91Ó°ÊÓ

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$\overline{)\mathrm{Î\pm }>}$in terms of the eigenvectors set, since the eigenvectors of form a complete, orthonormal set:

$\overline{)\mathrm{Î\pm }>}=\underset{m}{∑}{a}_m\overline{)e_m>}$

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:

$\hat{Q}\overline{)\mathrm{Î\pm }>}=\underset{m}{∑}{a}_m{q}_m\overline{)e_m>}$

Write $\hat{Q}$as:

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

then:

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

So:

$Q=\underset{n}{∑}{q}_n\overline{)e_n><e_n|}$

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