7. SPHERICAL HARMONICS 1

7. Spherical Harmonics by Christian Sch¨onenberger, Nov. 17, 2014

7.1 Angular momentum operator in spherical coordinates

Using vector calculus one can write the angular momentum operator in spherical coordinates. One first writes the gradient operator ∇~ in components: ∂ ∂ 1 ∂ 1 ∂ ∇~ = = ~e + ~e + ~e (1) ∂~r r ∂r φ rsin(θ) ∂φ θ r ∂θ

The three vectors ~er, ~eφ, ~eθ, are unit vectors of the spherical coordinate system. All three vectors depend on the angles φ and θ. This needs to be taken into account when are taken. The ~ ~ angular momentum is now obtained from L = r~er × (~/i)∇ as:  ∂ ∂ 1 ∂  L~ = ~ r(~e × ~e ) + (~e × ~e ) + (~e × ~e ) (2) i r r ∂r r θ ∂θ r φ sin(θ) ∂φ

Here, the first term is zero, ~er × ~eθ = ~eφ, and ~er × ~eφ = −~eθ. This leads to

 ∂ 1 ∂  L~ = ~ ~e − ~e (3) i φ ∂θ θ sin(θ) ∂φ

To extract from this equation the three components Lx, Ly, and Lz (in spherical coordinates), we have to express the unit vectors ~eφ and ~eθ in cartesian coordinates. This can be done as follows: 1 ∂ ~e = ~r = (−sin(φ), cos(φ), 0) φ rsin(θ) ∂φ (4) 1 ∂ ~e = ~r = (cos(θ)cos(φ), cos(θ)sin(φ), −sin(θ)) θ r ∂θ Adding this into equ. 3 yields the final result:

 ∂ 1 1  L = ~ −sin(φ) − cos(θ)cos(φ) x i ∂θ sin(θ) ∂φ  ∂ 1 1  L = ~ cos(φ) − cos(θ)cos(φ) (5) y i ∂θ sin(θ) ∂φ ∂ L = ~ z i ∂φ

2 7.2 Eigenvalues for Lz and L

The operator ∂ L = ~ (6) z i ∂φ from previous equation equ. 5 has the

χm(φ) = eimφ (7) with eigenvalues ~m. Since the functions should be unique, they must be 2π periodic in the angle φ. Hence, m must be an integer. One can in fact doubt this argument, because only the modulus is an observable. We see later below that this statement is almost correct. 7. SPHERICAL HARMONICS 2

Next, we look for the eigenfunctions of L2, which we call Y . These functions are defined by the equation 2 2 L Y = ~ λY (8) 2 2 where the eigenvalue of L has been abbreviated as ~ λ (~ has unit of angular momentum). Since 2 2 L and Lz commute, i.e. [L ,Lz] = 0, Y is at the same time an of Lz. The functions Y only depend on the two angles of the spherical coordinate system, the polar and azimuthal angle, θ and φ, respectively. They are called spherical harmonics (Kugelfunktionen in German) and they are distinguished by two quantum numbers, l and m, where l still needs to be derived. They can be decomposed in a product of two functions:

m m m Yl (θ, φ) = Ωl (θ)χ (φ) (9) m where the first part is the polar and the second the azimuthal function. Yl shall be normalized on the of unit radius. The goal is to derive that λ is given by l(l + 1) where l denotes the bounds for m given by the equation −l ≤ m ≤ l with m an integer. We have the following defining eigenvalue equations

m m LzYl = ~mYl (10) and 2 m 2 m L Yl = ~ λYl (11) Next we write 2 2 2 m 2 m (Lx + Ly + Lz)Yl = ~ λYl (12) and replace the Lz part with the respective eigenvalue leading to 2 2 m 2 2 m (Lx + Ly)Yl = ~ (λ − m )Yl (13) 2 2 Since the expectation value for Lj is larger than or equal zero, i.e. (Y,Lj Y ) = (LjY,LjY ) ≥ 0 ∀j, 2 2 we also have (Y, (Lx + Ly)Y ) ≥ 0 and therefore from equ. 13: λ ≥ m2 (14) This equation says that |m| is bound to some maximum value. Next, we define new operators: L+ := Lx + iLy (15)

L− := Lx − iLy (16) 2 Since [L ,Lj] = 0, we also have 2 [L ,L±] = 0 (17) One can easily show that the following also holds:

[Lz,L±] = ±~L± (18)

This follows simply by: [Lz,Lx + iLy] = [Lz,Lx] + i[Lz,Ly] = ~(iLy + Lx) = ~L+. Now let us look at the following combination:

2 m 2 m 2 m L (L±Yl ) = L±(L Yl ) = ~ λ(L±Yl ) (19) 2 m 2 In other words, since L and L± commute, the functions L±Yl are also eigenfunctions of L . How m do the functions L±Yl behave, when Lz is acting upon them? We write: m m Lz(L±Yl ) = ([Lz,L±] + L±Lz)Yl m = (±~L± + m~L±)Yl (20) m = ~(m ± 1)(L±Yl ) 7. SPHERICAL HARMONICS 3

m and hence, L±Yl is also eigenfunction of the operator Lz with the eigenvalue m ± 1. The operators L± are therefore so called ladder operators that increase or decrease the index m by one. Up to normalization factors N, N 0 we have: m+1 m Yl = N(L+Yl ) m−1 0 m (21) Yl = N (L−Yl )

We have already seen before in equ. 14 that |m| is bound to a maximum value. There exists therefore a maximum m and a minimum one, which we denote as mmax and mmin. In order for the ladder operators to terminate, we rquire

mmax mmin L+Yl = 0 and L−Yl = 0 (22)

2 Now we look at the product L+L− and L−L+: L+L− = (Lx + iLy)(Lx − iLy) = Lx − iLxLy + 2 2 2 2 2 2 2 iLyLx + Ly = Lx + Ly − i[Lx,Ly] = Lx + Ly + ~Lz = L − Lz + ~Lz. And similarly, we obtain: 2 2 m L−L+ = L − Lz − ~Lz. Applying these two results to the wavefunctions Yl yields:

mmin 2 2 mmin (L+L−)Y = ~ (λ − mmin + mmin)Y = 0 l l (23) mmax 2 2 mmax (L−L+)Yl = ~ (λ − mmax − mmax)Yl = 0 So we arrive at two algebraic equations:

mmin(mmin − 1) = λ (24) mmax(mmax + 1) = λ One can combine these two equations together and write

(mmax + mmin)(mmax − mmin + 1) = 0 (25)

Since mmax ≥ mmin, the second term cannot be zero. Hence, the first term must be zero, i.e. mmax + mmin = 0, or equally mmin = −mmax (26)

Due to the sequential generation with the , the difference mmax − mmin must be an integer. As a consequence mmax is ∈ N/2. We define mmax = l and call l the angular momentum . l is either an integer or a half integer. 1 The latter is for example realized by the of an electron, proton or neutron who all have a ‘self-momentum’ of l = 1/2. We confine ourselves to integer momentum, since only these are realized in the orbital motion, and discuss spin later. We have arrived at the very important result:

m = −l, −l + 1, . . . l − 1, l l ∈ N (27) m Just be surprised. We have derived this result without even knowing the details of the functions Yl . 2 2 Taken equ. 24, we further obtain for the eigenvalue of L /~ : λ = l(l + 1) (28) The final result reads: m m LzYl = ~mYl m = −l, l + 1, . . . l − 1, l 2 m 2 m (29) L Yl = ~ l(l + 1)Yl For a given angular momentum quantum number l, there are 2l + 1 different eigenfunctions which yield a definite momentum component Lz. The absolute value of the angular momentum, the length p of the vector in a classical picture, is ~ l(l + 1). The Lz component is a multiple of ~. Lx and Ly are not defined, that is have not a defined value, because they do not commute with Lz. The maximum p value of Lz is l~ ≤ ~ (l(l + 1). This is a consequence of the uncertainty relation. 1We now see that the argument in the bgining that m should be an integer was not fully correct. m can also be a half integer. 7. SPHERICAL HARMONICS 4

2 7.3 Eigenvalues for Lz and L

We have already derived Lx, Ly and Lz in spherical coordinates, see equ. 5. We obtain from these equations:  ∂ ∂   ∂ ∂  L = eiφ + icot(θ) L = e−iφ − icot(θ) (30) + ∂θ ∂φ − ∂θ ∂φ −l Since the ladder has to terminate at m = −l, we have L−Yl = 0. This can be expressed in the following differential equation: ∂Y −l l = lcot(θ)Y −l . (31) ∂θ l This equation has the solution −l l −ilφ Yl (θ, φ) = C(sin(θ)) e , (32) where C is the normalization constant. This constant is obtained by requiring: ZZ −l Yl (θ, φ)sin(θ)dθdφ = 1 . (33)

One obtains r (2l + 1)! 1 C = . (34) 4π l! 2l −l Now, we are in the position to generate all spherical waves starting with Yl and sequentially applying the raising operator L+. Each time L+ is used the obtained functions needs to be normalized, i.e. L+ does not protect the normalization. Using operator algebra, one can derive the respective normalization factor. This then yields:

m+1 1 m Yl = p L+Yl . (35) ~ (l − m)(l + m − 1)

The spherical functions can be found in many books. They are written in terms of so called associated m Legendre functions Pl (cos(θ)), which are obtained from the so called Lengendre polynomials: 1  d l P (x) = (x2 − 1)l . (36) l 2l l! dx The first five polynomials are P0(x) = 1

P1(x) = x

1 2 P2(x) = (3x − 1) 2 (37) 1 P (x) = (5x3 − 3x) 3 2 1 P (x) = (35x4 − 30x2 + 3) 4 8 The associates Legendre functions are then obtained as:

 d |m| P m(x) = (1 − x2)|m|/2 P (x) . (38) l dx l

m m −m Obviously the function Pl is even in the index m, i.e. Pl = Pl . Again, we give a few examples: 1 P 0(x) = (3x2 − 1) 2 3 1 p 2 (39) P2 (x) = 3x 1 − x 2 2 P2 (x) = 3(1 − x ) 7. SPHERICAL HARMONICS 5

Finally, the spherical harmonics are given by s 2l + 1 (l − |m|)! Y m(θ, φ) =  · eimφ P m(cos(θ)). (40) l 4π (l + |m|)! l

The factor  = (−1)m for m > 1 and  = 1 otherwise. For us, this factor is pure convention. We will not make use of it. It is best to visualize these functions on a sphere. It is a good idea to once google the term ‘spherical harmonics’ and look at the different representations! Here, some examples: 1 Y 0(θ, φ) = 0 4π r 3 Y 0(θ, φ) = cos(θ) 1 4π r 3 Y ±1(θ, φ) = ∓ sin(θ)e±iφ 1 8π (41) r 5 Y 0(θ, φ) = (3cos2(θ) − 1) 2 16π r 15 Y ±1(θ, φ) = ∓ sin(θ)cos(θ)e±iφ 2 8π r 15 Y ±2(θ, φ) = sin2(θ)e±2iφ 2 32π