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Operators and eigenvectors

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Show that operator Q can be written (Summation) q_n|q_n><q_n|
An operator Q has eigenvectors |q_n>,

Q|q_n> = q_n|q_n> n=1,2,3.....

Suppose that these eigenvectors |q_n> form a complete set. Show that in this case the operator Q can be written

Q= [(capital sigma, summation) with n below the S] q_n|q_n><q_n|

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Solution Summary

We give a straightforward derivation of the decomposition formula for an operator in terms of its eigenvectors and egivenvalues.

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In the bra-ket notation we can write down expressions like |k><k| which is are operators. If you let this act on a ket |z>, you get the ket vector |k><k|z>, if you let it act on a bra vector <z|, you get the bra vector <z|k><k|. For normalized |k>, this is then the projection on the one dimensional subspace spanned by |k>. ...

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