Bra-ket representation of projection operators

Question

Let me define (orthogonal) projection operator $P$ by its characterization: $P = P^2 = P^\dagger$

How can I show that $P$ can always be represented as $\sum_{i=1}^n |\psi_i\rangle\langle\psi_i|$ (where $\langle\psi_i|\psi_j\rangle = \delta_{ij}, n \le \mathrm{dim} \, \mathcal H$) in braket notation?

I only consider Hilbert spaces of finite dimention here (i.e. $\mathrm{dim} \, \mathcal H < \infty$).

It's easy to show that $\sum_i |\psi_i\rangle\langle\psi_i|$ satisfies the property, but I'm struggling to prove the converse.

Thanks in advance!

Answer 1

Operators that can be diagonalized look like \begin{align} P = \sum_i p_i \left | \psi _i \right > \left < \psi_i \right | \end{align} in their own eigenbasis. The spectral theorem tells us more than this; that $P$ is hermitian so it can be diagonalized by a unitary matrix. Therefore $\left < \psi_i | \psi_j \right > = \delta_{ij}$.

So far we have used $P^\dagger = P$ but not $P^2 = P$. As suggested above, you should use this to solve for the possible eigenvalues $p_i$.

Answer 2

Let us a assume a Hilbert space $\mathcal{H}$ with finite dimension $n$. The property $P =P^2$ implies that the eigenvalues of $P$ can only be $0$ or $1$. As $P$ is hermitian ($P^\dagger =P$), there exists an orthonormal basis of $\mathcal{H}$, $$\{|\psi_1\rangle, \ldots |\psi_r\rangle, |\psi_{r+1}\rangle, \ldots |\psi_n\rangle \}, \quad \langle \psi_i | \psi_j \rangle = \delta_{ij}, \quad \sum\limits_{i=1}^n |\psi_i \rangle \langle \psi_i | = \mathbf{1}_n,$$ of eigenvectors of $P$, with $P | \psi\rangle = |\psi_i \rangle $ for $i=1, \ldots r$ and $P |\psi_i \rangle =0$ for $i=r+1, \ldots n$ (assuming that the eigenvalue $1$ is $r$-fold degenerate). Consequently, the spectral representation of $P$ is given by $$P= \sum\limits_{i=1}^r |\psi_i \rangle \langle \psi_i |.$$ Obviously the choice of the vectors $\{ |\psi_i \rangle \}_{i=1}^r$ is not unique, any orthonormal basis of the $r$-dimensional subspace $P \mathcal{H}$ will do the job.