9 Angular Momentum I. Classical analogy, take. 9.1 Orbital Angular Momentum

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1 9 Angular Momentum I So far we haven t examined QM s biggest success atomic structure and the explanation of atomic spectra in detail. To do this need better understanding of angular momentum. In brief: we ll find that eigenfunctions of atomic problem have structure ψ(r) = u n (r)y lm (θ, φ) (1) where u n (r) is soln. to radial S.-eqn. we examined in Week 4, and Y lm (θ, φ) is eigenfunction of angular momentum operator L. Degeneracy of atomic states within a given shell corresponding to principal quantum number n explained by angular momentum quantum numbers l, m. 9.1 Orbital Angular Momentum Classical analogy, take 1

2 L = r ˆp (2) In other words ˆL x = y ˆp z z ˆp y ˆL y = z ˆp x xˆp z (3) ˆL z = xˆp y y ˆp x (4) Note that since [y, ˆp x ] = 0, etc., antisymmetry property r ˆp = ˆp r is ok, meaning there is no ambiguity in taking the classical expression and replacing r and p by their operator forms everywhere. Before solving H-atom, we need to derive formal properties of L. First summarize these, then prove a few of the interesting ones. Need definition of antisymmetric permutation symbol ɛ ijk = 1 if ijk are in cyclic order, e.g. 123, 312,etc. 1 if ijk are in anticyclic order, e.g. 321, 132, etc. 0 if any two indices are equal So the components of L in this notation are ˆL i = ɛ ijk x j ˆp k (6) with implied summation convention (if I write a two repeated indices below, it means sum over them, even if I leave out the Σ). Summary of useful relations involving L 1. ˆL i = ˆL i All components of L Hermitian. 2. [ ˆL i, ˆL j ] = i hɛ ijk ˆLk e.g. [ ˆL x, ˆL y ] = i h ˆL z 3. [ ˆL i, x j ] = i hɛ ijk x k e.g. [ ˆL y, x] = i hz 4. [ ˆL i, ˆp j ] = i hɛ ijk ˆp k e.g. [ ˆL y, ˆp x ] = i hˆp z 5. [ ˆL i, ˆp 2 ] = 0 (5) 2

3 6. [ ˆL i, V (r)] = 0 7. [ ˆL 2, ˆL i ] = 0 where ˆL 2 ˆL 2 x + ˆL 2 y + ˆL 2 z Major point will be that due to 5),6),7), the eigenfctns of central force problems can be chosen as simultaneous eigenfunctions of H = h2 2m 2 + V (r) and ˆL 2 and any component of L, e.g. ˆLz. Pf. of ˆL i = ˆL i : e.g., ˆL x = (y ˆp z z ˆp y ) = ˆp z y ˆp y z (7) = y ˆp z z ˆp y = ˆL x Pf. of [ ˆL i, ˆL j ] = i hɛ ijk ˆLk, e.g. [ ˆL z, ˆL y ] = [(xˆp y y ˆp x ), (z ˆp x xˆp z )] = [xˆp y, z ˆp x ] [xˆp y, xˆp z ] [y ˆp x, z ˆp x ] + [y ˆp x, xˆp z ] = ˆp y z[x, ˆp x ] y ˆp z [ˆp x, x] (8) = i h(ˆp y z y ˆp z ) = i h ˆL x Does it work? Check sign ɛ 321 = 1 OK! Pf. of [ ˆL i, ˆp 2 ] = 0: First need general relation [A, BC] = ABC BCA = ABC BAC + BAC BCA = [A, B]C + B[A, C] (9) Now evaluate using 4): [L i, ˆp 2 ] = j [L i, ˆp 2 j] 3

4 = j ( [ ˆLi, ˆp j ]ˆp j ˆp j [ ˆL i, ˆp j ] ) Pf. of [ ˆL 2, ˆL i ] = 0. Use (9): = i hɛ ijk ˆp k ˆp j + i hɛ ijk ˆp j ˆp k = 0 (10) [ ˆL 2, ˆL i ] = [ ˆL i, ˆL 2 ] = [ ˆL i, ˆL 2 x] [ ˆL i, ˆL 2 y] [ ˆL i, ˆL 2 z] = ([ ˆL i, ˆL x ] ˆL x + ˆL x [ ˆL i, ˆL x ]) + (x y) + (x z) = i hɛ i1k ( ˆL k ˆLx + ˆL x ˆLk ) +... = 0 (11) since ɛ ijk is antisymmetric. 9.2 Eigenfunctions of ˆL z For a central force problem, V (r) = V (r), ˆL2, ˆLz, and Π all commute with H, so we can find a complete set of eigenfctns. of all 4 ops. First construct eigenfctns of ˆL z in polar coordinates, Notice that x = r sin θ cos φ y = r sin θ sin φ z = cos θ. ψ φ r,θ = ψ x x φ + ψ y y φ + ψ z z φ = ψ x ψ ( )r sin θ sin φ + r sin θ sin φ y = x ψ y y ψ x (12) Mutiply by i h to find i h ψ φ = (xˆp y y ˆp x )ψ, or 4

5 L z = i h φ (13) Eigenfctns of ˆL z, i.e. soln. to ˆL z ψ = αψ is ψ e iαφ/ h (14) Require ψ be single-valued fctn. of position, i.e. when φ φ+2π, better get same value back again. Thus we find new quantum number α = m h, or Eigenvalues of ˆL z therefore m h. ψ e imφ, m = 0, ±1, ±2... (15) 9.3 L as generator of rotations Q: How does wave fctn. ψ(r) change when we rotate coordinate system to new coordinates r? Define rotation to be around axis ˆn, through angle θ. A: ψ = e iθˆn L/ h ψ (16) where U = e iθˆn L/ h is an operator to be understood in terms of its Taylor expansion, U = 1 iθˆn L/ h (iθˆn L/ h)2 +. Note θˆn L is a Hermitian operator, so U is unitary, U U = 1. The operator L is referred to as the generator of intfinesimal rotations, see below. Check in special case: rotate around ẑ, 1st by infinitesimal angle δφ: ψ (r, θ, φ) = ψ(r, θ, φ δφ) ψ(r, θ, φ) δφ ψ(r, θ, φ) (17) φ Now use representation of ˆL z we just worked out: 5

6 ψ = (1 δφ )ψ φ (18) = (1 iδφ ˆL z / h)ψ (19) Now rotate by finite angle φ 0 : apply above tranformation n 1 times for small φ 0 /n: Recall ψ = (1 iφ 0 ˆL z n h )n ψ (20) lim (1 + n x/n)n = e x (21) = ψ = e iφ 0 ˆL z / h Generalize to infinitesimal rotation δφ around ˆn to get Eq. (16) 9.4 Eigenvalues of ˆL 2 (22) Program is to find complete set of eigenfunctions of both ˆL z and ˆL 2. (Could pick any component of L, ˆL z is conventional): First note ˆL 2 ψ = aψ (23) ˆL z ψ = bψ. (24) (ψ, ˆL 2 ψ) = (ψ, ˆL 2 xψ) + (ψ, ˆL 2 yψ) + (ψ, ˆL 2 zψ) (25) 0 (26) since each term (ψ, ˆL 2 i ψ) = ( ˆL i ψ, ˆL i ψ) = dx ˆL i ψ 2. Now can figure out something about a and b: (ψ, L 2 ψ) = (ψ, L 2 xψ) + (ψ, L 2 yψ) + (ψ, L 2 zψ) (27) a(ψ, ψ) 0 0 b 2 (ψ, ψ) 6

7 So we see a b 2. Ladder operators for angular momentum Define Note ˆL + = ˆL, etc. (check!) ˆL + = ˆL x + i ˆL y (28) ˆL = ˆL x i ˆL y (29) Since [ ˆL 2, ˆL i ] = 0 and [ ˆL i, ˆL j ] = i hɛ ijk ˆLk, find [L 2, L ± ] = 0 (30) [ ˆL z, ˆL ± ] = ± h ˆL ±. (31) Now proceed à la harmonic oscillator case apply ˆL + to Eq.(23): ˆL + ( ˆL 2 )ψ = a ˆL + ψ = ˆL 2 ( ˆL + ψ) (32) so ˆL + ψ is an eigenfctn. of ˆL 2 with eigenvalue a. Now apply to Eq.(24): Rearrange to get ˆL + ˆLz ψ = b( ˆL + ψ) (33) (34) [ ˆL +, ˆL z ]ψ + ˆL z ( ˆL + ψ) = h ˆL + ψ + ˆL z ( ˆL + ψ) (35) ˆL z ( ˆL + ψ) = (b + h)( ˆL + ψ), (36) which = ˆL + ψ is eigenfctn of ˆL z with eigenvalue b + h. Label simult. e fctns of ˆL 2 and ˆL z by ψ ab, then in general. ˆL + ψ ab = ψ a,b+ h (37) ˆL ψ ab = ψ a,b h (38) So we begin to get the picture ˆL ± move us up and down the ladder of ˆL z quantum numbers. Recalling the SHO, need to find out where the top and bottom of the ladder are! 7

8 Know b 2 a from above, so sequence b, b± h, b±2 h must terminate above and below, i.e. there exist b max and b min for each choice of a: ˆL + ψ a,bmax = 0, ˆLz ψ a,bmax = b max ψ a,bmax (39) ˆL ψ a,bmin = 0, ˆLz ψ a,bmin = b min ψ a,bmin (40) Now apply ˆL 2 to ψ a,bmax to find a. Convenient to have form ˆL 2 = ˆL 2 x + ˆL 2 y + ˆL 2 z = ˆL ˆL+ + i[ ˆL y, ˆL x ] + L 2 z (41) = ˆL 2 = ˆL ˆL+ + ˆL 2 z + h ˆL z (42) and similarly ˆL 2 = ˆL + ˆL + ˆL 2 z h ˆL z (43) Therefore using (42) ˆL 2 ψ a,bmax = aψ a,bmax = (b 2 max + hb max )ψ a,bmax (44) and same argument applied to ψ a,bmin using (43) gives ˆL 2 ψ a,bmin = aψ a,bmin = (b 2 min hb min )ψ a,bmin (45) Summarize: Difference of (46) and (47) is a = b 2 max + hb max (46) a = b 2 min hb min (47) 0 = (b max b min )(b max + b min ) + h(b max + b min ) (48) which has soln. only when b max = b min. Now recall b is eigenvalue of ˆL z, showed b = m h, m = 0, ±1, ±2 so must have b max b min = n h (49) 8

9 for some integer n > 0. (48) and (49) together now say b max = n h 2 l h (50) where I ve defined 2l to be the integer n in (49). In terms of this l, plug back into (46) and find a = h 2 l(l + 1). (51) Allowed values of quantum numbers which follow from commutation relations: ˆL 2 : h 2 l(l + 1), l = 0, 1 2, 1, 3, 2, (52) 2 ˆL z : m h, m = l, m = l, l + 1, l 1, l (53) But note for orbital angular momentum specifically we showed m = integer, so only integer values of l are allowed. Half-integer values will be encountered when we discuss spin. 9