Angular Momentum

5.61 Spherical Harmonics page 1 ANGULAR MOMENTUM Now that we have obtained the general eigenvalue relations for angular momentum directly from the operators, we want to learn about the associated wave functions. Returning to spherical polar coordinates, we recall that the angular momentum operators are given by: ⎛ ∂ ∂ ⎞ L̂x = −i� ⎜ − sin φ − cot θ cos φ ⎟ ∂θ ∂φ ⎠ ⎝ ∂ ∂ ⎞ ⎛ L̂y = −i� ⎜ cos φ − cot θ sin φ ∂θ ∂φ ⎟⎠ ⎝ ∂ L̂z = −i� ∂φ Lˆ2 = Lˆ2x + Lˆ2Y + Lˆ2z 1 ∂2 ⎤ ∂ ⎛ ∂ ⎞ 2 2 ⎡ 1 ˆ ⇒ L = −� ⎢ ⎜ sin θ ⎟+ ∂θ ⎠ sin 2 θ ∂φ 2 ⎥⎦ ⎣ sin θ ∂θ ⎝ In terms of these, our original Schrödinger Equation for rigid rotations was L̂2 m m ˆ HYl = Yl = ElYl m 2I 2 −� ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ m ⇒ Y θ , φ ) = ElYl m (θ , φ ) ⎜ sin θ ⎟+ 2 ⎢ 2⎥ l ( 2I ⎣ sin θ ∂θ ⎝ ∂θ ⎠ sin θ ∂φ ⎦ 2 where l was the quantum number for L̂ and m was the quantum number for L̂z . 2 Taking what we learned in the last section about the eigenvalues of L̂ and L̂z we can say that at most we can have l = 0, 12 ,1, 23 , 2,... m = −l, −l + 1,..., l We will see that there is an additional restriction on the possible values of l in the present case, but these are the possible values for the quantum numbers. In terms of the quantum numbers, we have the eigenvalue relations Lˆ2Yl m = L2Yl m = � 2l ( l +1) Yl m L̂zYl m = �mYl m Now, the functions, Ylm , that satisfy these relations for rigid rotations are called Spherical Harmonics. It is possible to derive the spherical harmonics by solving the 2D differential equation above. McQuarrie goes through a fairly complete derivation and we outline that solution in the appendix to these notes (below). The result is that: 5.61 Spherical Harmonics Yl m (θ , φ ) = Alm Pl where Alm is a normalization constant and m page 2 ( cos θ ) eimφ m Pl ( x ) is an associated Legendre Polynomial. The first few Associated Legendre Polynomials are: P00 ( cos θ ) = 1 P10 ( cos θ ) = cos θ ( 3cos θ −1) P11 ( cos θ ) = sin θ P20 ( cos θ ) = P21 ( cos θ ) = 3cos θ sin θ P22 ( cos θ ) = 3sin 2 θ 1 2 2 There are a number of important features of the Spherical Harmonics we can recognize simply by inspecting these solutions: • The wavefunctions factorize into a product of a function of θ and a function of φ. Yl m (θ , φ ) ∝ f (θ ) g (φ ) This result is very reminiscent of the result we found for separable Hamiltonians, which is somewhat surprising because the Hamiltonian certainly did not appear at first sight to be separable into a Hamiltonian for θ and a Hamiltonian for φ: Ĥ = −� 2 2I ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ sin θ + ≠ Ĥ θ + Ĥ φ ⎟ ⎢ sin θ ∂θ ⎜ 2 2⎥ ∂ θ sin θ ∂ φ ⎝ ⎠ ⎣ ⎦ However, physically it makes some sense for the motion along θ and φ to separate: we haven’t applied any potential that links them together so particles should be free to move along θ and φ independently, just as particles in a separable potential in x and y can move independently along those axes. The θ ­ φ cross terms above reflect the curvature of the 2D surface the particles are moving on. • It is easy to verify that these functions are eigenstates of L̂z : ∂ m ∂ m m Yl = −i� Alm Pl ( cos θ ) eimφ = �mAlm Pl ( cos θ ) eimφ = �mYl m ∂φ ∂φ 2 m They are also eigenfunctions of L̂ , as can be proven for any given Yl 2 m (after some algebra) by computing Lˆ Y and verifying that the result is L̂zYl m = −i� l just � l ( l +1) Yl . 2 • m We can now see why half­integer values of l are not allowed here. Recall that φ is the angle in the x­y plane and it varies from 0 to 2π. What should happen to Yl m (θ , φ ) when φ → φ + 2π ? Of course, the value of the wavefunction should not change because by incrementing φ by 2π: we’ve 5.61 Spherical Harmonics page 3 just moved the particle around in a full circle. Thus, for the wave function to be single­valued, we must have: Yl m (θ , φ ) = Yl m (θ , φ + 2π ) ⇒ Alm Pl m ( cos θ ) eimφ = Alm Pl m ( cos θ ) eim(φ +2π ) ⇒ eimφ = eim(φ +2π ) ⇒ eimφ = eimφ eim 2π ⇒ 1 = eim 2π ⇒ m = an integer • thus, m must be an integer, no matter what value of l we choose. However, since the minimal value for m is l, l must also be an integer. This continuity argument is the reason why half integer values of l are not allowed for rigid rotations. Note that there are a few interesting algebraic features of the spherical harmonics: 1) the φ part of the wavefunction does not depend on l 2) The lth order Legendre polynomial always involves sums of products of sines and cosines such that the sum of the sine and cosine powers is less than or equal to l 3) for m ≠ 0 the spherical Harmonics are complex and Ylm * = Yl−m . Thus, one can obtain two real functions from each ±m pair via Rlm = • • 1 2 (Y m l + Yl− m ) I lm = 1 i 2 (Y m l − Yl− m ) These features are helpful in trying to identify, when given an arbitrary function of the angles, which spherical harmonics might contribute to that function. Typically, the spherical Harmonics are associated with letters as you have seen in your previous chemistry courses. Thus, l=0 is ‘s’, l=1 is ‘p’, l=2 is ‘d’ …. In the absence of a potential, as is the case for rigid rotations, the spherical Harmonics are 2l+1-fold degenerate: l m Yl m ' s 2l + 1 0 0 Y00 1 1 ­1, 0, 1 Y1−1 , Y10 , Y11 3 2 ­2, ­1, 0, 1, 2 Y2−2 , Y2−1 , Y20 , Y21 , Y22 5 3 ­3, ­2, ­1, 0, 1, 2, 3 Y −3 , Y −2 , Y −1 , Y 0 , Y 1 , Y 2 , Y 3 3 3 3 3 3 3 3 7 5.61 Spherical Harmonics page 4 As discussed previously, we should expect this degeneracy to be broken if we apply a potential that is not spherically symmetric. In the presence of a potential, we expect these levels to be split. Thus, to summarize, for the spherical Harmonics we have: Yl m (θ , φ ) = Alm Pl m ( cosθ ) eimφ Lˆ2Yl m = L2Yl m = � 2l ( l + 1) Yl m l = 0,1, 2,3... Lˆ Y m = �mY m m = −l , −l + 1,..., l z l l APPENDIX: SOLVING FOR THE SPHERICAL HARMONICS We need to solve the differential equation ⎡ 1 ∂ ⎛ ∂ ⎞ 1 ∂2 ⎤ ⎢ sin θ ∂θ ⎜ sin θ ∂θ ⎟ + sin 2 θ ∂φ 2 ⎥ Y (θ , φ ) = EY (θ , φ ) ⎝ ⎠ ⎣ ⎦ ˆ (θ , φ ) = EY (θ , φ ) for Rigid rotations. Rearranging the Equation, This is HY �2 − 2I ∂ ⎛ ∂ ⎞ 2IE 2 ⎤ ∂2 ⎡ ⎢sin θ ∂θ ⎜ sin θ ∂θ ⎟ + � 2 sin θ ⎥ Y (θ , φ ) = − ∂φ 2 Y (θ , φ ) ⎝ ⎠ ⎣ ⎦ only θ only φ We’ve separated the variables, just as in the 3D harmonic oscillator. ∴ Try Define Y (θ , φ ) = Θ (θ ) Φ (φ ) β≡ 2IE �2 as a solution ( note β ∝ E ) ∂ ⎛ ∂ ⎞ ∂2 ⎡ 2 ⎤ ⎢sin θ ∂θ ⎜ sin θ ∂θ ⎟ + β sin θ ⎥ Θ (θ ) Φ (φ ) = − ∂φ 2 Θ (θ ) Φ (φ ) ⎝ ⎠ ⎣ ⎦ Dividing by Θ (θ ) Φ (φ ) and simplifying 5.61 Spherical Harmonics sin θ ∂ Θ (θ ) ∂θ page 5 ∂ ⎞ 1 ∂2 ⎛ 2 sin Θ + sin = − Φ (φ ) θ θ β θ ⎜ ⎟ ( ) ∂θ ⎠ Φ (φ ) ∂φ 2 ⎝ only θ only φ Since θ and φ are independent variables, each side of the equation must be equal to a constant ≡ m2. ∂2 Φ (φ ) = − m 2 2 Φ (φ ) ∂φ 1 ⇒ and sin θ ∂ Θ (θ ) ∂θ ⇒ I ∂ ⎞ ⎛ 2 2 ⎜ sin θ ⎟ Θ (θ ) + β sin θ = m ∂θ ⎠ ⎝ First solve for Φ (φ ) using I ∂ 2 Φ (φ ) = − m 2 Φ (φ ) 2 ∂φ Solutions are Φ (φ ) = Am e Boundary conditions imφ and A− m e − imφ ⇒ quantization Φ (φ + 2π ) = Φ (φ ) Am eim(φ +2π ) = Am eimφ ⇒ ∴ This is only true if and A− m e − im(φ +2π ) = A− m e − imφ eim( 2π ) = 1 and e − im( 2π ) = 1 m = 0, ± 1, ± 2, ± 3,.... m is the “magnetic” quantum number ∴ Φ (φ ) = Am eimφ m = 0, ± 1, ± 2, ± 3,.... II 5.61 Spherical Harmonics ∫ Normalization: 2π 0 Φ ∗ (φ )Φ (φ ) d φ = 1 1 imφ e 2π Φ (φ ) = ⇒ page 6 m = 0, ± 1, ± 2, ± 3,... Now let’s look at Θ (θ ) . Need to solve II sin θ ∂ Θ (θ ) ∂θ Change variables: x = cos θ Since 0 ≤ θ ≤ π Also ⇒ ∂ ⎞ ⎛ 2 2 ⎜ sin θ ⎟ Θ (θ ) + β sin θ = m ∂θ ⎠ ⎝ dx Θ (θ ) = P ( x ) = dθ − sin θ −1 ≤ x ≤ +1 sin 2 θ = 1− cos2 θ = 1− x 2 This equation turns out to be Legendre’s equation in terms of Θ: sin θ d dθ ⎡ dΘ ⎤ ⎢sin θ dθ ⎥ + ⎣ ⎦ (β sin2 θ − m2 ) Θ (θ ) = 0 which we can re­write: sin 2 θ d 2Θ dΘ θ θ + sin cos + β sin 2 θ − m 2 Θ (θ ) = 0 2 dθ dθ ) ( Let x = cosθ and Θ(θ ) = P (x ). 12 dΘ dP dx dP = = − sin θ = − 1− x 2 dθ dx dθ dx ) d 2Θ d = dθ 2 dθ ) ( dP dx 1 2 dP ⎤ ⎡ d Θ ⎤ ⎡ dx ⎤ d ⎡ 2 = − 1− x ⎢ ⎢ dθ ⎥ ⎢ dθ ⎥ dx dx ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ( ⎡ ⎤ 1 2 d 2P ⎥ ⎢ x dP 2 = − sin θ ⎢ ⎥ 1 2 dx − 1− x dx 2 ⎥ ⎢ 1− x 2 ⎣ ⎦ ( = −x ) ( ) dP d 2P + 1− x 2 dx dx 2 ( ) Substituting these results into Legendre’s equation gives 5.61 Spherical Harmonics ( 1− x 2 ) 2 page 7 d 2P 2 dP 2 2 2 − 2 x 1− x dx + β 1− x − m P ( x ) = 0 dx ( ( ) ( ) ) Divide by (1− x 2 ) to obtain the Legendre equation in a convenient form: ( 1− x 2 ) d 2P dP ⎡ m2 ⎤ x β − 2 + − ⎢ ⎥ P(x) = 0 dx ⎢⎣ dx 2 1− x 2 ⎥⎦ The solutions to this equation are known, but very messy. They are called the associated Legendre polynomials, Pl m (x) . Note that they only depend on |m| because the equation depends on m2: Pl m (x )= P (cosθ ) 1 P (cos θ )= (3cos θ − 1) 2 P (cos θ )= 3cos θ sin θ P (cos θ )= 3sin θ m l ( ) P (cos θ )= cos θ P (cos θ )= sin θ P00 cos θ = 1 0 2 0 1 2 1 2 1 1 2 2 2 etc. So Θ (θ )= A P lm l m ( ( ) ) (cosθ ) where Alm is the normalization constant ⇒ π 2 m Alm2 ∫ ⎡ Pl cos θ ⎤ sin θ dθ = 1 0 ⎣ ⎦ ( ) So now putting it all together: ψ lm (r0 ,θ , φ )= Yl m (θ , φ )= Θl (θ )Φ m (φ ) m Yl m ( ( ) ) 1 2 ⎡⎛ 2l + 1⎞ l − m !⎤ ⎥ Pl m cos θ eimφ θ ,φ = ⎢⎜ ⎟ ⎢⎝ 4π ⎠ l + m !⎥ ⎣ ⎦ ( ) 1 ⎡⎛ 2l + 1⎞ l − m !⎤ 2 ⎥ Alm = ⎢⎜ ⎢⎝ 2 ⎟⎠ l + m !⎥ ⎣ ⎦ These functions are the spherical harmonics. ( ) 5.61 Spherical Harmonics page 8 SPHERICAL HARMONICS SUMMARY ( ) m () () Yl m θ , φ = Θ l θ Φ m φ ( ( ) ) 1 ⎡⎛ 2l + 1⎞ l − m !⎤ 2 m ⎥ Pl m cos θ eimφ Yl θ , φ = ⎢⎜ ⎟ ⎢⎝ 4π ⎠ l + m !⎥ ⎣ ⎦ l = 0, 1, 2,... m = 0, ± 1, ± 2, ± 3,... ± l ( ) ( ) Yl m ’s are the eigenfunctions to Ĥψ = Eψ for the rigid rotor problem. 1 Y00 = ⎛ 5 ⎞2 2 Y20 = ⎜ ⎟ ( 3cos θ −1) ⎝ 16π ⎠ 1 12 ( 4π ) 1 1 ⎛ 3 ⎞2 Y10 = ⎜ ⎟ cos θ ⎝ 4π ⎠ ⎛ 15 ⎞ 2 Y2±1 = ⎜ ⎟ sin θ cos θ e ±iφ ⎝ 8π ⎠ 1 2 1 2 ⎛ 3 ⎞ Y11 = ⎜ ⎟ sin θ eiφ ⎝ 8π ⎠ ⎛ 15 ⎞ 2 ±2iφ Y2±2 = ⎜ ⎟ sin θ e ⎝ 32π ⎠ 1 ⎛ 3 ⎞2 Y1−1 = ⎜ ⎟ sin θ e −iφ ⎝ 8π ⎠ Yl m ’s are orthonormal: Krönecker delta Note: Switch ⎧1 ⎩0 δ ll′ = ⎨ ∫∫ Y (θ ,φ )Y (θ ,φ )sin θ dθ dφ = δ m ′∗ l′ if l = l′ if l ≠ l′ m l ⎧1 ⎩0 δ mm′ = ⎨ if m = m′ if m ≠ m′ ll ′ δ mm′ normalization orthogonality l → J conventional for molecular rotational quantum number (e.g.: l ( l + 1) ⇒ J ( J + 1) J = 0, 1, 2,... .)