Physical Chemistry - Angular Momentum_lecture18-19

5.61 Angular Momentum Page 1


Angular Momentum
Since L̂2 and L̂z commute, they share common eigenfunctions. These functions are
extremely important for the description of angular momentum problems – they
determine the allowed values of angular momentum and, for systems like the Rigid
Rotor, the energies available to the system. The first things we would like to know
are the eigenvalues associated with these eigenfunctions. We will denote the
eigenvalues of L̂2 and L̂z by α and β, respectively so that:
L̂2Y β (θ , φ ) = α Y β (θ , φ )
α Lˆ Y β (θ , φ ) = β Y β (θ , φ )
α z α α

For brevity, in what follows we will omit the dependence of the eigenstates on θ
and φ so that the above equations become
Lˆ 2Yαβ = α Yαβ Lˆ zYαβ = β Yαβ
It is convenient to define the raising and lowering operators (note the similarity to
the Harmonic oscillator!):
L̂± ≡ L̂ x ± iLˆ y
Which satisfy the commutation relations:
⎡ L̂+ , L̂− ⎤ = 2�L̂z ⎡ L̂z , L̂± ⎤ = ± �L̂± ⎡ L̂± , L̂2 ⎤ = 0
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
These relations are relatively easy to prove using the commutation relations we’ve
already derived:
⎡ Lˆ x , Lˆ y ⎤ = i�Lˆ z ⎡ Lˆ y , Lˆ z ⎤ = i�Lˆ x ⎡ Lˆ z , Lˆ x ⎤ = i�Lˆ y ⎡ Lˆ 2 , Lˆ z ⎤ = 0
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
For example:
⎡ L̂z , L̂± ⎤ = ⎡ L̂z , L̂ x ⎤ ± i ⎡ L̂z , L̂ y ⎤
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
(
= i�L y ± i ( −i�L x ) = ± � L x ± iLy )
= ± �L̂±
The raising and lowering operators have a peculiar effect on the eigenvalue of L̂z :
Lˆ z (Lˆ ±Yαβ ) = ( ⎡⎣ Lˆ z , Lˆ ± ⎤⎦ + Lˆ ± Lˆ z )Yαβ = ( ± �Lˆ ± + Lˆ ± β )Yαβ = ( β ± � ) (Lˆ ±Yαβ )
Thus, L̂+ ( L̂− ) raises (lowers) the eigenvalue of L̂z by � , hence the names. Since
the raising and lowering operators commute with L̂2 they do not change the value
of α and so we can write
Lˆ ±Yαβ ∝ Yαβ ±�
and so the eigenvalues of L̂z are evenly spaced!

What are the limits on this ladder of eigenvalues? Recall that for the harmonic
oscillator, we found that there was a minimum eigenvalue and the eigenstates could

, 5.61 Angular Momentum Page 2


be created by successive applications of the raising operator to the lowest state.
There is also a minimum eigenvalue in this case. To see this, note that
Lˆ2 + Lˆ2 = Lˆ2 + Lˆ2 ≥ 0 x y x y

This result simply reflects the fact that if you take any observable operator and
square it, you must get back a positive number. To get a negative value for the
average value of L̂2x or L̂2y would imply an imaginary eigenvalue of L̂ x or L̂ y , which is
impossible since these operators are Hermitian. Besides, what would an imaginary
angular momentum mean? We now apply the above equation for the specific
wavefunction Yαβ :

∫ ( ) ∫ (
0 ≤ Yαβ * L̂2x + L̂2y Yαβ = Yαβ * L̂2 − L̂2z Yαβ )
= ∫ Yαβ * (α − β 2 ) Yαβ

=α −β2
Hence β 2 ≤ α and therefore − α ≤ β ≤ α . Which means that there are both
maximum and minimum values that β can take on for a given α. If we denote these
values by βmax and βmin, respectively, then it is clear that
Lˆ +Yαβmax = 0 Lˆ −Yαβmin = 0 .
We can then use this knowledge and some algebra tricks trick to determine the
relationship between α and βmax (or βmin). First note that:
⇒ Lˆ − Lˆ +Yαβ max = 0 Lˆ + Lˆ −Yαβ min = 0
We can expand this explicitly in terms of L̂ x and L̂ x :
( )
⇒ Lˆ2x + Lˆ2y − i( Lˆ y Lˆ x − Lˆ x Lˆ y ) Yαβ max = 0 ( )
Lˆ2x + Lˆ2y + i( Lˆ y Lˆ x − Lˆ x Lˆ y ) Yαβ min = 0
However, this is not the most convenient form for the operators, because we don’t
know what L̂ x or L̂ y gives when acting on Yαβ . However, we can re­write the same
expression in terms of L̂2 and L̂z :
(
Lˆ2x + Lˆ2y ± i( Lˆ y Lˆ x − Lˆ x Lˆ y ) )
L̂2 − L̂2z −i�Lˆ z
So then we have
( )
⇒ L̂2 − L̂2z − �L̂z Yαβ max = 0 ( L̂ − L̂ + �L̂ ) Yαβ = 0
2 2
z z
min



⇒ (α − β 2
max − �β max = 0 ) ( α − β + �β ) = 0
2
min min

⇒ α = β max ( β max + �) = β min ( β min − �)
⇒ β max = − β min ≡ �l