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Angular Momentum 1 Angular Momentum in Quantum Mechanics

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Angular Momentum 1 Angular Momentum in Quantum Mechanics J. Broida UCSD Fall 2009 Phys 130B QM II Angular Momentum 1 Angular momentum in Quantum Mechanics As is the case with most operators in quantum mechanics, we start from the clas- sical definition and make the transition to quantum mechanical operators via the standard substitution x x and p i~∇. Be aware that I will not distinguish a classical quantity such→ as x from the→− corresponding quantum mechanical operator x. One frequently sees a new notation such as ˆx used to denote the operator, but for the most part I will take it as clear from the context what is meant. I will also generally use x and r interchangeably; sometimes I feel that one is preferable over the other for clarity purposes. Classically, angular momentum is defined by L = r p . × Since in QM we have [xi,pj ]= i~δij it follows that [Li,Lj] = 0. To find out just what this commutation relation is, first recall that components6 of the vector cross product can be written (see the handout Supplementary Notes on Mathematics) (a b) = ε a b . × i ijk j k Here I am using a sloppy summation convention where repeated indices are summed over even if they are both in the lower position, but this is standard when it comes to angular momentum. The Levi-Civita permutation symbol has the extremely useful property that ε ε = δ δ δ δ . ijk klm il jm − im jl Also recall the elementary commutator identities [ab,c]= a[b,c] + [a,c]b and [a,bc]= b[a,c] + [a,b]c . Using these results together with [xi, xj ] = [pi,pj ] = 0, we can evaluate the com- mutator as follows: [L ,L ] = [(r p) , (r p) ] = [ε x p ,ε x p ] i j × i × j ikl k l jrs r s = εiklεjrs[xkpl, xrps]= εiklεjrs(xk[pl, xrps] + [xk, xrps]pl) 1 = ε ε (x [p , x ]p + x [x ,p ]p )= ε ε ( i~δ x p + i~δ x p ) ikl jrs k l r s r k s l ikl jrs − lr k s ks r l = i~ε ε x p + i~ε ε x p =+i~ε ε x p i~ε ε x p − ikl jls k s ikl jrk r l ikl ljs k s − jrk kil r l = i~(δ δ δ δ )x p i~(δ δ δ δ )x p ij ks − is jk k s − ji rl − jl ri r l = i~(δ x p x p ) i~(δ x p x p ) ij k k − j i − ij l l − i j = i~(x p x p ) . i j − j i But it is easy to see that ε L = ε (r p) = ε ε x p = (δ δ δ δ )x p ijk k ijk × k ijk krs r s ir js − is jr r s = x p x p i j − j i and hence we have the fundamental angular momentum commutation relation [Li,Lj]= i~εijkLk . (1.1a) Written out, this says that [Lx,Ly]= i~Lz [Ly,Lz]= i~Lx [Lz,Lx]= i~Ly . Note that these are just cyclic permutations of the indices x y z x. Now the total angular momentum squared is L2 = L L =→L L→, and→ therefore · i i 2 [L ,Lj] = [LiLi,Lj]= Li[Li,Lj] + [Li,Lj ]Li = i~εijkLiLk + i~εijkLkLi . But ε L L = ε L L = ε L L ijk k i kji i k − ijk i k where the first step follows by relabeling i and k, and the second step follows by the antisymmetry of the Levi-Civita symbol. This leaves us with the important relation 2 [L ,Lj ]=0 . (1.1b) Because of these commutation relations, we can simultaneously diagonalize L2 and any one (and only one) of the components of L, which by convention is taken to be L3 = Lz. The construction of these eigenfunctions by solving the differential equations is at least outined in almost every decent QM text. (The old book In- troduction to Quantum Mechanics by Pauling and Wilson has an excellent detailed description of the power series solution.) Here I will follow the algebraic approach that is both simpler and lends itself to many more advanced applications. The main reason for this is that many particles have an intrinsic angular momentum (called spin) that is without a classical analogue, but nonetheless can be described mathematically exactly the same way as the above “orbital” angular momentum. 2 In view of this generality, from now on we will denote a general (Hermitian) angular momentum operator by J. All we know is that it obeys the commutation relations [Ji, Jj ]= i~εijkJk (1.2a) and, as a consequence, 2 [J , Ji]=0 . (1.2b) Remarkably, this is all we need to compute the most useful properties of angular momentum. To begin with, let us define the ladder (or raising and lowering) operators J+ = Jx + iJy (1.3a) J = (J+)† = Jx iJy . − − Then we also have 1 1 Jx = (J+ + J ) and Jy = (J+ J ) . (1.3b) 2 − 2i − − Because of (1.2b), it is clear that [J 2, J ]=0 . (1.4) ± In addtion, we have [Jz, J ] = [Jz, Jx] i[Jz, Jy]= i~Jy ~Jx ± ± ± so that [Jz, J ]= ~J . (1.5a) ± ± ± Furthermore, 2 2 [Jz, J ]= J [Jz, J ] + [Jz, J ]J = 2~J ± ± ± ± ± ± ± and it is easy to see inductively that k k [Jz, J ]= k~J . (1.5b) ± ± ± It will also be useful to note 2 2 J+J = (Jx + iJy)(Jx iJy)= Jx + Jy i[Jx, Jy] − − − 2 2 ~ = Jx + Jy + Jz and hence (since J 2 + J 2 = J 2 J 2) x y − z 2 2 ~ J = J+J + Jz Jz . (1.6a) − − Similarly, it is easy to see that we also have 2 2 ~ J = J J+ + Jz + Jz . (1.6b) − 3 2 Because J and Jz commute they may be simultaneously diagonalized, and we β denote their (un-normalized) simultaneous eigenfunctions by Yα where 2 β ~2 β β ~ β J Yα = αYα and JzYα = βYα . Since Ji is Hermitian we have the general result J 2 = ψ J 2ψ = J ψ J ψ = J ψ 2 0 h i i h | i i h i | i i k i k ≥ 2 2 2 2 2 β ~2 2 β and hence J Jz = Jx + Jy 0. But Jz Yα = β Yα and hence we must have h i − h i h i h i≥ β2 α . (1.7) ≤ Now we can investigate the effect of J on these eigenfunctions. From (1.4) we have ± 2 β 2 β ~2 β J (J Yα )= J (J Yα )= α(J Yα ) ± ± ± so that J doesn’t affect the eigenvalue of J 2. On the other hand, from (1.5a) we also have± β ~ β ~ β Jz(J Yα ) = (J Jz J )Yα = (β 1)J Yα ± ± ± ± ± ± and hence J raises or lowers the eigenvalue ~β by one unit of ~. And in general, from (1.5b) we± see that k β k β ~ k β ~ k β Jz((J ) Yα ) = (J ) (JzYα ) k (J ) Yα = (β k)(J ) Yα ± ± ± ± ± ± so the k-fold application of J raises or lowers the eigenvalue of Jz by k units of ~. k β ± 2 This shows that (J ) Yα is a simultaneous eigenfunction of both J and Jz with corresponding eigenvalues± ~2α and ~(β k), and hence we can write ± k β β k (J ) Yα = Yα ± (1.8) ± where the normalization is again unspecified. β 2 ~2 ~ Thus, starting from a state Yα with a J eigenvalue α and a Jz eigenvalue β, we can repeatedly apply J+ to construct an ascending sequence of eigenstates with 2 Jz eigenvalues ~β, ~(β + 1), ~(β + 2), . , all of which have the same J eigenvalue ~2α. Similarly, we can apply J to construct a descending sequence ~β, ~(β 1), ~(β 2), . , all of which also have− the same J 2 eigenvalue ~2α. However, because− of (1.7),− both of these sequences must terminate. Let the upper Jz eigenvalue be ~βu and the lower eigenvalue be ~βl. Thus, by definition, − J Y βu = ~β Y βu and J Y βl = ~β Y βl (1.9a) z α u α z α − l α with βu βl J+Yα = 0 and J Yα = 0 (1.9b) − and where, by (1.7), we must have β2 α and β2 α . u ≤ l ≤ 4 By construction, there must be an integral number n of steps from βl to βu, so that − βl + βu = n . (1.10) (In other words, the eigenvalues of J range over the n intervals β , β + 1, z − l − l βl +2,..., βl + (βl + βu) = βu.) − Now, using− (1.6b) we have 2 βu βu 2 ~ βu J Yα = J J+Yα + (Jz + Jz)Yα . − β Then by (1.9b) and the definition of Yα , this becomes ~2 βu ~2 βu αYα = βu(βu + 1)Yα so that α = βu(βu + 1) . In a similar manner, using (1.6a) we have 2 βl βl 2 ~ βl J Yα = J+J Yα + (Jz Jz)Yα − − or ~2 βl ~2 βl αYα = βl(βl + 1)Yα so also α = βl(βl + 1) . Equating both of these equations for α and recalling (1.10) we conclude that n β = β = := j u l 2 where j is either integral or half-integral, depending on whether n is even or odd. In either case, we finally arrive at α = j(j + 1) (1.11) and the eigenvalues of J range from ~j to ~j in integral steps of ~.
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