Читать книгу Quantum Mechanics for Nuclear Structure, Volume 2 - Professor Kris Heyde - Страница 15

Representations of SU(2) can be constructed using a method due to Schwinger. Consider

j=12,m=12≔a+†∣0〉,j=12,m=−12≔a−†∣0〉;(1.109)

i.e. a+† creates a state (particle in a state) of spin-12 up and a−† creates a state of spin-12 down by action on the ‘vacuum’ ∣0〉 (which has no particles), where

ai,aj†=δij,{i,j}={+,−}.(1.110)

Then, defining

Jˆ+≔a+†a−,Jˆ−≔a−†a+,Jˆ0≔a+†a+−a−†a−2,(1.111)

it follows that

[Jˆ0,Jˆ±]=±Jˆ±,(1.112)

[Jˆ+,Jˆ−]=2Jˆ0,(1.113)

which define the structure developed for angular momentum and spin (here, ℏ≡1).

Although the elementary building blocks in the Schwinger representation have spin-12, they should not be regarded as fermions. These spin-12 ‘objects’ are designed to be combined to produce any desired value of total spin: the number of spin-12’s needed to produce a total spin of j will be 2j. These building blocks can be regarded as bosons. They can be visualised in terms of a two-dimensional harmonic oscillator:

a+,a+†=1,a−,a−†=1,(1.114)

Nˆ+=a+†a+,Nˆ−=a−†a−,(1.115)

∣n+〉=a+†n+n+!∣0〉,∣n−〉=a−†n−n−!∣0〉,(1.116)

Nˆ+∣n+〉=n+∣n+〉,Nˆ−∣n−〉=n−∣n−〉,(1.117)

a+†∣n+〉=n++1∣n++1〉,a−†∣n−〉=n−+1∣n−+1〉,(1.118)

a+∣n+〉=n+∣n+−1〉,a−∣n−〉=n−∣n−−1〉,(1.119)

a+∣0〉=0,a−∣0〉=0.(1.120)

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)

equating coefficients of powers of a+† in equations (1.162) and (1.164),

l+k=j+m′.(1.165)

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)