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CHM 532 Notes on Angular Momentum Eigenvalues and Eigenfunctions

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CHM 532 Notes on Angular Momentum Eigenvalues and Eigenfunctions CHM 532 Notes on Angular Momentum Eigenvalues and Eigenfunctions 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 coor- dinates 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) Proof: ˆ ˆ L+Yl,m = L−Yl,m = 0. (48) ˆ ˆ ˆ2 ˆ2 ˆ 2 hYl,m|L−L+|Yl,mi = hYl,m|L − Lz − h¯Lz|Yl,mi = [f(l) − m(m + 1)]¯h = 0. (49) ˆ ˆ ˆ2 ˆ2 ˆ 2 hYl,m|L+L−|Yl,mi = hYl,m|L − Lz +h ¯Lz|Yl,mi = [f(l) − m(m − 1)]¯h = 0. (50) Then f(l) = m(m + 1) = m(m − 1). (51) Solving for m in terms of m, we obtain m = −m. (52) 2 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.
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