# Multiply-degenerate time-independent Rayleigh-Schrödinger perturbation theory

Comments
Last updated 2015-11-25 23:48:28 SGT

I've been working on this while revising for my Quantum Mechanics final, and I kinda dislike how Sakurai handles the degenerate case (i.e. more than one eigenket with the same energy eigenvalue with respect to the unperturbed Hamiltonian). Cue a few hours of working it out on my own:

## Nondegenerate case

Sakurai gets a lot of flak for working with complementary projection operators, but in the general case it is, in my view, fully justified, so we'll mostly take his approach (with some modifications). Let's consider a perturbation problem with the Hamiltonian given by

where $\lambda$ is some small expansion parameter. Suppose the unperturbed Hamiltonian has energy eigenvalues $E_n^{(0)}$ with corresponding eigenkets $\ket{n^{(0)}}$. Then the (time-independent) Schrödinger's equation reads

where by the usual ansatz $E_n$ and $\ket{n}$ can be expanded in powers of $\lambda$ as

subject to the normalisation condition $\braket{n^{(0)}}{n} = 1$. Then writing $E_n = E_n^{(0)} + \lambda \Delta_n^{(1)}$, we can rewrite the Schrödinger equation in the form (1):

Observe that contracting on the left with $\bra{n^{(0)}}$ gives 0 on the left hand side; therefore the ket on the right hand side lives in the orthogonal complement to all eigenvectors with eigenvalue $E_n$. In this subspace of the total Hilbert space, $E_n^{(0)} - \hat{H_0}$ is nonsingular, and can be inverted. Let $P_n^{(0)}$ be this projection operator, and $Q_n^{(0)}$ be such that $P_n^{(0)} + Q_n^{(0)} = \mathbb{1}$. Explicitly,

In the nondegenerate case, $Q_n^{(0)} = \ket{n^{(0)}}\bra{n^{(0)}}$ and $P_n^{(0)} = \sum_{m \ne n} \ket{m^{(0)}}\bra{m^{(0)}}$. This will become handy later, but it pays to see how it works in the simple case.

In the nondegenerate case, Equation (1) yields the expression

If our normalisation condition is to hold to all orders of $\lambda$, then $\braket{n^{(0)}}{n^{(k)}} = 0$ for $k>0$. Expanding this in powers of $\lambda$ then gives us, in general,

That is, knowing $\ket{n^{(k-1)}}$ is sufficient to tell us $E^{(k)}$ immediately. In particular, we can find $E_n^{(1)}$ immediately as $\left< n^{(0)} | \hat{V} | n^{(0)}\right>$.

In order to find $\ket{n^{(1)}}$, we note that $\left(E_n^{(0)} - \hat{H_0} \right)Q_n^{(0)} = 0$. In the subspace that $P_n^{(0)}$ projects to, $\left(E_n^{(0)} - \hat{H_0} \right)$ is nondegenerate; let us write its Moore-Penrose pseudoinverse as $G_n^{(0)} = P_n^{(0)}G_n^{(0)}P_n^{(0)}$. Then we can rewrite Eq. 1 as (2):

Noting that $\ket{n} = \ket{n^{(0)}} + P_n^{(0)}\ket{n}$, this gives us

This is highly suggestive of an order-by-order expansion in powers of $\lambda$. Following through with this yields

The first-order correction is particularly simple:

This, together with our expression for $E_n^{(k)}$ in terms of $\ket{n^{(k-1)}}$, permits us to determine the perturbed energy levels and eigenstates in an iterative fashion (albeit not normalised properly) — with the proviso that the states are not degenerate with respect to the unperturbed Hamiltonian $\hat{H_0}$.

## Degeneracy

Now suppose that there was a set $D = \left\{ \ket{m^{(0)}} \right\}$ of eigenvectors of $\hat{H}_0$, all of which are degenerate with eigenvalue $E_D^{(0)}$. Then the projection operators as defined above now take the form

Clearly, most of the statements above still hold. However, $G_n^{(0)}$, which enters into our expressions for both perturbed energies and eigenkets at higher orders, is singular on $D$1. Equation 1 can now be contracted on the left by any vector in $D$. Projecting on the left by $Q_D^{(0)}$, and noting that $Q_D^{(0)} P_D^{(0)} = 0$ by orthogonality, we obtain now

Inserting Eq. 2 into this then gives us, after some manipulation (3):

Observe that to 0th order in $\lambda$, this is simply the secular equation:

That is to say, the first-order energy corrections are the eigenvalues of $\hat{V}$ on the degenerate subspace, and the corresponding 0th-order eigenkets are simultaneously eigenkets of $Q_D^{(0)}\hat{V} Q_D^{(0)}$ — i.e. of the restriction of $\hat{V}$ to the degenerate subspace. We can go further, however, by observing that this is of basically the same form as our original perturbation problem:

Then to first order in $\lambda$ we can define an effective Hamiltonian $\hat{H}_1 = Q_D^{(0)}\hat{V} Q_D^{(0)}$ acting on the degenerate subspace, and an effective perturbation $\hat{V}_1 = Q_D^{(0)} \hat{V} G_n^{(0)}\hat{V}Q_D^{(0)}$. We can moreover define $\Delta_n^{(1)} = E_n^{(1)} + \lambda\Delta_n^{(2)}$, such that Eq. 3 can be written, to leading order in $\lambda$,

It is instructive to compare this to Eq. 1, and upon so doing the generalisation is obvious. We define second-order projection operators and pseudoinverses such that

Then if the degeneracy is lifted to first order, we can immediately treat this (at least to first order in $\hat{V}_1$) as if it were a time-dependent nondegenerate perturbation problem with an effective unperturbed Hamiltonian $\hat{H}_1$ and an effective perturbation $\hat{V}_1$. At higher orders we must be careful to expand the additional terms in powers of $\lambda$ also; this makes things considerably messier. Nonetheless, this approach suffices to recover up to first-order corrections in the eigenkets (and, if we so choose, second-order corrections to the energy). Explicitly,

Interestingly enough, the expression for the second-order energy correction returned by this procedure is basically the same as the general nondegenerate case — a rough check of correctness.

## Multiple degeneracy

Once again, however, we see that if $\ket{n^{(0)}}$ is degenerate with respect to both $\hat{H}_0$ and $\hat{V}$, then not only is $G_n^{(1)}$ singular on the doubly degenerate subspace, but once again it is unclear what basis is physically preferred in the limit as $\lambda \to 0$. But now the generalisation is obvious: we set up a secular equation for $\hat{H}_2 = Q_n^{(1)} \hat{V}_1 Q_n^{(1)}$ on the doubly degenerate subspace projected onto by $Q_n^{(1)}$, and again receive (to first order) an effective perturbation problem that gives us now second-order corrections to the perturbed energies as the eigenvalues of the effective Hamiltonian.

1. Sakurai provides a much nicer argument, which I will reproduce here: say $\hat{H}_0$ and $\hat{A}$ are simultaneously diagonalisable. Choose eigenkets of $\hat{A}$ to span $D$. But if $\left[\hat{V},\hat{A}\right] \ne 0$, then even at zeroth order (i.e. as $\lambda \to 0$) the perturbed eigenkets are not eigenkets of $\hat{A}$; there exists some other preferred basis.

comments powered by Disqus