Question

1. (25) Consider an infinite potential well with the following shape: 0 a/4 3al4 a h2 where 4 Using the ground state wavefunction of the original infinite potential well as a trial function, 2πχ trial = 1-sin- find the approximation of the ground state energy for this system with the variational method. (Note, this question is simplified by considering the two components of the Hamiltonian, and V, on their own) b) If we had used the 1st excited state wavefunction(n 2) as our trial function, we would not be approximating the ground state energy of our new system. Why not? (hint: think of the centre of the well as the origin) c) For the same system, use 1st order perturbation theory to approximate the ground state energy Compare this answer with your result in part a. What does the difference tell you? d) In considering the magnitude of Vo, discuss whether or not it would be reasonable to use 2nd order perturbation theory to approximate the energies. e) One minor flaw of using perturbation theory is that it forces us to slightly ignore the Copenhagen interpretation (wave function collapse). Why is this the case?

Quantum Mechanics Problem

Answer 1

(a) psi_trial = squareroot 2/a sin pi x/a E = < psi/H/psi > Hamiltonian of the system is given by H = -h^2/2m d^2/dx^2 + 1/2 m w^2 x^2 zero for this case Thus, E = < psi/ - h^2/2m d^2/dx^2 + 1/2 mw^2 x^2/psi > rightarrow (1) solve its two parts separately < psi/-h^2/2m d^2/dx^2 Vertical-bar psi > = -h^2/2m integral^a_0 squareroot 2/a sin pi x/a (d^2/dx^2) squareroot 2/a sin (pi x/a) dx on differentiation -h^2/2m times 2/a integral^a_0 -sin (pi x/a) sin (pi x/a) (pi/a)^2 dx h^2/2m times (pi/a)^2 times 2/a integral^a_0 sin^2 (pi x/a) dx h^2/2m pi^2 2/a^3 [integral^a_0 (1/2-1/2 cos (2 pi x/a)) dx] h^2 pi^2 2/2 ma^3 [x/2 - 1/2 sin (2 pi x/a) (a/2 pi)]^a_0 < psi Vertical-bar -h^2/2m d^2/dx^2 Vertical-bar psi > = 2 pi^2 h^2/2m a^3 times a/2 < psi Vertical-bar -h^2/2m d^2/dx^2 Vertical-bar psi > = pi^2 h^2/2ma^2 rightarrow (2) \r\n H = k.E + p.E as potential inside well is zero thus v(x) = 0 E_0 = psi Vertical-bar -h^2/2m d^2/dx^2 Vertical-bar psi > + < psi Vertical-bar v(x) Vertical-bar psi > e_0 = pi^2 h^2/2ma^2

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