Question

Consider the dimensionless harmonic oscillator Hamiltonian, (where m = h̄ = 1).

with p=_ïda 2 2

Consider the orthogonal wave functions $\Psi _{0}(x) = e^{-x^{2}/2}$ and , which are eigenfunctions of H with eigenvalues 1/2 and 5/2, respectively.

Answer 1

Given { H^^psi_0 (x) = 1/2 e^-x^2/2 = 1/2 phi_0 (x) H^^psi_0 (x) = 5/2 psi_2 (x) phi_0 (x) = A_0 psi_0 (x) Normalisation condition for phi_0 (x) integral^infinity_-infinity phi^_0 (x) phi_0 (x) dx = 2 = > A^2_0 integral^infinity_-infinity e^-x^2 dx = 2 A^2-0 squareroot pi = 2 A_0 = 1/pi 1/4 since integral^infinity_-infinity x^2n e^-d^2 dx = squareroot n + 1/2/alpha^n + 1/2 \r\n Normalisation for phi_2 (2) integral^infinity_-infinity Vertical-bar phi^2_alpha (x) Vertical-bar ^2 dx = 1 = > A^2_2 integral^+ infinity_-infinity (1 - 2x^2)^2 e^-x^2 dx = 2 A^2_2 [integral^infinity_-infinity e^-x^2 dx + 4 integral^infinity_-infinity x^4 e^-x^2 dx - 2 integral^infinity_-infinity x^3 e^-x^2 A^2_2 [ + 4 3/2 2/2 squareroot pi + (-2) times 1/2 squareroot pi] = 2 A^2_2 [squareroot pi + 3 squareroot pi - squareroot pi] = 1 A_2 1/( pi_^1/4 A_2 = 1/(9 pi)^1/4 psi (x, 0) = B(phi_0 (x) + 2/3 phi_2 (x)) Normalisation of psi (x, 0): integral^infinity_-infinity psi (x, 0) psi (x, 0) dx =