� 8.04 Quantum Physics Lecture XXISummary: angular momentum derivation L = r × p (21-1)Lx = ypz − zpy, etc. (21-2)[x,py]=0, etc. (21-3)Angular momentum commutation relations [Lx,LyhLz]= i¯(21-4) [Li,Lj ]= i¯(21-5) h�ijkLk Levi-Civita symbol: +1 for even permutation of xyz �ijk = (21-6) −1 for odd permutation In general, no simultaneous eigenstates of Lx, Ly, Lz, L2 = L2 x + L2 y + L2 z, (21-7) [L2,Lx]=[L2,Ly]=[L2,Lz]=0, (21-8) simultaneous eigenstates of L2 and one component (Lz). Define, without loss of generality, simultaneous eigenstates |l,m� of L2 and Lz such that Lzl,m� = mh¯l,m� m magnetic quantum number (21-9) ||→ �� L2h2l ≥ 0 quantum number of |l,m� =¯l(l + 1)|l,m�→ total average momentum (21-10) �l�,m�|l,m� = δll� δmm� , → orthonormality (21-11) Raising and lowering operators L= Lx ± iLy = L†(21-12) ±± [L2,L±] = 0 (21-13) Note. L± preserves l. Massachusetts Institute of Technology XXI-1 � � 8.04 Quantum Physics Lecture XXIL±|l,m� = |l,m �± 1�, from [L±,Lz]= �¯± (21-14) hLNote. L± increases (lowers) magnetic quantum number by 1. �l,m �± 1|l,m �± 1� = �L±l,m|L±l,m� =¯h2(l � m)(l ± m + 1) (21-15) (21-16) |m|≤ l Since L+ increases m by 1 we need L+|l,mmax� = 0 for some mmax or �l,m� max +1|l,m� max +1� =¯h2(l − mmax)(l + mmax + 1) = 0 (21-17) mmax = l (21-18) L−|l,mmin� =0, for some mmin (21-19) �l,m� min − 1|l,m� min − 1� =¯h2(l + mmin)(l − mmin + 1) = 0 (21-20) (21-21) mmin = −l since mmax − mmin = integer (integer number of application of L+ onto |l,mmin�). We need mmax − mmin =2l = 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� = cn|n� (21-22) n cn = �n|ψ� (21-23) 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 |x0� of the position operator with eigenvalue x0, xˆ|x0� = xo|x0� (21-24) (In position space, these states are δ-functions.) We can expand the wavefunction in terms of the continuum of eigenstates, |ψ� = dxc(x)|x� (21-25) Massachusetts Institute of Technology XXI-2 � � � � 8.04 Quantum Physics Lecture XXIFigure 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�. (21-26) 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�, (21-27) �� ��� �� arbitrary state scalar coefficient x eigenstate ψ(x)= �xψ� projection of |ψ� vector onto . (21-28) |→ 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�, (21-29) ψ˜(k)= �k|ψ�. (21-30) 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 polla coordinates (i.e., states where the particle is found with certainty in a direction Massachusetts Institute of Technology XXI-3 � 8.04 Quantum Physics Lecture XXIspecified by θ, φ, and nowhere else): |ψ� = dΩc(θ,φ)|θ,φ� (21-31) � 2π � π = dφ sin θdθc(θ,φ)|θ,φ� (21-32) �02π �0 −1 = dφ d(cos θ)c(θ,φ)|θ,φ� (21-33) 0 −1 with the angular wavefunction Y (θ,φ)= c(θ,φ)= �θ,φ|ψ�, (21-34) expansion coefficients in terms of angular eigenstates. 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 Ylm(θ,φ)= �θ,φ|l,m� (21-35) Massachusetts Institute of Technology XXI-4 � � � � � � � � � � 8.04 Quantum Physics Lecture XXIwithout proof: by expressing Lz = xpy +ypx etc. in polar coordinates and substituting = ¯h∂ we obtain the following operator expressions: pi i∂xi h¯∂ Lz = ,i ∂φ∂∂ L± =¯he±iφ ±∂θ + i cot θ∂φ . The eigenequation for Lz becomes �θ,φ|Lz|l,m� =¯hm�θ,φ|l,m� =¯hmYlm(θ,φ) �θ,φ|Lz|l,m� = = ∂ h¯∂i ∂φl, m�|�θ, φ¯h ∂ Ylm(θ, φ)i ∂φYlm(θ,φ)= imYlm(θ,φ)∂φ This differential has the solution Ylm(θ,φ)= Plm(θ)e imφ (21-36) (21-37) (21-38) (21-39) (21-40) (21-41) (21-42) (21-43) The stretched state m = l is characterized by L+l,m = l� =0 or |∂∂ heiφ¯+ i cot θYll(θ,φ)=0,∂θ ∂φ ∂∂ e iφ + i cot θPll(θ)e ilφ =0,∂θ ∂φ ∂ ∂θ − l cot θPll(θ)e(l+1)φ =0, ∂ ∂θ − l cot θPll(θ)=0, the solution of which is Pll(θ) = (sin θ)l . Consequently, Yll(θ,φ)= Cll(sin θ)l e ilφ . (21-44) (21-45) (21-46) (21-47) (21-48) As for the HO, the eigenstates for mpolynomials (1 − n2)− m 2 �l−m(l + m)!d(u)=(−1)l+m 2)lP m l (l − u, for m ≥ 0 (21-52)(l − m)! 2ll! du (l − m)!P mP −m(u)=(−1)m (l + m)! (u) (21-53)ll The first spherical harmonics are: ⎫ 1 ⎪ Y00 = ⎬√4π l = 0 (21-54) ⎭ ⎫ �3 Y11 = − e iφ sin θ 8π 3 ⎬Y10 = cos θ 8πl = 1 (21-55)�3 Y1,−1 =+ e−iφ sin θ 8π ⎭ �⎫ Y22 = 15 e 2iφ sin2 θ 32π 15 Y21 = − e iφ sin θ cos θ 8π�⎬ 5 Y20 = (3 cos2 θ − 1) l = 2 (21-56) 16π 15 Y2,−1 = − 8πe−iφ sin θ cos θ Y2,−2 = 15 e−2iφ sin2 θ ⎭ 32π Massachusetts Institute of Technology XXI-6 � 8.04 Quantum Physics Lecture XXI(a) |Y00|2 (b) |Y11|2 (c) |Y10|2 Figure III: Distance of displayed curve from origin in given direction indicates value of |Ylm|2 . Geometric interpretation of quantum mechanical featuur 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 (L2 z)cl =(L2)cl, and Lx = Ly = 0. In QM, Lz and Lx do not commute, which implies a Heisenberg uncertainty betwwee 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 =1|L2|l,m = l� =¯h2l(l + 1) (21-57) > �l,m =1|Lz2�l,m = l (21-58) =¯h2l2 (21-59) Consequently, some angular momentum must be pointing in some other direction: L2 + L2 = L2 − L2 (21-60) xy z =¯h2l(l + 1) − h¯2l2 (21-61) = lh¯2 (21-62) = 0 (21-63) So there is angular momentum √lh¯pointing elsewhere. Massachusetts Institute of Technology XXI-7 � 8.04 Quantum Physics Lecture XXILet us analyze Lx, Ly in the stretched state m = l: �Lx�m=l = �l,m|Lx|l,m� (21-64) 1 = �l,m|2(L+ + L−)|l,m� (21-65) 1�� =2 �l,m|l,m �+1� + �l,m|l,m �− 1� (21-66) = 0 (21-67) since states with different quantum numbers are orthogonal. So we have �Lx� = �Ly� = 0. (Similarly for Ly.) Where, then, is the missing angular momentum? 1 �Lx2 �l=m =4�l,m|(L+ + L−)2|l,m� (21-68) 1 =4�l,mL2 + L+L−L+ + L2 −�l,m (21-69) |+ − + L1 =4�l,m|L+L− + L−L+�l,m (21-70) 1 =4�l,m− L2 +¯hLz + L2 − L2 hLz�l,m (21-71) |L2 zz − ¯1 =2�l,m = l|L2 − L2 z|l,m� (21-72) =1 l(l + 1)¯h2 − l2h¯2 (21-73) 2 = lh¯2 (21-74) 2 and similarly for �L2 y�: �L2 x� = �L2 y� = lh¯2 (21-75) 2 Even though �Lx� = �Ly� = 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 maximaall 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 = cm|l,m�z (21-76) m=−l Massachusetts Institute of Technology XXI-8 8.04 Quantum Physics Lecture XXIFigure 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 Lx and Ly components. Massachusetts Institute of Technology XXI-9