Spherical Harmonics
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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] = ih¯ ijkLk. (2) When using the position representation, the action of the angular mo- mentum on any state is given by a differential operator h¯ h¯ h~x|L~ |ψi = ~x × ∇h~ ~x|ψi = ~x × ∇~ ψ(~x). (3) i i Loosely, we can write h¯ L~ = ~x × ∇~ , (4) 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, h¯ L~ |~xi = −~x × ∇|~ ~xi. (5) i 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) 1 we can rewrite the derivatives using the chain rule, ∂r ∂rx ∂ry ∂rz ∂x ∂θ = ∂θx ∂θy ∂θz ∂y ∂φ ∂φx ∂φy ∂φz ∂z sin θ cos φ sin θ sin φ cos θ ∂x = r cos θ cos φ r cos θ sin φ −r sin θ ∂y . (7) −r sin θ sin φ r sin θ cos φ 0 ∂z We invert the matrix and find ∂ r sin2 θ cos φ sin θ cos θ cos φ − sin φ ∂ x 1 r 2 ∂y = r sin θ sin φ sin θ cos θ sin φ cos φ ∂θ . (8) r sin θ 2 ∂z r sin θ cos θ − sin θ 0 ∂φ Now the orbital angular momentum operators can be written in terms of spherical coordinates, h¯ ∂ ∂ ! L = − sin φ − cot θ cos φ , (9) x i ∂θ ∂φ h¯ ∂ ∂ ! L = cos φ − cot θ sin φ , (10) y i ∂θ ∂φ h¯ ∂ L = . (11) z i ∂φ It is useful to take the combinations h¯ ∂ ∂ ! L = e±iφ ±i − cot θ . (12) ± i ∂θ ∂φ Finally, " 1 ∂ ∂ ! 1 ∂2 # L~ 2 = −h¯2 sin θ + . (13) sin θ ∂θ ∂θ sin2 θ ∂φ2 The free-particle Hamiltonian is ~p2 −h¯2 −h¯2 ∂2 2 ∂ ! L~ 2 = (∂2 + ∂2 + ∂2) = + + . (14) 2m 2m x y z 2m ∂r2 r ∂r 2mr2 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 nh¯ for n ∈ Z. Therefore, Lz is also quantized as mh¯ 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. 2 2 Spherical Harmonics ~ 2 Now we look for eigenstates of L and Lz to find the Hilbert space for the orbital angular momentum. In any spherically symmetric systems, energy eigenstates can be given by produce wave functions of the form∗ m ψ(~x) = R(r)Yl (θ, φ). (15) In other words, we are parameterizing the position eigenbasis in terms of polar rather than Cartesian coordinates, h~x| = hr, θ, φ|. (16) The spherical harmonics are defined as the wave functions of angular mo- mentum eigenstates m Yl (θ, φ) = hθ, φ|l, mi. (17) Sakurai uses the notation h~n| and call them “direction eigenkets.” Clearly, the defintions of angular momemum eigenstates L~ 2|l, mi = l(l + 1)¯h2|l, mi, (18) Lz|l, mi = mh¯|l, mi, (19) translate to the differential equations " 1 ∂ ∂ ! 1 ∂2 # − h¯2 sin θ + Y m = l(l + 1)¯h2Y m, (20) sin θ ∂θ ∂θ sin2 θ ∂φ2 l l h¯ ∂ Y m = mhY¯ m. (21) i ∂φ l l The latter equation is easy to solve: the azimuth dependence of the spherical harmonics must be eimφ. But figuring out the polar angle dependence needs more work. The rest of the discussion here is on this issue. Both in the case of the harmonic oscillator and the Landau levels (energy levels of a charged particle in a uniform magnetic field), it was useful to write ∗This corresponds to the separation of variables in Hailton–Jacobi theory, where the action S is written as a sum of different pieces S(r) + S(θ, φ). Because the wave function corresponds to eiS/h¯ , it is natural for the wave function to be given by a product of different pieces. 3 down an equation for the ground state of the form a|0i = 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, −li = 0 because it cannot be lowered any more. Sakurai starts with |l, li instead. Of course you get the same result, but I find this way somewhat less confusing. In the position representation, we find h¯ ∂ ∂ ! 0 = hθ, φ|L |l, −li = e−iφ −i − cot θ Y −l(θ, φ) = 0. (22) − i ∂θ ∂φ l On the other hand, we know already that h¯ ∂ − lhY¯ −l = hθ, φ|L |l, −li = Y −l (23) l z i ∂φ l −l −l −ilφ and hence the azimuth dependence of Yl is Yl (θ, φ) = f(θ)e . Therefore, Eq. (22) becomes d ! −i + il cot θ f(θ) = 0. (24) dθ This equation is solved easily, by writing it as 1 df = l cot θdθ, (25) f 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 −l l −ilφ Yl (θ, φ) = c sin θe . (28) 4 The absolute value of c can be fixed by the normalization Z Z 1 Z 2π l 2 2 2l 1 = dΩ|Yl | = d cos θ dφ|c| sin θ. (29) −1 0 φ integral trivially gives a factor of 2π. The θ integral is most conveniently done using the variable x = cos θ, Z 1 1 = 2π|c|2 dx(1 − x2)l.. (30) −1 Writing 1−x2 = (1−x)(1+x), and further change the variable to x = −1+2t, Z 1 Z 1 1 = 2π|c|2 2dt (2t(2 − 2t))l = 2π|c|222l+1 dt tl(1 − t)l. (31) 0 0 The integral is nothing but the Beta function Z 1 Γ(p)Γ(q) B(p, q) = dt tp−1(1 − t)q−1 = . (32) 0 Γ(p + q) Therefore, Γ(l + 1)Γ(l + 1) l!l! 1 = 2π|c|222l+1 = 4π|c|222l . (33) Γ(2l + 2) (2l + 1)! We find s 1 (2l + 1)! |c| = . (34) 2ll! 4π 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 s 1 (2l + 1)! Y −l(θ, φ) = sinl θe−ilφ. (35) l 2ll! 4π −l m Now that we know Yl , we can keep acting L+ on it to obtain all Yl . Recall that the general discussion of angular momentum taught us that q q L+|l, mi = l(l + 1) − m(m + 1)|l, m + 1i = (l − m)(l + m + 1)|l, m + 1i. (36) 5 Then we find m 1 1 Yl (θ, φ) = q q h¯ (l + m)(l − m + 1) h¯ (l + m − 1)(l − m + 2) " !#l+m 1 1 h¯ iφ ∂ ∂ −l ··· q q e i − cot θ Yl (θ, φ) h¯ (2)(l + l − 1) h¯ (1)(l + l) i ∂θ ∂φ v u " !#l+m u (l − m)! ∂ ∂ = t eiφ + i cot θ Y −l(θ, φ) (37) (2l)!(l + m)! ∂θ ∂φ l Now the question is to bring it to a more usable form. The φ derivatives always give eigenvalues. But each time L+ acts, there is a factor of e+iφ and makes the eigenvalue of −i∂/∂φ increase one by one. Therefore, v s u ! 1 (2l + 1)!u (l − m)! d Y m(θ, φ) = t eimφ − (m − 1) cot θ l 2ll! 4π (2l)!(l + m)! dθ d ! d ! d ! − (m − 2) cot θ ··· + (l − 1) cot θ + l cot θ sinl θ. dθ dθ dθ v u ! 1 u(2l + 1)(l − m)! d = t eimφ − (m − 1) cot θ 2ll! 4π(l + m)! dθ d ! d ! d ! − (m − 2) cot θ ··· + (l − 1) cot θ + l cot θ sinl θ. dθ dθ dθ (38) Here we have cancelled factors ofh ¯ and i. The next trick we need is to write d ! 1 d + k cot θ = sink θ (39) dθ sink θ dθ Then v u ! 1 u(2l + 1)(l − m)! 1 d Y m(θ, φ) = t eimφ sin−(m−1) θ l 2ll! 4π(l + m)! sin−(m−1) θ dθ 1 d ! 1 d ! 1 d ! sin−(m−2) θ ··· sinl−1 θ sinl θ sinl θ sin−(m−2) θ dθ sinl−1 θ dθ sinl θ dθ 6 v u !l+m 1 u(2l + 1)(l − m)! 1 d = t eimφ sinm θ sin2l θ 2ll! 4π(l + m)! sin θ dθ v u !l+m 1 u(2l + 1)(l − m)! d = t eimφ sinm θ − sin2l θ 2ll! 4π(l + m)! d cos θ v l+m u l+m (−1) u(2l + 1)(l − m)! d = t eimφ(1 − x2)m/2 (1 − x2)2l.