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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,
\end{displaymath} (42)

and have the property

X\vert x'\rangle = x'\vert x'\rangle,   X\vert x''\rangle = x''\vert x''\rangle \dots,
\end{displaymath} (43)

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,
\end{displaymath} (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 Eq. (44) (for the eigenvalue $\xi''$) reads

\langle \xi''\vert\xi = \xi''^\ast \langle \xi''\vert.
\end{displaymath} (45)

If we left-multiply Eq. (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.
\end{displaymath} (46)

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

\xi' = \xi'^\ast,
\end{displaymath} (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,
\end{displaymath} (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'.
\end{displaymath} (49)

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