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In Griffiths' degenerate perturbation chapter, he mentions how finding symmetries of the original and perturbing Hamiltonians can simplify the process of first order degenerate perturbations. The claim was that given some hermitian operator $A$ such that it commutes with both the original and perturbing hamiltonian, we can try to find certain linear combinations of the degenerate hamiltonian that is also an eigenfunction of $A$ with distinct eigenvalues. These linear combinations are the 'good' linear combinations to use, since we can just use 'ordinary' perturbation theory.

One example from Griffiths was the cubical square well with $V(x,y,z)=0$ when $0<x,y,z<a$ and infinite elsewhere. He introduces the perturbation $H'=V_0$ for $0<x,y<a/2$ and wants to find the first order energy corrections for the 1st excited state, which is three-fold degenerate. Since the stationary states have the form $$\psi_{n_x,n_y,n_z}\propto\sin\left(\dfrac{n_x\pi}{a}x \right)\sin\left(\dfrac{n_y\pi}{a}y \right)\sin\left(\dfrac{n_z\pi}{a}z \right)$$ I tried to take advantage of the symmetry using the swap operator $P_{xy}$ which swaps the $x$ and $y$ value. Therefore, I have: $$P_{xy}\psi_{112}=\psi_{112}$$ $$P_{xy}(\psi_{121}+\psi_{211})=\psi_{121}+\psi_{211}$$ $$P{xy}(\psi_{121}-\psi_{211})=-(\psi_{121}-\psi_{211})$$ So I found three 'good' states: $\psi_a=\psi_{112}$, $\psi_b=\psi_{121}+\psi_{211}$ and $\psi_c=\psi_{121}-\psi_{211}$. The problem is that Griffths asserts that these states must also have different eigenvalues with $P_{xy}$, but the first two of these states have eigenvalue of 1. How do I go about this? If I choose to work in this basis, then does it mean I can guarantee the matrix elements $\langle \psi_a | H'|\psi_c\rangle = 0$ and $\langle \psi_b | H'|\psi_c\rangle = 0$ since these two states have different eigenvalues? This does not seem to be true, since Griffths answer shows that $\langle \psi_b | H'|\psi_c\rangle \neq 0$. What am I doing wrong?

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You have confused your notation with the notation in Griffiths. In Griffiths, $\psi_a,\ \psi_b$ and $\psi_c$ are $\psi_{112}, \ \psi_{121}$ and $\psi_{211}$. These are not eigenstates of the perturbation. On the other hand, the states you write in your answer are indeed eigenstate of the perturbation, and therefore are the ‘good states’.

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