determines the degenerate eigenvectors of A, these are not necessarily eigenvectors of B as well; in general, one has to diagonalise B to determine its eigenvectors in terms of those of A.

By repeating this procedure for all subspaces En, we obtain a basis of E , formed by

eiegenvectors common to A and B; this proves the theorem. The reader should be able to prove its reciprocal: if there is a basis of eigenvectors common to A and B, then these two observables commute.

A remark about notation: eigenvectors common to A and B will be denoted generi- cally by_|ui

n,pi,

A_|ui_{n,pi = a}n|uin,pi (2.4.25)

B_|ui_{n,pi = b}p|uin,pi, (2.4.26)

such that n and p respectively run over the eigenvalues of A and B, while i runs over the possible degeneracies that remain for a given pair (an, bp).

2.4.3.2 Complete sets of commuting observables Central question here: how can we label bases in E ?

There are many operators in E , Many bases can be used Let A be an observable with eigenvectors {|ui

ni} and corresponding eigenvalues an.

• If all an are not degenerate ) can use an to uniquely label the basis formed by

eigenvectors of A: _|ui_{ni ! |a}ni. That is, all eigensubspaces are one-dimensional.

=_{) A constitutes, by itself, a complete set of commuting observables (CSCO).} • If one or several an are degenerate ) specifying an is not sufficient to uniquely

label the whole basis formed by eigenvectors of A. That is, some eigensubspaces have dimensions greater than 1.

=_{) Choose another observable, B, such that [B, A] = 0, and with eigenvalues} bp. Theorem III guarantees the existence of a common basis to A and B in the

degenerate subspace of A.

=_{) if all b}p are non-degenerate, then the basis in E can be uniquely labelled by

{an, bp}, and {A, B} form a CSCO.

=_{) if some b}p are degenerate, then choose another observable, C, with eigen-

values cr, and such that [C, A] = [C, B] = [A, B] = 0.

=) the process may go on, until all subspaces of common eigenvectors are one- dimensional.

In summary:

• all A, B, C, . . . commute in pairs;

• specifying the eigenvalues of all A, B, C, . . . suffices to label a unique common eigenvector.

Alternatively: A set of observables A, B, C, . . . is a CSCO if there exists a unique or- thonormal basis of common eigenvectors (to within multiplicative factors).

Notes:

(1) If{A, B} is a CSCO, another CSCO can be obtained by adding to it any observable C which commutes with A and B. However, one is always interested in ‘minimal’ sets; that is, C is redundant.

(2) Let_{{A, B, C} be a CSCO with eigenvalues {a}n, bp, cr}; then basis is usually denoted

by|an, bp, cri, or, often dropping the commas, by |anbpcri.

(3) For a given physical system, there are several CSCO’s.

(4) In general, it is not a simple matter to determine all observables making up one CSCO for a given system. While several commuting observables can usually be identified from symmetry considerations, the set may still be incomplete; nonethe- less, the use of incomplete bases still provides considerable insights into the physical properties of the systems.

Further reading: CT, solved problems HII 11 and 12.