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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.,
$\displaystyle H_0 \psi_1$ $\textstyle =$ $\displaystyle E_1 \psi_1,$ (875)
$\displaystyle H_0 \psi_2$ $\textstyle =$ $\displaystyle E_2 \psi_2.$ (876)

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

\begin{displaymath}
(H_0+H_1) \psi_E = E \psi_E.
\end{displaymath} (877)

We can, in fact, solve this problem exactly. Since the eigenstates of $H_0$ form a complete set, we can write [see Eq. (865)]
\begin{displaymath}
\psi_E = \langle 1\vert E\rangle \psi_1 + \langle 2\vert E\rangle \psi_2.
\end{displaymath} (878)

It follows from (877) that
\begin{displaymath}
\langle i\vert H_0 + H_1\vert E\rangle = E \langle i\vert E\rangle,
\end{displaymath} (879)

where $i=1$ or $2$. Equations (875), (876), (878), (879), and the orthonormality condition
\begin{displaymath}
\langle i\vert j\rangle = \delta_{ij},
\end{displaymath} (880)

yield two coupled equations which can be written in matrix form:
$\displaystyle \left(\begin{array}{cc}E_1-E+e_{11}& e_{12}\  [0.5ex]
e_{12}^\as...
...gle\end{array}\right)=\left(
\begin{array}{c} 0\  [0.5ex]
0\end{array}\right),$     (881)

where
$\displaystyle e_{11}$ $\textstyle =$ $\displaystyle \langle 1\vert H_1\vert 1\rangle,$ (882)
$\displaystyle e_{22}$ $\textstyle =$ $\displaystyle \langle 2\vert H_1\vert 2\rangle,$ (883)
$\displaystyle e_{12}$ $\textstyle =$ $\displaystyle \langle 1\vert H_1\vert 2\rangle = \langle 2\vert H_1\vert 1\rangle^\ast.$ (884)

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

\begin{displaymath}
e_{11}= e_{22} = 0.
\end{displaymath} (885)

The solution of Eq. (881) (obtained by setting the determinant of the matrix to zero) is
\begin{displaymath}
E = \frac{(E_1+E_2)\pm\sqrt{(E_1-E_2)^2 + 4 \vert e_{12}\vert^2}}
{2}.
\end{displaymath} (886)

Let us expand in the supposedly small parameter
\begin{displaymath}
\epsilon = \frac{\vert e_{12}\vert}{\vert E_1-E_2\vert}.
\end{displaymath} (887)

We obtain
\begin{displaymath}
E \simeq \frac{1}{2} (E_1+E_2) \pm \frac{1}{2} (E_1-E_2)(1+2 \epsilon^2+\cdots).
\end{displaymath} (888)

The above expression yields the modification of the energy eigenvalues due to the perturbing Hamiltonian:
$\displaystyle E_1'$ $\textstyle =$ $\displaystyle E_1 + \frac{\vert e_{12}\vert^2}{E_1-E_2}+ \cdots,$ (889)
$\displaystyle E_2'$ $\textstyle =$ $\displaystyle E_2 - \frac{\vert e_{12}\vert^2}{E_1-E_2}+\cdots.$ (890)

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
$\displaystyle \psi_1'$ $\textstyle =$ $\displaystyle \psi_1+ \frac{e_{12}^\ast}{E_1-E_2} \psi_2+ \cdots,$ (891)
$\displaystyle \psi_2'$ $\textstyle =$ $\displaystyle \psi_2 - \frac{e_{12}}{E_1-E_2} \psi_1+ \cdots.$ (892)

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
\begin{displaymath}
\vert e_{12}\vert\ll \vert E_1-E_2\vert.
\end{displaymath} (893)

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.


next up previous
Next: Non-Degenerate Perturbation Theory Up: Time-Independent Perturbation Theory Previous: Improved Notation
Richard Fitzpatrick 2010-07-20