CHM 532 Notes on Angular Momentum Eigenvalues and Eigenfunctions

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In your textbooks, the eigenfunctions and eigenvalues of the angular momentum operators are determined using diﬀerential equations methods. A more powerful approach is to solve the angular momentum eigenfunction problem using operator methods analogous to the creation and annihilation operators we have used in the harmonic oscillator problem. These notes provide details about the operator approach.

1 Preliminaries

We begin by reviewing the angular momentum operators and their commutation relations. The detailed derivation of these preliminary results can be found in your textbooks. Recall that the angular momentum is a vector with three components, and there is a Hermitian operator associated with each component h¯ ∂ ∂ ! Lˆ = y − z , (1) x i ∂z ∂y h¯ ∂ ∂ ! Lˆ = z − x , (2) y i ∂x ∂z and h¯ ∂ ∂ ! Lˆ = x − y . (3) z i ∂y ∂x Additionally, it is convenient to deﬁne the operator corresponding to the square of the length of the angular momentum vector ˆ2 ˆ2 ˆ2 ˆ2 L = Lx + Ly + Lz. (4) Because the angular momentum is a conserved quantity for systems having spherically sym- metric potentials, it is important to express these operators as well in spherical polar coordinates h¯ ∂ ∂ ! Lˆ = − sin φ − cot θ cos φ , (5) x i ∂θ ∂φ

1 h¯ ∂ ∂ ! Lˆ = cos φ − cot θ sin φ , (6) y i ∂θ ∂φ h¯ ∂ Lˆ = , (7) z i ∂φ and 1 ∂ ∂ 1 ∂2 ! Lˆ2 = −h¯2 sin θ + . (8) sin θ ∂θ ∂θ sin2 θ ∂φ2 It is important to recognize that in spherical polar coordinates, all the angular momentum operators are independent of the coordinate r, and the eigenfunctions must be functions only of the angular coordinates; i.e. θ and φ. Using the operator expressions in either coordinate system (it is most easy to use Carte- sian coordinates), it is possible to demonstrate the following commutators (see the text for details) ˆ ˆ ˆ [Lx, Ly] = ih¯Lz, (9) ˆ ˆ ˆ [Ly, Lz] = ih¯Lx, (10) ˆ ˆ ˆ [Lz, Lx] = ih¯Ly, (11) and ˆ ˆ2 [Li, L ] = 0 (12) for i = x, y or z. From the commutation relations, it is clear that we can know only one component of the angular momentum and the square of the angular momentum vector at ˆ the same time. Because Lz has a particularly simple form in spherical polar coordinates, we 2 ˆ choose to know Lz and L simultaneously; i.e. we seek simultaneous eigenfunctions of Lz and Lˆ2.

2 The Eigenvalues

In the case of the harmonic oscillator, we discovered the eigenvalues of the Hamiltonian by introducing creation and annihilation operators. In analogy, we make the following deﬁni- tions:

Deﬁnition 1 The step-up operator is deﬁned by ˆ ˆ ˆ L+ = Lx + iLy. (13)

Deﬁnition 2 The step-down operator is deﬁned as the adjoint of the step-up operator; i.e.

ˆ ˆ† ˆ ˆ L− = L+ = Lx − iLy. (14)

2 We next introduce and prove a series of lemmas from which we can extract the eigenvalues of ˆ ˆ2 Lz and L . We let {Yl,m} represent the common complete orthonormal set of eigenfunctions ˆ ˆ2 of Lz and L with m and l respectively the quantum numbers associated with each operator. We write ˆ LzYl,m = mhY¯ l,m (15) and ˆ2 2 L Yl,m = f(l)¯h Yl,m (16) where f(l) is some function of the l quantum number the explicit expression for which is derived below. It is easily veriﬁed that the units ofh ¯ are units of angular momentum, and we have included factors ofh ¯ explicitly in Eqs.(15) and (16). The orthonormality condition is written hYl,m|Yl0,m0 i = δl,l0 δm,m0 (17) We now introduce and provide proofs for the necessary lemmas.

Lemma 1 ˆ ˆ ˆ ˆ LzL+ = L+(Lz +h ¯). (18) Proof: ˆ ˆ ˆ ˆ ˆ LzL+ = Lz(Lx + iLy) (19) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = LxLz + [Lz, Lx] + i(LyLz + [Lz, Ly]) (20) ˆ ˆ ˆ ˆ ˆ ˆ = LxLz + ih¯Ly + i(LyLz − ih¯Lx) (21) ˆ ˆ ˆ ˆ ˆ ˆ ˆ = (Lx + iLy)Lz +h ¯(Lx + iLy) = L+(Lz +h ¯). (22) 2

Lemma 2 ˆ ˆ ˆ ˆ LzL− = L+(Lz − h¯). (23) Proof: ˆ ˆ ˆ ˆ ˆ LzL− = Lz(Lx − iLy) (24) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = LxLz + [Lz, Lx] − i(LyLz + [Lz, Ly]) (25) ˆ ˆ ˆ ˆ ˆ ˆ = LxLz + ih¯Ly − i(LyLz − ih¯Lx) (26) ˆ ˆ ˆ ˆ ˆ ˆ ˆ = (Lx − iLy)Lz − h¯(Lx − iLy) = L−(Lz − h¯). (27) 2

Lemma 3 ˆ L+Yl,m = c+(l, m)Yl,m+1 (28) where c+(l, m) is a function only of the quantum numbers l and m.

3 ˆ ˆ2 ˆ Proof: We must show that L+Yl,m is an eigenfunction of L and Lz with the value of the z ˆ ˆ2 quantum number increased by one unit ofh ¯. Notice that L+ commutes with L so that the ˆ l quantum number must be unaﬀected by the action of L+. Next ˆ ˆ ˆ ˆ Lz(L+Yl,m) = L+(Lz +h ¯)Yl,m (29) ˆ ˆ = L+(m + 1)¯hYl,m = (m + 1)¯h(L+Yl,m). (30) 2 Lemma 4 ˆ L−Yl,m = c−(l, m)Yl,m−1 (31) where c−(l, m) is a function only of the quantum numbers l and m. ˆ ˆ2 ˆ Proof: We must show that L−Yl,m is an eigenfunction of L and Lz with the value of the z ˆ ˆ2 quantum number decreased by one unit ofh ¯. Notice that L− commutes with L so that the ˆ l quantum number must be unaﬀected by the action of L−. Next ˆ ˆ ˆ ˆ Lz(L−Yl,m) = L−(Lz − h¯)Yl,m (32) ˆ ˆ = L−(m − 1)¯hYl,m = (m − 1)¯h(L−Yl,m). (33) 2 Lemma 5 f(l) ≥ m2. (34) Proof: We begin with ˆ ˆ ˆ ˆ hYl,m|L−L+|Yl,mi = hL+Yl,m|L+Yl,mi ≥ 0. (35)

Next, ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ˆ2 ˆ ˆ ˆ ˆ L−L+ = (Lx − iLy)(Lx + iLy) = Lx + Ly + i(LxLy − LyLx) (36) ˆ2 ˆ2 ˆ ˆ2 ˆ2 ˆ = Lx + Ly − h¯Lz = L − Lz − h¯Lz. (37) Then ˆ ˆ 2 2 2 2 2 hYl,m|L−L+|Yl,mi = f(l)¯h − m h¯ − h¯ m = [f(l) − m(m + 1)]¯h ≥ 0 (38) or f(l) − m(m + 1) ≥ 0. (39) Next, ˆ ˆ ˆ ˆ hYl,m|L+L−|Yl,mi = hL−Yl,m|L−Yl,mi ≥ 0. (40) Also, ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 ˆ2 ˆ ˆ ˆ ˆ L+L− = (Lx + iLy)(Lx − iLy) = Lx + Ly − i(LxLy − LyLx) (41) ˆ2 ˆ2 ˆ ˆ2 ˆ2 ˆ = Lx + Ly +h ¯Lz = L − Lz +h ¯Lz. (42)

4 Then ˆ ˆ 2 2 2 2 2 hYl,m|L+L−|Yl,mi = f(l)¯h − m h¯ +h ¯ m = [f(l) − m(m − 1)]¯h ≥ 0 (43) or f(l) − m(m − 1) ≥ 0. (44) Adding Eqs.(39) and (44), we obtain

f(l) − m2 ≥ 0 (45)

or f(l) ≥ m2. (46) 2

An implication of Eq.(46) is that for a given l, there exits a maximum value of m which we denote m and a minimum value of m which we denote m. The next lemma relates m to m.

Lemma 6 m = −m (47)

m = −m. (52)

From Eq.(49), we have f(l) = m(m + 1). (53) We deﬁne l = m; i.e. the maximum value of m for a given l, and we can write

f(l) = l(l + 1). (54)

In the next lemma, we identify what values l can have.

Lemma 7 The quantum number l can be an integer or a 1/2 integer.

5 Proof: We have the following series of equations ˆ L−Yl,l = c−(l, l)Yl,l−1 (55) ˆ L−Yl,l−1 = c−(l, l − 1)Yl,l−2 (56) . . ˆ L−Y (l, −l + 1) = c−(l, −l + 1)Yl,−l. (57) We connect m = l to m = −l by an integer number k of operations. Then l − (−l) = 2l = k (58) or k l = (59) 2 an integer or a half integer. 2 We summarize the results of these lemmas with the following set of equations and statements: ˆ LzYl,m = mhY¯ l,m (60) ˆ2 2 L Yl,m = l(l + 1)¯h Yl,m (61) −l ≤ m ≤ l (62) where m and l can either be integers or half integers. Notice for a given l, there are gl = 2l+1 different m-states. It is interesting that the algebraic commutation relations for the angular momentum operators allows both integer and half integer quantum numbers for the angular momentum. When we solve for the eigenfunctions in a subsequent section of these notes, we find that the half integer quantum numbers are excluded. However, the algebra admits half integer quantum numbers that, as we can expect, correspond to the spin quantum number. The spin of particles is derivable from a relativistic treatment of quantum theory and cannot be understood from the Schrödinger equation alone. We close this section with two final lemmas that give expressions for c+(l, m) and c−(l, m). Lemma 8 q c+(l, m) = l(l + 1) − m(m + 1)¯h. (63) Proof: ˆ ˆ ˆ ˆ 2 hYl,m|L−L+|Yl,mi = hL+Yl,m|L+Yl,mi = c+(l, m) (64) = [l(l + 1) − m(m + 1)]¯h2 (65) or q c+(l, m) = l(l + 1) − m(m + 1)¯h. (66) 2

6 Lemma 9 q c−(l, m) = l(l + 1) − m(m − 1)¯h. (67) Proof: ˆ ˆ ˆ ˆ 2 hYl,m|L+L−|Yl,mi = hL−Yl,m|L−Yl,mi = c−(l, m) (68) = [l(l + 1) − m(m − 1)]¯h2 (69) or q c−(l, m) = l(l + 1) − m(m − 1)¯h. (70) 2

3 The Angular Momentum Eigenfunctions

In the algebraic solution for the harmonic oscillator Hamiltonian eigenfunctions, the ground state eigenfunction is determined by first applying the annihilation operator to the ground state function and solving the resulting differential equation. The other eigenfunctions are determined by applying products of the creation operator to the ground state wavefunction. To find the eigenfunctions of the angular momentum operators, we can use a similar method starting with the step-up and step-down operators. In what follows, we sketch the treatment for the angular momentum eigenfunctions, but do not give full details. The detailed treatment given in the text using differential equation methods for determining the eigenfunctions is in some ways preferable. We begin with the equation ˆ L+Yl,l(θ, φ) = 0 (71) where we have explicitly expressed the eigenfunction as a function of the angular variables in spherical polar coordinates, the most convenient coordinate system for this problem. To ˆ continue, we need an expression for L+ in spherical polar coordinates. Using Eqs.(5), (6) and (13), we obtain ∂ ∂ ! Lˆ =he ¯ iφ + i cot θ (72) + ∂θ ∂φ so that ∂ ∂ ! + i cot θ Y (θ, φ) = 0. (73) ∂θ ∂φ l,l We can solve Eq.(73) using separation of variables. We write

Yl,l(θ, φ) = Θl,l(θ)Φl(φ) (74)

and substitute the separated form into Eq.(73). After rearrangement, we ﬁnd

1 dΘ (θ) 1 dΦ l,l = = s (75) cot θΘl,l(θ) dθ iΦl(φ) dφ

7 where s is the separation constant. We ﬁrst solve the diﬀerential equation for the variable φ dΦ l = is dφ (76) Φl or isφ Φl(φ) = Ae . (77) From Eq.(7) we know that the operator corresponding to the z-component of the angular ˆ momentum depends only on φ, and Eq.(77) must be an eigenfunction of Lz. Using Eq.(7) h¯ ∂ Lˆ Φ (φ) = lh¯Φ (φ) = Φ (φ) = sh¯Φ (φ) (78) z l l i ∂φ l l so that the separation constant s = l. We next normalize Φl

Z 2π Z 2π 2 ∗ 2 −ilφ ilφ A Φl (φ)Φl(φ) dφ = A e e dφ (79) 0 0 = A22π = 1 (80) or 1 A = √ . (81) 2π The value of l is determined using the appropriate boundary condition. Because φ is a cyclic coordinate, we must have Φl(φ) = Φl(φ + 2π) (82) or 1 1 √ eilφ = √ eil(φ+2π) (83) 2π 2π so that eil2π = 1 (84) so that l must be an integer. For orbital angular momentum, the values of l are restricted to l = 0, 1, 2,.... As indicated previously, the algebraic solution to the angular momentum eigenvalue problem admits half integer eigenvalues, but the boundary conditions on the diﬀerential equation exclude the half integer solutions. We next solve the θ-dependent part of Eq.(75). After rearrangement we have

dΘ l,l = l cot θ dθ (85) Θl,l cos θ = l dθ (86) sin θ d sin θ = l . (87) sin θ

8 Integrating both sides of Eq.(87) l Θl,l(θ) = B sin θ (88) where B is a normalization constant. We normalize using the equation (the extra factor of sin θ comes from the Jacobian)

Z π 2 Θl,l(θ) sin θ dθ = 1. (89) 0 After some algebra, we ﬁnd

"(2l + 1)!#1/2 Θ (θ) = sinl θ. (90) l,l 2(2ll!)2

Combining with the solution to the φ-dependent part of Eq.(75) we ﬁnally obtain

"(2l + 1)!#1/2 1 Y (θ, φ) = sinl θeilφ. (91) l,l 4π 2ll!

To obtain expressions for Yl,m(θ, φ) for m 6= l, we apply the step-down operator to Eq.(91). The result is identical to the result of the diﬀerential equation approach to ﬁnding the eigenfunctions that is discussed in your textbooks. The complete, orthonormal set of functions {Yl,m(θ, φ)} are called spherical harmonics, and it can be shown that the general expression for the spherical harmonics is given by

"(2l + 1)(l − |m|)!#1/2 dl+|m| Y (θ, φ) = (1 − y2)|m|/2 (1 − y2)leimφ (92) l,m 4π(l + |m|)! dyl+|m| where sin θ = (1 − y2)1/2. (93)