- #1
mmwave
- 647
- 2
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?
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?