Principles of Quantum Mechanics · Problem 10.3.5

Symmetries and Their Consequences — R. Shankar

Physics — Quantum Mechanics · exchange-operator

Problem

Consider the exchange operator P12P_{12} whose action on the XX basis is P12∣x1,x2⟩=∣x2,x1⟩P_{12}|x_1, x_2\rangle = |x_2, x_1\rangle ((1)) Show that P12P_{12} has eigenvalues ±1\pm 1. (It is Hermitian and unitary.) ((2)) Show that its action on the basis ket ∣ω1,ω2⟩|\omega_1, \omega_2\rangle is also to exchange the labels 1 and 2, and hence that VS/A\mathbb{V}_{S/A} are its eigenspaces with eigenvalues ±1\pm 1. ((3)) Show that P12X1P12=X2P_{12}X_1P_{12} = X_2, P12X2P12=X1P_{12}X_2P_{12} = X_1 and similarly for P1P_1 and P2P_2. Then show that P12Ω(X1,P1;X2,P2)P12=Ω(X2,P2;X1,P1)P_{12}\Omega(X_1, P_1; X_2, P_2)P_{12} = \Omega(X_2, P_2; X_1, P_1). [Consider the action on ∣x1,x2⟩|x_1, x_2\rangle or ∣p1,p2⟩|p_1, p_2\rangle, for the functions of XX and PP, 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→P12HP12H \to P_{12}HP_{12} and U→P12UP12U \to P_{12}UP_{12}. Given this, show that any eigenstate of P12P_{12} continues to remain an eigenstate with the same eigenvalue as time passes, i.e., elements of VS/A\mathbb{V}_{S/A} never leave the symmetric or antisymmetric subspaces they start in.

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