Summary: angular momentum derivation

Transcription

1 Summary: angular momentum derivation L = r p L x = yp z zp y, etc. [x, p y ] = 0, etc. (-) (-) (-3) Angular momentum commutation relations [L x, L y ] = i hl z (-4) [L i, L j ] = i hɛ ijk L k (-5) Levi-Civita symbol: { ɛ ijk = + for even permutation of xyz for odd permutation (-6) In general, no simultaneous eigenstates of L x, L y, L z, L = L x + L y + L z, (-7) [L, L x ] = [L, L y ] = [L, L z ] = 0, (-8) simultaneous eigenstates of L and one component (L z ). Define, without loss of generality, simultaneous eigenstates l, m of L and L z such that L z l, m = mh l, m m magnetic quantum number (-9) L h l 0 quantum number of l, m = l(l + ) l, m (-0) total average momentum l, m l, m = δ ll δ mm, orthonormality (-) Raising and lowering operators Note. L ± preserves l. L ± = L x ± il y = L ± (-) [L, L ± ] = 0 (-3) XXI-

2 L ± l, m = l, m ±, from [L±, L z ] = hl ± (-4) Note. L ± increases (lowers) magnetic quantum number by. l, m ± l, m ± = L± l, m L ± l, m = h (l m)(l ± m + ) (-5) m l (-6) Since L + increases m by we need L + l, m max = 0 for some m max or l, m max + l, m max + = h (l m max )(l + m max + ) = 0 (-7) m max = l (-8) L l, m min = 0, for some m min (-9) l, m min l, m min = h (l + m min )(l m min + ) = 0 (-0) m min = l (-) since m max m min = integer (integer number of application of L + onto l, m min ). We need m max m min = l = integer. (l integer of half-integer.) State vector notation and wavefunctions In the decomposition of an arbitrary state ψ, in terms of energy eigenstates n, ψ = c n n (-) n c n = n ψ (-3) Similarly, we can calculate the projection of the state ψ onto the state where the particle is found with certainty at x and nowhere else, i.e., onto the eigenstate x 0 of the position operator with eigenvalue x 0, xˆ x 0 = x o x 0 (-4) (In position space, these states are δ-functions.) We can expand the wavefunction in terms of the continuum of eigenstates, ψ = dxc(x) x (-5) XXI-

3 Figure I: Decomposition of a state vector into basis vectors. where x is the position operator eigenstate with eigenvalue x, ˆx x = x x, and the expansion coefficients are given by, c(x) = x ψ. (-6) Since c(x)dx is the probability to find the particle within the interval [x, x + dx], we identify the expansion coefficients with the spatial wavefunction and write ψ = dx ψ(x) x, (-7) }{{}}{{}}{{} arbitrary state scalar coefficient x eigenstate ψ(x) = x ψ projection of ψ vector onto. (-8) position eigenstate x The wavefunction in position space ψ(x) is the set of expansion coefficients of the state ψ in terms of position eigenstates, it is the projection of the state ψ onto the position eigenstate where the particle is localized at x. Similarly, we can expand in terms of momentum eigenstates, ψ = dkφ (k) k, (-9) ψ (k) = k ψ. (-30) The wavefunction in momentum space is the set of expansion coefficients in terms of momentum eigenstates. Similarly, we have for eigenstates of angle θ, φ in polar coordinates (i.e., states where the particle is found with certainty in a direction XXI-3

4 specified by θ, φ, and nowhere else): ψ = dωc(θ, φ) θ, φ (-3) π π = dφ sin θdθc(θ, φ) θ, φ (-3) 0 π with the angular wavefunction 0 expansion coefficients in terms of angular eigenstates. = dφ d(cos θ)c(θ, φ) θ, φ (-33) 0 Y (θ, φ) = c(θ, φ) = θ, φ ψ, (-34) Figure II: Angles θ, φ in spherical coordinates. Wavefunction of angular momentum eigenstate l, m in angle representation The wavefunction corresponding to state l, m is Y lm (θ, φ) = θ, φ l, m (-35) XXI-4

5 without proof: by expressing L z = xp y +yp x etc. in polar coordinates and substituting = h we obtain the following operator expressions: p i i x i h L z =, i φ L ± = he ±iφ ± + i cot θ. θ φ (-36) (-37) The eigenequation for L z becomes θ, φ L z l, m = hm θ, φ l, m = hmy lm (θ, φ) h θ, φ L z l, m = θ, φ l, m i φ h = Ylm (θ, φ) i φ Y lm (θ, φ) = imy lm (θ, φ) φ (-38) (-39) (-40) (-4) (-4) This differential has the solution Y lm (θ, φ) = P lm (θ)e imφ The stretched state m = l is characterized by L + l, m = l = 0 or he iφ + i cot θ Y ll (θ, φ) = 0, θ φ e iφ + i cot θ P ll (θ)e ilφ = 0, θ φ θ l cot θ P ll(θ)e (l+)φ = 0, θ l cot θ P ll(θ) = 0, (-43) (-44) (-45) (-46) (-47) the solution of which is P ll (θ) = (sin θ) l. Consequently, Y ll (θ, φ) = C ll (sin θ) l e ilφ. (-48) As for the HO, the eigenstates for m < l can be found by applying L to Y ll : Y ll (θ, φ) = c(lˆ ) l m (sin θ) l e ilφ, (-49) XXI-5

6 where the operator Lˆ is given on p. XXI-5. These are the spherical harmonics, given by [ ] l + (l m)! Y lm (θ, φ) = ( ) m P m 4π (l + m)! l (cosθ)e imφ, for m 0 (-50) Y (θ, φ) = Y l, m lm, for m 0 (-5) where the P lm (cos θ) are the associated Legendre polynomials ( n ) m l m (l + m)! d P l m (u) = ( ) l+m (l u ) l, for m 0 (l m)! l l! du (-5) (l m)! P m P m (u) = ( ) m l l (u) (l + m)! (-53) The first spherical harmonics are: Y 00 = 4π l = 0 (-54) 3 Y = e iφ sin θ 8π 3 Y 0 = cos θ 8π l = (-55) 3 Y, = + e iφ sin θ 8π 5 Y = e iφ sin θ 3π 5 Y = e iφ sin θ cos θ 8π 5 Y 0 = (3 cos θ ) l = (-56) 6π 5 Y, = 8π e iφ sin θ cos θ 5 Y, = e iφ sin θ 3π XXI-6

7 (a) Y 00 (b) Y (c) Y 0 Figure III: Distance of displayed curve from origin in given direction indicates value of Y lm. Geometric interpretation of quantum mechanical feature of angular momentum Classically, we can prepare an object to have its angular momentu completely aligned along an axis, say, the z axis. Then we have classically (L z) cl = (L ) cl, and L x = L y = 0. In QM, L z and L x do not commute, which implies a Heisenberg uncertainty between them. Quantum mechanically, the largest z component of angular momentum in that we can produce for a given total angular momentum l is m = l, but l, m = L l, m = l = h l(l + ) (-57) > l, m = L z l, m = l (-58) = h l (-59) Consequently, some angular momentum must be pointing in some other direction: L x + Ly = L L z (-60) = h l(l + ) h l (-6) = lh (-6) = 0 (-63) So there is angular momentum lh pointing elsewhere. XXI-7

8 Let us analyze L x, L y in the stretched state m = l: L x m=l = l, m L x l, m (-64) = l, m (L + + L ) l, m (-65) = l, m l, m + + l, m l, m (-66) = 0 (-67) since states with different quantum numbers are orthogonal. So we have L x = L y = 0. (Similarly for L y.) Where, then, is the missing angular momentum? and similarly for L y : L x l=m = 4 l, m (L + + L ) l, m (-68) = l, m L + L + L L + + L L l, m (-69) = l, m L + L + L L + l, m 4 (-70) = l, m L L + hl z + L L hl z l, m (-7) 4 z z = l, m = l L L z l, m (-7) = l(l + ) h l h (-73) = l h (-74) L x = L y = l h (-75) Even though L x = L y = 0, some angular momentum is contained in the x- and y- components as uncertainty. Since l is constant, we can draw the following geometrical picture for angular momentum: Note. There is nothing special about the z-direction, we could prepare, e.g., a maximally oriented state m = l along x (or, in fact, any other direction) by a linear combination of l, m states, l l, m = l x = c m l, m z (-76) m= l XXI-8

9 Figure IV: For given state l, m, the angular momentum points somewhere along the circle that corresponds to the given m-value, but we cannot predict the direction, i.e., the L x and L y components. XXI-9