# Eigenvalues and eigenvectors

- Page ID
- 1668

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.

## Outside Links

mysbfiles.stonybrook.edu/~kli...08-S09/Ch4.pdf