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 \(\lambda=1\).

© Copyright 2021 Alexei Gilchrist