221B Lecture Notes - Hitoshi Murayama

1 221b Lecture NotesNotes on Spherical bessel Functions1 DefinitionsWe would like to solve the free Schr odinger equation h22m[1rd2dr2r l(l+ 1)r2]R(r) = h2k22mR(r).(1)R(r) is the radial wave function (~x) =R(r)Yml( , ). By factoring out h2/2mand defining =kr, we find the equation[1 d2d 2 l(l+ 1) 2+ 1]R( ) = 0.(2)The solutions to this equation are spherical bessel functions. Due to somereason, I don t see the integral representations I use below in books on math-emtical formulae, but I believe they are behavior at the origin can be studied by power expansion. AssumingR n, and collecting terms of the lowest power in , we getn(n+ 1) l(l+ 1) = 0.(3)There are two solutions,n=lor l 1.(4)The first solution gives a positive power, and hence a regular solution at theorigin, while the second a negative power, and hence a singular solution atthe is easy to check that the following integral representations solve theabove equation Eq. (2):h(1)l( ) = ( /2)ll! i +1ei t(1 t2)ldt,(5)andh(2)l( ) =( /2)ll!

2 I 1ei t(1 t2)ldt.(6)1By acting the derivatives in Eq. (2), one finds[1 d2d 2 l(l+ 1) 2+ 1]h(1)l( )= ( /2)ll! i 1(1 t2)l[l(l+ 1) 2+2(l+ 1)it t2 l(l+ 1) 2+ 1]dt= ( /2)ll!1i i 1ddt[ei t(1 t2)l+1]dt.(7)Therefore only boundary values contribute, which vanish both att= 1 andt=i for =kr >0. The same holds forh(2)l( ).One can also easily see thath(1) l( ) =h(2)l( ) by taking the complexconjugate of the expression Eq. (5) and changing the variable fromtto integral representation Eq. (5) can be expanded in powers of 1/ .For instance, forh(1)l, we change the variable fromttoxbyt= 1 +ix, andfindh(1)l( ) = ( /2)ll! 0ei (1+ix)xl( 2i)l(1 x2i)lidx= i( /2)ll!ei ( 2i)ll k=0lCk 0e x ( x2i)kxldx= iei l k=0( i)l k(l+k)!2kk!(l k)!1 k.(8)Similarly, we findh(2)l( ) =ie i l k=0il k(l+k)!2kk!(l k)!1 k.(9)Therefore bothh(1,2)lare singular at = 0 with power l combinationjl( ) = (h(1)l+h(2)l)/2 is regular at = 0. This can beseen easily as follows. Becauseh(2)lis an integral fromt= 1 toi , whileh(1)lfromt= +1 toi , the differencd between the two corresponds to anintegral fromt= 1 tot=i and coming back tot= +1.

3 Because theintegrand does not have a pole, this contour can be deformed to a straightintegral fromt= 1 to +1. Therefore,jl( ) =12( /2)ll! 1 1ei t(1 t2)ldt.(10)2In this expression, 0 can be taken without any problems in the integraland hencejl l, , regular. The other linear combinationnl= (h(1)l h(2)l)/2iis of course singular at = thath(1)l( ) =jl( ) +inl( )(11)is analogous toei = cos +isin .(12)It is useful to see some examples for , j1=sin 2 cos ,j2=3 2 3sin 3 2cos ,n0= cos , n1= cos 2 sin , n2= 3 2 3cos 3 2sin ,h(1)0= iei , h(1)1= i(1 2 i )ei h(1)2= i(3 2 3 3i 2)ei .h(2)0=ie i , h(2)1=i(1 2+i )e i h(2)2=i(3 2 3+3i 2)e i .(13)2 Power Series ExpansionEq. (5) can be used to obtain the power series expansion. We first split theintegration region into two parts,h(1)l( ) = ( /2)ll! i +1ei t(1 t2)ldt= ( /2)ll![ i 0 10]ei t(1 t2)ldt.(14)The first term can be expanded in a power series by a change of variable,t=i / ,the first term = ( /2)ll!

4 0e (1 + 2 2)lid = i1l!2l l+1 0e ( 2+ 2)ld = i1l!2l l+1 0e l n=llCn 2n 2l 2nd = i1l!2l l+1l n=ll!n!(l n)! 2n (2l 2n+ 1)1 Note that my notation fornldiffers from Sakurai s by a sign as seen in Eq. ( )on page 409. I m sorry for that, but I stick with my convention, which was taken i12l l+1l n=l(2l 2n)!n!(l n)! 2n.(15)On the other hand, the second term can be expanded asthe second term =( /2)ll! 10ei t(1 t2)ldt=( /2)ll! 10 n=0inn! ntn(1 t2)ldt=( /2)ll! n=0inn! 10 ntn(1 t2)ldt=( /2)ll! n=0inn! n 10x(n 1)/2(1 x)l12dx=12( /2)ll! n=0inn! n (n2+12) (l+ 1) (n2+l+32).(16)At this point, it is useful to separate the sum to evenn= 2kand oddn= 2k+ 1,the second term=12( 2)l( k=0( 1)k(2k)! 2k (k+12) (k+l+32)+ k=0i( 1)k(2k+ 1)! 2k+1 (k+ 1) (k+l+ 2))= l2l+1( k=0( 1)k(2k)! 2k (k+12) (k+l+32)+ k=0i( 1)k(2k+ 1)! 2k+1k!(k+l+ 1)!)(17)Becauseh(1)l( ) =jl( ) +inl( ), we findjl( ) = l2l+1 k=0( 1)k(2k)! (k+12) (k+l+32) 2k(18)nl( ) = 12l l+1l n=l(2l 2n)!n!(l n)! 2n+ l2l+1 k=0( 1)kk!

5 (2k+ 1)!(k+l+ 1)! 2k+1.(19)The expression forjlcan be simplified using the identity (n+12) =(n 12)(n 32) 3212 (12) =(2n 1)!!2n ,jl( ) = l2l+1 k=0( 1)k(2k)!(2k 1)!! /2k(2k+ 2l+ 1)!! /2k+l+1 2k4= l k=0( 1)k(2k)!1(2k+ 2l+ 1)(2k+ 2l 1) (2k+ 1) 2k= l k=0( 1)k(2k)!(2k+ 2l)(2k+ 2l 2) (2k+ 2)(2k)!(2k+ 2l+ 1)! 2k= l k=0( 1)k2l(k+l)!k!(2k+ 2l+ 1)! 2k= (2 )l k=0( 1)k(k+l)!k!(2k+ 2l+ 1)! 2k.(20)To write out the first three terms,= l(2l+ 1)!![1 22(2l+ 3)+ 48(2l+ 5)(2l+ 3) ].(21)It suggests that the leading term is a good approximation when 2l1 , the expression fornl( ) can also be simplified,nl( ) = 12l l+1l n=l(2l 2n)!n!(l n)! 2n+ l2l+1 k=0( 1)kk!(2k+ 1)!(k+l+ 1)! 2k+1= 12l l+1(l n=l(2l 2n)!n!(l n)! 2n 12 k=0( 1)kk!(2k+ 1)!(k+l+ 1)! 2k+2l+2)= 12l l+1 l n=l(2l 2n)!n!(l n)! 2n+12 n=l+1( 1)n+l(n l 1)!(2n 2l 1)!n! 2n = 12l l+1 l n=l(2l 2n)!n!(l n)! 2n+ n=l+1( 1)n+l(n l)!(2n 2l)!n! 2n = 12l l+1l n=l( 1)nn! (2l 2n+ 1) (l n+ 1) 2n.(22)To write out the first three terms,= (2l 1)!

6 ! l+1[1 + 22(2l 1)+ 48(2l 1)(2l 3)+ ].(23)It suggests that the leading term is a good approximation when 2l1 Asymptotic BehaviorEqs. (8,9) give the asymptotic behaviors ofh(1)lfor :h(1)l iei ( i)l= iei( l /2) .(24)By taking linear combinations, we also findjl sin( l /2) ,(25)nl cos( l /2) .(26)These expressions are good approximations when l2. As seen in thenext section, there are better approximations when l LargelBehaviorStarting from the integral from Eq. (10), we use the steepest descent methodto find the largelbehavior. Changing the variablet=l ,jl( ) =12( /2)ll! 1 1ei t(1 t2)ldt=12( /2)ll! 1/l 1/lel(log(1 l2 2)+i )ld =12( /2)ll! 1/l 1/lexp[ l+ l2 2+llog2l(l l2 2) 2 l 2 l2 22(l l2 2)( il l2 2l )2+O( )3 ld '12( /2)l 2 l lle le le l2 2(2l(l l2 2) 2)l(2 (l l2 2)l 2 l2 2)1/2l=12 e l2 2(l l2 2 )l(l l2 2 l2 2)1/2.(27)This expression works very well as long asl 1 and from the integral from Eq. (5), we use the steepest descentmethod to find the largelbehavior.]

7 Change the variablet=il ,h(1)l( ) = ( /2)ll! i +1ei t(1 t2)ldt= il( /2)ll! i/lexp[ l l2 2+llog2l(l+ l2 2) 2 l 2 l2 22(l+ l2 2)( l+ l2 2l )2+O( )3 ' ie l2 2(l+ l2 2 )l(l+ l2 2 2 l2 2)1/2.(28)Note that there are actually two saddle points, =l l2 2l .(29)In the above calculation, we picked the saddle point with the negative signwith the steepest descent, while the other saddle point is what we picked forjl( ). Therefore,nl( )' 1 e l2 2(l+ l2 2 )l(l+ l2 2 l2 2)1/2.(30)This expression again works very well as long asl 1 and l(but nottoo close).On the other hand, for l 1, the saddle points above becomecomplex. The contribution to theh(1)l( ) is given by the saddle point =l i 2 l2l , and henceh(1)l( )'1 ei 2 l2(l i 2 l2 )l(l i 2 l2i 2 l2)1/2.(31)This works very well as long asl 1 and l(but not too close).jl(nl)is given by the real (imaginary) part ofh(1)l( ). In practice, this form worksremarkably well even forl= l=100exactlarge llarge jl( )Figure 1: Comparison of the large behavior and the largelbehavior ofjl( )to the exact result.]

8 The largelbehavior is a very good approximation for 105>100 =l, while the large behavior is still a poor approximationunless > O(l2).It is interesting to note that this asymptotic behavior ofh(1)l( ) is whatyou expect from the semi-classical approximation for the free-particle wavefunction. The classical action for a free particle isS(r) = h rl/k k2 l2r 2dr = h (kr)2 l2 2larctan kr lkr+l(32)and henceeiS(r)/ h=ei (kr)2 l2 l i (kr)2 l2kr l,(33)which agrees with the largelbehavior above except for the last factor whichcomes from the lowest-order quantum 'l 1, two saddle points collide and I don t know what to Recursion FormulaeStarting from Eq. (5), we take the derivativedd h(1)l=l h(1)l ( /2)ll! i +1ei tit(1 t2)ldt.(34)The second term can be integrated by parts, and gives=l h(1)l+ 2( /2)ll! i +1ei t(1 t2)l+1dt=l h(1)l h(1)l+1.(35)8In fact, other functionsjl,nl, andh(2)lall satisfy the same relation which canbe easily checked. Referring to all of them generically aszl( ), we find therecursion formulaz l=l zl zl+1.

9 (36)Becausezl( ) satisfies the differential equation Eq. 2, we can combine itwith the above recursion relation and find0 =(d2d 2+2 dd l(l+ 1) 2+ 1)zl=zl 2l+ 3 zl+1+zl+2.(37)Relabelingltol 1, we obtainzl 1+zl+1=2l+ 1 zl.(38)Finally, combining the two recursion relations, we also obtainz l=zl 1 l+ 1 zl.(39)6 Plane Wave ExpansionThe non-trivial looking formula we used in the classeikz= l=0(2l+ 1)iljl(kr)Pl(cos )(40)can be obtained quite easily from the integral representation Eq. (10). Thepoint is that one can keep integrating it in parts. By integratingei tfactorand differentiating (1 t2)lfactor, the boundary terms att= 1 alwaysvanish up tol-th time because of the (1 t2)lfactor. Therefore,jl=12( /2)ll! 1 11(i )lei t( ddt)l(1 t2)ldt.(41)Note that the definition of the Legendre polynomials isPl(t) =12l1l!dldtl(t2 1)l.(42)9 Using this definition, the spherical bessel function can be written asjl=121il 1 1ei tPl(t)dt.(43)Then we use the fact that the Legendre polynomials form a complete set oforthogonal polynomials in the intervalt [ 1,1].

10 Noting the normalization 1 1Pn(t)Pm(t)dt=22n+ 1 n,m,(44)the orthonormal basis isPn(t) (2n+ 1)/2, and hence n=02n+ 12Pn(t)Pn(t ) = (t t ).(45)By multipyling Eq. (43) byPl(t )(2l+ 1)/2 and summing overn, l=02il2l+ 12Pl(t )jl( ) = 1 1ei t n=02l+ 12Pl(t )Pl(t)dt=ei t .(46)By setting =krandt = cos , we prove Eq. (40).If the wave vector is pointing at other directions than the positivez-axis, the formula Eq. (40) needs to be generalized. NotingY0l( , ) = (2l+ 1)/4 Pl(cos ), we findei~k ~x= 4 l=0iljl(kr)l m= lYm l( ~k, ~k)Yml( ~x, ~x)(47)7 Delta-Function NormalizationAn important consequence of the identity Eq. (47) is the innerproduct of twospherical bessel functions. We start with d~xei~k ~xe i~k ~x= (2 )3 (~k ~k ).(48)Using Eq. (47) in the of this equation, we find d~xei~k ~xe i~k ~x= l,m l ,m (4 )2 d ~xdrr2Ym l( ~k)Yml( ~x)Ym l ( ~x)Ym l ( ~k )jl(kr)jl (k r)= l,m(4 )2 drr2jl(kr)jl(k r)Ym l( ~k)Yml( ~k ).(49)10On the other hand, the of Eq. (48) is(2 )3 (~k ~k ) = (2 )31k2 (k k ) ( ~k ~k )= (2 )31k2sin (k k ) ( ) ( ).