# Eigenvalues and eigenvectors

In general, the ket $$X\vert A\rangle$$ is not a constant multiple of $$\vert A\rangle$$. However, there are some special kets known as the eigenkets of operator $$X$$. These are denoted

$\vert x'\rangle, \vert x''\rangle, \vert x'''\rangle \ldots,$

and have the property

$X\vert x'\rangle = x'\vert x'\rangle, X\vert x''\rangle = x''\vert x''\rangle \dots,$

where $$x'$$, $$x''$$, $$\ldots$$ are numbers called eigenvalues. Clearly, applying $$X$$ to one of its eigenkets yields the same eigenket multiplied by the associated eigenvalue.

Consider the eigenkets and eigenvalues of a Hermitian operator $$\xi$$. These are denoted

$\xi \vert\xi'\rangle = \xi' \vert\xi' \rangle, \label{44}$

where $$\vert\xi'\rangle$$ is the eigenket associated with the eigenvalue $$\xi'$$. Three important results are readily deduced:

(i) The eigenvalues are all real numbers, and the eigenkets corresponding to different eigenvalues are orthogonal. Since $$\xi$$ is Hermitian, the dual equation to Equation \ref{44} (for the eigenvalue $$\xi''$$ reads

$\langle \xi''\vert\xi = \xi''^\ast \langle \xi''\vert. \label{45}$

If we left-multiply Equation \ref{44} by $$\langle \xi''\vert$$, right-multiply the above equation by $$\vert\xi'\rangle$$, and take the difference, we obtain

$(\xi' - \xi''^\ast) \langle \xi''\vert\xi'\rangle = 0. \label{46}$

Suppose that the eigenvalues $$\xi'$$ and $$\xi''$$ are the same. It follows from the above that

$\xi' = \xi'^\ast, \label{47}$

where we have used the fact that $$\vert\xi'\rangle$$ is not the null ket. This proves that the eigenvalues are real numbers. Suppose that the eigenvalues $$\xi'$$ and $$\xi''$$ are different. It follows that

$\langle \xi''\vert\xi'\rangle = 0, \label{48}$

which demonstrates that eigenkets corresponding to different eigenvalues are orthogonal.

(ii) The eigenvalues associated with eigenkets are the same as the eigenvalues associated with eigenbras. An eigenbra of $$\xi$$ corresponding to an eigenvalue $$\xi'$$ is defined

$\langle \xi'\vert\xi = \langle \xi'\vert\xi'. \label{49}$

(iii) The dual of any eigenket is an eigenbra belonging to the same eigenvalue, and conversely.

### Contributors

http://en.wikipedia.org/wiki/Bra-ket_notation