# Why does degenerate perturbation theory work? [duplicate]

As is well-known, the first-order correction to the $$n$$th unperturbed eigenstates is given by $$|\psi^n_1 \rangle = \sum_m \frac{\langle m| H_1 |n \rangle}{\varepsilon_n - \varepsilon_m},$$ where I have used conventional notation and have for now assumed the unperturbed Hamiltonian $$H_0$$ is nondegenerate, so that $$\varepsilon_n - \varepsilon_m \neq 0$$. But if it's degenerate, then as is well-known we must take as our unperturbed states $$|n \rangle$$ those states which diagonlize $$H_1$$ in the corresponding subspace. My problem is why on earth this should do the trick. The crux, according to my book, is that if $$|n,r \rangle$$ denotes the various eigenstates which diagonalize $$H_1$$ in the degenerate subspace of $$\varepsilon_n$$ with $$r$$ indexing the different degenerate eigenstates, then $$\langle n,r|H_1 |n,r'\rangle = 0$$ for $$r \neq r'$$. Why should this be so? I might argue that the point is that $$H_1$$ as restricted to the $$\varepsilon_n$$ eigenspace is still a Hermitian operator, and can therefore be diagonalized, but it's not even an operator on this restricted subsapce (it has components outside the subspace). So how does all this work?

• closely related and possible duplicate of this question. Commented Sep 6, 2023 at 23:15
• Yes, it is definitely related, but I suppose here I am interested in a proof of why the diagonalization works, not just the assertion that it works. Your answer there is very helpful in setting the context, thank you @ZeroTheHero My comments on Prof. Steane's answer may also help in clarifying my concern here.
– EE18
Commented Sep 7, 2023 at 0:42
• It is a necessary condition. You assume that you can write your state as a series, where a term $\psi_0$ - the zero-order term- appears. But these are undefined if $H_0$ has degeneracies. Thus, you seek conditions on these states. And, as any book demonstrates, a necessary condition these states have to obey is to diagonalize the perturbation in the degenerate subspace(s). If $H_1$ restricted to the subspace is non-degenerate, you have a condition which uniquely fixes the $\psi_0$ (apart from phase and normalization). If not, you have to go one order higher... Commented Sep 7, 2023 at 6:24
• Regarding the last point, see e.g. this. IIRC, an example is discussed in Barton Zwiebach's book. That being said, two more things: 1. Give an exact reference. 2. H1 restricted to the relevant degenerate subspace is hermitian and in particular a linear operator. – Commented Sep 7, 2023 at 6:32
• @TobiasFünke I suppose I am trying to do is to prove why it is necessary here. (1) I am referring to Ballentine here, and I am not sure he does enough to prove it to me. (2) Yes, after further consideration I agree. One argues that $H_1$ as restricted to the degenerate eigenspace is itself still Hermitian, so that by the relevant spectral theorem as applied to the restriction of $H_1$ to this subspace, we can take an orthonormal eigenbasis (of $H_1$) of this eigenspace (of $H_0$) and then extend it to a basis of $\mathcal{H}$ by appending to the these the bases of all other $H_0$ eigenspaces
– EE18
Commented Sep 7, 2023 at 13:03

In perturbation theory the unperturbed states you should use are the limit of the perturbed states in the limit where the perturbation goes to zero. The diagonalization method is a good way to find those states.

The reason for the claim in the first sentence above is that if each unperturbed state (also called zero order state) is not close to its perturbed version, then the assumptions of perturbation theory have broken down.

• I'd suggest to add a note specifying that the limit of the perturbed states as required, will also be 1) orthogonal to every other such limit states 2) and also be first order correction orthogonal. The diagonalisation accomplishes all three at once. Commented Sep 7, 2023 at 0:24
• Yes @naturallyInconsistent, that is what I am concerned about. My concern with this question is (1) why the states which diagonlize $H_1$ in their own subspace are necessarily orthogonal (or can be taken to be as such -- I think this has to do with $H_1$ being Hermitian in the subspace) and (2) why this diagonalization guarantees the first order correction is also orthogonal to these states (when I traced the derivation, all I found out is that the components along these other degenerate unperturbed states were indeterminate, $0=0$).
– EE18
Commented Sep 7, 2023 at 0:40
• @EE18 It will become obvious upon hindsight. We are trying to diagonalise not the entire $H_1$, but rather only its influence in the degenerate subspace of $H_0$. Now, once you have a diagonalised perturbation matrix, then it is clear that the standard basis itself accomplishes orthogonality on both within themselves and on the perturbation. Commented Sep 7, 2023 at 3:34

Remember what the goal is: diagonalize the full Hamiltonian. Diagonalizing is done by a unitary transformation \begin{align} U^\dagger H U&\to H_d\, ,\\ H&= H_0+\epsilon H_1+\epsilon^2 H_2+\ldots \end{align} where $$H_d$$ is diagonal. The matrix $$U$$ is the matrix of exact eigenvectors and encapsulates the change of basis from the unperturbed to the perturbed basis. In perturbation theory, we find the perturbed eigenstates order by order, so we "build up" $$U$$ order by order.

I'm going to illustrate the issue with a $$2\times 2$$ example but the argument works for any dimension.

Your Hamiltonian is of the form $$H=E_0\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)+\epsilon \left(\begin{array}{cc}a&b\\ b^*&c\end{array}\right)+\epsilon^2 H_2$$ Because the term in $$\epsilon^0$$, i.e. the unperturbed Hamiltonian, $$H_0= E_0\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$$ is proportional to the identity matrix, $$H_0$$ commutes with any unitary matrix $$U$$. As a result, the change of basis needed to bring $$H$$ to diagonal form does not depend on $$H_0$$: $$U^\dagger H_0 U=H_0$$ to order $$\epsilon$$ for any $$U$$. The change of basis, to order $$\epsilon$$, is instead determined by $$\epsilon \left(\begin{array}{cc}a&b\\ b^*&c\end{array}\right)$$ i.e. there is no change of basis to order $$\epsilon^0$$, and biggest contribution to the change of basis is of order $$\epsilon$$. Since all the terms in $$H_1$$ are of the same size, the change of basis to first order is an exact diagonalization of $$H_1$$ that will bring your $$H$$ to something like $$H=E_0\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)+\epsilon \left(\begin{array}{cc}\tilde{a}&0\\ 0&\tilde c\end{array}\right)+\epsilon^2 \tilde{H}_2$$ i.e. the change of basis will diagonalize the first order perturbation.

So "why" does perturbation theory works in the degenerate case? Because it happens to coincide with the exact diagonalization procedure, at least to order $$\epsilon$$ in the perturbation. It coincides because the term $$H_0$$, being proportional to the identity matrix, commutes with any matrix $$U$$ that produces a change of basis.

Once you have the basis to order $$\epsilon$$, and assuming the diagonal entries are different, you can then proceed with regular perturbation theory since the denominators like $$\tilde{a}-\tilde{c}\ne 0$$.

• Thank you for this answer. I can't quite tell though -- did the end get cut off?
– EE18
Commented Sep 7, 2023 at 14:46
• oups. yes an edit remained out of place. Commented Sep 7, 2023 at 14:53