# Projectors

## Alexei Gilchrist

Projectors are linear operators with the property that the square of the operator is the operator.

### 1 Subspaces

Given a vector space, a subset of vectors that form a vector space is called a subspace (so it must be closed, have inverses etc). Since a vector space has to have a null vector $$|\textbf{0}\rangle$$, all subspaces must contain this null vector.

Definition: Direct-sum

For subspaces $$\mathbb{V}_1$$ and $$\mathbb{V}_2$$ of $$\mathbb{V}$$, then the direct-sum $$\mathbb{V}_1\oplus\mathbb{V}_2$$ is all linear combinations of vectors from $$\mathbb{V}_1$$ and $$\mathbb{V}_2$$.

Definition: Orthogonality for subspaces

Two subspaces are orthogonal if every vector from one subspace is orthogonal to every vector in the other subspace.

Definition: Orthogonal complement

The orthogonal complement, $$\mathbb{V}_1^\perp$$, to a vector subspace $$\mathbb{V_1}$$ in $$\mathbb{V}$$, is every vector $$|v\rangle\in\mathbb{V}$$ for which $$\langle v| u\rangle=0$$ holds for all $$|u\rangle\in\mathbb{V}_1$$.

For any subspace $$\mathbb{V}_1$$ in $$\mathbb{V}$$, then $$\mathbb{V}_1\oplus\mathbb{V}_1^\perp = \mathbb{V}$$.

One useful decomposition to bear in mind is that every vector $$|x\rangle\in\mathbb{V}$$ can be uniquely written as $$|x\rangle = |v\rangle+|v^\perp\rangle$$ where $$|v\rangle\in\mathbb{V}_1$$ for a given subspace $$\mathbb{V}_1$$, and $$|v^\perp\rangle\in\mathbb{V}_1^\perp$$ (to be proved).

### 2 Projectors

Definition: Projector

Say $$\mathbb{H}_a$$ is a subspace of $$\mathbb{H}$$ and $$\mathbb{H}_b = \mathbb{H}_a^\perp$$ is the orthogonal complement. For any state $$|\psi\rangle=|\psi_a\rangle+|\psi_b\rangle$$ where $$|\psi_a\rangle\in\mathbb{H}_a$$ and $$|\psi_b\rangle\in\mathbb{H}_b$$ (and of course $$\langle \psi_a| \psi_b\rangle=0$$), the projector $$\mathbb{P}_a$$ onto subspace $$\mathbb{H}_a$$ is defined by the action on such states
$\mathbb{P}_a|\psi\rangle = |\psi_a\rangle$

Projectors are Hermitian. For some other vector $$|\phi\rangle=|\phi_a\rangle+|\phi_b\rangle$$ where $$|\phi_a\rangle\in\mathbb{H}_a$$ and $$|\phi_b\rangle\in\mathbb{H}_b$$, in addition to $$|\psi\rangle$$ above. Then

\begin{align*}(|\phi\rangle, \mathbb{P}_a |\psi\rangle) &= (|\phi_a\rangle+|\phi_b\rangle, |\psi_a\rangle) = (|\phi_a\rangle,|\psi_a\rangle) \\ (\mathbb{P}_a|\phi\rangle, |\psi\rangle) &= (|\phi_a\rangle,|\psi_a\rangle+|\psi_b\rangle) = (|\phi_a\rangle,|\psi_a\rangle)\end{align*}

The signature feature of projectors is $$\mathbb{P}^2=\mathbb{P}$$:

$\mathbb{P}_a^2|\psi\rangle = \mathbb{P}_a\mathbb{P}_a|\psi\rangle = \mathbb{P}_a|\psi_a\rangle = |\psi_a\rangle$
Obviously this extends to any positive interger power.

Since projectors are Hermitian, their eigenvalues must be real, and since $$\mathbb{P}^2=\mathbb{P}$$, the eigenvalues must either be 1 or 0: say $$|\lambda\rangle$$ is an eigenvector of projector $$\mathbb{P}$$ with eigenvalue $$\lambda$$, then

\begin{align*}\mathbb{P}\mathbb{P}|\lambda\rangle &= \mathbb{P}|\lambda\rangle = \lambda |\lambda\rangle \\ \mathbb{P}\mathbb{P}|\lambda\rangle &= \lambda\mathbb{P}|\lambda\rangle = \lambda^2 |\lambda\rangle.\end{align*}
So we must have $$\lambda^2 = \lambda$$ and this is only satisfied if $$\lambda=0$$ or $$\lambda=1$$.