Why does degenerate perturbation theory work?

Question

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?

Answer 1

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.

Answer 2

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 $\tilde{a}-\tilde{c}\ne 0$.