Angular momentum. Quantum mechanics. Orbital angular momentum

Transcription

1 Angular momentum 1 Orbital angular momentum Consider a particle described by the Cartesian coordinates (x, y, z r and their conjugate momenta (p x, p y, p z p. The classical definition of the orbital angular momentum of such a particle about the origin is L = r p, giving L x = y p z z p y, (1 L y = z p x x p z, (2 L z = x p y y p x. (3 Let us assume that the operators (L x, L y, L z L which represent the components of orbital angular momentum in quantum mechanics can be defined in an analogous manner to the corresponding components of classical angular momentum. In other words, we are going to assume that the above equations specify the angular momentum operators in terms of the position and linear momentum operators. Note that L x, L y, and L z are Hermitian, so they represent things which can, in principle, be measured. Note, also, that there is no ambiguity regarding the order in which operators appear in products on the right-hand sides Eqs. (1 (3, since all of the products consist of operators which commute. The fundamental commutation relations satisfied by the position and linear momentum operators are [x i, x j ] = 0, (4 [p i, p j ] = 0, (5 [x i, p j ] = i h δ ij, (6 where i and j stand for either x, y, or z. Consider the commutator of the operators L x and L z : [L x, L y ] = [(y p z z p y, (z p x x p z ] = y [p z, z] p x + x p y [z, p z ] = i h ( y p x + x p y = i h L z.

2 Orbital angular momentum The cyclic permutations of the above result yield the fundamental commutation relations satisfied by the components of an angular momentum: These can be summed up more succinctly by writing [L x, L y ] = i h L z, (8 [L y, L z ] = i h L x, (9 [L z, L x ] = i h L y. (10 L L = i h L. (11 The three commutation relations (8 (10 are the foundation for the whole theory of angular momentum in quantum mechanics. Whenever we encounter three operators having these commutation relations, we know that the dynamical variables which they represent have identical properties to those of the components of an angular momentum (which we are about to derive. In fact, we shall assume that any three operators which satisfy the commutation relations (8 (10 represent the components of an angular momentum. Suppose that there are N particles in the system, with angular momentum vectors L i (where i runs from 1 to N. Each of these vectors satisfies Eq. (11, so that L i L i = i h L i. (12 However, we expect the angular momentum operators belonging to different particles to commute, since they represent different degrees of freedom of the system. So, we can write L i L j + L j L i = 0, (13 for i j. Consider the total angular momentum of the system, L = N i=1 L i. It is clear from Eqs. (12 and (13 that L L = L i i=1 = i h L j = j=1 L i = i h L. i=1 L i L i i=1 (L i L j + L j L i i,j=1

3 Orbital angular momentum L 2 y Consider the magnitude squared of the angular momentum vector, L 2 L 2 + Lz 2. The commutator of L 2 and L z is written It is easily demonstrated that [L 2, L z ] = [L 2 x, L z ] + [L 2 y, L z ] + [L 2 z, L z ]. (15 x + so [L 2 x, L z ] = i h (L x L y + L y L x, (16 [L 2 y, L z ] = +i h (L x L y + L y L x, (17 [L 2 z, L z ] = 0, (18 [L 2, L z ] = 0. (19 Since there is nothing special about the z-axis, we conclude that L 2 also commutes with L x and L y. It is clear from Eqs. ( 8 (10 and (19 that the best we can do in quantum mechanics is to specify the magnitude of an angular momentum vector along with one of its components (by convention, the z-component. It is convenient to define the shift operators L + and L : Note that L + = L x + i L y, (20 L = L x i L y. (21 [L +, L z ] = h L +, (22 Note, also, that both shift operators commute with L 2. [L, L z ] = + h L, (23 [L +, L ] = 2 h L z. (24

4 Eigenfunctions of orbital angular momentum Eigenfunctions of orbital angular momentum In Cartesian coordinates, the three components of orbital angular momentum can be written L x L y L z = i h y z z y = i h z x x z = i h x y y x using the Schrödinger representation. Transforming to standard spherical polar coordinates, 26 we obtain x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ, L x = i h sin ϕ θ L y = i h cos ϕ θ L z = i h ϕ. L 2 = Lx 2 + Ly 2 + Lz 2 + cot θ cos ϕ ϕ cot θ sin ϕ ϕ (27 (28 (29 (30 (31 (32 (33 from the above eqs. (30, (31, (32 and (34, we can obtain: L 2 = h 2 1 sin θ θ sin θ θ sin 2 θ ϕ 2. (34

5 Eigenfunctions of orbital angular momentum Quantum The eigenvalue problem for L 2 takes the form where Equation ψ(r, θ, ϕ L 2 ψ = λ h 2 ψ, (35 is the wave-function, and λ is a numbeṙ Let us write (r, θ, ϕ = R(r Y(θ, ϕ. (36 (35 reduces to 1 sin θ θ sin θ θ sin 2 θ ϕ 2 Y + λ Y = 0, (37 where use has been made of Eq. (34 As is well-known, square integrable solutions to this equation only exist when λ takes the values l (l + 1, where l is an integer. These solutions are known as spherical harmonics, and can be written Yl m (θ, ϕ = 2 l + 1 (l m! 4π (l + m! ( 1m e i m ϕ Pl m (cos ϕ, (38 where m is a positive integer lying in the range 0 m l. Here, P m l (ξ is an associated Legendre function satisfying the equation We define [ d (1 ξ 2 dpm l dξ dξ ] m2 1 ξ 2Pm l + l (l + 1 P m l = 0. Y m l = ( 1 m (Y m l, (40 (41 which allows m to take the negative values l m < 0. The spherical harmonics are orthogonal functions, and are properly normalized with respect to integration over the entire solid angle: π 2π 0 0 Y m l (θ, ϕ Y m l (θ, ϕ sin θ dθ dϕ = δ ll δ mm. (42

6 Eigenfunctions of orbital angular momentum The spherical harmonics also form a complete set for representing general functions of θ and ϕ. By definition, L 2 Y m l = l (l + 1 h 2 Y m l, (43 where l is an integer. It follows from Eqs. (32 and (38 that L z Y m l = m h Y m l, (44 where m is an integer lying in the range l m l. Thus, the wave-function (r, θ, ϕ = R(r Y m l (θ, φ, where R is a general function, has all of the expected features of the wave-function of a simultaneous eigenstate of L 2 and L z belonging to the quantum numbers l and m.