The operatorsL+,L−andLZdo not changeℓ. That is, after we operate, the new state is still an
eigenstate ofL^2 with the same eigenvalue,ℓ(ℓ+ 1).
The eigenvalue ofLzis changed when we operate withL+orL−.
Lz(L±Yℓm) = L±LzYℓm+ [Lz,L±]Yℓm
= m ̄h(L±Yℓm)± ̄hL±Yℓm= (m±1) ̄h(L±Yℓm)(This should remind you of the raising and lowering operators in the HOsolution.)
From the above equation we can see that (L±Yℓm) is an eigenstate ofLz.
L±Yℓm=C±(ℓ,m)Yℓ(m±1)These operators raise or lower thezcomponent of angular momentum by one unit of ̄h.
SinceL†±=L∓, its easy to show that the following is greater than zero.
〈L±Yℓm|L±Yℓm〉≥ 0
〈Yℓm|L∓L±Yℓm〉≥ 0WritingL+L−in terms of our chosen operators,
L∓L± = (Lx∓iLy)(Lx±iLy) =L^2 x+L^2 y±iLxLy∓iLyLx
= L^2 x+L^2 y±i[Lx,Ly] =L^2 −L^2 z∓ ̄hLzwe can derive limits on the quantum numbers.
〈Yℓm|(L^2 −L^2 z∓ ̄hLz)Yℓm〉≥ 0
(ℓ(ℓ+ 1)−m^2 ∓m) ̄h^2 ≥ 0
ℓ(ℓ+ 1)≥m(m±1)We know that the eigenvalueℓ(ℓ+ 1) ̄h^2 is greater than zero. We can assume that
ℓ≥ 0because negative values just repeat the same eigenvalues ofℓ(ℓ+ 1) ̄h^2.
The condition thatℓ(ℓ+ 1)≥m(m±1) then becomes a limit onm.
−ℓ≤m≤ℓNow,L+raisesmby one andL−lowersmby one, and does not changeℓ. Sincemis limited to be
in the range−ℓ≤m≤ℓ, the raising and lowering must stop form=±ℓ,
L−Yℓ(−ℓ)= 0
L+Yℓℓ= 0