As far as I gather, before a perturbation is applied, the eigenspace associated with the degenerate energy is multidimensional but after applying the perturbation this space 'splits' into different eigenspaces. If we use Non-degenerate perturbation theory, then we would end up dividing by zero when calculating the coefficients for the linear combination of the perturbed state. The way to get around this is apparently to diagonalise the perturbation Hamiltonian with a basis of the unperturbed degenerate eigenstates, and this is what I fail to understand.

Why is it that diagonalising the perturbation Hamiltonian will solve the problem of dividing by zero? Also how can I understand visually what is happening to the linear transformation on the eigenspace before and after diagonalising the perturbation Hamiltonian?

Another problem I have is understanding the process of calculating the energy and state corrections after finding a basis in which the perturbation Hamiltonian is diagonalised. Every example or explanation I have found is incredibly abstract. I would love to see a simple example with matrices and vectors if possible if anybody has links to good resources or can be bothered to go through one in an answer. I find matrix/vector representations to be much more intuitive.

Lastly, another question that I have just thought of: Are these eigenstates that we find to diagonalise the perturbation Hamiltonian with precisely the eigenstates of the new total Hamiltonian? And in that case, are they regarded as the first order corrections to the space of eigenstates whatever that would mean?


2 Answers 2


This is a good question, on an issue that seem to confuse many students. I will try to clarify to the best of my ability the issue with degenerate perturbation theory.

When we have a degeneracy in the Hamiltonian, a problem is that there is no 'true' preferred basis to work with. Any rotation within a degenerate sub-space is allowed without changing the fact that the Hamiltonian is still diagonal.

Let's consider, for example, the angular-momentum Hamiltonian $\mathcal{H} = \frac{\omega}{\hbar}{\bf L}^2$, and consider a system with total angular momentum of $l=1$. We approach the problem and decide to diagonalize it in the usual basis of $|l=1, l_z = -1\rangle$, $|l=1, l_z = 0\rangle$ and $|l=1, l_z = +1\rangle$. However all this states have identical energy of $E = 2\hbar\omega$. Therefore, the basis $$|l=1, l_z = -1\rangle, \frac{1}{\sqrt{2}}(|l=1, l_z = +1\rangle \pm |l=1, l_z = 0\rangle)$$ is also legitimate and diagonalize the Hamiltonian.

When we apply perturbation theory, we always seek corrections to a specific state with specific energy. But in the degenerate case we face a problem - what is the correct state to start working from? What is the 'true' basis state to which we need to seek perturbative corrections? This question does not arise in the case of a nondegenerate subspace of the Hamiltonian.

So, here, the perturbation itself determines this for us. Perturbation theory is a tool to help us analyze the Hamiltonian with the perturbation, so it might be (and indeed, in many cases) that the full Hamiltonian is non degenerate. The thing that lifts the degeneracy is the perturbation. So we look in each degenerate subspace of the Hamiltonian, and diagonalie the perturbation within this subspace - this will tell us what is the correct unperturbed basis to work with. This is another important point - the state that we find at the first step are still eigenstates of the unperturbed Hamiltonian. They do not include the perturbation!

Now, once we have the correct basis, we can proceed in a similar manner to the nondegenerate perturbation theory: the first correction to the energy is just $\langle \psi^0 | V | \psi^0 \rangle$ with $|\psi^0\rangle$ the (correct) eigenstates of the unperturbed Hamiltonian, and the first order correction to the state itself comes from summing over matrix elements of the perturbation with states outside the degenerate subspace.

In the example above, let's say we added a perturbation $\omega_0 L_x$ with $\omega_0 \ll \omega$. Then it is clear that the correct basis to work with is the $L_x$ basis.

Lastly, it might be that the perturbation will not lift the degeneracy, or only partially lift it (for example, $\omega_0/{\hbar}L_z^2$ will leave $|l=1,l_z=\pm 1\rangle$ degenerate). In that case, we still have the freedom to choose the basis, and the perturbation theory will work for every choice.


From an operational point of view, the problem is that terms of the form

$$\frac{\langle n^{(0)} | \hat V | m^{(0)} \rangle}{E_n - E_m}$$

arise when summing over all $|m\rangle \neq |n\rangle$. If $|m^{(0)}\rangle$ and $|n^{(0)}\rangle$ are degenerate, then $E_n=E_m$ and we are dividing by zero, as you say.

If we choose the basis of the degenerate subspace such that $\hat V$ is diagonalized, however, then there are no such terms because $\langle n^{(0)} | \hat V | m^{(0)} \rangle=0$ for all $|n^{(0)}\rangle \neq |m^{(0)}\rangle$.

  • $\begingroup$ This is very much after this has been answered, but I still find myself flummoxed. All that has been shown is that, if the correct basis is chosen, we get $0/0$ (indeed, if we trace back the derivation for the correction to the eigenstate that you quote above, we see that we get a $0=0$ indeterminate equation), which essentially tells us that we've got to do something else to show that the term is 0. So how do we know that $\langle n^{(0)} | m^{(1)} \rangle=0$ for all $|n^{(0)}\rangle$ in this degenerate subsapce? $\endgroup$
    – EE18
    Commented Mar 20, 2021 at 12:45
  • 1
    $\begingroup$ @1729_SR I don't know what resource you're looking at, so I can't say exactly where the crucial step comes. However, the idea is that the entire procedure is modified so that the sum is not over all states $|m^{(0)}\rangle\neq |n^{(0)}\rangle$, but rather over all states $|m^{(0)}\rangle \notin \mathcal D_n$, where $\mathcal D_n$ is the degenerate subspace containing $|n^{(0)}\rangle$. If you look at the non-degenerate case, the $|n^{(0)}\rangle$ term is removed from the sum; for the same reason the whole subspace is removed in the degenerate case. $\endgroup$
    – J. Murray
    Commented Mar 20, 2021 at 15:04
  • $\begingroup$ I'm looking at the books by Townsend and Shankar, wherein each just handwave things away by saying "now the numerator and denominator are each 0, so the sum is just over all unperturbed eigenkets outside the degenerate subspace". I have consulted Sakurai (3rd edition) who actually makes a concerted effort to treat the case rigorously, but equation (5.87) there in Sakurai seems to imply that $\langle n^{(0)} | m^{(1)} \rangle \neq 0$ necessarily. I'm therefore at a loss. Sorry if this is taking us off on a tangent -- I can post a full question if you'd prefer. $\endgroup$
    – EE18
    Commented Mar 20, 2021 at 17:05
  • $\begingroup$ @1729_SR Let's discuss this in chat. $\endgroup$
    – J. Murray
    Commented Mar 20, 2021 at 17:19

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