Question

A particle of mass m is moving in the three-dimensional harmonic oscillator potential V(x, y, z) = 1/2 mω²(x² + y² + z²). A weak perturbation is applied in the form of the function AV = kxyz + \(\frac{k^2}{h \omega} x^2 y^2 z^2\), where k is a small constant. Note the same constant k appears in both terms.

(a) Calculate the shift in the ground state energy to second order in k.

(b) Using an argument that does not depend on perturbation theory, what is the expectation value of x in the ground state of this system?

Answer 1

(a) The ground state of a particle in the potential well of three-dimensional harmonic oscillator is

While the perturbation \(\Delta\)V’ = kxyz does not give rise to first order correction, it is to be considered for second order perturbation in order to calculate the energy shift accurate to k². Its perturbation Hamiltonian has matrix elements

where n = n₁ + n₂ + n₃ and so the second order energy correction is

Therefore the energy shift of the ground state accurate to k² is

(b) V + AV is not changed by the inversion x → -x, y → -y, .i.e.,

Furthermore the wave function of the ground state is not degenerate, so \(\psi\)(--2, -y, z) = \(\psi\)(x, y, z) and, consequently,

where we have applied the transformation x'= -x, y'= -y, z’ = z.

Hence (x) = 0. In the same way we find (y) = 0, (z) = 0. Thus (x) = 0.