This is a nice little problem for a Wednesday morning (I didn't actually solve the problem this morning, I think it was sometime very late a couple nights ago, but it's been sitting in my notebook
and it's begging to be posted)! We will start by constructing our general eigenstate in the following form:
$$|\psi^{0}_g\rangle \ = \ \displaystyle\sum_{j} \alpha_j |\psi^{0}_j\rangle$$
Where each of the $|\psi_j^{0}\rangle$ state vectors are the degenerate state vectors for the eigenvalue $E^{0}$, for a system with $j$-fold degeneracy. Now, we know that for a small perturbation, we have:
$$H \ = \ H^{0} \ + \ \lambda H'$$
And our goal is to find the new eigenstates of the perturbed Hamiltonian. Thus, we want to solve the equation:
$$H|\psi'\rangle \ = \ E|\psi'\rangle \ \Rightarrow \ (H^{0} \ + \ \lambda H')(|\psi^{0}\rangle \ + \ \lambda |\psi^{1}\rangle \ + \ ...) \ = \ (E^{0} \ + \ \lambda E^{1} \ + \ ...)(|\psi^{0}\rangle \ + \ \lambda |\psi^{1}\rangle \ + \ ...)$$
Since our general state is an eigenstate of some $E^{0}$, we are allowed to plug it into our equation as $|\lambda^{0}\rangle$. We are also only looking for the first-order corrections, so we
will truncate the sums at the first-order terms in the Taylor expansion:
$$(H^{0} \ + \ \lambda H')(|\psi^{0}_g\rangle \ + \ \lambda |\psi^{1}_g\rangle) \ = \ (E^{0} \ + \ \lambda E^{1})(|\psi^{0}_g\rangle \ + \ \lambda |\psi^{1}_g\rangle)$$
Let's do a bit of expanding!
$$H^{0} |\psi^{0}_g\rangle \ + \ \lambda H^{0} |\psi^1_g\rangle \ + \ \lambda H' |\psi^0_g\rangle \ + \ \lambda^2 H' |\psi_g^1\rangle \ = \ E^{0} |\psi^{0}_g\rangle \ + \ \lambda E^{0} |\psi^1_g\rangle \ + \ \lambda E^{1} |\psi^0_g\rangle \ + \ \lambda^2 E^{1} |\psi_g^1\rangle$$
We then group terms using $\lambda$:
$$H^{0} |\psi^{0}_g\rangle \ + \ (H^{0} |\psi^1_g\rangle \ + \ H' |\psi^0_g\rangle) \lambda \ + \ \lambda^2 H' |\psi_g^1\rangle \ = \ E^{0} |\psi^{0}_g\rangle \ + \ (E^{0} |\psi^1_g\rangle \ + \ E^{1} |\psi^0_g\rangle) \lambda \ + \ \lambda^2 E^{1} |\psi_g^1\rangle$$
$$H^{0} |\psi^{0}_g\rangle \ = \ E^{0} |\psi^{0}_g\rangle \ \Rightarrow \ (H^{0} |\psi^1_g\rangle \ + \ H' |\psi^0_g\rangle) \lambda \ + \ \lambda^2 H' |\psi_g^1\rangle \ = \ (E^{0} |\psi^1_g\rangle \ + \ E^{1} |\psi^0_g\rangle) \lambda \ + \ \lambda^2 E^{1} |\psi_g^1\rangle$$
Now, we take the inner product of both sides with the $|\psi_j^0\rangle^{\dagger}$:
$$(\langle \psi^0_j | H^{0} |\psi^1_g\rangle \ + \ \langle \psi^0_j | H' |\psi^0_g\rangle) \lambda \ + \ \lambda^2 \langle \psi^0_j | H' |\psi_g^1\rangle \ = \ (\langle \psi^0_j | E^{0} |\psi^1_g\rangle \ + \ \langle \psi^0_j | E^{1} |\psi^0_g\rangle) \lambda \ + \ \lambda^2 \langle \psi^0_j | E^{1} |\psi_g^1\rangle$$
$$\langle \psi^0_j | H^{0} |\psi^1_g\rangle \ = \ E^{0} \langle \psi^0_j | \psi^1_g\rangle \ \Rightarrow \ \langle \psi^0_j | H' |\psi^0_g\rangle \lambda \ + \ \lambda^2 \langle \psi^0_j | H' |\psi_g^1\rangle \ = \ \langle \psi^0_j | E^{1} |\psi^0_g\rangle \lambda \ + \ \lambda^2 \langle \psi^0_j | E^{1} |\psi_g^1\rangle$$
Now, by the uniqueness of power series representations, we have:
$$\langle \psi^0_j | H' |\psi^0_g\rangle \ = \ \langle \psi^0_j | E^{1} |\psi^0_g\rangle \ = \ E^{1} \langle \psi^0_j | \displaystyle\sum_{i} \alpha_i |\psi^{0}_i\rangle \ = \ E^{1} \alpha_j$$
We just changed the variable we sum over to $i$ because we already used $j$. Now, we have:
$$\displaystyle\sum_i \alpha_i \langle \psi^0_j | H' | \psi_i^0 \rangle \ = \ E^1 \alpha_j \ \Rightarrow \ \displaystyle\sum_{i} \alpha_i W_{ji} \ = \ E^{1} \alpha_j$$
This is equivalent to the system of eigenvalue equations for the $W$ matrix (which is just the $H'$ matrix expressed in the basis of energy eigenstates). Basically, this equation is saying
that the $W$ matrix acting upon the general eigenstate must be equivalent to multiplying the eigenstate by $E^1$, thus the first-order perturbed energies are simply the eigenstates of the $W$ matrix.
Now, let's consider the "good" states. The goal of finding these good states is essentially to minimize our work, so we don't have to solve a system of eigenvalue equations. Consider our general vector once more:
$$|\psi_g^0\rangle \ = \ \displaystyle\sum_j \alpha_j |\psi_j^0\rangle$$
Now, the only condition we imposed upon $|\psi_j^0\rangle$ is that it is an eigenstate of $H^0$. Note that in a system with degeneeracy linear combinations of degenerate states corresponding to the same energy are also eigenstates with that energy.
Thus, there are a bunch of different bases in which we can express our general state vector $|\psi_g^0\rangle$, but we have no idea which one to choose! Well, we can actually choose a basis that makes our calculations easier, which Griffiths calls the "good" states!
Essentially, we want to choose the basis of eigenstates of the $W$ matrix. If we do this, then we get:
$$\displaystyle\sum_i \alpha_i \langle \psi^0_j | H' | \psi_i^0 \rangle \ = \ \displaystyle\sum_i V \alpha_i \langle \psi^0_j | \psi_i^0 \rangle \ = \ V \alpha_j \langle \psi^0_j | \psi_j^0 \rangle \ = \ \alpha_j \langle \psi^0_j | H' | \psi^0_j \rangle \ = \ E^{1} \alpha_j$$
Where $V$ is the eigenvalue. We also know that the eigenvalues of $H'$ are the first order energies, $E^1$, so, since for the "good" states, all terms in the sum become $0$ except for $\langle \psi^0_j | H' | \psi^0_j \rangle$, we simply get:
$$\langle \psi^0_j | H' | \psi^0_j \rangle \ = \ E^1$$
Where each $|\psi_j^0\rangle$ is a "good" state (eigenvector of the $W$ matrix). Thus, instead of solving a large system of equations, if we choose a basis that sends all terms of the sum to $0$ except for one, we can simply evaluate the inner products of this form to get the first-order energies. This is much less messy thsan solving a massive
system of linear equations. For a $j$-fold degenerate system, we will have $j$ eigenstates, so we will get $j$ new, perturbed energies, which is the "lifting" of energies that Griffiths discussed in the beginning of the chapter.