Projection Operators

Vectors in $\mathbb{R}^3$

$\vec{a}\cdot\vec{b}=a^xb^x+a^yb^y+a^zb^z=a^ib^i=ab\cos\theta = a_{||}b = ab_{||}$.

They pick out the components of a vector to a subspace of $\mathcal{V}$.

$P^\dagger P=P$.

$P^2=P$
$P^\dagger = P$
$P_1+P_2$ is a projection operator iff $P_1P_2=P_2P_1=0$ (Orthogonal projections)

$\mathcal{W}^\perp = \{|w^\perp\rangle:\langle w^\perp|w\rangle \forall |w\rangle\in\mathcal{W}\}$.

$|v\rangle = |v_w\rangle + |v_w^\perp\rangle$
$\mathcal{V} = \mathcal{W}\oplus\mathcal{W}^\perp$, $\oplus$ is called the direct sum.
If $P_w$ projects onto $\mathcal{W}$
- $P_w|v\rangle=|w\rangle\in\mathcal{W}$
- $P_w|w\rangle = |w\rangle \forall |w\rangle\in\mathcal{W}$
then $P^\perp_w=\mathbb{I}-P_w$ projects onto $\mathcal{W}$.
Then if $\{P_i\}$ is a set of orthogonal projections, $P_iP_j=P_i\delta^i_j$, then $P=\sum_i P_i$ projects onto the `span of $P_i$’.

If $\{P_i\}$ is an orthonormal basis for $\mathcal{V}$ then

$P_i=|i\rangle\langle i|$ projects onto the $i$-th basis direction
$P_i+P_j = |i\rangle\langle i| + |j\rangle\langle j|$ projects onto the subspace spanned by the $i$ and $j$-th basis vectors
$\sum_i P_i = \mathbb{I}$

Proof.

$P_i|a\rangle = |i\rangle\langle i|a\rangle = a^i|i\rangle$.
$(P_i+P_j)|a\rangle = |i\rangle\langle i|a\rangle + |j\rangle\langle j|a\rangle = a^i|i\rangle + a^j|j\rangle$.
$\left[\sum_i P_i\right]|a\rangle = \left[\sum_i P_i|a\rangle\right] = \sum_i a^i|i\rangle = |a\rangle$.

$P_i=|i\rangle\langle i|$. $\sum_n P_n = \mathbb{I}$ iff $\{|i\rangle\}$ is orthonormal basis (spanning, linearly independent, orthogonal, normal).

$\mathbb{I}|a\rangle = \sum_n P_n |a\rangle = a^n|n\rangle$.
$A = \mathbb{I}A\mathbb{I} = \sum_n P_n A \sum_n P_m = \sum_n\sum_m |n\rangle\langle m|\langle n|A|m\rangle$.
$A^i_j = \langle i|A|j\rangle = \sum_n\sum_m\langle i|n\rangle\langle m|j\rangle\langle n|A|m\rangle = \sum_n\sum_m\delta^i_n\delta^m_j\langle n|A|m\rangle = \langle i|A|j\rangle$.

$A = \sum_n A^n P_n$