"Good states" in degenerate perturbation theory

Question

So, we recently covered degenerate (time independent) perturbation theory in my quantum 2 course (undergrad). I feel pretty ok about obtaining the energy correction values (find the eigenvalues of $H'$ from the "secular equation" is at least how I understood it in a procedural sense). However, I'm really having a problem understanding what Griffith's (textbook for class) and a vast majority of resources I have found mean by "good states". So, supposing a first order energy correction "lifts" a doubly degenerate eigenvalue into 2 different eigenvalues, how do I go about finding the wavefunctions for the system with the hamiltonian $H=H_0+H'$? I understand that the "good states" are the eigenvectors/eigenfunct.s/eigenstates of $H'$, and, since they are linear combos of the eigenstates of H₀ then they are also an eigenstate of H₀. What I don't understand, is what this means/how it can be used/ what the significance of these "good states" are. I'm under the impression that they have something to do with finding the first order approximation of the eigenfunctions for the H=H₀+H' system, but, other than that I'm fairly lost on how this is actually carried out. To rephrase... consider a doubly degenerate energy level of some system with hamiltonian, H₀, where |n,k>=|k> (only considering a single n/energy level so can just use k) for all k=1,2 are the eigenstates of this level/energy eigenvalue. Now suppose we have a system with hamiltonian,H=H₀+H', where H' is a small perturbation. Now say we find the first order energy corrections to the energy eigenvalue of H₀ which, upon correction, gives us the energy spectrum of the corresponding states for the H=H₀+H' system post-perturbation. (maps E for this level in unperturbed system to two E levels in the perturbed system). Now, suppose I want to find the wavefunctions for these definite energy states in the H=H₀+H' system. How is this done?

Answer 1

Zeroth order in perturbation theory

The starting point for perturbation theory is zeroth order, when we ignore the perturbation and we focus on the free Hamiltonian, $H_0$. In order for degenerate perturbation theory to be relevant, it must be the case that the free Hamiltonian has at least one degenerate energy level. Any remnant symmetry can be used to classify the states after the degeneracy is lifted.

To make this concrete, suppose at energy $E$, there is a 2-dimensional degenerate subspace. Let's say that $|a\rangle$ and $|b\rangle$ form a basis for this subspace. Then, there are an infinite number of states that have energy $E$, of the form $c_1 | a\rangle + c_2 |b\rangle$, where $|c_1|^2 + |c_2|^2 = 1$. More concretely, for all pairs of complex numbers $c_1$ and $c_2$ obeying the normalization condition, \begin{equation} H_0 \left(c_1 | a \rangle + c_2 | b \rangle \right) = E\left(c_1 | a \rangle + c_2 | b \rangle\right) \end{equation} In other words, all states in this two dimensional space have the same energy $E$, and all of them are eigenstates of the free Hamiltonian.

First order in perturbation theory

Now, we go to first order in perturbation theory, and add the perturbation Hamiltonian, $H_1$. Typically, the result of the perturbation is to lift the degeneracy of the states with energy $E$. This means that the energy level $E$ will be split into two energy levels, one with energy $E + \delta E_1$ and one with energy $E + \delta E_2$, with $\delta E_1 \neq \delta E_2$.

As a result, general state of the form $c_1 |a \rangle + c_2 | b \rangle$, which was an eigenstate of the free Hamiltonian $H_0$ with energy $E$, will not be an eigenstate of the full Hamiltonian $H_0 + H_1$.

However, there will be a special pair of eigenstates of the free Hamiltonian, which we will call $|A \rangle$ and $|B \rangle$, which are also eigenstates of $H_0 + H_1$, with energies $E+\delta E_1$ and $E + \delta E_2$, respectively. These are the so-called "good states." Let me go through some properties of them, and then talk about how to find them.

Properties of "good" and "bad" states

Let's write out some properties of these special states $|A \rangle$ and $|B \rangle$, as well as the not-so-special states $|a\rangle$ and $|b\rangle$, in equations.

• First, since $|A\rangle$ and $|B\rangle$ are members of the degenerate subspace of $H_0$ with energy $E$, they can be expanded in terms of $|a\rangle$ and $|b \rangle$ \begin{eqnarray} |A \rangle &=& \alpha_1 | a \rangle + \alpha_2 | b \rangle \\ | B \rangle &=& \beta_1 | a \rangle + \beta_2 | b \rangle \end{eqnarray} where $\alpha_{1, 2}$ and $\beta_{1,2}$ are complex numbers (you can think of them as specific choices of $c_1$ and $c_2$). You can also express $|a \rangle$ and $|b\rangle$ in terms of $|A\rangle$ and $B\rangle$ \begin{eqnarray} |a \rangle &=& \gamma_1 | A \rangle + \gamma_2 | B \rangle \\ | b \rangle &=& \lambda_1 | A \rangle + \lambda_2 | B \rangle \end{eqnarray}
• Second, $|A\rangle$ and $|B\rangle$ are eigenstates of the free Hamiltonian $H_0$ with energy $E$, meaning \begin{eqnarray} H_0 | A \rangle &=& E | A \rangle \\ H_0 | B \rangle &=& E | B \rangle \end{eqnarray} Of course, $|a \rangle$ and $|b\rangle$ are also eigenstates of $H_0$ \begin{eqnarray} H_0 | a \rangle &=& E | a \rangle \\ H_0 | b \rangle &=& E | b \rangle \end{eqnarray}
• Third, $A\rangle$ and $|B\rangle$ are eigenstates of $H_0 + H_1$, but now with different energies \begin{eqnarray} \left(H_0 + H_1\right) | A \rangle &=& \left(E + \delta E_1\right) | A \rangle \\ \left(H_0 + H_1\right) | B \rangle &=& \left(E + \delta E_2\right) | B \rangle \end{eqnarray} However, $|a \rangle$ and $| b \rangle$ are not eigenstates of $H_0 + H_1$, since \begin{eqnarray} \left(H_0 + H_1\right) | a \rangle &=& \gamma_1 \left(E + \delta E_1\right) | A \rangle + \gamma_2 \left(E + \delta E_2\right) | B \rangle \\ \left(H_0 + H_1\right) | b \rangle &=& \lambda_1 \left(E + \delta E_1\right) | A \rangle + \lambda_2 \left(E + \delta E_2\right) | B \rangle \end{eqnarray}

Since we want to find eigenstates of the full Hamiltonian, $H_0 + H_1$, we clearly are going to want to work with $|A\rangle$ and $|B\rangle$ instead of $|a \rangle$ and $|b\rangle$ (or any of the other infinite number of states that are eigenstates of $H_0$ with energy $E$). However, finding $|A\rangle$ and $|B\rangle$ can be tricky.

Finding good states

I won't attempt to give a general algorithm for constructing the good states (although, it can be done). Instead, I will just point out that often, degeneracies in the spectrum of the free Hamiltonian arise because there is a symmetry. The perturbation Hamiltonian tends to break symmetries.

To be concrete, let's suppose that $H_0$ is rotationally symmetric. This means that it commutes with all three components of the angular momentum, $L_i$, where $i=\{x, y, z\}$. We have \begin{equation} [H_0, L_i] = 0 \end{equation} This is the origin of labeling states of Hydrogen by $|n, \ell, m \rangle$; the $\ell, m$ quantum numbers are eigenvalues associated with $L^2$ and $L_z$, respectively.

Now, typically, the perturbation Hamiltonian breaks the symmetry, so some of the symmetry operators will not commute with $H_1$. Let's consider the Stark effect, where we add an electric field in the $z$ direction. Then $L_x$ and $L_y$ will not commute with $H_1$. As a result, we expect states with different $\ell$ values (the eigenvalue of $L^2$) to have different energies -- the loss of symmetry has lifted the degeneracy that was previously present. However, $L_z$ does commute with $H_1$ (since the system is still symmetric with respect to rotations about the $z$ axis). This means that $L_z$ satisfies the property \begin{equation} [H_0, L_z] = [H_1, L_z] = 0 \end{equation} This means we can simultaneously diagonalize the free Hamiltonian, $H_0$, the full Hamiltonian, $H_0 + H_1$, and the $z$-component of the angular momentum operator $L_z$. Eigenstates of $L_z$ are easy to find. And, since these states are eigenstates of both $H_0$ and $H_0+H_1$, they will satisfy all the properties of the $|A\rangle$ and $|B\rangle$ states given above -- the eigenstates of $L_z$ are the good states.

In general, if we can find a Hermitian operator $O$ that commutes with $H_0$ and $H_1$, then eigenstates of $O$ are good states for degenerate perturbation theory.