The two-state system

Consider the simplest possible non-trivial quantum mechanical system. In such a system, there are only two independent eigenstates of the unperturbed Hamiltonian: i.e.,

$$H_0\,\psi_1 = E_1\,\psi_1,$$ (854)

$$H_0\,\psi_2 = E_2\,\psi_2.$$ (855)

It is assumed that these states, and their associated eigenvalues, are known. We also expect the states to be orthonormal, and to form a complete set.

Let us now try to solve the modified energy eigenvalue problem

$$(H_0+H_1)\,\psi_E = E\,\psi_E.$$ (856)

We can, in fact, solve this problem exactly. Since the eigenstates of $H_0$ form a complete set, we can write [see Eq. (844)]

$$\psi_E = \langle 1\vert E\rangle\,\psi_1 + \langle 2\vert E\rangle\,\psi_2.$$ (857)

It follows from (856) that

$$\langle i\vert H_0 + H_1\vert E\rangle = E\,\langle i\vert E\rangle,$$ (858)

where $i=1$ or $2$. Equations (854), (855), (857), (858), and the orthonormality condition

$$\langle i\vert j\rangle = \delta_{ij},$$ (859)

yield two coupled equations which can be written in matrix form:

$$\left(\begin{array}{cc}E_1-E+e_{11}& e_{12}\\ [0.5ex] e_{12}^\as... ...gle\end{array}\right)=\left( \begin{array}{c} 0\\ [0.5ex] 0\end{array}\right),$$ (860)

where

$$e_{11} = \langle 1\vert H_1\vert 1\rangle,$$ (861)

$$e_{22} = \langle 2\vert H_1\vert 2\rangle,$$ (862)

$$e_{12} = \langle 1\vert H_1\vert 2\rangle = \langle 2\vert H_1\vert 1\rangle^\ast.$$ (863)

Here, use has been made of the fact that $H_1$ is an Hermitian operator.

Consider the special (but not uncommon) case of a perturbing Hamiltonian whose diagonal matrix elements are zero, so that

$$e_{11}= e_{22} = 0.$$ (864)

The solution of Eq. (860) (obtained by setting the determinant of the matrix to zero) is

$$E = \frac{(E_1+E_2)\pm\sqrt{(E_1-E_2)^2 + 4\,\vert e_{12}\vert^2}} {2}.$$ (865)

Let us expand in the supposedly small parameter

$$\epsilon = \frac{\vert e_{12}\vert}{\vert E_1-E_2\vert}.$$ (866)

We obtain

$$E \simeq \frac{1}{2}\,(E_1+E_2) \pm \frac{1}{2}\,(E_1-E_2)(1+2\,\epsilon^2+\cdots).$$ (867)

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

$$E_1' = E_1 + \frac{\vert e_{12}\vert^2}{E_1-E_2}+ \cdots,$$ (868)

$$E_2' = E_2 - \frac{\vert e_{12}\vert^2}{E_1-E_2}+\cdots.$$ (869)

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

$$\psi_1' = \psi_1+ \frac{e_{12}^\ast}{E_1-E_2}\,\psi_2+ \cdots,$$ (870)

$$\psi_2' = \psi_2 - \frac{e_{12}}{E_1-E_2}\,\psi_1+ \cdots.$$ (871)

Thus, the modified energy eigenstates consist of one of the unperturbed eigenstates, plus a slight admixture of the other. Now our expansion procedure is only valid when $\epsilon\ll 1$. This suggests that the condition for the validity of the perturbation method as a whole is

$$\vert e_{12}\vert\ll \vert E_1-E_2\vert.$$ (872)

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