Quantum Mechanics for Nuclear Structure, Volume 2. Professor Kris Heyde

      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

are independent functions. We require u and

to satisfy (again, ℏ≡1)

      Thus, we deduce the realisation

      It follows from equations (1.148), (1.149) and (1.147) that

      and

      It should be noted that

are elements of a complex function space which is formally developed in section 1.17 under the title of the Bargmann representation, i.e. this function space is known as Bargmann space. These bases are irreducible (unlike Cartesian tensors).

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)

      The functions ϕlm(r⃗) are proportional to the spherical harmonics, Ylm(θ,ϕ) (see table 1.1). This follows from the relationship between Cartesian coordinates and spherical polar coordinates:

      x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθ,(1.201)

      whence

ϕ1,±1=∓(x±iy)=∓rsinθe±iϕ=r8π3Y1,±1(θ,ϕ).(1.202)

      Similarly,

ϕ1,0(r⃗)=2rcosθ=r8π3Y1,0(θ,ϕ).(1.203)

      ϕ2,±2(r⃗)=r2sin2θe±2iϕ=r232π15Y2,±2(θ,ϕ).(1.204)

      ϕ2,±1(r⃗)=∓2r2cosθsinθe±iϕ=r232π15Y2,±1(θ,ϕ).(1.205)