Orbital angular momentum and the spherical harmonics

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1 Orbital angular momentum and the spherical harmonics March 8, 03 Orbital angular momentum We compare our result on representations of rotations with our previous experience of angular momentum, defined for a point particle as L = x p or, for a quantum system as the operator relationship ˆL = ˆx ˆp Notice that since ˆL i = ε ijk ˆx j ˆp k there is no ordering ambiguity: ˆx j and ˆp k commute as long as j k, and the cross product insures this. Computing commutators, we have [ˆLi, ˆL m ] = [ε ijk ˆx j ˆp k, ε mnsˆx n ˆp s ] Therefore, with = ε ijk ε mns [ˆx j ˆp k, ˆx n ˆp s ] = ε ijk ε mns ˆx j [ˆp k, ˆx n ˆp s ] + [ˆx j, ˆx n ˆp s ] ˆp k = ε ijk ε mns ˆx j [ˆp k, ˆx n ] ˆp s + ˆx n [ˆx j, ˆp s ] ˆp k = ε ijk ε mns i δ knˆx j ˆp s + i δ jsˆx n ˆp k = i ε ijk ε mksˆx j ˆp s + ε ijk ε mnj ˆx n ˆp k = i δ im δ js δ is δ jm ˆx j ˆp s δ im δ kn δ in δ km ˆx n ˆp k = i δ imˆx j ˆp j ˆx m ˆp i δ imˆx k ˆp k + ˆx i ˆp m = i ˆx m ˆp i + ˆx i ˆp m ε imn ˆLi = ε imn ε ijk ˆx j ˆp k = δ mj δ nk δ mk δ nj ˆx j ˆp k = ˆx m ˆp n ˆx n ˆp m this becomes [ˆLi, ˆL m ] = i ε imn ˆLn and we see that ˆL m satisfies the angular momentum commutation relations and must therefore admit l, m representations satisfying ˆL z l, m = m l, m ˆL l, m = l l + l, m

2 along with raising and lowering operators, ˆL ±. However, in this case we have an explicit coordinate representation for the operators. For the z-component, x ˆL 3 α = x ˆx ˆp ˆx ˆp α = i x y y x Similarly, the x- and y-components are x ˆL α = i y z z y x ˆL α = i z x x z so we may construct the raising and lowering operators, [ x ˆL ± α = i y z z y = i ± i z x x z [ ±iz x z y i x ± iy z ] ] This exposes an essential asymmetry between spinors and vectors. We have seen that 3-vectors may be represented a matrices in a complex, -dim spinor representation, there does not exist a similar representation of spinors using 3-dim coordinates. Thus, since orbital angular momentum operators may be written in a coordinate representation, we will see that they only admit integer j representations, so the states l, m only exist for integer l. To see this in detail, we need to change from Cartesian to spherical coordinates. Changing to spherical coordinates Here we rewrite ˆL z, ˆL ± and ˆL in spherical coordinates. The coordinate transformation and its inverse are given by and r = x + y + z x θ = tan + y r ϕ = tan y x We also need the derivative operators, x = r cos ϕ y = r sin ϕ z = r cos θ x i. Using the chain rule, we have x = r x r + θ x θ + ϕ x ϕ = r y y r + θ y θ + ϕ y ϕ = r z z r + θ z θ + ϕ z ϕ

3 Computing the partial derivatives, we start with the differential of r, dr = x r dx + y r dy + z r dz and read off Next, for θ, we take the differential of tan θ, r x = x r r y = y r r z = z r x + y tan θ = z cos θ dθ = x x + y z dx + y x + y z dy x + y z dz Then, since we have cos θ = z r dθ = xz x + y r dx + yz x + y r dy x + y r dz and read off the partials, θ x = xz x + y r θ y θ z = = x + y yz x + y Finally, we compute the differential of tan ϕ = y x, and use cos ϕ = and once again read off the partials r r x x +y cos ϕ dϕ = x dy y x dx dϕ = x x + y dy y x + y dx ϕ x ϕ y ϕ z = y x + y x = x + y = 0 3

4 Now, returning to the chain rule expansions, we substitute to find x = x r y z r + xz x + y r θ y x + y ϕ = cos ϕ r + r cos θ cos ϕ θ sin ϕ r ϕ = y r r + yz x + y r θ + x x + y ϕ = sin ϕ r + r cos θ sin ϕ θ + r = z r r x + y r θ = cos θ r r and we may substitute into the orbital angular momentum operators. θ cos ϕ 3 Orbital angular momentum operators in spherical coordiates Carrying out the coordinate substitutions, for ˆL 3 we have i x y y = i r cos ϕ sin ϕ x r + r cos θ sin ϕ θ + r +i r sin ϕ cos ϕ r + r cos θ cos ϕ θ r For the raising operator, we have while the lowering operator is = i ϕ ˆL + = z x + iz y x + iy z = r cos θ +r cos θ cos ϕ r + r cos θ cos ϕ θ r i sin ϕ r + i r cos θ sin ϕ θ + i r r cos ϕ + i sin ϕ cos θ r r θ ϕ sin ϕ ϕ cos ϕ ϕ cos ϕ sin ϕ = cos ϕ + i sin ϕ e iϕ r cos θ r + cos θe iϕ θ + eiϕ sin θ θ +i cos ϕ + i sin ϕ cos θ ϕ = e iϕ θ + icos θ ϕ ˆL = z x + iz y + x iy z ϕ ϕ 4

5 = r cos θ +r cos θ cos ϕ r + r cos θ cos ϕ θ r i sin ϕ r + i r cos θ sin ϕ θ + i r +r cos ϕ i sin ϕ cos θ r r θ sin ϕ ϕ cos ϕ ϕ Collecting these, Finally, since = re iϕ cos θ r re iϕ cos θ r e iϕ cos θ θ e iϕ sin θ cos θ + ie iϕ θ ϕ = e iϕ cos θ + ie iϕ θ ϕ ˆL + = e iϕ θ + icos θ ϕ ˆL = e iϕ θ + icos θ ϕ ˆL + ˆL = L L 3 + L 3 we have L = ˆL + ˆL + L 3 L 3 = e iϕ θ + icos θ = cos θ ϕ θ + i cos θ sin θ cos θ i θ +i cos θ ϕ + i ϕ = cos θ θ [ e iϕ θ + icos θ ϕ ϕθ + i θϕ cos θ sin θ +i cos θ = θ ϕ θ + cos θ sin θ sin θ + + cos θ sin θ + θ sin θ cos θ sin θ ϕ ϕ ϕ ϕ ] + i + i ϕ ϕ ϕ This last equation establishes the relationship between the spherical harmonics and the angular momentum states, because the Laplace equation in spherical coordinates is = r r + r r r + θ θ r sin θ ϕ = r r + r r r ˆL 5

6 and we know that the general solution for f r, θ, ϕ is given in terms of spherical harmonics, where the spherical harmonics satisfy θ f r, θ, ϕ = θ Y l m θ, ϕ l l=0 m= l A l r Y l m θ, ϕ + sin θ ϕ Y l m θ, ϕ + l l + Y l m θ, ϕ = 0 for integer l and m = l, l +,... + l, while the eigenstates of ˆL satisfy precisely the same equation, x ˆL l, m = l l + x l, m with ˆL given above. This shows that orbital angular momentum only describes integer j states. 4 Spherical harmonics We can now use the quantum formalism to find the spherical harmonics, Y l m θ, ϕ = θ, ϕ l, m. For any state α, we know the effect of ˆL z is given by so for an eigenstate, This is trivially integrated to give θ, ϕ ˆL z α = i θ, ϕ α ϕ θ, ϕ ˆL z l, m = m θ, ϕ l, m θ, ϕ ˆL z l, m = i θ, ϕ l, m ϕ i θ, ϕ l, m = m θ, ϕ l, m ϕ θ, ϕ l, m = e imϕ θ, ϕ l Furthermore, we know that the raising operator will anihilate the state with the highest value of m, ˆL + l, m = l = 0 In a coordinate basis, this translates to a differential equation, Setting x l = f l θ, we have 0 = θ, ϕ ˆL + l, l = e iϕ θ + icos θ = e iϕ θ + icos θ This is solved by f l = sin l θ, so we have, for m = l e ilϕ θ, ϕ l, l ϕ e ilϕ θ, ϕ l ϕ 0 = f l θ l cos θf l Y l l θ, ϕ = A ll e ilϕ sin l θ 6

7 Now we can find all other Y l m θ, ϕ by acting with the lowering operator, θ, ϕ ˆL l, m = l l + m m θ, ϕ l, m Inserting the coordinate expression for θ, ϕ ˆL l, m and solving for the next lower state, we have θ, ϕ l, m = thereby defining all Y l m θ, ϕ recursively. e iϕ l l + m m θ + icos θ e iϕ e imϕ = l l + m m θ + mcos θ ϕ θ, ϕ l, m θ l, m 7

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