# Projection Operators

## Vectors in $$\mathbb{R}^3$$

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

## Definition

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

$$P^\dagger P=P$$.

## Properties

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

## Definition of Perpendicular Subspace

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

### Properties

• $$|v\rangle = |v_w\rangle + |v_w^\perp\rangle$$
• $$\mathcal{V} = \mathcal{W}\oplus\mathcal{W}^\perp$$, $$\oplus$$ 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 $$P_i$$’.

## Theorem

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.

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

## Completeness Relation

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

### Names

• Resolution of the Identity
• Completeness Relation

## Inserting Identity

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

## Diagonal Matrix

$$A = \sum_n A^n P_n$$

## Gram-Schmidt Orthogonalization

Created: 2023-06-25 Sun 02:24

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