5.61
Spherical Harmonics
page 1
ANGULAR MOMENTUM
Now that we have obtained the general eigenvalue relations for angular
momentum directly from the operators, we want to learn about the associated
wave functions. Returning to spherical polar coordinates, we recall that the
angular momentum operators are given by:
⎛
∂
∂ ⎞
L̂x = −i� ⎜ − sin φ
− cot θ cos φ ⎟
∂θ
∂φ ⎠
⎝
∂
∂ ⎞
⎛
L̂y = −i� ⎜ cos φ
− cot θ sin φ
∂θ
∂φ ⎟⎠
⎝
∂
L̂z = −i�
∂φ
Lˆ2 = Lˆ2x + Lˆ2Y + Lˆ2z
1 ∂2 ⎤
∂ ⎛
∂ ⎞
2
2 ⎡ 1
ˆ
⇒ L = −� ⎢
⎜ sin θ
⎟+
∂θ ⎠ sin 2 θ ∂φ 2 ⎥⎦
⎣ sin θ ∂θ ⎝
In terms of these, our original Schrödinger Equation for rigid rotations was
L̂2 m
m
ˆ
HYl = Yl = ElYl m
2I
2
−� ⎡ 1 ∂ ⎛
∂ ⎞
1 ∂2 ⎤ m
⇒
Y θ , φ ) = ElYl m
(θ , φ )
⎜ sin θ
⎟+ 2
⎢
2⎥ l (
2I ⎣ sin θ ∂θ ⎝
∂θ ⎠ sin θ ∂φ ⎦
2
where l was the quantum number for L̂ and m was the quantum number for L̂z .
2
Taking what we learned in the last section about the eigenvalues of L̂ and L̂z
we can say that at most we can have
l = 0, 12 ,1, 23 , 2,...
m = −l, −l + 1,..., l
We will see that there is an additional restriction on the possible values of l in
the present case, but these are the possible values for the quantum numbers. In
terms of the quantum numbers, we have the eigenvalue relations
Lˆ2Yl m = L2Yl m = � 2l ( l +1) Yl m
L̂zYl m = �mYl m
Now, the functions, Ylm , that satisfy these relations for rigid rotations are
called Spherical Harmonics. It is possible to derive the spherical harmonics by
solving the 2D differential equation above. McQuarrie goes through a fairly
complete derivation and we outline that solution in the appendix to these notes
(below). The result is that:
5.61
Spherical Harmonics
Yl m (θ ,φ ) = Alm Pl
where Alm is a normalization constant and
m
page 2
( cos θ ) eimφ
m
Pl ( x ) is an associated Legendre
Polynomial. The first few Associated Legendre Polynomials are:
P00 ( cos θ ) = 1
P10 ( cos θ ) = cos θ
( 3cos θ −1)
P11 ( cos θ ) = sin θ
P20 ( cos θ ) =
P21 ( cos θ ) = 3cos θ sin θ
P22 ( cos θ ) = 3sin 2 θ
1
2
2
There are a number of important features of the Spherical Harmonics we can
recognize simply by inspecting these solutions:
• The wavefunctions factorize into a product of a function of θ and a
function of φ.
Yl m (θ , φ ) ∝ f (θ ) g (φ )
This result is very reminiscent of the result we found for separable
Hamiltonians, which is somewhat surprising because the Hamiltonian
certainly did not appear at first sight to be separable into a Hamiltonian
for θ and a Hamiltonian for φ:
Ĥ =
−� 2
2I
⎡ 1 ∂ ⎛
∂ ⎞
1 ∂2 ⎤
sin
θ
+
≠ Ĥ θ + Ĥ φ
⎟
⎢ sin θ ∂θ ⎜
2
2⎥
∂
θ
sin
θ
∂
φ
⎝
⎠
⎣
⎦
However, physically it makes some sense for the motion along θ and φ to
separate: we haven’t applied any potential that links them together so
particles should be free to move along θ and φ independently, just as
particles in a separable potential in x and y can move independently along
those axes. The θ ­ φ cross terms above reflect the curvature of the 2D
surface the particles are moving on.
• It is easy to verify that these functions are eigenstates of L̂z :
∂ m
∂
m
m
Yl = −i�
Alm Pl ( cos θ ) eimφ = �mAlm Pl ( cos θ ) eimφ = �mYl m
∂φ
∂φ
2
m
They are also eigenfunctions of L̂ , as can be proven for any given Yl
2 m
(after some algebra) by computing Lˆ Y and verifying that the result is
L̂zYl m = −i�
l
just � l ( l +1) Yl .
2
m
• We can now see why half­integer values of l are not allowed here. Recall
that φ is the angle in the x­y plane and it varies from 0 to 2π. What
should happen to Yl
m
(θ , φ )
when φ → φ + 2π ? Of course, the value of the
wavefunction should not change because by incrementing φ by 2π: we’ve
5.61
Spherical Harmonics
page 3
just moved the particle around in a full circle. Thus, for the wave
function to be single­valued, we must have:
Yl m (θ , φ ) = Yl m (θ , φ + 2π )
⇒ Alm Pl m ( cos θ ) eimφ = Alm Pl m ( cos θ ) eim(φ +2π )
⇒ eimφ = eim(φ +2π )
⇒ eimφ = eimφ eim 2π
⇒ 1 = eim 2π
⇒ m = an integer
thus, m must be an integer, no matter what value of l we choose. However,
since the minimal value for m is l, l must also be an integer. This
continuity argument is the reason why half integer values of l are not
allowed for rigid rotations.
• Note that there are a few interesting algebraic features of the spherical
harmonics: 1) the φ part of the wavefunction does not depend on l 2) The
lth order Legendre polynomial always involves sums of products of sines
and cosines such that the sum of the sine and cosine powers is less than
or equal to l 3) for m ≠ 0 the spherical Harmonics are complex and
Ylm * = Yl−m . Thus, one can obtain two real functions from each ±m pair via
Rlm =
1
2
(Y
m
l
+ Yl− m
)
I lm =
1
i 2
(Y
m
l
− Yl− m
)
These features are helpful in trying to identify, when given an arbitrary
function of the angles, which spherical harmonics might contribute to that
function.
• Typically, the spherical Harmonics are associated with letters as you have
seen in your previous chemistry courses. Thus, l=0 is ‘s’, l=1 is ‘p’, l=2 is ‘d’
….
• In the absence of a potential, as is the case for rigid rotations, the
spherical Harmonics are 2l+1-fold degenerate:
l
m
Yl m ' s
2l + 1
0
0
Y00
1
1
­1, 0, 1
Y1−1 , Y10 , Y11
3
2
­2, ­1, 0, 1, 2
Y2−2 , Y2−1 , Y20 , Y21 , Y22
5
3 ­3, ­2, ­1, 0, 1, 2, 3 Y −3 , Y −2 , Y −1 , Y 0 , Y 1 , Y 2 , Y 3
3
3
3
3
3
3
3
7
5.61
Spherical Harmonics
page 4
As discussed previously, we should expect this degeneracy to be broken if
we apply a potential that is not spherically symmetric. In the presence of
a potential, we expect these levels to be split.
Thus, to summarize, for the spherical Harmonics we have:
Yl m (θ , φ ) = Alm Pl
m
( cosθ ) eimφ
Lˆ2Yl m = L2Yl m = � 2l ( l + 1) Yl m
l = 0,1, 2,3...
Lˆ Y m = �mY m
m = −l , −l + 1,..., l
z l
l
APPENDIX: SOLVING FOR THE SPHERICAL HARMONICS
We need to solve the differential equation
⎡ 1 ∂ ⎛
∂ ⎞
1 ∂2 ⎤
⎢ sin θ ∂θ ⎜ sin θ ∂θ ⎟ + sin 2 θ ∂φ 2 ⎥ Y (θ , φ ) = EY (θ , φ )
⎝
⎠
⎣
⎦
ˆ (θ , φ ) = EY (θ , φ ) for Rigid rotations. Rearranging the Equation,
This is HY
�2
−
2I
∂ ⎛
∂ ⎞ 2IE 2 ⎤
∂2
⎡
⎢sin θ ∂θ ⎜ sin θ ∂θ ⎟ + � 2 sin θ ⎥ Y (θ , φ ) = − ∂φ 2 Y (θ , φ )
⎝
⎠
⎣
⎦
only θ
only φ
We’ve separated the variables, just as in the 3D harmonic oscillator.
∴
Try
Define
Y (θ , φ ) = Θ (θ ) Φ (φ )
β≡
2IE
�2
as a solution
( note β ∝ E )
∂ ⎛
∂ ⎞
∂2
⎡
2 ⎤
⎢sin θ ∂θ ⎜ sin θ ∂θ ⎟ + β sin θ ⎥ Θ (θ ) Φ (φ ) = − ∂φ 2 Θ (θ ) Φ (φ )
⎝
⎠
⎣
⎦
Dividing by Θ (θ ) Φ (φ ) and simplifying
5.61
Spherical Harmonics
sin θ ∂
Θ (θ ) ∂θ
page 5
∂ ⎞
1 ∂2
⎛
2
sin
Θ
+
sin
=
−
Φ (φ )
θ
θ
β
θ
⎜
⎟ ( )
∂θ ⎠
Φ (φ ) ∂φ 2
⎝
only θ
only φ
Since θ and φ are independent variables, each side of the equation must be
equal to a constant ≡ m2.
∂2
Φ (φ ) = − m 2
2
Φ (φ ) ∂φ
1
⇒
and
sin θ ∂
Θ (θ ) ∂θ
⇒
I
∂ ⎞
⎛
2
2
⎜ sin θ
⎟ Θ (θ ) + β sin θ = m
∂θ ⎠
⎝
First solve for Φ (φ ) using
I
∂ 2 Φ (φ )
= − m 2 Φ (φ )
2
∂φ
Solutions are Φ (φ ) = Am e
Boundary conditions
imφ
and A− m e − imφ
⇒
quantization
Φ (φ + 2π ) = Φ (φ )
Am eim(φ +2π ) = Am eimφ
⇒
∴
This is only true if
and A− m e − im(φ +2π ) = A− m e − imφ
eim( 2π ) = 1 and
e − im( 2π ) = 1
m = 0, ± 1, ± 2, ± 3,....
m is the “magnetic” quantum number
∴
Φ (φ ) = Am eimφ
m = 0, ± 1, ± 2, ± 3,....
II
5.61
Spherical Harmonics
∫
Normalization:
2π
0
Φ ∗ (φ )Φ (φ ) d φ = 1
1 imφ
e
2π
Φ (φ ) =
⇒
page 6
m = 0, ± 1, ± 2, ± 3,...
Now let’s look at Θ (θ ) . Need to solve II
sin θ ∂
Θ (θ ) ∂θ
Change variables: x = cos θ
Since 0 ≤ θ ≤ π
Also
⇒
∂ ⎞
⎛
2
2
⎜ sin θ
⎟ Θ (θ ) + β sin θ = m
∂θ ⎠
⎝
dx
Θ (θ ) = P ( x )
= dθ
− sin θ
−1 ≤ x ≤ +1
sin 2 θ = 1− cos2 θ = 1− x 2
This equation turns out to be Legendre’s equation in terms of Θ:
sin θ
d
dθ
⎡
dΘ ⎤
⎢sin θ dθ ⎥ +
⎣
⎦
(β sin2 θ − m2 ) Θ (θ ) = 0
which we can re­write:
sin 2 θ
d 2Θ
dΘ
θ
θ
+
sin
cos
+ β sin 2 θ − m 2 Θ (θ ) = 0
2
dθ
dθ
)
(
Let x = cosθ and Θ(θ ) = P (x ).
12
dΘ dP dx
dP
=
= − sin θ
= − 1− x 2
dθ dx dθ
dx
)
d 2Θ d
=
dθ 2 dθ
)
(
dP
dx
1 2 dP ⎤
⎡ d Θ ⎤ ⎡ dx ⎤ d ⎡
2
=
−
1−
x
⎢
⎢ dθ ⎥ ⎢ dθ ⎥ dx
dx ⎥⎦
⎣
⎦ ⎣
⎦
⎣
(
⎡
⎤
1 2 d 2P ⎥
⎢
x
dP
2
= − sin θ ⎢
⎥
1 2 dx − 1− x
dx 2 ⎥
⎢ 1− x 2
⎣
⎦
(
= −x
)
(
)
dP
d 2P
+ 1− x 2
dx
dx 2
(
)
Substituting these results into Legendre’s equation gives
5.61
Spherical Harmonics
(
1− x 2
)
2
page 7
d 2P
2 dP
2
2
2 − 2 x 1− x dx + β 1− x − m P ( x ) = 0
dx
( (
)
(
)
)
Divide by (1− x 2 ) to obtain the Legendre equation in a convenient form:
(
1− x 2
)
d 2P
dP ⎡
m2 ⎤
x
β
−
2
+
−
⎢
⎥ P(x) = 0
dx ⎢⎣
dx 2
1− x 2 ⎥⎦
The solutions to this equation are known, but very messy. They are called the
associated Legendre polynomials, Pl
m
(x) .
Note that they only depend on |m|
because the equation depends on m2:
Pl
m
(x )= P (cosθ )
1
P (cos θ )= (3cos θ − 1)
2
P (cos θ )= 3cos θ sin θ
P (cos θ )= 3sin θ
m
l
( )
P (cos θ )= cos θ
P (cos θ )= sin θ
P00 cos θ = 1
0
2
0
1
2
1
2
1
1
2
2
2
etc.
So Θ
(θ )= A
P
lm l
m
(
(
)
)
(cosθ )
where Alm is the normalization constant
⇒
π
2
m
Alm2 ∫ ⎡ Pl cos θ ⎤ sin θ dθ = 1
0 ⎣
⎦
(
)
So now putting it all together:
ψ lm (r0 ,θ , φ )= Yl m (θ , φ )= Θl (θ )Φ m (φ )
m
Yl
m
(
(
)
)
1
2
⎡⎛ 2l + 1⎞ l − m !⎤
⎥ Pl m cos θ eimφ
θ ,φ = ⎢⎜
⎟
⎢⎝ 4π ⎠ l + m !⎥
⎣
⎦
( )
1
⎡⎛ 2l + 1⎞ l − m !⎤ 2
⎥
Alm = ⎢⎜
⎢⎝ 2 ⎟⎠ l + m !⎥
⎣
⎦
These functions are the spherical harmonics.
(
)
5.61
Spherical Harmonics
page 8
SPHERICAL HARMONICS SUMMARY
( )
m
() ()
Yl m θ , φ = Θ l θ Φ m φ
(
(
)
)
1
⎡⎛ 2l + 1⎞ l − m !⎤ 2
m
⎥ Pl m cos θ eimφ
Yl θ , φ = ⎢⎜
⎟
⎢⎝ 4π ⎠ l + m !⎥
⎣
⎦
l = 0, 1, 2,...
m = 0, ± 1, ± 2, ± 3,... ± l
( )
(
)
Yl m ’s are the eigenfunctions to Ĥψ = Eψ for the rigid rotor problem.
1
Y00 =
⎛ 5 ⎞2
2
Y20 = ⎜
⎟ ( 3cos θ −1)
⎝ 16π ⎠
1
12
( 4π )
1
1
⎛ 3 ⎞2
Y10 = ⎜
⎟ cos θ
⎝ 4π ⎠
⎛ 15 ⎞ 2
Y2±1 = ⎜ ⎟ sin θ cos θ e ±iφ
⎝ 8π ⎠
1
2
1
2
⎛ 3 ⎞
Y11 = ⎜ ⎟ sin θ eiφ
⎝ 8π ⎠
⎛ 15 ⎞
2
±2iφ
Y2±2 = ⎜
⎟ sin θ e
⎝ 32π ⎠
1
⎛ 3 ⎞2
Y1−1 = ⎜ ⎟ sin θ e −iφ
⎝ 8π ⎠
Yl m ’s are orthonormal:
Krönecker delta
Note: Switch
⎧1
⎩0
δ ll′ = ⎨
∫∫ Y (θ ,φ )Y (θ ,φ )sin θ dθ dφ = δ
m ′∗
l′
if l = l′
if l ≠ l′
m
l
⎧1
⎩0
δ mm′ = ⎨
if m = m′
if m ≠ m′
ll ′
δ mm′
normalization
orthogonality
l → J conventional for molecular rotational quantum number
(e.g.: l ( l + 1) ⇒ J ( J + 1)
J = 0, 1, 2,... .)