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

(584) | |||

(585) |

It is assumed that these states, and their associated eigenvalues, are known. Because 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

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

(587) |

Right-multiplication of Equation (586) by and yields two coupled equations, which can be written in matrix form:

Here,

(589) | |||

(590) | |||

(591) |

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

(592) |

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

(593) |

Let us expand in the supposedly small parameter

(594) |

We obtain

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

(596) | |||

(597) |

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

(598) | |||

(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 . This suggests that the condition for the validity of the perturbation expansion is

(600) |

In other words, when we say that needs to be small compared to , what we really mean is that the above inequality needs to be satisfied.