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

$$\vert x'\rangle, \vert x''\rangle, \vert x'''\rangle, ~\ldots,$$ (42)

and have the property

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

$$\xi\, \vert\xi'\rangle = \xi' \,\vert\xi' \rangle,$$ (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 (44) (for the eigenvalue $\xi''$ ) reads

$$\langle \xi''\vert\,\xi = \xi''^{\,\ast}\, \langle \xi''\vert.$$ (45)

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

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

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

$$\xi' = \xi'^{\,\ast},$$ (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

$$\langle \xi''\vert\xi'\rangle = 0,$$ (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

$$\langle \xi'\vert\,\xi = \langle \xi'\vert\,\xi'.$$ (49)

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