[L 2, L z ] = 0 It means we can find the common set of eigen function for L 2 and L z. Suppose we have eigen function α, m > such that,

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1 Angular Momentum For any vector operator V = {v x, v y, v z } For infinitesimal rotation, [V i, L z ] = i hɛ ijk V k For rotation about any axis, [V i, L z ]δφ j = i hɛ ijk V k δφ j We know, [V i, n.l]dφ n = i h( n V ) i dφ n [L, L z ] = 0 It means we can find the common set of eigen function for L and L z. Suppose we have eigen function α, m > such that, L α, m >= α α, m > L z α, m >= hm α, m > (assuming eigen values of L z are integer multiple of h.) To find eigen values of L : Lets invent, L + = L x + il y Also,L = (L + ) Now, L = L x il y [L z, L ± ] = [L z, L x ] ± i[l z, L y ] = i hl y ± i(i hl x ) = i hl y ± hl x Again, = ± hl ± L [L + α, m >] = L + L α, m >

2 So, = αl + α, m > L z [L + α, m >] = [L + L z + [l z, L + ]] α, m > = m hl + α, m >] + hl + α, m > = (m + ) hl + α, m > L ± α, m >= C ± (α, m) α, m ± > C is introduced to account normalization. In addition,m can t be arbitrary. It should be fixed within some limit. Here, Therefore, < α, m L L z α, m > =< α, m L x + L y α, m > = L x α, m > + L y α, m > whichis > 0 < α, m L L z α, m >= α m h > 0 The value of m is constrained by the this equation.m can t be arbitrarily positive or negative. It should follow that, Here, L + l, m max >= 0 L L + = L x + L y + i[l x + L y ] = L L z + i(i hl z ) So, Therefore, = L L z hl z L L z hl z l, m max >= 0 or, (α h m h m) l, m max >= 0 α = h m max (m max + )

3 calling m max = l Thus we obtain, = h L z l, m >= hm l, m > L l, m >= h l, m > This gives the condition for m max.similarly, we can find that,m min = l Hence, the range for m is l to +l. Now, For, l = 0, m = 0 l =, m =, l =, m =, 0, l = 3, m = 3,,, 3 We have l + eigen vectors for any given l. L z should give only integer value.in order to generalize integer value we invent J which describes rotations such that J has eigen function j, m j >. Now, and, J j, m j >= h j(j + ) j, m j > J z j, m j >= hm j j = o,,...n ; m j = j, j +,... + j Quantum mechanics tells us that there might be rotations of space that can t be defined by r p which exist in classical mechanics. Lets evaluate C ± : C ± = L + l, m > =< l, m L L + l, m > = h [ m m] or, C + = h m(m + ) Similarly, C = h m(m ) 3

4 And, In spherical coordinates, and from HW Set 9: x = rsinθcosφ, y = rsinθsinφ, z = rcosθ ( L + = he iφ θ + icotθ ) φ ( L = he iφ θ + icotθ ) φ L = h ( sin θ sin θ θ + θ sin θ ). φ We know that from which we can deduce that L + l, l = 0 Y l l (θ, φ) = ( ) l (l + )! 4π l l! sinl θe ilφ. All other Y m l s can be deduced from this by repeatedly applying L. An arbitrary state with given angular momentum quantum numbers l, m can be written where for m > 0, Y m l ψ(r, θ, φ) = R(r)Y m l (θ, φ) (θ, φ) = ( ) m l + (l m)! 4π(l + m)! Pl m (cosθ)e imφ and for m < 0, Y m l (θ, φ) = ( ) m l + (l m )! 4π(l + m)! P m l (cosθ)e imφ For a particle in a rotationally symmetric potential V (r) the Hamiltonian in spherical coordinates is H = h m + V (r) 4

5 with = ( r r r r + r sin θ Therefore, we can write the Hamiltonian as θ sin θ θ + r sin θ = L mr h D r m + V (r) where we define D r = r r r r. We are interested in eigenstates of the Hamiltonian ). φ or, h D r ψ m Assume ψ = R(r)Yl m (θ, φ). Also Hψ E,l,m = Eψ E,l,m + L ψ + V (r) = Eψ mr mr L Yl m = h m Y mr l (θ, φ) The radial equation is therefore, [ h h ] m D rr(r) + + V (r) R(r) = ER(r) mr Let R(r) = U(r) r Then, [ d dr r m ] [V (r) E] U(r) = 0. h For now, let s restrict ourselves to the case V (r) = 0 (it could be an arbitrary constant, but we can always set it to zero by redefining E). In this case, the Schrödinger equation reads [ d k dr r with k = m [V E]. h Now lets look at the limiting { cases: sinkr When r 0, for l = 0, U = coskr 5 ] U(r) = 0

6 We can ignore the nd possibility since it would lead to an overall wave function ψ = R(r)Y 0 0 (θ, φ) r at the origin but we know that the Laplacian of this is a delta function at the origin. So it cannot solve the Schrödinger equation unless the potential itself is a delta function. For l 0, r dominates over the potential term, so Its solutions are, U r U = 0. U r l+ U r l Again, the nd solution cannot be correct at the origin since the full wave function ψ = R(r)Yl m (θ, φ) could not be normalized: ψ = r dr d cos θ = r dr R(r) dφ R(r) Y m l (θ, φ) r dr r l+ which explodes at the origin. We already know what the solution for a free particle looks like in cartesian coordinates: ψ( r) = i p r e h (π h) 3/ To solve the radial equation in spherical coordinates, U = U r k U, we define ρ = kr so that U(r) = V (kr) = V (ρ). Then, U = k v = k U U ρ 6

7 or Introduce d l = ρ + l+ ρ d ( v dρ + and d l = ) v = 0. ρ + l+ ρ ρ. This yields d l d l = + ρ ρ d l d l = (l + )(l + ) + ρ ρ Now we can rewrite the equation for v: d l d l v l(ρ) = v l (ρ) Multiplying from the left with d l and regrouping the ladder operators we find ( ) d l d ld l v (l + )(l + ) l(ρ) = + d ρ ρ l v l(ρ) = d l v l(ρ) which is the equation for v l+ (ρ). We can therefore conclude that the ladder operator d l acting on v l (ρ) turns it into (a multiple of) the radial eigenstate for the next higher l. Thus, if we can start with the solution for l = 0, we can produce all higher-l solutions by repeated application of the ladder operator. We already { know that for l = 0 the solutions are sinρ v 0 (ρ) = cosρ If we want to find v l (ρ) then,we have to carry out the following: v l (ρ) = d l d l...d 0v 0 (ρ). We can rewrite this if we remind ourselves that we are really after the radial solutons { R(r) = U(r)/r = v(ρ)/r: sinkr = ksinkr = kj R E,l=0 (r) = r kr 0 (kr) coskr = kcoskr = kn r kr 0 (kr) In general { R E,l (r) = v l(ρ) jl (ρ) = k r n l (rho) where j l is the spherical Bessel function and n l is the Neumann function. We can show (using the expression involving the ladder operator) that ( ) l j l (ρ) = ( ρ) l j 0 (ρ) ρ ρ 7

8 ( ) l n l (ρ) = ( ρ) l n 0 (ρ) ρ ρ While we know that only the spherical Bessel functions can contribute at the origin, this more general solution is useful if we have to piece together a solution for a potential that has one value near the origin and a different value for r > a (particle in a spherical box of radius a). In that case, the j l contribute in the center and we have to use the continuity of the wave function and its derivative to match to a combination of Bessel and Neumann functions at r a. For a truly free particle, any value of E > 0 is allowed and there is exactly one solution for each combination E, l, m. On the other hand, if the particle is locked inside a rigid sphere (V (r) = for r > a), we have to require that R E,l (r) = 0 for r = a. This leads to only certain values of k and therefore E being permissible we once again get quantized energy eigenstates. For instance, for l = 0, we have to require ka = nπ with integer n. Finally, we can make the connection with the cartesian form of the free particle wave function - since the eigenstates in spherical coordinates must form a complete basis, we should be able to express the plane wave as a linear combination of solutions in spherical coordinates. For simplicity, let s assume that the wave travels along the z-direction, p = pẑ. Then ψ( r) = ipz e h (π h) 3/ = (π h) e ipr cos θ 3/ h which is already expressed in spherical coordinates. Since there is no φ dependence, we can assume that only Yl 0 s can contribute. Indeed, the plane wave can be written as ψ( r) = Σ l c l kj l (kr)y 0 l (θ, φ). 8