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

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      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)