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1.8: Eigenvalues and Eigenvectors

  1. Last updated
    22:49, 15 Nov 2014
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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

 

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

 

 

and have the property

 

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

 

$\displaystyle \xi\, \vert\xi'\rangle = \xi' \,\vert\xi' \rangle,$ (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

 

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

 

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

 

 

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

 

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

 

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

 

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

 

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

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