Section 5.6 Projection Operators
A special class of operators, called projection operators, are particularly useful for finding the component of a vector along a particular direction and for changing basis.
Definition 5.3. Projection Operator.
A operator \(P_v\) that gives the same result when acting on any vector twice in succession as when it acts just once, i.e.
\begin{equation}
P_v^2=P_v\text{,}\tag{5.6.1}
\end{equation}
is called a projection operator.
Bra-Ket Representation of Projection Operators.
Given any normalized vector \(|v\rangle\text{,}\) that is, a vector satisfying
\begin{equation}
\langle v | v \rangle = 1\text{,}\tag{5.6.2}
\end{equation}
we can show that a projection operator
\(P_v\) can be constructed by taking the
outer product, see
Section 3.3, of the vector with its Hermitian adjoint.
\begin{equation}
P_v = |v \rangle \langle v|\text{.}\tag{5.6.3}
\end{equation}
The proof is a straightforward calculation using the norm condition
(5.6.2)
\begin{align}
P_v^2
\amp = (|v \rangle \langle v|)(|v \rangle \langle v|)\notag\\
\amp = |v \rangle(\langle v|v \rangle )\langle v|\notag\\
\amp = |v \rangle \langle v|\notag\\
\amp = P_v\text{.}\tag{5.6.4}
\end{align}
Using Projection Operators to Implement Projections.
Using methods analogous to
(5.6.4), we show that
\(P_v\) takes
\(|v\rangle\) to itself, that is,
\begin{align}
P_v |v\rangle \amp = (|v \rangle \langle v|)|v\rangle\notag\\
\amp = |v\rangle(\langle v|v\rangle)\notag\\
\amp = |v\rangle\text{;}\notag
\end{align}
and takes any vector \(w\) that is orthogonal to \(v\text{,}\) (i.e. \(\langle v|w \rangle = 0\)), to zero,
\begin{align}
P_v |w\rangle \amp = (|v \rangle \langle v|)|w\rangle\notag\\
\amp = |v\rangle(\langle v|w\rangle)\notag\\
\amp = 0\text{.}\notag
\end{align}
This is what we mean when we say \(P_v\) projects any vector along \(|v\rangle\text{.}\)
Activity 5.2. Matrix Representations of Projection Operators.
Find the matrix representation of the projection operators corresponding to the vectors:
\begin{align*}
\vert + \rangle \amp \doteq \begin{pmatrix}1\\0\end{pmatrix} ,\\
\vert + \rangle_y \amp \doteq
\frac{1}{\sqrt 2}\begin{pmatrix}1\\i\end{pmatrix} ,\\
\vert + \rangle_9
\amp \doteq \begin{pmatrix}a\\be^{i\phi}\end{pmatrix} .
\end{align*}
Hint 1.
The answer should be a square matrix. If you are getting a positive, real number, you are incorrectly taking the inner-product rather than the outer product. Check the order of your matrices.
Hint 2.
Don’t forget to complex conjugate when you turn the ket into a bra.
