Show that projection operators are idempotent: ${\hat{\mathbf{P}}}^{\mathbf{2}}\mathit{=}\hat{\mathbf{P}}$. Determine the eigenvalues of $\hat{\mathbf{P}}$ , and characterize its eigenvectors.

Begin by defining a$\mathit{n}\mathbf{-}\mathit{d}\mathit{i}\mathit{m}\mathit{e}\mathit{n}\mathit{s}\mathit{i}\mathit{o}\mathit{n}\mathit{a}\mathit{l}$vector written in orthonormal$\overline{)\mathbf{i}\mathbf{>}}$ basis. Then, an arbitrary vector can be expressed as:

$\overline{)\mathbf{\psi }\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}}{\mathit{c}}_{\mathbf{1}}\overline{)\mathbf{i}\mathbf{>}}$

Now, denote a projector onto $\mathit{m}$ – state as localid="1658117337914" ${\hat{\mathbf{P}}}_{\mathbf{m}}$. Thus the latter operator can be written in the following form:

localid="1658117894083" ${\hat{\mathbf{P}}}_{\mathbf{m}}\mathbf{=}\overline{)\mathbf{m}\mathbf{>}\mathbf{<}\overline{)\mathbf{m}}}$

Acting with the projector operator on the arbitrary vector, obtain the following expression:

localid="1658119307238"

From which see that the action of the projector ${\hat{P}}_m$is equivalent to pulling out the -th component of the vector, which in some textbooks is taken as the definition. Now, let's observe how the operator ${{\hat{P}}_m}^2$acts on the same vector localid="1658118048579" $\overline{)\psi >}$.

$$\begin{gathered} {{\hat{P}}_m}^2\overline{)\psi >=}{\hat{P}}_m\left({\hat{P}}_m\overline{)\psi >)} \\ {{\hat{P}}_m}^2\overline{)\psi >=}{\hat{P}}_m\right({\hat{C}}_m\overline{)m>)} \\ {{\hat{P}}_m}^2\overline{)\psi >=}{c}_m{\hat{P}}_m\overline{)m}> \\ {{\hat{P}}_m}^2\overline{)\psi >=}{c}_m\overline{)m}> \\ {{\hat{P}}_m}^2\overline{)\psi >=}{\hat{P}}_m\overline{)m}> \end{gathered}$$

${{\hat{P}}_m}^2={\hat{P}}_m$

To find the associated eigenvalues and eigenvectors, let's see how the projector ${\hat{P}}_m$acts on all elements of the $\overline{)i>}$basis.

localid="1658118608775"

See that if i=m then localid="1658118689334" ${\hat{\mathrm{P}}}_m\overline{)\mathrm{m}}>=1.\overline{)\mathrm{m}>}$and if$i\ne m$then ${\hat{P}}_m\overline{)i>=0.\overline{)m>}}$. It follows that any vector proportional to the vector which the projector projects upon is an eigenvector with eigenvalue 1, and every other vector which is perpendicular to it (in this case, all other elements of the basis) are eigenvectors as well with eigenvalue 0. As is independent eigenvectors, conclude that you have found all of them, since the dimension of the space is n.

A more general approach to the problem would be to use the idempotency property. Now assume for an arbitrary projector $\hat{P}$the following eigenvector and eigenvalue:

$$\begin{gathered} \hat{P}\overline{)\varphi >=\lambda }\overline{)\varphi >} \\ {\hat{P}}^2\overline{)\varphi >=}\hat{P}\left(\hat{P}\overline{)\varphi }\right) \\ \hat{P}\lambda \overline{)\varphi >=\lambda \hat{P}}\overline{)\varphi >} \\ ={\lambda }^2\overline{)\varphi >} \end{gathered}$$

But since ${\hat{P}}^2=\hat{P}$it follows that ${\lambda }^2=\lambda$therefore ${\lambda }_1=1$ or ${\lambda }_2=0$ . As for the eigenvectors, follow the same reasoning as in the upper paragraph. Thus It follows that any vector proportional to the vector which the projector projects upon is an eigenvector with eigenvalue 1, and every other vector which is perpendicular to it (in this case, all other elements of the basis) are eigenvectors as well with eigenvalue 0.