The raising and lowering operators changemin integer steps, so, starting fromm=−ℓ, there will
be states in integer steps up toℓ.
m=−ℓ,−ℓ+ 1,...,ℓ− 1 ,ℓHaving the minimum at−ℓand the maximum at +ℓwith integer steps only works ifℓis an integer
or a half-integer. There are 2ℓ+ 1 states with the sameℓand different values ofm. We now know
what eigenstates are allowed.
The eigenstates ofL^2 andLzshould be normalized
〈Yℓm|Yℓm〉= 1.The raising and lowering operators acting onYℓmgive
L±Yℓm=C±(ℓ,m)Yℓ(m±1)The coefficientC±(ℓ,m) can be computed.
〈L±Yℓm|L±Yℓm〉 = |C(ℓ,m)|^2 〈Yℓ(m±1)|Yℓ(m±1)〉
= |C(ℓ,m)|^2
〈L±Yℓm|L±Yℓm〉 = 〈Yℓm|L∓L±Yℓm〉
= 〈Yℓm|(L^2 −L^2 z∓ ̄hLz)Yℓm〉
= (ℓ(ℓ+ 1)−m^2 ∓m) ̄h^2
|C(ℓ,m)|^2 =(
ℓ(ℓ+ 1)−m^2 ∓m)
̄h^2
C±(l,m) = ̄h√
ℓ(ℓ+ 1)−m(m±1)We now have the effect of the raising and lowering operators in termsof the normalized eigenstates.
L±Yℓm= ̄h√
ℓ(ℓ+ 1)−m(m±1)Yℓ(m±1)14.5 Examples
14.5.1 The Expectation Value ofLz
What is the expectation value ofLzin the stateψ(~r) =R(r)(
√
2
3 Y^11 (θ,φ)−i√
1
3 Y^1 −^1 (θ,φ))?The radial part plays no role. The angular part is properly normalized.
〈ψ|Lz|ψ〉 =〈√
2
3
Y 11 −i√
1
3
Y 1 − 1 |Lz|√
2
3
Y 11 −i√
1
3
Y 1 − 1
〉
=
〈√
2
3
Y 11 −i√
1
3
Y 1 − 1 |
√
2
3
̄hY 11 +i√
1
3
̄hY 1 − 1〉
=
(
2
3
−
1
3
)
̄h=1
3
̄h