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7.2: Two-State System

Let us start 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

$\displaystyle H_0 \,\vert 1\rangle$ $\displaystyle =$ $\displaystyle E_1 \,\vert 1\rangle,$ (584)
$\displaystyle H_0 \,\vert 2\rangle$ $\displaystyle =$ $\displaystyle E_2 \,\vert 2\rangle.$ (585)

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


$\displaystyle (H_0 + H_1) \,\vert E\rangle = E\,\vert E\rangle.$ (586)



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


$\displaystyle \vert E\rangle = \langle 1\vert E\rangle \vert 1\rangle + \langle 2\vert E\rangle \vert 2\rangle.$ (587)



Right-multiplication of Equation (586) by $ \langle 1\vert$ and $ \langle 2\vert$ yields two coupled equations, which can be written in matrix form:


$\displaystyle \left( \begin{array}{c c} E_1 -E + e_{11} & e_{12} \\ e_{12}^{\,\...
...ngle\end{array} \!\right)= \left(\!\begin{array}{c}0\\ 0 \end{array}\! \right).$ (588)




$\displaystyle e_{11}$ $\displaystyle =$ $\displaystyle \langle 1\vert\,H_1\, \vert 1\rangle,$ (589)
$\displaystyle e_{22}$ $\displaystyle =$ $\displaystyle \langle 2 \vert\,H_1\, \vert 2\rangle,$ (590)
$\displaystyle e_{12}$ $\displaystyle =$ $\displaystyle \langle 1\vert\,H_1\,\vert 2\rangle.$ (591)

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


$\displaystyle e_{11} = e_{22} = 0,$ (592)



the solution of Equation (588) (obtained by setting the determinant of the matrix equal to zero) is


$\displaystyle E = \frac{(E_1+E_2) \pm \sqrt{(E_1-E_2)^{\,2} + 4\,\vert e_{12}\vert^{\,2}}}{2}.$ (593)



Let us expand in the supposedly small parameter


$\displaystyle \epsilon = \frac{\vert e_{12}\vert}{\vert E_1-E_2\vert}.$ (594)



We obtain


$\displaystyle E\simeq \frac{1}{2} \,(E_1+E_2) \pm \frac{1}{2}\,(E_1-E_2)\,(1+2\,\epsilon^2 + \cdots).$ (595)



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

$\displaystyle E_1'$ $\displaystyle =$ $\displaystyle E_1 + \frac{\vert e_{12}\vert^{\,2}}{E_1-E_2} + \cdots,$ (596)
$\displaystyle E_2'$ $\displaystyle =$ $\displaystyle E_2 - \frac{\vert e_{12}\vert^{\,2}}{E_1-E_2} + \cdots.$ (597)

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

$\displaystyle \vert 1\rangle'$ $\displaystyle =$ $\displaystyle \vert 1\rangle + \frac{e_{12}^{~\ast}}{E_1-E_2}\, \vert 2\rangle + \cdots,$ (598)
$\displaystyle \vert 2\rangle'$ $\displaystyle =$ $\displaystyle \vert 2\rangle - \frac{e_{12}}{E_1-E_2}\, \vert 1\rangle +\cdots.$ (599)

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 Equation (595) only converges if $ 2\,\vert\epsilon\vert<1$ . This suggests that the condition for the validity of the perturbation expansion is


$\displaystyle \vert e_{12}\vert < \frac{\vert E_1-E_2\vert}{2}.$ (600)



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.