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Two interacting spin 1/2 particles in a square well

Two identical spin-1/2 particles are confined to an infinite one-dimensional square well of width a with infinite potential barriers at x=0 and x=a. The potential is V(x)=0 for 0 <= x <= a. Suppose that the particles interact weakly by the potential V_1(x)=Kdelta(x_1 - x_2), where x_1 and x_2 are the positions of the two particles , K is a constant, and delta is the Dirac delta function. This represents a very short-range weak force between the two particles.

a) Using first-order perturbation theory, find the perturbation to the energy of the lowest-energy singlet state.

b) Show that the first-order perturbation to the energy of the lowest-energy triplet state is zero.

c) What is the physical reason for the answer in part (b)?

Solution Preview

Let's denote the unperturbed eigenstates for a single spinless (scalar) particle as |n>. The wavefunction of these states is given by:

<x|n> = psi_n(x) = sqrt(2/a) sin(n pi x/a)

If we have a spin 1/2 particle then the eigenstates can be written as a tensor product:

|n>|s>

where |s> denotes the spin part of the state. The energy doesn't depend on the spin so, the energy levels are two fold degenerate. If we have two spin 1/2 paticles then we could denote a general eigenstate as:

|n1,n2>|s1,s2>

where we've combined the two spin and two configuration parts into a single kets for each. The first entry always refers to one particle, the second to the other. Now, for spin 1/2 particles, the state is required to be anti-symmetric under interchange of the two particles. This means that the above state is not a valid state. Instead we have a state of the form:

|space>|spin>

where either |spin> is ...

Solution Summary

We consider two spin 1/2 particles in a square well. We show in detail how the degeneracy between triplet and singlet states is lifted due to a perturbation V(x)=Kdelta(x_1 - x_2), where x_1 and x_2 are the positions of the two particles.

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