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)

Definition of Perpendicular Subspace

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


  • \(|v\rangle = |v_w\rangle + |v_w^\perp\rangle\)
  • \(\mathcal{V} = \mathcal{W}\oplus\mathcal{W}^\perp\), \(\oplus\) 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
  • \(\sum_i P_i = \mathbb{I}\)


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

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).


  • 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

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:24