Example of degenerate perturbation theory

Suppose we have a three state basis and an

{\hat{H}}^{(0)}

whose eigenstates,

| 1^{(0)} ⟩

| 2^{(0)} ⟩

and

| 3^{(0)} ⟩

, have energies

E_{1}^{(0)}

E_{2}^{(0)}

and

E_{3}^{(0)}

(all initially assumed to be diﬀerent). A representation of this system is

| 1^{(0)} ⟩ = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), | 2^{(0)} ⟩ = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), | 3^{(0)} ⟩ = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}), {\hat{H}}^{(0)} = (\begin{matrix} E_{1}^{(0)} & 0 & 0 \\ 0 & E_{2}^{(0)} & 0 \\ 0 & 0 & E_{3}^{(0)} \end{matrix}) .

{\hat{H}}^{(1)} = a (\begin{matrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) .

E_{1}^{(1)} = E_{2}^{(1)} = E_{3}^{(1)} = a, | 1^{(1)} ⟩ = \frac{a}{E_{1}^{(0)} - E_{2}^{(0)}} | 2^{(0)} ⟩, | 2^{(1)} ⟩ = \frac{a}{E_{2}^{(0)} - E_{1}^{(0)}} | 1^{(0)} ⟩, | 3^{(1)} ⟩ = 0

E_{1}^{(2)} = - E_{2}^{(2)} = \frac{a^{2}}{E_{1}^{(0)} - E_{2}^{(0)}}, E_{3}^{(2)} = 0

We note that because

{\hat{H}}^{(1)}

is already diagonal in the

| 3^{(0)} ⟩

space, the ﬁrst-order shift in energy is exact and there is no change to that eigenvector. In this case it is straightforward to obtain the eigenvalues of

{\hat{H}}^{(0)} + {\hat{H}}^{(1)}

exactly:

E_{1} = \frac{1}{2} (2 a + E_{1}^{(0)} + E_{2}^{(0)} - \sqrt{4 a^{2} + {(E_{2}^{(0)} - E_{1}^{(0)})}^{2}}), E_{2} = \frac{1}{2} (2 a + E_{1}^{(0)} + E_{2}^{(0)} + \sqrt{4 a^{2} + {(E_{2}^{(0)} - E_{1}^{(0)})}^{2}})

and

E_{3} = E_{3}^{(0)} + a

, and so we can check the expansion to order

a^{2}

Now consider the case where

E_{2}^{(0)} = E_{1}^{(0)}

. We note that

| 1^{(0)} ⟩

and

| 2^{(0)} ⟩

are just two of an inﬁnite set of eigenstates with the same energy

E_{1}^{(0)}

, since any linear combination of them is another eigenstate. Our results for the third state are unchanged, but none of those obtained for the ﬁrst two still hold. Instead we have to work in a new basis,

| {1^{'}}^{(0)} ⟩

and

| {2^{'}}^{(0)} ⟩

which diagonalises

{\hat{H}}^{(1)}

. By inspection we see that this is

| {1^{'}}^{(0)} ⟩ = \frac{1}{\sqrt{2}} (| 1^{(0)} ⟩ + | 2^{(0)} ⟩) and | {2^{'}}^{(0)} ⟩ = \frac{1}{\sqrt{2}} (| 1^{(0)} ⟩ - | 2^{(0)} ⟩) .

Then

{\hat{H}}^{(1)} | {1^{'}}^{(0)} ⟩ = 2 a | {1^{'}}^{(0)} ⟩

and

{\hat{H}}^{(1)} | {2^{'}}^{(0)} ⟩ = 0

. Hence

E_{1^{'}}^{(1)} = 2 a, E_{2^{'}}^{(1)} = 0, | {1^{'}}^{(1)} ⟩ = | {2^{'}}^{(1)} ⟩ = 0, E_{1^{'}}^{(2)} = E_{2^{'}}^{(2)} = 0

In this case because

{\hat{H}}^{(1)}

doesn’t mix states 1 & 2 with 3, diagonalising it in the subspace is actually equivalent to solving the problem exactly. We can check our results against the exact eigenvalues for

E_{2}^{(0)} = E_{1}^{(0)}

and see that they are correct, except that we made the “wrong” choice for our labelling of

| {1^{'}}^{(0)} ⟩

and

| {2^{'}}^{(0)} ⟩

3.2 Example of degenerate perturbation theory