Two-State System

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

$$H_0 \,\vert 1\rangle = E_1 \,\vert 1\rangle,$$ (7.2)

$$H_0 \,\vert 2\rangle = E_2 \,\vert 2\rangle.$$ (7.3)

It is assumed that these states, and their associated eigenvalues, are known. Because $H_0$ is, by definition, an Hermitian operator, its two eigenkets are mutually orthogonal 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

$$(H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$$ (7.4)

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

$$\vert E\rangle = \langle 1\vert E\rangle \vert 1\rangle + \langle 2\vert E\rangle \vert 2\rangle.$$ (7.5)

Substituting the previous expansion into Equation (7.4), and then right-multiplying by either $\langle 1\vert$ or $\langle 2\vert$ , we get two coupled equations that can be written in matrix form:

$$\left( \begin{array}{c c} E_1 -E + e_{11}, & e_{12} \\ e_{12}^{\,... ...ngle\end{array} \!\right)= \left(\!\begin{array}{c}0\\ 0 \end{array}\! \right).$$ (7.6)

Here,

$$e_{11} = \langle 1\vert\,H_1\, \vert 1\rangle,$$ (7.7)

$$e_{22} = \langle 2 \vert\,H_1\, \vert 2\rangle,$$ (7.8)

$$e_{12} = \langle 1\vert\,H_1\,\vert 2\rangle$$ (7.9)

are the so-called matrix elements of the perturbing Hamiltonian (with respect to the unperturbed eigenstates). In the special (but common) case of a perturbing Hamiltonian whose diagonal matrix elements are zero, so that

$$e_{11} = e_{22} = 0,$$ (7.10)

the non-trivial solution of Equation (7.6) (obtained by setting the determinant of the matrix equal to zero [92]) is

$$E = \frac{(E_1+E_2) \pm \sqrt{(E_1-E_2)^{\,2} + 4\,\vert e_{12}\vert^{\,2}}}{2}.$$ (7.11)

Let us expand in the supposedly small parameter

$$\epsilon = \frac{\vert e_{12}\vert}{\vert E_1-E_2\vert}.$$ (7.12)

We obtain

$$E\simeq \frac{1}{2} \,(E_1+E_2) \pm \frac{1}{2}\,(E_1-E_2)\,(1+2\,\epsilon^{\,2} + \cdots).$$ (7.13)

The previous expression yields the modifications to the energy eigenvalues caused by the perturbing Hamiltonian:

$$E_1' = E_1 + \frac{\vert e_{12}\vert^{\,2}}{E_1-E_2} + \cdots,$$ (7.14)

$$E_2' = E_2 - \frac{\vert e_{12}\vert^{\,2}}{E_1-E_2} + \cdots.$$ (7.15)

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

$$\vert 1\rangle' =\vert 1\rangle + \frac{e_{12}^{~\ast}}{E_1-E_2}\, \vert 2\rangle + \cdots,$$ (7.16)

$$\vert 2\rangle' =\vert 2\rangle - \frac{e_{12}}{E_1-E_2}\, \vert 1\rangle +\cdots.$$ (7.17)

Thus, the modified energy eigenstates consist of one of the unperturbed eigenstates with a slight admixture of the other. Actually, the series expansion on the right-hand side of Equation (7.13) only converges when $2\,\vert\epsilon\vert<1$ [59]. This suggests that the condition for the validity of the perturbation expansion is

$$\vert e_{12}\vert < \frac{\vert E_1-E_2\vert}{2}.$$ (7.18)

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