Answer to Question #202902 in Quantum Mechanics for Swagata Ghosh

Question

Consider Harmonic oscillator Hamiltonian in 2-D

(P_x)^2/2+(p_y) ^2/2+x²/2+y^2/2+lamda×x×y

Find the ground state energy and energy of first exited state

Answer 1

The Harmonic Oscillator Hamiltonian is given by,

$H = \frac{p^2}{2m} + \frac{1}{2} m^2 x^2$

The differential equation to be solved is given as -

$[\frac{(\bar{h})^2}{ 2m}] \frac{d^2 u }{ dx^2 }+ (0.5) m ^2 x^2 u = E u$

The energy eigenvalues are given by,

E_n = (n + \frac{1}{2}) hbar 

for n = 0, 1, 2, ...There are a countably infinite number of solutions with equal energy spacing.

The ground state wave function is given as :

$u_o (x) = (\frac{m }{ hbar})^\frac{1}{4}e^{-m x^2 / 2 hbar}$

This is a Gaussian (minimum uncertainty) distribution.

The first excited state is an odd parity state, with a first-order polynomial multiplying the same Gaussian.

$u_1 (x) = (\frac{m }{ hbar})^\frac{1}{4}\frac{2m}{2 hbar}e^{-2m / 2 hbar}$

The second excited state is even parity, with a second order polynomial multiplying the same Gaussian.

$u_2 (x) =C(1- \frac{2mx^2}{ hbar})e^{-2mx^2/ 2 hbar}$