Question

Two particles, each of mass M, are bound in a one-dimensional harmonic oscillator potential V = \(\frac 12\)kx² and interact with each other through an attractive harmonic force \(F_{12} = - K(x_1 - x_2)\). You may take K to be small.

(a) What are the energies of the three lowest states of this system?

(b) If the particles are identical and spinless, which of the states of (a) are allowed?

(c) If the particles are identical and have spin 1/2, what is the total spin of each of the states of (a)?

Answer 1

The Hamiltonian of the system is

The system can be considered as consisting of two independent harmonic oscillators of angular frequencies \(\omega _1\) and \(\omega _2\) given by

where n, m = 0,1,2,... , and \(\psi^{(k)}_n\) is the nth eigenstate of a harmonic oscillator of spring constant k.

(a) The energies of the three lowest states of the system are

(b) If the particles are identical and spinless, the wave function must be symmetric with respect to the interchange of the two particles. Thus the states \(|00\rangle\), \(|10\rangle\) are allowed, while the state \(|01\rangle\) is not allowed.

(c) If the particles are identical with spin 1/2, the total wave function, including both spatial and spin, must be antisymmetric with respect to an interchange of the two particles. As the spin function for total spin S = 0 is antisymmetric and that for S = 1 is symmetric, we have