# 3.1: Time-Independent Degenerate Perturbation Theory


We have seen how we can find approximate solutions for a system whose Hamiltonian is of the form

$\hat{H} = \hat{H}_0 + \hat{V} \nonumber$

When we assumed that $$\hat{H}$$ and $$\hat{H}_0$$ possess discrete, non-degenerate eigenvalues only. This led to a mixing of states where

$|\phi_0 \rangle = |n_0 \rangle + \sum_{k \neq 0} \frac{V_{k0}}{(E_0 − E_k)} |n_k \rangle \nonumber$

Clearly, if $$E_0 = E_k$$ this diverges. As do the higher order energy shifts (see 2.4). Thus for the degenerate case we cannot associate a particular perturbed state $$|\phi_0 \rangle$$ with a particular unperturbed state $$|n_0 \rangle$$: we need to take a different approach. In fact, the approximation we make is completely different: we assume that the small perturbation only mixes those states which are degenerate. We then solve the problem exactly for that subset of states.

Assume that $$\hat{H}_0$$ possesses $$N$$ degenerate eigenstates $$|m \rangle$$ with eigenvalue $$E_{deg}$$. It may also possesses non-degenerate eigenstates, which can be treated separately by non-degenerate perturbation theory. We write a perturbed eigenstate $$|\phi_j \rangle$$ as an linear expansion in the unperturbed degenerate eigenstates only:

$|\phi_j \rangle = \sum_i |m_i \rangle \langle m_i | \phi_j \rangle = \sum_i c_{ji} | m_i \rangle \nonumber$

Where $$i$$ here runs over degenerate states only. The TISE now becomes:

$[\hat{H}_0 + \hat{V} ] |\phi_j \rangle = [\hat{H}_0 + \hat{V} ] \sum_i c_{ji} |m_i \rangle = E_j \sum_i c_{ji} |m_i \rangle \nonumber$

but we know that for all degenerate eigenstates $$\hat{H}_0|m_i \rangle = E_{deg} |m_i \rangle$$. So we obtain:

$\sum_i c_{ji} \hat{V} | m_i \rangle = (E_j - E_{deg}) \sum_i c_{ji} | m_i \rangle \nonumber$

premultiplying by some unperturbed state $$\langle m_k|$$ gives

$\sum_i c_{ji} \left[ \langle m_k | \hat{V} | m_i \rangle - \delta_{ik} (E_j - E_{deg}) \right] = 0 \nonumber$

We can get a similar equation from each unperturbed state $$|m_k \rangle$$. We thus have an eigenvalue problem: the eigenvector has elements $$c_{ji}$$ and the eigenvalues are $$\Delta E_j = E_j − E_{deg}$$. Writing the matrix elements between the $$i^{th}$$ and $$k^{th}$$ unperturbed degenerate states as $$V_{ik} \equiv \langle m_i |\hat{V} |m_k \rangle$$ we recover the determinantal equation:

$\begin{vmatrix} V_{11} − \Delta E_j & V_{12} & ... & V_{1N} \\ V_{21} & V_{22} − \Delta E_j & ... & V_{2N} \\ ... & ... & ... & ... \\ V_{N1} & V_{N2} & ... & V_{NN} − \Delta E_j \end{vmatrix} = 0 \nonumber$

The $$N$$ eigenvalues obtained by solving this equation give the shifts in energy due to the perturbation, and the eigenvectors give the perturbed states $$|\phi \rangle$$ in the unperturbed, degenerate basis set $$|m \rangle$$.

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