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. 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 .

2 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).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.

3 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. 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.

4 ( )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.( )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 .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.

5 ( )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~.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).

6 , all of which also have the sameJ2eigenvalue~2 . 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).In a similar manner, using ( ) we haveJ2Y l =J+J Y l + (J2z ~Jz)Y l or~2 Y l =~2 l( l+ 1)Y l so also = l( l+ 1).Equating both of these equations for and recalling ( ) we conclude that u= l=n2:=jwherejis either integral or half-integral, depending on whethernis even or either case, we finally arrive at =j(j+ 1)( )and the eigenvalues ofJzrange from ~jto~jin integral steps of~.

7 We can now label the eigenvalues ofJzby~minstead of~ , where the integer orhalf-integermranges from jtojin integral steps. Thus our eigenvalue equationsmay be writtenJ2 Ymj=j(j+ 1)~2 YmjJzYmj=m~Ymj.( )We say that the statesYmjare Angular Momentum eigenstates with Angular mo-mentumjandz-component of Angular momentumm. Note that ( ) is nowwrittenJ+Yjj= 0 andJ Y jj= 0.( )5 Since (J ) =J , using equations ( ) we havehJ Ymj|J Ymji=hYmj|J J Ymji=hYmj|(J2 J2z ~Jz)Ymji=~2[j(j+ 1) m2 m]hYmj|Ymji=~2[j(j+ 1) m(m 1)]hYmj| know thatJ Ymjis proportional toYm 1j. So if we assume that theYmjarenormalized, then this equation implies thatJ Ymj=~pj(j+ 1) m(m 1)Ym 1j.( )If we start at the top stateYjj, then by repeatedly applyingJ , we can construct allof the statesYmj. Alternatively, we could equally well start fromY jjand repeatedlyapplyJ+to also construct the us see if we can find a relation that defines theYmj. SinceYjjis definedbyJ+Yjj= 0, we will only define our states up to an overall normalization ( ), we haveJ Yjj=~pj(j+ 1) j(j 1)Yj 1j=~p2j Yj 1jorYj 1j=~ 11 2jJ we have(J )2 Yjj=~2p2jpj(j+ 1) (j 1)(j 2)Yj 2j=~2p(2j)2(2j 1)Yj 2jorYj 2j=~ 21p(2j)(2j 1)2(J ) once more should do it:(J )3 Yjj=~3p(2j)(2j 1)2pj(j+ 1) (j 2)(j 3)Yj 3j=~3p(2j)(2j 1)(2)(3)(2j 2)Yj 3jorYj 3j=~ 31p2j(2j 1)(2j 2)3!

8 (J ) thatm=j 3 so that 3! = (j m)! and 2j 3 = 2j (j m) =j+m, itis easy to see we have shown thatYmj=~m js(j+m)!(2j)!(j m)!(J )j mYjj.( )6 And an exactly analogous argument starting withY jjand applyingJ+repeatedlyshows that we could also writeYmj=~ m js(j m)!(2j)!(j+m)!(J+)j+mY jj.( )It is extremely important to realize that everything we havedone up to thispoint depended only on the commutation relation ( ), andhence applies to bothinteger and half-integer Angular momenta. While we will return in a later section todiscuss spin (including the half-integer case),for the rest of this section we restrictourselves to integer values of Angular Momentum , and hence we will be discussingorbital Angular next thing we need to do is to actually construct the Angular momentumwave functionsYml( , ). (Since we are now dealing with orbital Angular momen-tum, we replacejbyl.) To do this, we first need to writeLin spherical way to do this is to start fromLi= (r p)i= ijkxjpkwherepk= i~( / xk),and then use the chain rule to convert from Cartesian coordinatesxito sphericalcoordinates (r, , ).

9 Usingx=rsin cos y=rsin sin z=rcos so thatr= (x2+y2+z2)1/2 = cos 1z/r = tan 1y/xwe have, for example, x= r x r+ x + x =xr r+xzr3sin yx2cos2 = sin cos r+cos cos r sin rsin with similar expressions for / yand / z. Then using terms such asLx=ypz zpy= i~ y z z y we eventually arrive atLx= i~ sin cot cos ( )Ly= i~ cos cot sin ( )Lz= i~ .( )7 However, another way is to start from the gradient in spherical coordinates (seethe section on vector calculus in the handoutSupplementary Notes on Mathematics) = r r+ 1r + 1rsin .ThenL=r p= i~r = i~r( r ) so that (since r, and areorthonormal)L= i~r r r r+ r 1r + r 1rsin = i~ 1sin If we write the unit vectors in terms of their Cartesian components (again, see thehandout on vector calculus) = (cos cos ,cos sin , sin ) = ( sin ,cos ,0)thenL= i~ x sin cot cos + y cos cot sin + z which is the same as we had in ( ).

10 Using these results, it is now easy to write the ladder operatorsL =Lx iLyin spherical coordinates:L = ~e i icot .( )To find the eigenfunctionsYml( , ), we start from the definitionL+Yll= 0. Thisyields the equation Yll +icot Yll = can solve this by the usual approach of separation of variables if we writeYll( , ) =T( )F( ). Substituting this and dividing byTFwe obtain1 Tcot T = i1F F .Following the standard argument, the left side of this is a function of only, andthe right side is a function of only. Since varying won t affect the right side,and varying won t affect the left side, it must be that both sides are equalto aconstant, which I will callk. Now the equation becomesdFF=ikd 8which has the solutionF( ) =eik (up to normalization). ButYllis an eigenfunc-tion ofLz= i~( / ) with eigenvaluel~, and hence so isF( ) (sinceT( ) justcancels out). This means that i~ eik =k~eik :=l~eik and therefore we must havek=l, so that (up to normalization)Yll=eil T( ).