Consider the eigenvalue equation of eq262 for a three-dimensional infinite cubical well, where

where .

From eq278, the ground state is , while the first excited state is characterised by , and . It is obvious, from eq279, that the first excited state is degenerate. Let’s assume that the potential in the well is perturbed such that first-order correction to is

We shall analyse the degenerate subspace of the first excited state. The first step is to compute the matrix elements and substitute them in eq274, with the assumption that :

Finding the non-trivial solutions of the above linear homogeneous equation,

The roots are , and , which means that the perturbation lifts the degeneracy of the unperturbed states. To find the exact eigenstates , we substitute the roots back in eq281. For the first root, we have

Comparing both sides of the above equation, can be any number but and . So, the normalised wavefunction associated with is . For the second root, we have

Comparing both sides, since , and ( and are not equal to zero because that will result in being a zero eigenfunction). The normalised wavefunction associated with is . Similarly, for , we have .

Finally, by substituting , and in eq273, we find that they diagonalise . We call these linearly combined states that diagonalise “good” states.