The two-state system

Let us begin by considering time-independent perturbation theory, in which the modification to the Hamiltonian, $H_1$, has no explicit dependence on time. It is usually assumed that the unperturbed Hamiltonian, $H_0$, is also time-independent.

Consider the simplest non-trivial system, in which there are only two independent eigenkets of the unperturbed Hamiltonian. These are denoted

$$H_0 \vert 1\rangle = E_1 \vert 1\rangle,$$ (588)

$$H_0 \vert 2\rangle = E_2 \vert 2\rangle.$$ (589)

It is assumed that these states, and their associated eigenvalues, are known. Since $H_0$ is, by definition, an Hermitian operator, its two eigenkets are orthonormal and form a complete set. The lengths of these eigenkets are both normalized to unity. Let us now try to solve the modified energy eigenvalue problem

$$(H_0 + H_1) \vert E\rangle = E \vert E\rangle.$$ (590)

In fact, we can solve this problem exactly. Since, the eigenkets of $H_0$ form a complete set, we can write

$$\vert E\rangle = \langle 1\vert E\rangle \vert 1\rangle + \langle 2\vert E\rangle \vert 2\rangle.$$ (591)

Right-multiplication of Eq. (590) by $\langle 1\vert$ and $\langle 2\vert$ yields two coupled equations, which can be written in matrix form:

$$\left( \begin{array}{c c} E_1 -E + e_{11} & e_{12} \\ e_{12... ...\right)= \left(\!\begin{array}{c}0\\ 0 \end{array}\! \right).$$ (592)

Here,

$$e_{11} = \langle 1\vert H_1 \vert 1\rangle,$$ (593)

$$e_{22} = \langle 2 \vert H_1 \vert 2\rangle,$$ (594)

$$e_{12} = \langle 1\vert H_1\vert 2\rangle.$$ (595)

In the special (but common) case of a perturbing Hamiltonian whose diagonal matrix elements (in the unperturbed eigenstates) are zero, so that

$$e_{11} = e_{22} = 0,$$ (596)

the solution of Eq. (592) (obtained by setting the determinant of the matrix equal to zero) is

$$E = \frac{(E_1+E_2) \pm \sqrt{(E_1-E_2)^2 + 4 \vert e_{12}\vert^2}}{2}.$$ (597)

Let us expand in the supposedly small parameter

$$\epsilon = \frac{\vert e_{12}\vert}{\vert E_1-E_2\vert}.$$ (598)

We obtain

$$E\simeq \frac{1}{2} (E_1+E_2) \pm \frac{1}{2}(E_1-E_2)(1+2 \epsilon^2 + \cdots).$$ (599)

The above expression yields the modifications to the energy eigenvalues due to the perturbing Hamiltonian:

$$E_1' = E_1 + \frac{\vert e_{12}\vert^2}{E_1-E_2} + \cdots,$$ (600)

$$E_2' = E_2 - \frac{\vert e_{12}\vert^2}{E_1-E_2} + \cdots.$$ (601)

Note that $H_1$ causes the upper eigenvalue to rise, and the lower eigenvalue to fall. It is easily demonstrated that the modified eigenkets take the form

$$\vert 1\rangle' = \vert 1\rangle + \frac{e_{12}^{ \ast}}{E_1-E_2} \vert 2\rangle + \cdots,$$ (602)

$$\vert 2\rangle' = \vert 2\rangle - \frac{e_{12}}{E_1-E_2} \vert 1\rangle +\cdots.$$ (603)

Thus, the modified energy eigenstates consist of one of the unperturbed eigenstates with a slight admixture of the other. Note that the series expansion in Eq. (599) only converges if $2 \vert\epsilon\vert<1$. This suggests that the condition for the validity of the perturbation expansion is

$$\vert e_{12}\vert < \frac{\vert E_1-E_2\vert}{2}.$$ (604)

In other words, when we say that $H_1$ needs to be small compared to $H_0$, what we really mean is that the above inequality needs to be satisfied.