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221A Lecture Notes
Spherical Harmonics
1 Oribtal Angular Momentum
The orbital angular momentum operator is given just as in the classical mechanics, L~ = ~x × ~p. (1)
From this definition and the canonical commutation relation between the po- sition and momentum operators, it is easy to verify the commutation relation among the components of the angular momentum,
[Li, Lj ] = i¯hijkLk. (2)
When using the position representation, the action of the angular mo- mentum on any state is given by a differential operator
〈~x|L~|ψ〉 = ~x × ¯h i ∇〈~ ~x|ψ〉 = ~x × ¯h i ∇~ψ(~x). (3)
Loosely, we can write ~L = ~x × ¯h i
but you have to be careful on what this differential operator is acting on. If you act on a position ket instead of a bra,
L^ ~|~x〉 = −~x × ¯h i ∇|~ ~x〉. (5)
Whenever I write the orbital angular momentum operator as a differential operator in this note, it is understood that it acts on a position bra instead of ket. With this caveat in mind, we can rewrite the orbital angular momentum operator in the polar coordinates. Following the usual definitions
x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, (6)
we can rewrite the derivatives using the chain rule,
∂r ∂θ ∂φ
=
∂rx ∂ry ∂rz ∂θx ∂θy ∂θz ∂φx ∂φy ∂φz
∂x ∂y ∂z
sin θ cos φ sin θ sin φ cos θ r cos θ cos φ r cos θ sin φ −r sin θ −r sin θ sin φ r sin θ cos φ 0
∂x ∂y ∂z
. (7)
We invert the matrix and find
∂x ∂y ∂z
= 1 r sin θ
r sin^2 θ cos φ sin θ cos θ cos φ − sin φ r sin^2 θ sin φ sin θ cos θ sin φ cos φ r sin θ cos θ − sin^2 θ 0
∂r ∂θ ∂φ
. (8)
Now the orbital angular momentum operators can be written in terms of spherical coordinates,
Lx = ¯h i
( − sin φ
∂θ − cot θ cos φ
∂φ
) , (9)
Ly = ¯h i
( cos φ
∂θ − cot θ sin φ
∂φ
) , (10)
Lz = ¯h i
∂φ
It is useful to take the combinations
L± = ¯h i
e±iφ
( ±i
∂θ
− cot θ
∂φ
)
. (12)
Finally,
~L^2 = −¯h^2
[ 1 sin θ
∂θ
( sin θ
∂θ
)
sin^2 θ
∂^2
∂φ^2
]
. (13)
The free-particle Hamiltonian is
~p^2 2 m
−¯h^2 2 m
(∂ x^2 + ∂ y^2 + ∂ z^2 ) = −¯h^2 2 m
( ∂^2 ∂r^2
r
∂r
)
L~^2
2 mr^2
Looking at the expression for Lz , you can see that it is of the same form as the momentum operator of a particle on a circle, whose eigenvalues are quantized as n¯h for n ∈ Z. Therefore, Lz is also quantized as m¯h for m ∈ Z. An immediate consequence is that no half-odd values are allowed for Lz , and hence half-odd j cannot be obtained for orbital angular momentum. Only integer j is possible.
down an equation for the ground state of the form a| 0 〉 = 0. It was useful because it gave us a linear differential equation instead of a quadratic one (e.g., Schr¨odinger equation). The linear differential equations are far easier to solve. We can take the same strategy for the spherical harmonics.
2.1 Derivation of Spherical Harmonics
We know from the general representation theory of angular momenta that L−|l, −l〉 = 0 because it cannot be lowered any more. Sakurai starts with |l, l〉 instead. Of course you get the same result, but I find this way somewhat less confusing. In the position representation, we find
0 = 〈θ, φ|L−|l, −l〉 = ¯h i
e−iφ
( −i
∂θ
− cot θ
∂φ
) Y (^) l− l(θ, φ) = 0. (22)
On the other hand, we know already that
− l¯hY (^) l− l= 〈θ, φ|Lz |l, −l〉 = ¯h i
∂φ
Y (^) l− l (23)
and hence the azimuth dependence of Y (^) l− lis Y (^) l −l(θ, φ) = f (θ)e−ilφ. Therefore, Eq. (22) becomes (^) (
−i d dθ
) f (θ) = 0. (24)
This equation is solved easily, by writing it as
1 f
df = l cot θdθ, (25)
and integrating both sides to
log f = l log sin θ + const. (26)
Therefore, f (θ) = c sinl^ θ (27)
with an overall normalization constant c. Putting things together, we find
Y (^) l− l(θ, φ) = c sinl^ θe−ilφ. (28)
The absolute value of c can be fixed by the normalization
1 =
∫ dΩ|Y (^) ll |^2 =
∫ (^1) − 1
d cos θ
∫ (^2) π 0
dφ|c|^2 sin^2 l^ θ. (29)
φ integral trivially gives a factor of 2π. The θ integral is most conveniently done using the variable x = cos θ,
1 = 2π|c|^2
∫ (^1) − 1
dx(1 − x^2 )l.. (30)
Writing 1−x^2 = (1−x)(1+x), and further change the variable to x = −1+2t,
1 = 2π|c|^2
∫ (^1) 0
2 dt (2t(2 − 2 t))l^ = 2π|c|^222 l+
∫ (^1) 0
dt tl(1 − t)l. (31)
The integral is nothing but the Beta function
B(p, q) =
∫ (^1) 0
dt tp−^1 (1 − t)q−^1 = Γ(p)Γ(q) Γ(p + q)
Therefore,
1 = 2π|c|^222 l+^ Γ(l + 1)Γ(l + 1) Γ(2l + 2) = 4π|c|^222 l^ l!l! (2l + 1)!
We find
|c| =
2 ll!
√ (2l + 1)! 4 π
The phase of c is fixed by picking a convention. The commonly used convention (the same as Sakurai’s) is not to have an additional phase factor
Y (^) l− l(θ, φ) =
2 ll!
√ (2l + 1)! 4 π sinl^ θe−ilφ. (35)
Now that we know Y (^) l− l, we can keep acting L+ on it to obtain all Y (^) lm. Recall that the general discussion of angular momentum taught us that
L+|l, m〉 =
√ l(l + 1) − m(m + 1)|l, m + 1〉 =
√ (l − m)(l + m + 1)|l, m + 1〉. (36)
2 ll!
√√ √√ (2l + 1)(l − m)! 4 π(l + m)! eimφ^ sinm^ θ
( 1 sin θ
d dθ
)l+m sin^2 l^ θ
2 ll!
√√ √√ (2l + 1)(l − m)! 4 π(l + m)!
eimφ^ sinm^ θ
( − d d cos θ
)l+m sin^2 l^ θ
(−1)l+m 2 ll!
√√ √√ (2l + 1)(l − m)! 4 π(l + m)! eimφ(1 − x^2 )m/^2 dl+m dxl+m^ (1 − x^2 )^2 l. (40)
In the last line, we used the variable x = cos θ.
2.2 Legendre Polynomials
Here are a few new notations. Legendre polynomials are defined by
Pl(x) = (−1)l 2 ll!
dl dxl^ (1 − x^2 )l^ (41)
P (^) lm (x) = (1 − x^2 )m/^2 dm dxm^ Pl(x). (42)
The definition Eq. (42) works only for m > 0. Another way to write P (^) lm is just putting them together,
P (^) lm (x) =
(−1)l 2 ll! (1 − x^2 )m/^2
dl+m dxl+m^ (1 − x^2 )l. (43)
This expression allows you to pick m < 0. P (^) lm is called associated Legendre polynomials, and P (^) lm (x) = Pl(x). The definitions look complicated, but they are just polynomials! Pl is a polyno- mial of order l. P (^) lm has this funny factor (1 − x^2 )m/^2 with a fractional power, but we will set x = cos θ in the end, and (1 − x^2 )m/^2 = sinm^ θ. Remember θ is the polar angle, and we only consider 0 ≤ θ ≤ π so that sin θ ≥ 0. Therefore, P (^) lm is a polynomial of order m in sin θ and l − m in cos θ. Now let us prove a surprising identity
P (^) l− m(x) = (−1)m^ (l − m)! (l + m)!
P (^) lm (x). (44)
Let us start with (^) dxdl+l+mm (1 − x^2 )l^ for m > 0 which appears in P (^) lm (x). We expand it out and see how it can be related to d l−m dxl−m^ (1^ −^ x (^2) )l
dl+m dxl+m^ (1 − x^2 )l^ = dl+m dxl+m^ (1 − x)l(1 + x)l
l∑+m
r=
l+mCr
( dr dxr^ (1 − x)l
) ( dl+m−r dxl+m−r^ (1 + x)l
)
. (45)
Even though the sum extends for 0 ≤ r ≤ l + m, the term in the first parentheses vanishes for r > l while the second for l + m − r > l. Therefore, the sum is taken only for m ≤ r ≤ l. Then
dl+m dxl+m^ (1 − x^2 )l
=
∑^ l r=m
l+mCr
(−1)rl! (l − r)!
(1 − x)l−r^ l! (r − m)!
(1 + x)r−m
l∑−m
s=
l+mCm+s
(−1)m+sl! (l − m − s)! (1 − x)l−m−s^ l! s! (1 + x)s. (46)
In the last line, I rewrote it with r = m + s. Now I multiply it by (1 − x^2 )m and divide by it,
dl+m dxl+m^ (1 − x^2 )l
=
(1 − x^2 )m
l∑−m s=
l+mCs+s
(−1)m+sl! (l − m − s)! (1 − x)l−s^
l! s! (1 + x)m+s
(1 − x^2 )m
l∑−m
s=
(l + m)! (m + s)!(l − s)!
(−1)m+sl! (l − m − s)! (1 − x)l−s^ l! s! (1 + x)m+s
(1 − x^2 )m
l∑−m
s=
(l + m)! s!(l − m − s)!
(−1)m+sl! (l − s)! (1 − x)l−s^ l! (m + s)! (1 + x)m+s
(l + m)! (l − m)!
(1 − x^2 )m
l∑−m
s=
(l − m)! s!(l − m − s)! (−1)m+s ( (−1)s^ ds dxs^ (1 − x)l−s
) ( dl−m−s dxl−m−s^ (1 + x)m+s
)
(l + m)! (l − m)!
(−1)m (1 − x^2 )m
l∑−m
s=
l−mCs
( ds dxs^
(1 − x)l−s
) ( dl−m−s dxl−m−s^
(1 + x)m+s
)
(l + m)! (l − m)!
(−1)m (1 − x^2 )m
dl−m dxl−m^ (1 − x^2 )l. (47)
Multiplying both sides of the equation by (−1) l 2 ll! (1^ −^ x
(^2) )m/ (^2) , we find
P (^) lm (x) = (−1)l 2 ll! (1 − x^2 )m/^2 dl+m dxl+m^ (1 − x^2 )l
(−1)l+l 2 l+ll!l! (−1)l+m
∫ (^1) − 1 dx(1 − x^2 )l
( dl+m dxl+m^ (−1)m(x^2 )m^ dl+m dxl+m^ (−1)l(x^2 )l
)
22 ll!l!
∫ (^1) − 1 dx(1 − x^2 )l
( dl+m dxl+m^ (x^2 )m^ (2l)! (l − m)! xl−m
)
22 ll!l!
∫ (^1) − 1 dx(1 − x^2 )l(l + m)! (2l)! (l − m)!
Change the variable to x = −1 + 2t, ∫ (^1) − 1
dxP (^) lm (x)P (^) lm (x) =
22 ll!l!
(2l)!(l + m)! (l − m)!
∫ (^1) 0
2 dt(2t(2 − 2 t))l
l!l!
(2l)!(l + m)! (l − m)!
∫ (^1) 0 dttl(1 − t)l
=
l!l!
(2l)!(l + m)! (l − m)!
Γ(l + 1)Γ(l + 1) Γ(2l + 2) =
(2l + 1)!
(2l)!(l + m)! (l − m)! =
2 l + 1
(l + m)! (l − m)!
2.3 More Conventional Expression
Compared to Eq. (40), we see that P (^) lm gives Y (^) lm ,
Y (^) lm = (−1)m
√√ √√ (2l + 1)(l − m)! 4 π(l + m)! eimφP (^) lm (cos θ). (53)
A special case of this is
Y (^) l^0 =
√ (2l + 1) 4 π
Pl(cos θ). (54)
When m < 0, an alternative expression is obtained by using the identity Eq. (44),
Y (^) lm =
√√ √√ (2l + 1)(l − |m|)! 4 π(l + |m|)! eimφP (^) l| m|(cos θ). (55)
In particular, this expression shows a relation
Y (^) lm = (−1)m(Y (^) l− m)∗. (56)
2.4 Orthonormality and Completeness
You can verify the orthonormality of spherical harmonics explicitly. Here and below, Ω refers to the polar coordinates θ, φ, and the integration volume is dΩ = d cos θdφ. ∫ dΩY (^) l m(Ω)(Y m ′ l′^ (Ω)) ∗
= (−1)m+m ′
√√ √√ (2l + 1)(l − m)! 4 π(l + m)!
√√ √√ (2l′^ + 1)(l′^ − m′)! 4 π(l′^ + m′)! ∫ (^1) − 1 d cos θ
∫ (^2) π 0 dφP (^) lm (cos θ)P m ′ l′^ (cos^ θ)e imφe−im′φ. (57)
Because φ integral vanishes unless m = m′,
= 2 πδm,m′
√√ √√ (2l + 1)(l − m)! 4 π(l + m)!
√√ √√ (2l′^ + 1)(l′^ − m)! 4 π(l′^ + m)!
∫ (^1) − 1
dxP (^) lm (x)P (^) lm′ (x)
= 2 πδm,m′
√√ √√ (2l + 1)(l − m)! 4 π(l + m)!
√√ √√ (2l′^ + 1)(l′^ − m)! 4 π(l′^ + m)!
2 l + 1
(l + m)! (l − m)! δl,l′
= 2 πδm,m′^ δl,l′ (2l + 1)(l − m)! 4 π(l + m)!
2 l + 1
(l + m)! (l − m)! δl,l′ = δm,m′^ δl,l′^. (58)
This is an explicit verification of the expected orthonormality
δm,m′^ δl,l′^ = 〈l′, m′|l, m〉 =
∫ dΩ〈l′, m′|θ, φ〉〈θ, φ|l, m〉 =
∫ dΩ(Y m ′ l′^ ) ∗Y m l.^ (59) The completeness relation is
δ^2 (Ω − Ω′) = 〈θ, φ|θ′φ′〉
∑ l,m
〈θ, φ|l, m〉〈l, m|θ′φ′〉
=
∑ l,m
Y (^) lm (θ, φ)(Y (^) lm (θ′, φ′))∗. (60)
Here,
δ^2 (Ω − Ω′) = δ(cos θ − cos θ′)δ(φ − φ′) = δ(θ − θ′) sin θ δ(φ − φ′). (61)
Then you look at x, y, z dependences to identify a particular orbital. Note that the dependence on r should not be taken seriously; it is supposed to be multiplied by a radial wave function anyway. We multiplied spherical harmonics by rl^ just to make to expressions become polynomials in x, y, z. From these expressions, it is clear that the pz orbital corresponds to Y 10 , while px to (Y 11 + Y 1 − 1 )/
2 and py to (Y 11 − Y 1 − 1 )/i
- dx (^2) −y 2 corresponds to (Y 22 + Y 2 − 2 )/
2, dyz to (Y 21 − Y 2 − 1 )/i
I’m sure you have seen “shapes” of spherical harmonics in textbooks. It is important to understand what they actually are. In many cases, what is shown is a surface given by points
r = |Y (^) lm (θ, φ)|^2. (74)
In other words, the distance of the surface from the origin along a direction is determined by the probability of finding the particle along that direction. They actually do not represent the “shapes” of the wave function. They just show along which direction the probability is big or small. In certain cases, though, these plots do represent the “shapes” of the actual wave function. Remember that the actual wave function has the radial wave function on top of the spherical harmonics. Suppose the radial wave function is a smoothly decaying function, say, e−r/a^0. Now you try to draw a surface of constant probability density in three dimensions. Then along the directions where |Y (^) lm |^2 is larger, the constant probability is attained even at higher r; but along the directions with small |Y (^) lm |^2 , you need to go closer to the origin to get the same probability density. Then the plot mentioned above can approximate the “shape” of the actual wave function. But if you want to interpret the plots in this manner, it obviously depends on the details of the radial wave function, and what value you chose for the surface of constant probability density. Just presenting the “shapes” of, say, pz orbitals independent of n (the principal quantum number) is therefore misleading.
2.6 Mathematica
It is useful to know that Mathematica has built-in commands for the spher- ical harmonics SphericalHarmonicY[l,m,θ,φ] for Y (^) lm (θ, φ), the Legendre polynomials LegendreP[n,x] for Pn(x), and the associated Legendre poly- nomials LegendreP[n,m,x] for P (^) nm (x).
