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Eigenvalues and Eigenvectors

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

$\displaystyle \vert x'\rangle, \vert x''\rangle, \vert x'''\rangle, ~\ldots,$ (1.45)

and have the property

$\displaystyle X\,\vert x'\rangle = x'\,\vert x'\rangle,$   $\displaystyle X\,\vert x''\rangle = x''\,\vert x''\rangle,$   $\displaystyle X\,\vert x'''\rangle = x'''\,\vert x'''\rangle,$   $\displaystyle \dots,$ (1.46)

where $ x'$ , $ x''$ , $ x'''$ , $ \ldots$ are complex 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 an Hermitian operator $ \xi$ . These are denoted

$\displaystyle \xi\, \vert\xi'\rangle = \xi' \,\vert\xi' \rangle,$ (1.47)

where $ \vert\xi'\rangle$ is the eigenket associated with the eigenvalue $ \xi'$ . Three important results are readily deduced:
  1. The eigenvalues are all real numbers, and the eigenkets corresponding to different eigenvalues are orthogonal. Because $ \xi$ is Hermitian, the dual equation to Equation (1.47) (for the eigenvalue $ \xi''$ ) reads

    $\displaystyle \langle \xi''\vert\,\xi = \xi''^{\,\ast}\, \langle \xi''\vert.$ (1.48)

    If we left-multiply Equation (1.47) by $ \langle \xi''\vert$ , right-multiply the previous equation by $ \vert\xi'\rangle$ , and take the difference, then we obtain

    $\displaystyle (\xi' - \xi''^{\,\ast})\, \langle \xi''\vert\xi'\rangle = 0.$ (1.49)

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

    $\displaystyle \xi' = \xi'^{\,\ast},$ (1.50)

    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

    $\displaystyle \langle \xi''\vert\xi'\rangle = 0,$ (1.51)

    which demonstrates that eigenkets corresponding to different eigenvalues are orthogonal.

  2. 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

    $\displaystyle \langle \xi'\vert\,\xi = \langle \xi'\vert\,\xi'.$ (1.52)

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


next up previous
Next: Observables Up: Fundamental Concepts Previous: Outer Product
Richard Fitzpatrick 2016-01-22