Let $\hat{\mathbf{Q}}$be an operator with a complete set of orthonormal eigenvectors:localid="1658131083682" $\hat{\mathbf{Q}}{\overline{)\mathbf{e}}}_{\mathbf{n}}\mathbf{>}\mathbf{=}{\mathit{q}}_{\mathbf{n}}\overline{){\mathbf{e}}_{\mathbf{n}}}$(n=1,2,3,....) Show that$\hat{\mathbf{Q}}$can be written in terms of its spectral decomposition:$\hat{\mathbf{Q}}\mathbf{=}\mathbf{\sum }_{\mathbf{n}}{\mathit{q}}_{\mathbf{n}}\overline{){\mathbf{e}}_{\mathbf{n}}\mathbf{>}\mathbf{<}{\mathbf{e}}_{\mathbf{n}}\mathbf{|}}$

Hint: An operator is characterized by its action on all possible vectors, so what you must show is that$\hat{\mathbf{Q}}\mathbf{=}\left\{\sum _n{q}_n\overline{)e_n><e_n|}\right\}$ for any vector $\overline{)\mathbf{\alpha }\mathbf{>}}$.

The underlying vector space on which the operator functions is further decomposed canonically by the spectral decomposition, which is provided by the spectral theorem. Every real, symmetric matrix is diagonalizable according to the spectral theorem for symmetric matrices, which was established by Augustin-Louis Cauchy.

Consider an operator$\hat{Q}$with a complete, orthonormal set of eigenvectors, that is:

$\hat{Q}{\overline{)e_m>=q_m\overline{)e}}}_m>$

Where$q_m$is the eigenvalue. Write the operator as a spectral decomposition operator.

First, write any vector$\overline{)\alpha >}$in terms of the eigenvectors set, since the eigenvectors of form a complete, orthonormal set:

$\overline{)\alpha >}=\sum _m{a}_m\overline{)e_m>}$

Here $a_m$is the coefficient of the basis vector $\overline{)e_m>}$. Now apply $\hat{Q}$on this equation and get the result as:

$\hat{Q}\overline{)\alpha >}=\sum _m{a}_m{q}_m\overline{)e_m>}$

Write $\hat{Q}$as:

$\hat{Q}=\sum _n{q}_n\overline{)e><e_n|}$

then:

role="math" localid="1658132420035" $$\begin{gathered} \hat{Q}\overline{)\alpha >=\left[\sum _n{q}_n\overline{)e_n><e_n|}\right]\sum _m{a}_m\overline{)e_m>}} \\ =\sum _{n,m}{q}_n{a}_m{\delta }_{nm}\overline{)e}><e_m| \\ =\sum _m{a}_m{q}_m{\overline{)e}}_m> \end{gathered}$$

So:

$Q=\sum _n{q}_n\overline{)e_n><e_n|}$

Thus, the operator can be written as a spectral decomposition operator $Q=\sum _n{q}_n\overline{)e_n><e_n|}$.