Angular Momentum 1 Angular momentum in Quantum …

1 J. BroidaUCSD Fall 2009 Phys 130B QM IIAngular Momentum1 Angular Momentum in Quantum MechanicsAs 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 thestandard substitutionx xandp i~ . Be aware that I will not distinguisha classical quantity such asxfrom the corresponding Quantum mechanical operatorx. One frequently sees a new notation such as xused to denote the operator, butfor the most part I will take it as clear from the context what is meant.

2 I will alsogenerally usexandrinterchangeably; sometimes I feel that one is preferable overthe other for clarity , Angular Momentum is defined byL=r in QM we have[xi,pj] =i~ ijit follows that [Li,Lj]6= 0. To find out just what this commutation relation is, firstrecall that components of the vector cross product can be written (see the handoutSupplementary Notes on Mathematics)(a b)i= I am using a sloppy summation convention where repeatedindices are summedover even if they are both in the lower position, but this is standard when it comesto Angular Momentum .

3 The Levi-Civita permutation symbol has the extremelyuseful property that ijk klm= il jm im recall the elementary commutator identities[ab,c] =a[b,c] + [a,c]band [a,bc] =b[a,c] + [a,b] these results together with [xi,xj] = [pi,pj] = 0, we can evaluate the com-mutator as follows:[Li,Lj] = [(r p)i,(r p)j] = [ iklxkpl, jrsxrps]= ikl jrs[xkpl,xrps] = ikl jrs(xk[pl,xrps] + [xk,xrps]pl)1= ikl jrs(xk[pl,xr]ps+xr[xk,ps]pl) = ikl jrs( i~ lrxkps+i~ ksxrpl)= i~ ikl jlsxkps+i~ ikl jrkxrpl= +i~ ikl ljsxkps i~ jrk kilxrpl=i~( ij ks is jk)xkps i~( ji rl jl ri)xrpl=i~( ijxkpk xjpi) i~( ijxlpl xipj)=i~(xipj xjpi).

4 But it is easy to see that ijkLk= ijk(r p)k= ijk krsxrps= ( ir js is jr)xrps=xipj xjpiand hence we have the fundamental Angular Momentum commutation relation[Li,Lj] =i~ ijkLk.( )Written out, this says that[Lx,Ly] =i~Lz[Ly,Lz] =i~Lx[Lz,Lx] =i~ that these are just cyclic permutations of the indicesx y z the total Angular Momentum squared isL2=L L=LiLi, and therefore[L2,Lj] = [LiLi,Lj] =Li[Li,Lj] + [Li,Lj]Li=i~ ijkLiLk+i~ ijkLkLi= kjiLiLk= ijkLiLkwhere the first step follows by relabelingiandk, and the second step follows by theantisymmetry of the Levi-Civita symbol.

5 This leaves us withthe important relation[L2,Lj] = 0.( )Because of these commutation relations, we can simultaneously diagonalizeL2and anyone(and only one) of the components ofL, which by convention is takento beL3=Lz. The construction of these eigenfunctions by solving the differentialequations is at least outined in almost every decent QM text.(The old bookIn-troduction to Quantum Mechanicsby Pauling and Wilson has an excellent detaileddescription of the power series solution.) Here I will follow the algebraic approachthat is both simpler and lends itself to many more advanced applications.

6 Themain reason for this is that many particles have an intrinsicangular Momentum (calledspin) that is without a classical analogue, but nonetheless can be describedmathematically exactly the same way as the above orbital Angular view of this generality, from now on we will denote a general (Hermitian) Angular Momentum operator byJ. All we know is that it obeys the commutationrelations[Ji,Jj] =i~ ijkJk( )and, as a consequence,[J2,Ji] = 0.( )Remarkably, this is all we need to compute the most useful properties of begin with, let us define theladder(orraising and lowering)operatorsJ+=Jx+iJyJ = (J+) =Jx iJy.

7 ( )Then we also haveJx=12(J++J ) andJy=12i(J+ J ).( )Because of ( ), it is clear that[J2,J ] = 0.( )In addtion, we have[Jz,J ] = [Jz,Jx] i[Jz,Jy] =i~Jy ~Jxso that[Jz,J ] = ~J .( )Furthermore,[Jz,J2 ] =J [Jz,J ] + [Jz,J ]J = 2~J2 and it is easy to see inductively that[Jz,Jk ] = k~Jk .( )It will also be useful to noteJ+J = (Jx+iJy)(Jx iJy) =J2x+J2y i[Jx,Jy]=J2x+J2y+~Jzand hence (sinceJ2x+J2y=J2 J2z)J2=J+J +J2z ~Jz.( )Similarly, it is easy to see that we also haveJ2=J J++J2z+~Jz.( )3 BecauseJ2andJzcommute they may be simultaneously diagonalized, and wedenote their (un-normalized) simultaneous eigenfunctions byY whereJ2Y =~2 Y andJzY =~ Y.

8 SinceJiis Hermitian we have the general resulthJ2ii=h |J2i i=hJi |Ji i=kJi k2 0and hencehJ2i hJ2zi=hJ2xi+hJ2yi 0. ButJ2zY =~2 2Y and hence we musthave 2 .( )Now we can investigate the effect ofJ on these eigenfunctions. From ( ) wehaveJ2(J Y ) =J (J2Y ) =~2 (J Y )so thatJ doesn t affect the eigenvalue ofJ2. On the other hand, from ( ) wealso haveJz(J Y ) = (J Jz ~J )Y =~( 1)J Y and henceJ raises or lowers the eigenvalue~ by one unit of~. And in general,from ( ) we see thatJz((J )kY ) = (J )k(JzY ) k~(J )kY =~( k)(J )kY so thek-fold application ofJ raises or lowers the eigenvalue ofJzbykunits of~.

9 This shows that (J )kY is a simultaneous eigenfunction of bothJ2andJzwithcorresponding eigenvalues~2 and~( k), and hence we can write(J )kY =Y k ( )where the normalization is again , starting from a stateY with aJ2eigenvalue~2 and aJzeigenvalue~ ,we can repeatedly applyJ+to construct an ascending sequence of eigenstates withJzeigenvalues~ ,~( + 1),~( + 2), .. , all of which have the sameJ2eigenvalue~2 . Similarly, we can applyJ to construct a descending sequence~ ,~( 1),~( 2), .. , all of which also have the sameJ2eigenvalue~2.

10 However, becauseof ( ), both of these sequences must the upperJzeigenvalue be~ uand the lower eigenvalue be ~ l. Thus, bydefinition,JzY u =~ uY u andJzY l = ~ lY l ( )withJ+Y u = 0 andJ Y l = 0( )and where, by ( ), we must have 2u and 2l .4By construction, there must be an integral numbernof steps from lto u, sothat l+ u=n.( )(In other words, the eigenvalues ofJzrange over thenintervals l, l+ 1, l+ 2,.., l+ ( l+ u) = u.)Now, using ( ) we haveJ2Y u =J J+Y u + (J2z+~Jz)Y u .Then by ( ) and the definition ofY , this becomes~2 Y u =~2 u( u+ 1)Y u so that = u( u+ 1).