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

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      The states ∣n+〉, ∣n−〉 can be written in the combined form ∣n+n−〉 which, from

      [a−,a+†]=[a−,a+]=[a−†,a+†]=[a−†,a+]=0,(1.121)

      obey

∣n+n−〉=a+†n+a−†n−n+!n−!∣00〉,(1.122)

      Nˆ+∣n+n−〉=n+∣n+n−〉,Nˆ−∣n+n−〉=n−∣n+n−〉,(1.123)

      a+†∣n+n−〉=n++1∣n++1,n−〉,a−†∣n+n−〉=n−+1∣n+,n−+1〉,(1.124)

      a+∣n+n−〉=n+∣n+−1,n−〉,a−∣n+n−〉=n−∣n+,n−−1〉,(1.125)

      a+∣00〉=0,a−∣00〉=0.(1.126)

      Then, from equations (1.111) and (1.115), i.e.

      Jˆ+=a+†a−,Jˆ−=a−†a+,Jˆ0=12a+†a+−a−†a−=12(Nˆ+−Nˆ−),(1.127)

      together with

      Nˆ≔Nˆ++Nˆ−=a+†a++a−†a−(1.128)

      and

      Jˆ2≔Jˆ02+12(Jˆ+Jˆ−+Jˆ−Jˆ+),(1.129)

      we obtain:

Jˆ+∣n+n−〉=n−(n++1)∣n++1,n−−1〉,(1.130)

      Jˆ−∣n+n−〉=n+(n−+1)∣n+−1,n−+1〉,(1.131)

Jˆ0∣n+n−〉=12(n+−n−)∣n+n−〉,(1.132)

      Nˆ∣n+n−〉=(n++n−)∣n+n−〉,(1.133)

      and

Jˆ2∣n+n−〉=Jˆ02∣n+n−〉+12Jˆ+Jˆ−∣n+n−〉+12Jˆ−Jˆ+∣n+n−〉=14(n+−n−)2∣n+n−〉+12n+(n−+1)∣n+n−〉+12n−(n++1)∣n+n−〉=n++n−2n++n−2+1∣n+n−〉=n2n2+1∣n+n−〉,(1.134)

      where

      n=n++n−.(1.135)

      Evidently, by making the associations

      n+↔j+m,n−↔j−m,(1.136)

      we obtain

      n=2j;(1.137)

      and from equations (1.130)–(1.132) and (1.134)

      Jˆ+∣n+n−〉=(j−m)(j+m+1)∣n++1,n−−1〉,(1.138)

      Jˆ−∣n+n−〉=(j+m)(j−m+1)∣n+−1,n−+1〉,(1.139)

      Jˆ0∣n+n−〉=m∣n+n−〉,(1.140)

      Jˆ2∣n+n−〉=j(j+1)∣n+n−〉,(1.141)

      respectively. Thus, by comparison with

      Jˆ+∣jm〉=(j−m)(j+m+1)∣jm+1〉,(1.142)

      Jˆ−∣jm〉=(j+m)(j−m+1)∣jm−1〉,(1.143)

      Jˆ0∣jm〉=m∣jm〉,(1.144)

      Jˆ2∣jm〉=j(j+1)∣jm〉,(1.145)

      we can assert that

      ∣n+n−〉≡∣jm〉,(1.146)

      and from equation (1.122)

∣jm〉≡a+†j+ma−†j−m(j+m)!(j−m)!∣00〉.(1.147)

      Two special cases of note are: m=+j, i.e.

∣jj〉≡a+†2j(2j)!∣00〉;(1.148)

      and m=−j, i.e.

∣j,−j〉≡a−†2j(2j)!∣00〉.(1.149)

      Consider then the rotation of the states ∣j=12,m=12〉≡∣12,12〉 and ∣j=12,m=−12〉≡∣12,−12〉:

      Dy(β)12,12↔cosβ2−sinβ2sinβ2cosβ210=cosβ2sinβ2,(1.150)

      Dy(β)12,−12↔cosβ2−sinβ2sinβ2cosβ201=−sinβ2cosβ2;(1.151)

      i.e.

      Dy(β)12,−12=cosβ212,12+sinβ212,−12,(1.152)

      Dy(β)12,−12=−sinβ212,12+cosβ212,−12.(1.153)

      Then from

      12,12=a+†∣0〉,12,−12=a−†∣0〉,(1.154)

      we have

      Dy(β)12,12=Dy(β)a+†Dy−1(β)Dy(β)∣0〉,(1.155)

      Dy(β)12,−12=Dy(β)a−†Dy−1(β)Dy(β)∣0〉;(1.156)

      whence

      Dy(β)a+†Dy−1(β)≡a+†′=cosβ2a+†+sinβ2a−†,(1.157)

      Dy(β)a−†Dy−1(β)≡a−†′=−sinβ2a+†+cosβ2a−†.(1.158)

      Thus,

      Dy(β)∣jm〉≔a+†′j+ma−†′j−m(j+m)!(j−m)!∣00〉.(1.159)

∴Dy(β)∣jm〉=cosβ2a+†+sinβ2a−†j+m−sinβ2a+†+cosβ2a−†j−m(j+m)!(j−m)!∣00〉.(1.160)

      The right-hand side of equation (1.160) can be expanded using the binomial theorem:

      Dy(β)∣jm〉=1(j+m)!(j−m)!∑l(j+m)!l!(j+m−l)!a+†cosβ2la−†sinβ2j+m−l×∑k(j−m)!k!(j−m−k)!(−a+†sinβ2)ka−†cosβ2j−m−k∣00〉.(1.161)

∴Dy(β)∣jm〉=(j+m)!(j−m)!∑l,k(−1)kcosβ2j−m+l−ksinβ2j+m−l+kl!(j+m−l)!k!(j−m−k)!×a+†l+ka−†2j−l−k∣00〉,(1.162)

      and comparing with

      Dy(β)∣jm〉=∑m′∣jm′〉dm′m(j)(β),(1.163)

      i.e.

Dy(β)∣jm〉=∑m′dm′m(j)(β)a+†j+m′a−†j−m′(j+m′)!(j−m′)!∣00〉,(1.164)

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