two fold degeneracy perturbation question

[mmwave]

consider a two fold degeneracy such that
H Psi_a = E Psi_a and H Psi_b = E Psi_b and <Psi_a | Psi_b> = 0

All of the above are the unperturbed states, Hamiltonian and eigenvalue. Notice the two states share the eigen value E.

Form the linear combination of the two states

Psi = a * Psi_a + b * Psi_b

Clearly H Psi = E Psi and the eigenvalue of the linear combination is still the value E.

Now a perturbation to the hamiltonian H will split the energy level into two values E+ and E- and two states Psi+ and Psi-.

Finally the question:

I am told that if the perturbation is "dialed down to zero" then Psi+ and Psi- reduce to two different linear combinations of Psi_a and Psi_b. What does this mean?

Shouldn't the states be simply Psi_a and Psi_b? Why would there be two unique linear combinations instead of just the two eigenstates themselves?

[lethe]

Originally posted by mmwave
Shouldn't the states be simply Psi_a and Psi_b? Why would there be two unique linear combinations instead of just the two eigenstates themselves?

why would you think it would be Psi_a and Psi_b? you know that if the perturbation is nonzero, than the eigenstates are not necessarily Psi_a, Psi_b.

any linear combination is an eigenstate of the unperturbed hamiltonian. the perturbation just chooses a particular linear combination that diagonalize the perturbation as well.

[mmwave]

Originally posted by lethe
why would you think it would be Psi_a and Psi_b? you know that if the perturbation is nonzero, than the eigenstates are not necessarily Psi_a, Psi_b.



Because the question comes when the perturbation is removed. It seems to me that the system should go back to where it was with eigenstates Psi_a and Psi_b. I think there may be an ustated assumption I am ignorant of here.

[lethe]

Originally posted by mmwave
Because the question comes when the perturbation is removed. It seems to me that the system should go back to where it was with eigenstates Psi_a and Psi_b. I think there may be an ustated assumption I am ignorant of here.

when the perturbation is very very small, you get two orthogonal eigenstates that have very very close to the same eigenvalue.

when the perturbation is zero, the two orthogonal eigenstates have exactly the same eigenvalue. they are now degenerate.

all you have done is choose a basis for the degenerate eigenspace, when there was no good choice before you turned on the perturbation.