Question

Two particles of mass m are placed in a rectangular box of sides a > b > c in the lowest energy state of the system compatible with the conditions below. The particles interact with each other according to the potential V = A\(\delta\)(r₁ - r₂). Using first order perturbation theory to calculate the energy of the system under the following conditions:

(a) Particles not identical.

(b) Identical particles of spin zero.

(c) Identical particles of spin one-half with spins parallel.

Answer 1

(a) The unperturbed system can be treated as consisting of two separate single-particle systems and the wave function as a product of two singleparticle wave functions:

The lowest energy state wave function is thus

First order perturbation theory gives an energy correction

(b) For a system of spin-0 particles, the total wave function must be symmetric for interchange of a pair of particles. Hence the lowest energy state is

which is the same as that in (a). The energy to first order perturbation is also

(c) For a system of spin-; particles the total wave function must be antisymmetric. As the spins are parallel, the spin wave function is symmetric and so the spatial wave function must be antisymmetric. As \(\frac 1{a^2} < \frac 1{b^2}<\frac 1{c^2}\), the lowest energy state is

where \(\psi\)₁₁₁(r) and \(\psi\)₂₁₁(r) are the ground and first excited single-particle states respectively. The unperturbed energy is