238 CHAPTER15. IDENTICALPARTICLES
andsz isthe spinindex,ifthe particleshave anintrinisicspin. Then thestates
satisfyingtheInterchangeHypothesiswouldsatisfy
PEψ(z 1 ,z 2 )=eiδψ(z 1 ,z 2 ) (15.14)Its easy to see that PE is an Hermitian operator(why?), and eq. (15.14) isan
eigenvalueequation. Then,bysamelogicweusedtodeducetheeigenstatesofparity,
weapplyeq. (15.12)twicetoget
PEPEψ(z 1 ,z 2 ) = PEψ(z 2 ,z 1 )
= ψ(z 1 ,z 2 ) (15.15)Thenapplyeq. (15.14)twice,tofind
PEPEψ(z 1 ,z 2 ) = !iδPEψ(z 1 ,z 2 )
=(
!iδ) 2
ψ(z 1 ,z 2 ) (15.16)Comparing(15.15)and(15.16),wededucethattheonlypossibleeigenvaluesare
eiδ=± 1 (15.17)Therefore,theonly2-particlephysicalstateswhichsatisfytheInterchangeHypothesis
arethesymmetricstatesψS(z 1 ,z 2 ),withtheproperty
ψS(z 1 ,z 2 )=+ψ(z 2 ,z 1 ) (15.18)andtheantisymmetricstatesψA(z 1 ,z 2 )
ψA(z 1 ,z 2 )=−ψA(z 2 ,z 1 ) (15.19)Allofthiseasilygeneralizestosystemscontaininganynumberofidenticalpar-
ticles. SupposethereareNsuchparticles,andthe corresponding wavefunction is
denotedψ(z 1 ,z 2 ,...,zN). DenotebyPEijtheoperatorwhichexchangesthei-thposi-
tionandspinwiththej-thpositionandspin,i.e.
PEijψ(z 1 ,z 2 ,..,zi,..,zj,...,zN)=ψ(z 1 ,z 2 ,..,zj,..,zi,...,zN) (15.20)ThenthestateswhichareallowedbytheInterchangehypothesisarethesymmetric
states,satisfying
ψS(z 1 ,z 2 ,..,zi,..,zj,...,zN)=+ψS(z 1 ,z 2 ,..,zj,..,zi,...,zN) (15.21)andtheantisymmetricstates,satisfying
ψA(z 1 ,z 2 ,..,zi,..,zj,...,zN)=−ψA(z 1 ,z 2 ,..,zj,..,zi,...,zN) (15.22)