Consider the exchange operator P12 whose action on the X basis is
P12∣x1,x2⟩=∣x2,x1⟩
((1)) Show that P12 has eigenvalues ±1. (It is Hermitian and unitary.)
((2)) Show that its action on the basis ket ∣ω1,ω2⟩ is also to exchange the labels 1 and 2, and hence that VS/A are its eigenspaces with eigenvalues ±1.
((3)) Show that P12X1P12=X2, P12X2P12=X1 and similarly for P1 and P2. Then show that P12Ω(X1,P1;X2,P2)P12=Ω(X2,P2;X1,P1). [Consider the action on ∣x1,x2⟩ or ∣p1,p2⟩, for the functions of X and P, assume they are given by power series and consider any term in the series. If you need help, peek into the discussion leading to Eq. (11.2.22).]
((4)) Show that the Hamiltonian and propagator for two identical particles are unaffected under H→P12HP12 and U→P12UP12. Given this, show that any eigenstate of P12 continues to remain an eigenstate with the same eigenvalue as time passes, i.e., elements of VS/A never leave the symmetric or antisymmetric subspaces they start in.