Then, for a particular choice of m′,
l=j+m′−k(1.166)
and
dm′m(j)(β)=∑k(no negative factorials)(−1)k(j+m)!(j−m)!(j+m′)!(j−m′)!(j+m′−k)!(m−m′+k)!k!(j−m−k)!×cosβ22j−2k+m′−m×sinβ22k+m−m′.(1.167)
1.8 A spinor function basis for SU(2)
The Schwinger representation and its associated basis leads directly to a spinor function basis for SU(2):
where u and
Thus, we deduce the realisation
It follows from equations (1.148), (1.149) and (1.147) that
and
It should be noted that
1.9 A spherical harmonic basis for SO(3)
The use of spinor functions as a basis for SU(2) and the relations for Jˆ0, Jˆ± given in equations (1.172)–(1.174) lead to the consideration of a functional representation of the ∣lm〉 for (ℏ≡1)
Lˆ0=Lˆz=xˆpˆy−yˆpˆx↔−ix∂∂y+iy∂∂x;(1.177)
Lˆ+=Lˆx+iLˆy=yˆpˆz−zˆpˆy+izˆpˆx−ixˆpˆz,(1.178)
∴Lˆ+↔−iy∂∂z+iz∂∂y+z∂∂x−x∂∂z;(1.179)
Lˆ−↔−iy∂∂z+iz∂∂y−z∂∂x+x∂∂z;(1.180)
where the postion representation has been used. Evidently, Lˆ0, Lˆ± in the form given by equations (1.177), (1.179) and (1.180) leave the degree of a polynomial in x, y and z unchanged. Therefore, we consider the space of homogeneous polynomials in x, y and z, i.e.
f(x,y,z)=(ax+by+cz)l,(1.181)
where a, b, and c are complex numbers.
We start with the homogeneous polynomials ϕlm=−l(r⃗), r⃗≔(x,y,z), that satisfy the so-called ‘lowest weight’ conditions:
Lˆ0ϕl,−l(r⃗)=−lϕl,−l(r⃗)(1.182)
and
Lˆ−ϕl,−l(r⃗)=0.(1.183)
Then, consider
Lˆ0(ax+by+cz)l=−ix∂∂y+iy∂∂x(ax+by+cz)l=−ixl(ax+by+cz)l−1b+iyl(ax+by+cz)l−1a=l(ax+by+cz)l−1(−ibx+iay),(1.184)
and the right-hand side fulfils equation (1.182), i.e.
Lˆ0(ax+by+cz)l=−l(ax+by+cz)l,(1.185)
provided a=1,b=−i,c=0. Thus,
ϕl,−l(r⃗)=(x−iy)l.(1.186)
Evidently,
Lˆ−ϕl,−l(r⃗)=−iy∂∂z+iz∂∂y−z∂∂x+x∂∂z(x−iy)l=izl(−i)(x−iy)l−1−zl(x−iy)l−1=0.(1.187)
We can construct the ϕlm(r⃗) using (ℏ≡1)
Lˆ+ϕlm(r⃗)=(l−m)(l+m+1)ϕl,m+1(r⃗).(1.188)
For l = 1: from
ϕ1,−1(r⃗)=x−iy,(1.189)
Lˆ+ϕ1,−1(r⃗)=−iy∂∂z+iz∂∂y+z∂∂x−x∂∂z(x−iy)=iz(−i)+z=2z≔2ϕ1,0(r⃗);(1.190)
∴ϕ1,0(r⃗)=2z.(1.191)
Then,
Lˆ+ϕ1,0(r⃗)=2−iy∂∂z+iz∂∂y+z∂∂x−x∂∂zz=2(−iy−x)=−2(x+iy)≔2ϕ1,1(r⃗);(1.192)
∴ϕ1,1(r⃗)=−(x+iy).(1.193)
For l = 2: from
ϕ2,−2(r⃗)=(x−iy)2,(1.194)
Lˆ+ϕ2,−2(r⃗)=−iy∂∂z+iz∂∂y+z∂∂x−x∂∂z(x−iy)2=iz2(x−iy)(−i)+z2(x−iy)=4z(x−iy)≔2ϕ2,−1(r⃗);(1.195)
∴ϕ2,−1(r⃗)=2z(x−iy).(1.196)
Then,
Lˆ+ϕ2,−1(r⃗)=−iy∂∂z+iz∂∂y+z∂∂x−x∂∂z2z(x−iy)=−iy2(x−iy)+iz2z(−i)+z2z−x2(x−iy)=−2(x−iy)(x+iy)+4z2=−2(x2+y2)+4z2≔6ϕ2,0(r⃗);(1.197)
∴ϕ2,0(r⃗)=23(−x2−y2+2z2).(1.198)
Similarly,
ϕ2,1(r⃗)=−2z(x+iy),(1.199)
ϕ2,2(r⃗)=(x+iy)2.(1.200)
