Question

We've been learning in class that when approaching a perturbed Hamiltonian with degeneracies you should try to choose a basis made of eigenstates of both the unperturbed Hamiltonian and the perturbation. But at that point, why do you need perturbation theory? Don't you now have the exact solutions to the Schrodinger equation?

Answer 1

Choosing a basis of eigenstates of both the unperturbed Hamiltonian and the perturbation doesn't provide exact solutions to the Schrödinger equation, but simplifies the problem. Perturbation theory is still needed to evaluate state mixing and energy level corrections, especially in complex systems without exact solutions.

Step-by-step explanation:

When learning about perturbation theory and dealing with a degenerate perturbed Hamiltonian, choosing a basis of eigenstates of both the unperturbed Hamiltonian and the perturbation can be very helpful.

However, even when such a convenient basis is found, it doesn't mean that we have the exact solutions to the Schrödinger equation. Perturbation theory is still needed to determine how the eigenstates mix and evolve due to the perturbation, as well as to calculate the corrections to the energy levels, especially for complex systems where exact solutions are not feasible.

Solving the time-independent Schrödinger equation for a quantum particle in a box is one of the simpler quantum mechanics problems, which provides insights into quantized energy states resulting from standing wave conditions.

However, for more complex or realistic systems, such as those with varying potentials or interactions, exact solutions are often unattainable, and perturbation theory becomes a valuable tool for approximating solutions and understanding the behavior of quantum systems.