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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 27 "Clebsch-Gordan coefficien
ts" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 91 "The
 calculation of vector coupling coefficients is based on representing \+
J_total -sqared = " }{TEXT 291 1 "J" }{TEXT -1 185 "^2 in the basis of
 the single-particle angular momentum eigenstates and diagonalization.
 We make use of the matrix elements of the raising and lowering operat
ors to assemble the matrix." }}{PARA 0 "" 0 "" {TEXT 19 62 "<j,m+1| J+
 |j,m> = <j,m| J- | j,m+1> = -h_*sqrt((j-m)*(j+m+1))" }}{PARA 0 "" 0 "
" {TEXT -1 31 "We are using the decomposition " }{TEXT 263 1 "J" }
{TEXT -1 9 "_tot^2 = " }{TEXT 264 2 "J1" }{TEXT -1 7 "^2  +  " }{TEXT 
265 2 "J2" }{TEXT -1 9 "^2  +  2 " }{TEXT 266 2 "J1" }{TEXT -1 5 "_z *
 " }{TEXT 267 2 "J2" }{TEXT -1 5 "_z + " }{TEXT 270 2 "J1" }{TEXT -1 
4 "_ * " }{TEXT 269 3 "J2+" }{TEXT -1 5 "  +  " }{TEXT 268 2 "J1" }
{TEXT -1 1 "\000" }{TEXT 271 2 "+ " }{TEXT -1 3 "*  " }{TEXT 272 3 "J2
_" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "Whe
n assembling the matrix representation we keep in mind that the 5 term
s on the RHS of the above equation separate into two groups:" }}{PARA 
0 "" 0 "" {TEXT -1 120 "the first three terms will contribute only on \+
the diagonal, and the last two terms only one line away from the diago
nal." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "S
uppose we want to couple " }{TEXT 274 2 "J1" }{TEXT -1 9 "=3/2 and " }
{TEXT 273 2 "J2" }{TEXT -1 33 "=1. We work in units where h-bar " }
{TEXT 19 6 "h_ = 1" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 46 "restart; interface(warnlevel=0): with(linalg):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "j1:=3/2; j2:=1;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 90 "Normally, we are interested in picking an
 allowed valueof total angular momentum to which " }{TEXT 19 2 "j1" }
{TEXT -1 5 " and " }{TEXT 19 2 "j2" }{TEXT -1 32 " couple (i.e., out o
f the range " }{TEXT 19 18 "|j1-j2| .. (j1+j2)" }{TEXT -1 2 ")." }}
{PARA 0 "" 0 "" {TEXT -1 56 "However, the way things will work in the \+
approach where " }{TEXT 275 1 "J" }{TEXT -1 26 "^2 is diagonalized in \+
the " }{TEXT 19 13 "|j1 j2 m1 m2>" }{TEXT -1 45 " basis, we will not u
se the specification of " }{TEXT 19 1 "J" }{TEXT -1 55 ", but rather f
ind the CG coefficients for all possible " }{TEXT 19 1 "J" }{TEXT -1 
22 " values. To leave the " }{TEXT 19 1 "z" }{TEXT -1 21 "-projection \+
of total " }{TEXT 276 1 "J" }{TEXT -1 69 " as unrestricted as possible
 we determine the maximum allowed length:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "J:=j1+j2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Ou
r task is to represent " }{TEXT 277 1 "J" }{TEXT -1 10 "^2 in the " }
{TEXT 19 13 "|j1 j2 m1 m2>" }{TEXT -1 12 " basis with " }{TEXT 19 2 "j
1" }{TEXT -1 5 " and " }{TEXT 19 2 "j2" }{TEXT -1 87 " specified throu
gh their quantum numbers. This means that the matrix representation of
 " }{TEXT 278 1 "J" }{TEXT -1 21 "^2 can be labeled by " }{TEXT 19 2 "
m1" }{TEXT -1 5 " and " }{TEXT 19 2 "m2" }{TEXT -1 11 ". However, " }
{TEXT 19 2 "m1" }{TEXT -1 5 " and " }{TEXT 19 2 "m2" }{TEXT -1 64 " do
 not vary independently, but are restricted by the condition " }{TEXT 
19 11 "m1 + m2 = M" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 53 "The
refore, we now pick a projection out of the range " }{TEXT 19 7 "M=-J.
.J" }{TEXT -1 163 ". Let us pick anyone but the largest (or lowest) so
 that we will get some coupling. The reason for avoiding the largest p
rojection is that only one combination of " }{TEXT 19 1 "z" }{TEXT -1 
16 "-projections of " }{TEXT 280 2 "J1" }{TEXT -1 5 " and " }{TEXT 
279 2 "J2" }{TEXT -1 20 " works in that case." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 11 "M:=-J+j2+1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
20 "The solution of the " }{TEXT 290 1 "J" }{TEXT -1 47 "^2 eigenvalue
/eigenvector problem will give us:" }}{PARA 0 "" 0 "" {TEXT -1 59 "(i)
 through the eigenvalues the allowed quantum numbers of " }{TEXT 259 
1 "J" }{TEXT -1 10 "^2, i.e., " }{TEXT 19 12 "h_^2 J (J+1)" }{TEXT -1 
31 ", with the possible choices of " }{TEXT 19 1 "J" }{TEXT -1 37 " th
at are compatible with the chosen " }{TEXT 19 1 "M" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 55 "(ii) in anticipation of half-integer spin
s we pick the " }{TEXT 19 1 "M" }{TEXT -1 45 " value above such that i
t is compatible with " }{TEXT 19 1 "J" }{TEXT -1 20 ", i.e,. for integ
er " }{TEXT 19 1 "J" }{TEXT -1 17 " we pick integer " }{TEXT 19 1 "M" 
}{TEXT -1 23 ", and for half-integer " }{TEXT 19 1 "J" }{TEXT -1 22 " \+
we pick half-integer " }{TEXT 19 1 "M" }{TEXT -1 1 "." }}{PARA 0 "" 0 
"" {TEXT -1 34 "(iii) the eigenvector for a given " }{TEXT 258 1 "J" }
{TEXT -1 125 "^2-eigenvalue gives us the set of coupling coefficients \+
that says how the uncoupled states combine to form an eigenvector of \+
" }{TEXT 260 1 "J" }{TEXT -1 3 "^2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 57 "Now the question is: how do we find the \+
correct range of " }{TEXT 19 7 "(m1,m2)" }{TEXT -1 40 " values by use \+
of restrictions (such as " }{TEXT 19 7 "M=m1+m2" }{TEXT -1 40 ") ? Thi
s will determine the matrix size." }}{PARA 0 "" 0 "" {TEXT -1 66 "We d
etermine by a counting method the matrix size in the variable " }
{TEXT 19 2 "ic" }{TEXT -1 39 ", and the configurations are stored in \+
" }{TEXT 19 2 "cf" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 186 "ic:=0: for m1 from -j1 to j1 do: for m2 from -j2 to \+
j2 do: if m1+m2=M then ic:=ic+1: cf[ic]:=[m1,m2]: print(`combo `,ic,` \+
involves [m1,m2]= `,cf[ic],` with m1+m2=M: `,M); fi; od: od: ic;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Now we understand that " }{TEXT 
19 2 "j2" }{TEXT -1 59 " (the smaller of the two) steps through all al
lowed values:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Jsq:=matri
x(ic,ic,0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The matrix is a sq
uare matrix: we let the row and column indeces go through all allowed \+
values of " }{TEXT 19 5 "m1,m2" }{TEXT -1 29 ", i.e., each index goes \+
from " }{TEXT 19 1 "1" }{TEXT -1 4 " to " }{TEXT 19 2 "ic" }{TEXT -1 
15 ". We call them " }{TEXT 19 5 "m1,m2" }{TEXT -1 5 " and " }{TEXT 
19 7 "m1p,m2p" }{TEXT -1 41 " for row and column indices respectively.
" }}{PARA 0 "" 0 "" {TEXT -1 66 "We also need indeces to refer to the \+
matrix entries (ranging from " }{TEXT 19 1 "1" }{TEXT -1 4 " to " }
{TEXT 19 2 "ic" }{TEXT -1 9 ") called " }{TEXT 19 3 "im2" }{TEXT -1 5 
" and " }{TEXT 19 4 "im2p" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
128 "To understand the diagonal matrix entries is straightforward. The
y are simply the sum of the diagonal elements of the operators " }
{TEXT 19 20 "J1^2+J2^2+ 2*J1z*J2z" }{TEXT -1 46 ", i.e., of the eigenv
alues of these operators." }}{PARA 0 "" 0 "" {TEXT -1 76 "The off-diag
onal elements require one to understand the following structure:" }}
{PARA 0 "" 0 "" {TEXT -1 5 "when " }{TEXT 19 8 "m2=m2p+1" }{TEXT -1 
17 " it follows from " }{TEXT 19 15 "m1=M-m2=M-m2p-1" }{TEXT -1 5 " an
d " }{TEXT 19 14 "m1p=M-m2p=m1+1" }{TEXT -1 6 " that " }{TEXT 19 8 "m1
=m1p-1" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 17 "Vice versa: whe
n " }{TEXT 19 8 "m2=m2p-1" }{TEXT -1 6 " then " }{TEXT 19 8 "m1=m2p+1
" }{TEXT -1 83 ". This allows J1+J2-  and J1-J2+ to have entries just \+
above and below the diagonal." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 59 "for irow from 1 to ic do: m1:=cf[irow][1]: m2:=cf[irow][2]:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "for icol from 1 to ic do: m1p:=cf[i
col][1]: m2p:=cf[icol][2]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "if m2
=m2p then Jsq[irow,icol]:=j1*(j1+1)+j2*(j2+1)+2*m1*m2: " }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 128 "elif m1=m1p-1 and m2=m2p+1 then Jsq[irow,icol
]:=sqrt((j2-m2+1)*(j2+m2)*(j1+m1+1)*(j1-m1)): print(m1,m1p,m2,m2p,Jsq[
irow,icol]); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "elif m1=m1p+1 and
 m2=m2p-1 then Jsq[irow,icol]:=sqrt((j1-m1+1)*(j1+m1)*(j2+m2+1)*(j2-m2
)): print(m1,m1p,m2,m2p,Jsq[irow,icol]); " }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "fi: od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 11 "print(Jsq);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "To find out \+
the allowed total angular momentum for which the chosen " }{TEXT 19 1 
"M" }{TEXT -1 89 " sublevel can be realized we calculate the eigenvalu
es and keep in mind that they follow " }{TEXT 19 12 "h_^2*J*(J+1)" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eigenvalue
s(Jsq);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Among the three possib
le eigenvalues of " }{TEXT 282 1 "J" }{TEXT -1 60 "^2 we do find the p
reviously selected maximum configuration " }{TEXT 19 7 "J=j1+j2" }
{TEXT -1 25 ", as well as the minimum " }{TEXT 19 7 "J=j1-j2" }{TEXT 
-1 21 ", and one in-between." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 8 "J*(J+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "abs(j1-j2)
*(abs(j1-j2)+1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "We obtain the
 mixing coefficients that tell us the following:" }}{PARA 0 "" 0 "" 
{TEXT -1 38 "the uncoupled basis states labeled by " }{TEXT 19 7 "[j1 \+
m1]" }{TEXT -1 5 " and " }{TEXT 19 7 "[j2 m2]" }{TEXT -1 19 " can be c
oupled to " }{TEXT 19 5 "[J M]" }{TEXT -1 27 " values specified by fix
ed " }{TEXT 19 1 "M" }{TEXT -1 19 " and by calculated " }{TEXT 19 1 "J
" }{TEXT -1 30 " (from the diagonalization of " }{TEXT 283 1 "J" }
{TEXT -1 101 "^2). The meaning of the eigenvector entries is that the \+
coeffcients tell how much of the product of  " }{TEXT 19 7 "[j1 m1]" }
{TEXT -1 5 " and " }{TEXT 19 7 "[j2 m2]" }{TEXT -1 23 " states is need
ed with " }{TEXT 19 5 "m1,m2" }{TEXT -1 37 " given by the specificatio
n in table " }{TEXT 19 2 "cf" }{TEXT -1 75 ". Thus the first eigenvect
or component is the admixing coefficient for the " }{TEXT 19 8 "(m1, m
2)" }{TEXT -1 17 " values given in " }{TEXT 19 5 "cf[1]" }{TEXT -1 61 
", the second eigenvector entry is the one for the product of " }
{TEXT 19 7 "[j1 m1]" }{TEXT -1 6 " with " }{TEXT 19 7 "[j2 m2]" }
{TEXT -1 6 " with " }{TEXT 19 2 "m1" }{TEXT -1 5 " and " }{TEXT 19 2 "
m2" }{TEXT -1 10 " given in " }{TEXT 19 5 "cf[2]" }{TEXT -1 6 ", etc.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "vec:=eigenvects(Jsq);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 53 "The eigenvector corresponding to the chosen value of " }
{TEXT 19 1 "J" }{TEXT -1 49 " when normalized gives the coupling coeff
icients." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "nops([vec]);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "for i from 1 to nops([vec
]) do: if vec[i][1]=J*(J+1) then myvec:=op(vec[i][3]); print(`unnormal
ized eigenvector: `,myvec); fi: od:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "CG:=map(combine,normalize(myvec));" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 125 "The normalization is important to maintain uni
tarity, and thus a probabilistic interpretation. If we prepare the sys
tem in a " }{TEXT 19 14 "[J, M, j1, j2]" }{TEXT -1 30 " eigenstate, an
d then measure " }{TEXT 19 4 "j1_z" }{TEXT -1 5 " and " }{TEXT 19 4 "j
2_z" }{TEXT -1 113 ", then the square of the CG coefficient will provi
de us the probabilities to measure a particular set of allowed " }
{TEXT 19 8 "(m1, m2)" }{TEXT -1 24 " values. First we check:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "add(CG[i]^2,i=1..vectdim(CG)
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "print(cf);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "CG[1]^2,CG[2]^2,CG[3]^2;" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "The statement is that when we are
 in the " }{TEXT 19 30 "|J=5/2, M=-1/2, j1=3/2, j2=1 >" }{TEXT -1 40 "
 eigenstate, then the probabilities are:" }}{PARA 0 "" 0 "" {TEXT -1 
17 "10% to be in the " }{TEXT 19 19 "(m1,m2) = (-3/2, 1)" }{TEXT -1 
12 " combination" }}{PARA 0 "" 0 "" {TEXT -1 17 "60% to be in the " }
{TEXT 19 19 "(m1,m2) = (-1/2, 0)" }{TEXT -1 12 " combination" }}{PARA 
0 "" 0 "" {TEXT -1 17 "30% to be in the " }{TEXT 19 19 "(m1,m2) = (1/2
, -1)" }{TEXT -1 13 " combination." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT 285 11 "Exercise1 :" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "Go through the calculation for the other allowed " }
{TEXT 19 1 "M" }{TEXT -1 21 " values and the same " }{TEXT 19 1 "J" }
{TEXT -1 14 " value, i.e., " }{TEXT 19 7 "J=j1+j2" }{TEXT -1 65 ". Cal
culate the probabilities to find the system in a particular " }{TEXT 
19 7 "(m1,m2)" }{TEXT -1 60 " configuration and display graphically us
ing arrows for the " }{TEXT 19 7 "(j1,m1)" }{TEXT -1 5 " and " }{TEXT 
19 7 "(j2,m2)" }{TEXT -1 34 " angular momentum vectors how the " }
{TEXT 19 5 "(J M)" }{TEXT -1 19 " state comes about." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 286 11 "Exercise2 :" }}{PARA 
0 "" 0 "" {TEXT -1 18 "Look at the other " }{TEXT 19 1 "J" }{TEXT -1 
29 " values for which the set of " }{TEXT 19 7 "(j1 m1)" }{TEXT -1 5 "
 and " }{TEXT 19 9 "(j2 M-m1)" }{TEXT -1 27 " configurations couples t
o " }{TEXT 19 5 "(J M)" }{TEXT -1 98 " and construct the probabilities
 as well as the graphical representations discussed in exercise 1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 287 11 "Exercise
 3:" }}{PARA 0 "" 0 "" {TEXT -1 28 "Pick another combination of " }
{TEXT 19 2 "j1" }{TEXT -1 5 " and " }{TEXT 19 2 "j2" }{TEXT -1 52 ", a
nd repeat the calculations for exercises 1 and 2." }}{PARA 0 "" 0 "" 
{TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 284 41 "A program for C
lebsch-Gordan coefficients" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 73 "One can arrive at a general formula given, e.g., \+
in William J. Thompson: " }{TEXT 257 16 "Angular Momentum" }{TEXT -1 
141 ", Wiley 1994, eq. 7.51. One introduces first the related (3j) coe
fficient for coupling three angular momenta to zero. It involves a sum
 over " }{TEXT 19 1 "k" }{TEXT -1 37 " (which is carried out as a sum \+
over " }{TEXT 19 3 "2*k" }{TEXT -1 52 " to allow half-integer angular \+
momentum projection)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "Th
reeJ:=proc(j1,m1,j2,m2,j3,m3) local k,tk,tkmin,tkmax,res,phas,n1,n2,d1
,d2,d3,term;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "if m1+m2+m3 <> 0 th
en RETURN(0) fi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "tkmin:=2*max(0,
j2-j1-m3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "tkmax:=2*min(j3-m3,j3
-j1+j2); res:=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "phas:=(-1)^(tkm
in/2+j2+m2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "for tk from tkmin t
o tkmax by 2 do: k:=tk/2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "n1:=(j
2+j3+m1-k)!; #print(\"n1= \",n1);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
34 "n2:=(j1-m1+k)!; #print(\"n2= \",n2);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "d1:=k!*(j3-j1+j2-k)!; #print(\"d1= \",d1);" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 45 "d2:=(j3-m3-k)!; #print(\"d2= \",d2,k,j1,j
2,m3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "d3:=(k+j1-j2+m3)!; #print
(\"d3= \",d3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "term:=phas*n1*n2/
(d1*d2*d3);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "phas:=-phas;" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "res:=res+term; od;" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 147 "simplify(res)*(-1)^(j1-j2-m3)*sqrt((j3+j1-j2)!*
(j3-j1+j2)!*(j1+j2-j3)!*(j3-m3)!*(j3+m3)!/((j1+j2+j3+1)!*(j1-m1)!*(j1+
m1)!*(j2-m2)!*(j2+m2)!)); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 91 "CGC:=(j1,m1,j2,m2,J)->combine(ThreeJ(j1,m1,j2,m2,J,-m1-m2)*sqr
t(2*J+1)*(-1)^(j1-j2+m1+m2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "
For our example we have " }{TEXT 19 5 "j1 j2" }{TEXT -1 18 " and the r
ange of " }{TEXT 19 2 "m2" }{TEXT -1 5 " and " }{TEXT 19 2 "m1" }
{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "j1,j2,[-1,
0,1],[M-(-1),M,M-1]; #the matrix was done by stepping with m2 [-1,0,1]
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "CG[1];" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "CGC(j1,M+1,j2,-1,J);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 5 "%-%%;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
6 "CG[2];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "CGC(j1,M,j2,0,
J);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(%-%%);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "CGC(j1,M-1,j2,1,J);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "CG[3];" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 5 "%-%%;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "I
n a nutshell the meaning of the calculation is:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 19 81 "|JM j1 j2> = |5/2, -1/2, \+
3/2, 1> = add( CG[M-m2,m2]* |j1,M-m2> |j2 m2> ,m2=-1..1)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "J,M,j1,j2;" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 170 "There are sum rules for CG-coefficients given in Thom
pson's book (or other books on group theory/angular momentum algebra),
 which can be checked by explicit calculations." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 11 "Application" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 319 "One useful applic
ation of CG coefficients or (3j) symbols is in the evaluation of angul
ar integrals over three spherical harmonics. W.J. Thompson lists it as
 eq. (7.102) and shows how it follows as a special case of an integral
 over three Wigner rotation matrices. We start with a definition of th
e spherical harmonics." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(orthopoly);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 53 "Plm:=proc(theta,l::nonnegint,m::integer) \+
local x,y,f;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "x:=cos(theta);" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "if m>0 then f:=subs(y=x,diff(P(l,y)
,y$m));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "else f:=subs(y=x,P(l,y))
; fi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "(-1)^m*sin(theta)^m*f; end
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Plm(theta,3,2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "For the spherical harmonics we don
't need the Plm's with negative argument." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 52 "Y:=proc(theta,phi,l::nonnegint,m::integer) local m1
;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "m1:=abs(m); if m1>l then RETUR
N(\"|m\} has to be <= l for Y_lm\"); fi;" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 78 "exp(I*m*phi)*Plm(theta,l,m1)*(-1)^m*sqrt((2*l+1)*(l-m
1)!/(4*Pi*(l+m1)!)); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "Y(theta,phi,3,0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Y(
theta,phi,3,1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "No:=(l,
m)->int(int(evalc(Y(theta,phi,l,m)*conjugate(Y(theta,phi,l,m))),phi=0.
.2*Pi)*sin(theta),theta=0..Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 8 "No(1,0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Now suppose we
 are interested in the following integral:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 141 "AI:=(J,M,j1,m1,j2,m2)->int(int(evalc(Y(theta,phi,j
1,m1)*Y(theta,phi,j2,m2)*conjugate(Y(theta,phi,J,M))),phi=0..2*Pi)*sin
(theta),theta=0..Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "There ar
e restrictions on the angular momentum parameters: (i) triangle rule o
n " }{TEXT 19 9 "(j1,j2,J)" }{TEXT -1 7 "; (ii) " }{TEXT 19 7 "j1+j2+J
" }{TEXT -1 23 " = even integer; (iii) " }{TEXT 19 7 "m1+m2=M" }{TEXT 
-1 14 ". For example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "AI
(2,1,1,0,1,1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "In terms of (3j
) symbols the integral can be calculated as:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 128 "AI3j:=(J,M,j1,m1,j2,m2)->simplify((-1)^M*sqrt((
2*j1+1)*(2*j2+1)*(2*J+1)/(4*Pi))*ThreeJ(j1,m1,j2,m2,J,-M)*ThreeJ(j1,0,
j2,0,J,0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "AI3j(2,1,1,0
,1,1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 289 11 "Exercise 4:" }}{PARA 
0 "" 0 "" {TEXT -1 187 "Verify the short-cut formula by comparing it a
gainst the symbolic computation of the integral. Go to large values of
 the angular momentum parameters and check its computational advantage
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 294 "Integrals of this type arise in atomic p
hysics in the computation of multipole expansions of the potential of \+
a charge distribution, and in the computation of dipole matrix element
s (e.g., in the radiative decay) in which the Cartesian components of \+
the position vector are represented through " }{TEXT 19 8 "r*Y(1,m)" }
{TEXT -1 8 ", e.g., " }{TEXT 19 14 "x+I*y=r*Y(1,1)" }{TEXT -1 1 "." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 2 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }