U and U² Perturbation

A particle is moving in the three-dimensional harmonic oscillator with potential energy V(r). A weak perturbation δV is applied:

(i)

(ii)

The same small constant U occurs in both terms of δV. Use perturbation theory to calculate the change in the ground state energy to order O(U²).

SOLUTION

The result from first-order perturbation theory is obtained by taking the integral of the perturbation δV with the ground state wave function ѱ_g:

(1)

(2)

The ground state energy is E₀ = 3hω/2. The first term in δV has odd parity and integrates to zero in the above expression. The second term in δV has even parity and gives a nonzero contribution. In this problem it is easiest to keep the eigenfunctions in the separate basis x, y, z of rather than to combine them into r. In one dimension the average of x² = ⟨0|x²|0⟩, so we have

(3)

(4)

where x²₀ = h/mω. This is probably the simplest way to leave the answer. This completes the discussion of first-order perturbation theory.

The other term Uxyz in δV contributes an energy of O(U²) in secondorder perturbation theory. The excited state must have the symmetry of xyz, which means it is the state ѱ_ex(r) = ѱ₁(x) ѱ₁(y) ѱ₁(z). This has three quanta excited, so it has an energy E_ex = 9hω/2:

(5)

(6)

(7)

Now we combine the results from first- and second-order perturbation theory:

(8)