8.04 Quantum Physics Lecture XX Angular momentum The eigenequation associated with angular momentum reads ˆL2Y (θ, φ) = 2mr 2EL(r)Y (θ, φ) = const · Y (θ, φ) (20-1) where 2mr2EL is the eigenvalue, and ˆL2 = −¯h2 � ∂2 ∂θ2 + cot θ ∂ ∂θ + 1 sin θ ∂2 ∂θ2 � (20-2) Similar to the HO problem, we can proceed in two ways. We can either: 1. solve the differential equation using some Taylor expansion. 2. we can take a more abstract operator approach. Here we will do the latter. (For the direct approach see Gasiorowicz, supplement 7-B, or F&T.) We analyze the commutation relations for the angular momentum operator Lˆ=ˆr × pˆ(20-3) Note. that since waves in orthogonal directions are independent, we have no Heisenbeer uncertainty restriction on, say x and py, and consequently the commutator is zero, [x,py] = 0. Let us calculate the commutator between different components of L: omit operator symbol [Lx,Ly]=[ypz − zpy,zpx − xpz] (20-4) = y[pz,z]px + x[z,py]py (20-5) h¯= ypx hxpy+ i¯(20-6) i = ih¯(xpy − ypx) (20-7) = i¯(20-8) hLz [Lx,LyhLz]= i¯(20-9) [Ly,LzhLx]= i¯(20-10) [Lz,LxhLy]= i¯(20-11) The fact that the different components of angular momentum do not commute means that it is not possible to find simultaneous eigenstates of, say, Lx and Lz, unless Lz = 0 for that state (see previous lecture). Massachusetts Institute of Technology XX-1 8.04 Quantum Physics Lecture XXWhat about L2? [Lz, L2]=[Lz,L2 ]+[Lz,L2] (20-12) xy= Lx[Lz,Lx]+[Lz,Lx]Lx + Ly[Lz,Ly]+[Lz,Ly]Ly (20-13) = i¯+ i¯hLyhLx(20-14) hLxLy hLyLx − i¯Lx − i¯Ly = 0 (20-15) This implies that one can find simultaneous eigenstates of L2 and one component of L2 and one component of L, e.g., Lz, but not of all components: Proof. (Direct proof by contradiction) For a simultaneous eigenstate |n� of Lx and Ly with Lx|n� = l1|n�, (20-16) Ly|n� = l2|n�. (20-17) we have [Lx,Ly]|n� =0= Lz|n� (20-18) and 1 l2|n� = Ly|n� = ih¯[Lz,Lx]|n� =0 → l2 = 0 (20-19) and similarly l1 = 0. Only for L = 0 can we have simultaneous eigenstates of Lx, Ly, Lz. In general, we can only have simultaneous eigenstates of L2 and Lz (or Lx or Ly, Lz by convention). Let us denote such an eigenstate by |l,m� with Lz|l,m� = mh¯|l,m� (20-20) L2|l,m� =¯h2l(l + 1)|l,m� (20-21) The reason for the strange definition of the quantum number l (or L2 eigenvalue h¯2l(l + 1)) will become apparent later. m, l are dimensionless numbers, since L = r × p has units of ¯h. We assume that the simultaneous eigenstates of L2 and Lz are normalized, �l�,m�|l,m� = δll� δmm� orthonormality for → angular momentum (20-22) eigenstates Massachusetts Institute of Technology XX-2 = �¯ ± 8.04 Quantum Physics Lecture XXRaising and lowering operators for angular momentum It is useful to define the following non-Hermitian operators (20-23) (20-24) (20-25) L± = Lx ± iLy L†= L+ − L†= L+− L+ and Lare Hermitian conjugate of each other (reminiscent of ˆa = xxˆ0 + i ppˆ0 ,− aˆ† = xˆ− i pˆ). To understand similar significance of these operators, let us analyze x0 p0 their commutation relations: [L2,L±]=0 (20-26) since [L2,Lx]=0, [L2,Ly] = 0. [L+,L−]=[Lx + iLy,Lx − iLy] (20-27) = −i[Lx,Ly]+ i[Ly,Lx] (20-28) = −2i[Lx,Ly] (20-29) = −2ii¯(20-30) hLz = 2¯hLz (20-31) [L+,L−] = 2¯hLz (20-32) [L±,Lz]=[Lx ± iLy,Lz] (20-33) =[Lx,Lz] ± i[Ly,Lz] (20-34) = −i¯hLx(20-35) hLy ± i(i¯) = �¯hLyhLx − i¯(20-36) = �h¯(Lx ± Ly) (20-37) hL(20-38) [L±,LzhL±]= �¯(20-39) We also note that L+L− =(Lx + iLy)(Lx − iLy) (20-40) = L2 x + L2 y − iLxLy + LyLx (20-41) = L2 − L2 z − i[Lx,Ly] (20-42) = L2 − L2 z +¯hLz (20-43) and similarly L= L2 hLz.−L+ − L2 z − ¯ Massachusetts Institute of Technology XX-3 | ±�. 8.04 Quantum Physics Lecture XX L+L− L−L+ = = L2 − L2 z + ¯hLz L2 − L2 z − ¯hLz As for the HO, we now proceed to analyze the range of allowed values for l, m: Since L2 = L2 + L2 + L2 and Lx, Ly, Lz are Hermitian operators, we have xyz �l,m|L2 |l,m� = �L†(l,m)|Lx(l,m)� = �Lx(l,m)|Lx(l,m)�≥ 0, (20-44) xxsimilarly for y, z, and consequently �l,m|L2|l,m�≥ 0 or 0 ≤�l,m|L2|l,m� =¯h2l(l + 1)�l,m|l,m� =¯h2l(l + 1). (20-45) Consequently, we can choose l ≥ 0. we define l� := −(l + 1), then (If l ≤−1, l(l +1) = −l�(l� + 1) and l� ≥ 0.) To understand the operators L±, let us define a new state |ψ±� := L±|l,m�, (20-46) and act on it with L2 . L2|ψ±� = L2L±|l,m� (20-47) = L±L2|l,m� (20-48) =¯h2l(l + 1)L±|l,m� (20-49) =¯h2l(l + 1)|ψ±�, (20-50) so |ψ±� is an eigenstate of L2 with the same quantum number l. Also we have Lz|ψ±� = LzL±|l,m� (20-51) =(L±Lz ± ¯±)|l,m� (20-52) hL=(mh¯± h¯)L±|l,m� (20-53) =(m ± 1)¯hL±|l,m� (20-54) =(m ± 1)¯hψ(20-55) This means that L±|l,m� is also an eignenstate of Lz, but with an eigenvalue (m±1)¯h that differs from the original one by one. Since m is the quantum number associated with the z component of angular momentum, we call m the azimuthal (or magnettic quantum number, while l is the quantum number associated with total angular momentum. L+ (L−) raises (lowers) the magnetic quantum number by one, while preserving the total angular momentum l. Let us calculate the length of |l,m �± 1� := L±|l,m�, (20-56) Massachusetts Institute of Technology XX-4 �� � � �� � ������������������ � 8.04 Quantum Physics Lecture XX l,m ± 1l,m ± 1� = �l,mthe unnormalized state vector.L�L±l,m� (20-57) ||L2 − L2 hLzl,m� (20-58) z � ¯|h2 = �l,m|=¯h2l(l + 1) − h¯2 2 2)�l,mm � ¯|l,m� (20-59)ml(l + 1) − m(m ± 1) (20-60) =¯h2=¯h2(l � m)(l ± m + 1) (20-61) Since the length squared of any vector must be non-negative, it follows that l(l + 1) − m(m ± 1) ≥ 0. (20-62) Consequently, 11 m(m ± 1) = m 2 ± m + (20-63) 4 − 4�211 − (20-64) m ±=24�211 + l = l + (20-65) ≤ l2 2 − 4 orm ±12≤+121= l + (20-66)2since l ≥ 0, m ≤ l, for m > 0 (20-67) and also − m ≤ l, for m ≤ 0. (20-68) Therefore, m is bounded both from above and from below: −l ≤ m ≤ l , l ≥ 0. (20-69) Since |ψ+� = L+|l,m� is also an eigenstate of L2 and Lz, but with new eigenvalue m� = m + 1, the bound on m is only consistent with this fact if L+|l,m� = 0 for some m. Consequently, with L+|l,m� = |(20-70)l,m +1� l,m +1l,m +1� (l − m)(l + m + 1). (20-72) 0= �|(20-71)=¯h2 Massachusetts Institute of Technology XX-5 8.04 Quantum Physics Lecture XXmmax = l (20-73) Similarly, for ketψ− = L−|l,m� we have mmin = −l. (20-74) Thus, we have a ladder of eigenvalues spaced by one, and connected by the raising and lowering operators L+ and L− m = −l, −l +1,...,l − 1,l , l ≥ 0 (20-75) This is only possible is l is integer or half integer. It turns out that half-integer Figure I: Ladder of eigenvalues for fixed l. values of l have no simple spatial representation, and correspond to an internal form of angular momentum called spin of the particle. Here we will restrict ourselves to orbital angular momentum, which requires l to be an integer. Massachusetts Institute of Technology XX-6