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# 11.3: 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: that is,

\begin{aligned} \label{e12.21} H_0\,\psi_1 &= E_1\,\psi_1,\\[0.5ex] H_0\,\psi_2&=E_2\,\psi_2.\label{e12.22}\end{aligned} 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 $\label{e12.23} (H_0+H_1)\,\psi_E = E\,\psi_E.$ We can, in fact, solve this problem exactly. Because the eigenstates of $$H_0$$ form a complete set, we can write [see Equation ([e12.13a])]

$\label{e12.24} \psi_E = \langle 1|E\rangle\,\psi_1 + \langle 2|E\rangle\,\psi_2.$ It follows from Equation ([e12.23]) that

$\label{e12.25} \langle i|H_0 + H_1|E\rangle = E\,\langle i|E\rangle,$ where $$i=1$$ or $$2$$. Equations ([e12.21]), ([e12.22]), ([e12.24]), ([e12.25]), and the orthonormality condition $\langle i|j\rangle = \delta_{ij},$ yield two coupled equations that can be written in matrix form:

\begin{aligned} \label{e12.27} \left(\begin{array}{cc}E_1-E+e_{11},& e_{12}\\[0.5ex] e_{12}^\ast, & E_2-E+e_{22}\end{array}\right)\left( \begin{array}{c}\langle 1|E\rangle\\[0.5ex] \langle 2|E\rangle\end{array}\right)=\left( \begin{array}{c} 0\\[0.5ex] 0\end{array}\right),\end{aligned} where \begin{aligned} e_{11} &= \langle 1|H_1|1\rangle,\\[0.5ex] e_{22}&=\langle 2|H_1|2\rangle,\\[0.5ex] e_{12}&=\langle 1|H_1|2\rangle = \langle 2|H_1|1\rangle^\ast.\end{aligned} 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 $e_{11}= e_{22} = 0.$ The solution of Equation ([e12.27]) (obtained by setting the determinant of the matrix to zero) is $E = \frac{(E_1+E_2)\pm\sqrt{(E_1-E_2)^{\,2} + 4\,|e_{12}|^{\,2}}} {2}.$ Let us expand in the supposedly small parameter $\epsilon = \frac{|e_{12}|}{|E_1-E_2|}.$ We obtain $E \simeq \frac{1}{2}\,(E_1+E_2) \pm \frac{1}{2}\,(E_1-E_2)(1+2\,\epsilon^{\,2}+\cdots).$ The previous expression yields the modification of the energy eigenvalues due to the perturbing Hamiltonian: \begin{aligned} E_1' &= E_1 + \frac{|e_{12}|^{\,2}}{E_1-E_2}+ \cdots,\\[0.5ex] E_2' &= E_2 - \frac{|e_{12}|^{\,2}}{E_1-E_2}+\cdots.\end{aligned} 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 \begin{aligned} \psi_1' &=\psi_1+ \frac{e_{12}^\ast}{E_1-E_2}\,\psi_2+ \cdots,\\[0.5ex] \psi_2'&= \psi_2 - \frac{e_{12}}{E_1-E_2}\,\psi_1+ \cdots.\end{aligned} 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 $|e_{12}|\ll |E_1-E_2|.$ 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 previous inequality must be satisfied.
