Angular Momentum We review some basic classical mechanics. Consider a classical point particle of mass µ whose position at time t is denoted by ~r(t) moving in a specified central potential V (r) (the potential depends only on r = |~r|) in three dimensions. The energy is given by 1 d~r d~r E = µ · + V (r) . (1) 2 dt dt dV The force is given by F~ = −∇~ V = −rˆ and points along the radial direction since V dr depends only on r; its magnitude depends on the magnitude of ~r. This is a central force. Clearly the torque ~r × F~ vanishes and hence, angular momentum is conserved. Since the angular momentum, ~r × ~p, is perpendicular to ~r, the constancy of the angular momentum implies that the radius vector of the particle, ~r lies in a plane perpendicular to the angular momentum. Thus angular momentum conservation reduces three-dimensional motion to motion in a plane. For the gravitational potential, closed orbits are ellipses. Choose the plane of motion to be the xy-plane and use polar coordinates. 1 E = µ (r ˙2 + r2φ˙2) + V (r) (2) 2 Note that in elementary parlance φ˙ is the angular velocity ω and the angular part of the 1 2 2 1 2 kinetic energy is 2 µr ω = 2 Iω where we have used the definition of moment of inertia. We re-express the angular part of the kinetic energy in terms of the conserved quantity, the orbital angular momentum. The magnitude of the angular momentum is given by µr2 φ˙ (or Iω). More explicitly1 ~r × ~p = µ ~r × ~v = µ rrˆ × (r ˙rˆ + rφ˙φˆ)=zµr ˆ 2φ˙ i.e., the angular momentum is orthogonal to the plane. Since L = µr2φ˙ the angular part of the kinetic energy can be rewritten2 as 1 L2 E = µ r˙2 + + V (r) . (3) 2 2µr2 This separation important result and occurs in free particle Hamiltonian in quantum mechanics. The energy describes a particle moving in one dimension (with the one - dimensional coordinate r restricted to non-negative values) in an effective potential given 1The velocity consists of the radial velocityr ˙ alongr ˆ plus a tangential velocity rφ˙ along φˆ. 2Substituting for φ˙ from the expression for L 2 1 1 L L2 µr2φ˙2 = µr2 = . 2 2 µr2 2µr2 1 by L2 V (r) = V (r) + . eff 2µr2 This allows us to provide a qualitative description of the motion. Draw figures in class. Angular momentum commutation relations We will need the commutation relations for coordinates and momenta of a particle in three dimensions. Consider a particle in three dimensions with the coordinate and momentum operators along the three directions denoted by rl and pj = − i~∂/∂rj . The notation r1 = x, r2 = y, and r3 = z allows a compact way of representing the relations. The commutation relations are [ri, rj] = 0 , [pi, pj] = 0 and [ri, pj] = i~δij (4) which state that the coordinates can be measured simultaneously, the momenta along the three directions can be measured simultaneously, so can a coordinate and a momentum component along a different direction. We will use x, y, and z or 1 and 2 and 3 for the three directions. We wish to find the commutation relations for the angular momentum operators de- fined in the text and in the lecture. Since L~ = ~r × ~p, we have Lx = ypz − zpy, Ly = zpx − xpz and Lz = xpy − ypx . Note that the order of the operators does not matter since x commutes with py etc., but it is conventional to place the differential operators to the right. Check that these are Hermitian operators. Finding the commutation relations is tedious; a little thought can reduce the labor. We obtain [Lx, Ly] = i~Lz , [Ly, Lz] = i~Lx , and [Lz, Lx] = i~Ly . Angular momentum in spherical coordinates We wish to write Lx, Ly, and Lz in terms of spherical coordinates. Recall that the gradient operator in spherical coordinates (See Griffiths “Classical Electrodynamics”) is 1 1 ∇~ =r∂ ˆ + θˆ ∂ + φˆ ∂ . r r θ r sin θ φ 2 You should be able to write this down from a simple geometrical picture of spherical coordinates. Using L~ = ~r × ~p → −i~~r × ∇~ and orthonormality of the unit vectorsr ˆ, θˆ, and φˆ we find 1 L~ = (−i~) φˆ ∂ − θˆ ∂ . θ sin θ φ We use the expressions for the unit vectors (obtained from geometry) xˆ =r ˆ sin θ cos φ + θˆ cos θ cos φ − φˆ sin φ yˆ =r ˆ sin θ sin φ + θˆ cos θ sin φ + φˆ sin φ zˆ =r ˆ cos θ − θˆ sin θ . Given L~ in spherical coordinates above and the cartesian unit vectors we obtain Lz using Lz =z ˆ· L~ . This is the simplest of the three components and only one term survives. We find Lz = −i~ ∂/∂φ . Give a simple geometrical interpretation: the angular momentum along z in classi- cal mechanics arises from rotation about the z=axis, i.e., from temporal variations in φ. The linear momentum along x in classical mechanics arises from time variation of the position x. Since in quantum mechanics px → −i~∂/∂x, one might naively expect Lz → −i~ ∂/∂φ and this expectation is checked by the calculations outlined above. Computing the scalar products yields ~ ∂ ∂ Lx = −i − sin φ ∂θ − cos φ cot θ ∂φ , ~ ∂ ∂ Ly = −i cos φ ∂θ − sin φ cot θ ∂φ , ~ ∂ Lz = −i ∂φ . A very useful operator is the total angular momentum operator defined by 2 ~ ~ 2 2 2 L = L · L = Lx + Ly + Lz . One can verify that 2 2 2 [L , Lx] = [L , Ly] = [L , Lz]=0 . (5) 2 The commutation relations imply that only two of the set of four operators{Lx, Ly, Lz, L } can have definite values (with no uncertainty) in a quantum mechanical state simultane- 2 ously; conventionally they are chosen to be L and Lz. Given the cartesian components we can compute an explicit expression for L2 in spher- ical polar coordinates. A straightforward calculation yields ∂2 ∂ 1 ∂2 L~ · L~ = −~2 + cot θ + (6) ∂θ2 ∂θ sin2 θ ∂φ2 1 ∂ ∂ 1 ∂2 = −~2 sin θ + . (7) sin θ ∂θ ∂θ sin2 θ ∂φ2 3 This is precisely the derivative combinations that appear in the Laplacian in spherical coordinates, We recall that the Laplacian in spherical polar coordinates is given by 1 ∂ ∂ 1 ∂ ∂ 1 ∂2 ∇2 = r2 + sin θ + . (8) r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2 Thus the kinetic energy operator (denoted by H0) in three-dimensional, single-particle quantum mechanics can be written as follows: Note that 2 ~p · ~p ~ 2 H0 = = − ∇ 2µ 2µ Substituting for the Laplacian we have 2 2 2 ~ 1 ∂ 2 ∂ ~ 1 ∂ ∂ 1 ∂ H0 = − r − sin θ + (9) 2µ r2 ∂r ∂r 2µ r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2 ~2 1 ∂ ∂ Lˆ2 = r2 + . (10) 2µ r2 ∂r ∂r 2µr2 We have restored the hat on L to remind you that it is an operator. This is an important identity and you should be familiar with it. The angular part Lˆ2 of the kinetic energy operator is . The classical calculation would have naturally 2µr2 suggested this form. 2 Since L and Lz commute they share a set of eigenvectors, i.e., we can find a set of functions f(θ,φ) such that L2 f(θ, φ) = λ ~2 f(θ, φ) Lz f(θ, φ) = ν ~ f(θ, φ) . Of course, the eigenvalues are not the same; they do not even have the same dimensions. We have denoted the corresponding eigenvalues by λ ~2 and ν ~. This way the dimensions are taken care of and λ and ν are numbers. We have to determine the eigenvalues and eigenvectors. Now it is easy to obtain the eigenfunctions of Lz: Lz fθ,φ) = ν ~f(r,θ,φ) . Since Lz = −i~ ∂/∂φ (as you remember!) θ is effectively a constant and we consider functions of φ only and set f(θ,φ) = g(φ). Canceling a factor of ~ we have d −i g(φ) = νφ ⇒ g(φ) ∝ eiνφ . dφ 4 Since under φ → φ + 2π the system returns back to the original position and we expect nothing to be changed. Therefore, we demand that ν is an integer which will be denoted by m. (This is a slippery argument and I will just let it slip by.) Thus the eigenfunctions of Lz are exp(imφ) with eigenvalue m~. Please remember this! We determine the eigenvalues and the eigenfunctions of L2 next. Canceling the con- stant factor ~2 the eigenvalue equation reads 1 ∂ ∂ 1 ∂2 − sin θ + f(θ,φ) = λf(θ,φ) . sin θ ∂θ ∂θ sin2 θ ∂φ2 m First the result: the eigenfunctions are the spherical harmonics denoted by Y` (θ,φ) and the eigenvalues are `(` + 1) ~2 where ` is a non-negative integer. That is to say λ = `(` + 1). The superscript m indicates the eigenvalue of Lz, m~. We use separation of variables to solve the equation. Let f(θ,φ) = Θ(θ)Φ(φ) and multiply the equation by sin2 θ and divide by ΘΦ and obtain 1 d2Φ sin θ d dΘ − = sin θ + λ sin2 θ . Φ dφ2 Θ dθ dθ The left-hand side and right-hand side are respectively functions of φ and θ only and so they have to be equal to the same constant for the equation to be satisfied for all values of the variables. We denote the separation constant by m2 we obtain Φ(φ) = eimφ. This is consistent with our result for the eigenfunctions of Lz. The other equation becomes d dΘ sin θ sin θ + `(` + 1) sin2 θ Θ − m2Θ=0 .