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Eigenvalues and Eigenvectors

In general, the ket $ X,円\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

[画像:$\displaystyle \vert x'\rangle, \vert x''\rangle, \vert x'''\rangle, ~\ldots,$] (42)

and have the property

[画像:$\displaystyle X,円\vert x'\rangle = x',円\vert x'\rangle,~~~X,円\vert x''\rangle = x'',円\vert x''\rangle, ~~\dots,$] (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

[画像:$\displaystyle \xi,円 \vert\xi'\rangle = \xi' ,円\vert\xi' \rangle,$] (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

[画像:$\displaystyle \langle \xi''\vert,円\xi = \xi''^{,円\ast},円 \langle \xi''\vert.$] (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

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

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

[画像:$\displaystyle \xi' = \xi'^{,円\ast},$] (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

[画像:$\displaystyle \langle \xi''\vert\xi'\rangle = 0,$] (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

[画像:$\displaystyle \langle \xi'\vert,円\xi = \langle \xi'\vert,円\xi'.$] (49)

(iii) 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 2013年04月08日