If I begin with the Hamiltonian:

$$\begin{align} \hat{H} = -\frac{\hbar^{2}}{2mR^{2}} \partial_{\phi}^{2} + g\delta(\phi) \end{align} $$

And look at the unperturbed portion $(g=0)$, I find that the general solution is

$$\begin{align} |\psi\rangle = Ae^{\pm ik \phi} \end{align} $$

Normalizing, and solving the eigenstates and eigen-energies for its configuration (ie a particle on a ring):

$$\begin{align} |\psi_{n}\rangle &= \frac{1}{\sqrt{2\pi R}}e^{\pm in \phi} \\ \mathcal{E}_{n} &=\frac{ n^{2} \hbar^{2}}{2mR^{2}} \end{align} $$

And we have $2n+1$ number of degeneracies indexed by $n$ where $n \in \mathbb{Z}$

When I know look at the perturbed state $(g \delta(\phi))$ I construct the matrix: $$\begin{align} \begin{pmatrix} \langle k|g\delta(\phi) |k\rangle & \langle k|g\delta(\phi)|k'\rangle \\ \langle k'|g\delta(\phi)|k'\rangle & \langle k'|g\delta(\phi)|k'\rangle \end{pmatrix} = \begin{pmatrix} \frac{g}{2\pi} & \frac{g}{2\pi}\\ \frac{g}{2\pi } & \frac{g}{2\pi } \end{pmatrix} \end{align}$$ where $k,k'$ are the matrix indexes.

Taking the determinant then gives the following eigenvectors with their eigen-values:

$$\begin{align} |\eta_{1} \rangle = \frac{1}{\sqrt{2}} \left( |k\rangle + |k'\rangle \right) \qquad \mathrm{for}\ \mathcal{E} = \frac{g}{\pi} \\ |\eta_{2} \rangle = \frac{1}{\sqrt{2}} \left( |k\rangle - |k'\rangle \right) \qquad \mathrm{for}\ \mathcal{E} = 0 \end{align}$$

So now my question is really, what do I do with this information to find the perturbed energies (keeping it second order) and eigenstates? Or importantly, what is the necessity so speak to have these eigenvectors?

Is really what I did was solve for the perturbed "portions" of the states $\eta_{1,2}$ in the basis of $k$ and $k'$ and then I can use that as the ket when solving for the perturbed state equations ie:

$$\begin{align} E_{\eta_{1}}^{(1)} =\langle \eta_{1}| g \delta(\phi)|k \rangle \end{align}$$



1 Answer 1


The eigenstates $\vert\eta_1\rangle$ and $\vert \eta_2\rangle$ now become your "new" basis.

Remember the game is ultimately to label states using energy, so we look for eigenstates because those have definite energy. The difficulty in degenerate perturbation theory is to find a useful "new" basis since any combination of the degenerate states is also an eigenstates of the unperturbed Hamiltonian. By diagonalizing the first order perturbation, you select a useful combination of the original basis states. By "useful" I mean you can now use $\vert\eta_1\rangle$ and $\vert \eta_2\rangle$ to continue with the normal nondegenerate approach since these "new" states now have different energy and (presumably) enough of an energy difference to apply the usual perturbation using those.

  • $\begingroup$ Ah, I see. I was trying to find the right words for describing everything. Thank you for shedding a new light. One last question to further expand off of this, when you say I can then use my new basis for non-degenerate perturbation theory, I can then go about our perturbation equation $E^{(1)}_{n} = \langle \psi_{n}| V |\psi_{n}\rangle $ (from griffiths) and replace my new basis into these? Or is it really that I'm evaluating each energy for each new basis (in my case): $\endgroup$
    – iron2man
    Commented Feb 26, 2017 at 15:07
  • $\begingroup$ $E^{(1)}_{\eta_{1},n} = \langle \eta_{1}| V |\psi_{n}\rangle $ and $E^{(1)}_{\eta_{2},n} = \langle \eta_{2}| V |\psi_{n}\rangle $ $\endgroup$
    – iron2man
    Commented Feb 26, 2017 at 15:08
  • 2
    $\begingroup$ You do everything in the "new" basis; it is an "improved" original basis in that it accounts for the zeroeth and first order perturbation, so there is no need to recalculate $E^{(1)}_n$ since you already found the first order correction to be $g/\pi$ and $0$ (in your case). All matrix elements are now of the type $\langle \eta_i \vert V \vert \eta_k\rangle$. $\endgroup$ Commented Feb 26, 2017 at 15:10
  • $\begingroup$ @Countto10 is this for me? $\endgroup$ Commented Feb 26, 2017 at 16:14
  • 1
    $\begingroup$ @Countto10 I wish my students were so appreciative! Be well! $\endgroup$ Commented Feb 26, 2017 at 16:23

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