Projection Operators \bgroup\color{black}$\vert j\rangle\langle j\vert$\egroup and Completeness

Now we move on a little with our understanding of operators. A ket vector followed by a bra vector is an example of an operator. For example the operator which projects a vector onto the $j^{th}$ eigenstate is

\begin{displaymath}\bgroup\color{black} \vert j\rangle\langle j\vert \egroup\end{displaymath}

First the bra vector dots into the state, giving the coefficient of \bgroup\color{black}$\vert j\rangle$\egroup in the state, then its multiplied by the unit vector \bgroup\color{black}$\vert j\rangle$\egroup, turning it back into a vector, with the right length to be a projection. An operator maps one vector into another vector, so this is an operator.

The sum of the projection operators is 1, if we sum over a complete set of states, like the eigenstates of a Hermitian operator.

\bgroup\color{black}$\displaystyle \sum\limits_i \vert i\rangle \langle i\vert=1 $\egroup
This is an extremely useful identity for solving problems. We could already see this in the decomposition of \bgroup\color{black}$\vert\psi\rangle$\egroup above.

\begin{displaymath}\bgroup\color{black}\vert\psi\rangle=\sum\limits_i \vert i\rangle \langle i\vert\psi\rangle .\egroup\end{displaymath}

The same is true for definite momentum states.

\bgroup\color{black}$\displaystyle \int\limits_{-\infty}^\infty \vert p\rangle\langle p\vert dp=1 $\egroup

We can form a projection operator into a subspace.

\begin{displaymath}\bgroup\color{black}P=\sum\limits_{subspace} \vert i\rangle\langle i\vert \egroup\end{displaymath}

We could use this to project out the odd parity states, for example.

Jim Branson 2013-04-22