Projection operators
The most general spin state of a spin-1/2 particle is given by,
\[\begin{equation} \mid \psi \rangle = c_+ \mid +\mathbf{z}\rangle + c_- \mid -\mathbf{z} \rangle \label{eq1} \end{equation}\]where, \(c_-\) and \(c_+\) can be complex numbers, and \(\mid +\mathbf{z}\rangle\) and \(\mid -\mathbf{z}\rangle\) are basis vectors. We can find \(c_-\) and \(c_+\) by taking inner products of \eqref{eq1} with \(\langle +\mathbf{z}\mid\) and \(\langle -\mathbf{z}\mid\).
\[\begin{equation} \begin{aligned} \langle +\mathbf{z} \mid \psi \rangle &= c_{+} \langle +\mathbf{z}\mid +\mathbf{z} \rangle + c_{-}\langle +\mathbf{z} \mid -\mathbf{z} \rangle = c_{+} \\ \langle -\mathbf{z} \mid \psi \rangle &= c_{+} \langle -\mathbf{z} \mid +\mathbf{z} \rangle + c_{-}\langle -\mathbf{z} \mid -\mathbf{z} \rangle = c_{-} \end{aligned} \label{eq2} \end{equation}\]By combining eqns. \eqref{eq1} and \eqref{eq2}, we can write,
\[\begin{equation} \mid \psi \rangle = \mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \psi \rangle + \mid -\mathbf{z} \rangle \langle -\mathbf{z} \mid \psi \rangle. \label{eq3} \end{equation}\]Since \(\langle \pm \mathbf{z}\mid \psi \rangle\) actually means \(\langle \pm \mathbf{z}\mid \mid \psi \rangle\), we can write eqn. \eqref{eq3} as,
\[\begin{equation} \begin{aligned} \mid \psi \rangle &= \mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \mid \psi \rangle + \mid -\mathbf{z} \rangle \langle -\mathbf{z} \mid \mid \psi \rangle \\ &= \left(\mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid + \mid -\mathbf{z} \rangle \langle -\mathbf{z} \mid \right) \psi \\ \end{aligned} \label{eq4} \end{equation}\]What’s going on here? The strange looking thing inside the bracket in eqn. \eqref{eq4} is an operator acting on the ket \(\mid \psi \rangle\). Since it is leaving the ket unchanged, it is an identity operator.
\[\begin{equation} \boxed{\mid +\mathbf{z}\rangle\langle+\mathbf{z}| + \mid -\mathbf{z} \rangle \langle-\mathbf{z}| = \hat{I}} \label{eq5} \end{equation}\]In general, in an \(n-\) dimensional vector space, the identity operator is given by,
\[\begin{aligned} \hat{I} &= \mid 1\rangle\langle1\mid + \mid 2 \rangle \langle2 \mid + \mid 3 \rangle \langle3 \mid + \cdots \mid n \rangle \langle n \mid \\ &=\sum_{i}^{n}\mid i \rangle \langle i \mid. \end{aligned}\]We can consider the identity operator to be composed of two operators \(\hat{P}_{+}\) and \(\hat{P}_{-}\):
\[\begin{equation} \hat{P}_{+} = \mid +\mathbf{z}\rangle\langle+\mathbf{z} \mid \text{ and }\, \hat{P}_{-} = \mid -\mathbf{z}\rangle\langle -\mathbf{z} \mid \label{eq6} \end{equation}\]We have,
\[\begin{equation*} \hat{P}_{+}\mid \psi \rangle = \left(\mid +\mathbf{z}\rangle\langle+\mathbf{z} \mid \right) \mid \psi \rangle = \mid +\mathbf{z}\rangle\langle+\mathbf{z} \mid \psi \rangle. \end{equation*}\]That is, the operator \(\hat{P_+}\) projects out the \(\mid +\mathbf{z}\rangle\) component of \(\mid \psi \rangle\). Similarly, \(\hat{P}_{-}\) projects out the \(\mid -\mathbf{z}\rangle\) component of \(\mid \psi \rangle\). That’s why they are called projection operators.
Let’s now look at some of the properties of the projection operators. We have,
\[\begin{equation} \begin{aligned} \hat{P}_{+} \mid +\mathbf{z}\rangle &= \left(\mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \right) \mid +\mathbf{z} \rangle = \mid +\mathbf{z} \rangle \\ \hat{P}_{-} \mid +\mathbf{z}\rangle &= \left(\mid -\mathbf{z}\rangle \langle -\mathbf{z} \mid \right) \mid +\mathbf{z} \rangle = 0 \end{aligned} \label{eq8} \end{equation}\]That is, the basis state \(\mid +\mathbf{z} \rangle\) is an eigenstate of the operator \(\hat{P_+}\) with an eigen value 1, and \(\mid -\mathbf{z} \rangle\) an eigenstate with an eigenvalue 0. Similalrly, \(\mid +\mathbf{z} \rangle\) and \(\mid -\mathbf{z} \rangle\) are eigenstates of \(\hat{P_{-}}\) with eigenvalues 0 and 1 respectively.
Also,
\[\begin{align*} \hat{P}_{+}^{2} &= \hat{P}_{+} \hat{P}_{+} \\ &= \mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \\ &= \mid +\mathbf{z}\rangle \langle +\mathbf{z} \mid \\ &= \hat{P}_{+} \end{align*}\]Similarly, \(\hat{P_-}^{2} = \hat{P_-}\). In general,
\[\begin{equation*} \boxed{\hat{P}_{\pm}^{n} = \hat{P}_{\pm}}. \end{equation*}\]Matrix representation of projection operators
The matrix representation of an operator \(\hat{A}\) in \(S_{z}\) basis is given by (more details here):
\[\begin{equation*} \hat{A} \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{cc} \langle+\mathbf{z}|\hat{A}|+\mathbf{z}\rangle & \langle+\mathbf{z}|\hat{A}|-\mathbf{z}\rangle \\ \langle-\mathbf{z}|\hat{A}|+\mathbf{z}\rangle & \langle-\mathbf{z}|\hat{A}|-\mathbf{z}\rangle \end{array}\right) \end{equation*}\]For the projection operator \(\hat{P}_{+}\)
\[\begin{equation} \hat{P}_{+} \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{cc} \langle+\mathbf{z}\mid\hat{P}_{+}\mid+\mathbf{z}\rangle & \langle+\mathbf{z}\mid\hat{P}_{+}\mid-\mathbf{z}\rangle \\ \langle-\mathbf{z}\mid\hat{P}_{+}\mid+\mathbf{z}\rangle & \langle-\mathbf{z}\mid\hat{P}_{+}\mid-\mathbf{z}\rangle \end{array}\right) \label{eq9} \end{equation}\]By substituting \(\hat{P}_{+} = \mid +\mathbf{z}\rangle \langle+\mathbf{z} \mid\), we get,
\[\begin{equation} \begin{aligned} \langle+\mathbf{z} \mid \hat{P}_{+} \mid +\mathbf{z}\rangle &= \langle+\mathbf{z} \mid \left( \mid +\mathbf{z}\rangle \langle+\mathbf{z} \mid \right) \mid +\mathbf{z}\rangle \\ &=\langle+\mathbf{z} \mid \mid +\mathbf{z}\rangle \langle+\mathbf{z} \mid \mid +\mathbf{z}\rangle \\ &= \langle+\mathbf{z} \mid +\mathbf{z}\rangle \langle+\mathbf{z} \mid +\mathbf{z}\rangle \\ &= 1 \end{aligned} \label{eq10} \end{equation}\]Similarly, we can verify that,
\[\begin{equation} \begin{aligned} \langle+\mathbf{z} \mid \hat{P}_{+} \mid -\mathbf{z}\rangle = \langle-\mathbf{z} \mid \hat{P}_{+} \mid +\mathbf{z}\rangle = \langle-\mathbf{z} \mid \hat{P}_{+} \mid -\mathbf{z}\rangle = 0 \end{aligned} \label{eq11} \end{equation}\]Substituting eqns. \eqref{eq10} and \eqref{eq11} in \eqref{eq9}, we get:
\[\begin{equation} \hat{P}_{+} \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right). \label{eq12} \end{equation}\]Let’s now find the matrix representation of \(\hat{P}_{-}\).
We have,
\[\begin{equation} \hat{P}_{-} \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{cc} \langle+\mathbf{z}\mid\hat{P}_{-}\mid+\mathbf{z}\rangle & \langle+\mathbf{z}\mid\hat{P}_{-}\mid-\mathbf{z}\rangle \\ \langle-\mathbf{z}\mid\hat{P}_{-}\mid+\mathbf{z}\rangle & \langle-\mathbf{z}\mid\hat{P}_{-}\mid-\mathbf{z}\rangle \end{array}\right) \label{eq13} \end{equation}\]By substituting \(\hat{P}_{-} = \mid -\mathbf{z}\rangle \langle-\mathbf{z} \mid\), we can verify that,
\[\begin{equation} \langle - \mathbf{z}\mid\hat{P}_{-}\mid-\mathbf{z}\rangle = 1, \end{equation}\]and all other terms are zero.
Therefore,
\[\begin{equation} \hat{P}_{-} \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right). \end{equation}\]Also,
\[\begin{equation} \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) + \left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right), \end{equation}\]which is in agreement with eqn. \eqref{eq5}.
We can also check the action the action of \(\hat{P_+}\) and \(\hat{P_+}\) on the basis states \(\mid +\mathbf{z}\rangle\) and \(\mid -\mathbf{z}\rangle\).
\[\begin{equation} \begin{aligned} \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) \left(\begin{array}{c} 1 \\ 0 \end{array}\right) &= \left(\begin{array}{c} 1 \\ 0 \end{array}\right) \\ \left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right) \left(\begin{array}{c} 0 \\ 1 \end{array}\right) &= \left(\begin{array}{c} 0 \\ 0 \end{array}\right), \end{aligned} \end{equation}\]which, again, is in agreement with eqn. \eqref{eq8}.