wang
(Wang)
#1
Given thatJ 1 andJ 2 are self-adjoint, the operatorsJ±are not self-adjoint, but are instead
the adjoints of one another,
(J+)†=J− (J−)†=J+ (5.90)
In terms of these operators, the angular momentum algebra assumes the form,
[J 3 ,J±] =± ̄hJ± ⇔ J 3 J±=J±(J 3 ± ̄h)
[J+,J−] = +2 ̄hJ 3 (5.91)
SinceJ^2 commutes withJ±, the valuejorλ(j) remains unchanged upon applyingJ±to a
state|λ(j),m〉. Them-value will be lowered or raised, since
J 3 J±|j,m〉= ̄h(m±1)J±|j,m〉 (5.92)
5.7.3 Constructing the spectrum
Our goal is to realize the angular momentum algebra of a Hilbert space. Since λ(j) and
m label different eigenvalues of self-adjoint operators, the corresponding eigenstates are
orthogonal, and we may normalize them by requiring
〈j,m|j′,m′〉=δj,j′δm,m′ (5.93)
We shall now show that, givenj, the range for the eigenvaluemmust be bounded. To do
so, expressJ^2 in terms ofJ 3 andJ±,
J^2 =
1
2
(J+J−+J−J+) +J 32
= J+J−+J 32 − ̄hJ 3
= J−J++J 32 + ̄hJ 3 (5.94)
Now, both the operatorsJ+J−andJ−J+are positive, since for allj,m,
〈j,m|J+J−|j,m〉=||J−|j,m〉||^2 = ̄h^2 [λ(j)−m^2 +m]≥ 0
〈j,m|J−J+|j,m〉=||J+|j,m〉||^2 = ̄h^2 [λ(j)−m^2 −m]≥ 0 (5.95)
Clearly, givenλ(j), the values ofmare bounded from above and from below. But since
successive application ofJ+would raisem indefinitely, there must be a state with value
m= m+which is annihilated byJ+. Similarly, since successive application ofJ−would
lowermindefinitely, there must be a state with valuem=m−which is annihilated byJ−.
By construction, we will havem+≥m−, and
J+|j,m+〉= 0 〈j,m+|j,m+〉= 1
J−|j,m−〉= 0 〈j,m−|j,m−〉= 1 (5.96)
