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--- adjoints_and_hermitian_operators [2022/09/22 19:08] – [Hermitian and Skew-Hermitian Operators] admin
+++ adjoints_and_hermitian_operators [2022/10/06 00:45] (current) – [Projection Operators] admin
@@ Line 102: / Line 102: @@
 where $V^{\perp}$ is the orthogonal complement of $V$ in $\mathcal{H}$.
-In order to completely define the operator, we need to say what it does to vectors that are neither in $V$ or $V^{\perp}$.  However, since $\mathcal{H} = V\oplus V^{\perp}$, we can write any vector $\ket{\psi}$ as $\ket{\psi} = a\ket{\phi} + b\ket{\phi^{\perp}}$, where $\ket{\psi}\in V$ and $\ket{\psi^{\perp}}\in V^{\perp}$.  Then, if we want $\hat{P}_V$ to be a //linear// operator, we must have
+In order to completely define the operator, we need to say what it does to vectors that are neither in $V$ or $V^{\perp}$.  However, since $\mathcal{H} = V\oplus V^{\perp}$, we can write any vector $\ket{\psi}$ as $\ket{\psi} = a\ket{\phi} + b\ket{\phi^{\perp}}$, where $\ket{\phi}\in V$ and $\ket{\phi^{\perp}}\in V^{\perp}$.  Then, if we want $\hat{P}_V$ to be a //linear// operator, we must have
 \[\hat{P}_V \ket{\psi} = \hat{P}_V \left ( a\ket{\phi} + b\ket{\phi^{\perp}} \right ) = a\hat{P}_V\ket{\phi} + b\hat{P}_V \ket{\phi^{\perp}} = a\ket{\phi}.\]
 In other words, the projection operator $\hat{P}_V$ simply returns the component of the vector $\ket{\psi}$ that lies in the subspace $V$.
@@ Line 110: / Line 110: @@
 Another trivial example of a projector is $\hat{0}$ which just multiplies a vector by the scalar $0$, i.e.
 \[\hat{0}\ket{\psi} = 0 \ket{\psi} = \boldsymbol{0},\]
-for all vectors $\ket{\psi}\in\mathcal{H}$.  This is the projector onto the zero-dimensional subspace that just consists of the zero vector $\boldymbol{0}$.  This subspace is the orthogonal complement of $\mathcal{H}$.
+for all vectors $\ket{\psi}\in\mathcal{H}$.  This is the projector onto the zero-dimensional subspace that just consists of the zero vector $\boldsymbol{0}$.  This subspace is the orthogonal complement of $\mathcal{H}$.
 We already know that, if $\ket{\phi_1}, \ket{\phi_2},\cdots$ is an orthonormal basis for $\mathcal{H}$ then
@@ Line 127: / Line 127: @@
 where $\ket{\phi_1},\ket{\phi_2} \in V$ and $\ket{\phi_1^{\perp}},\ket{\phi_2^{\perp}} \in V^{\perp}$.  Then,
 \begin{align*}
-   \sand{\psi_1}{\hat{P}_V}{\psi_2} & = \left ( \right ) \hat{P} \left ( \right ) \\
+   \sand{\psi_1}{\hat{P}_V}{\psi_2} & = \left ( \bra{\phi_1} + \bra{\phi_1^{\perp}} \right ) \hat{P}_V \left (\ket{\phi_2} + \ket{\phi_2^{\perp}} \right ) \\
-   & = \sand{}{}{} + \sand{}{}{} + \sand{}{}{} + \sand{}{}{} \\\
+   & = \sand{\phi_1}{\hat{P}_V}{\phi_2} + \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}}.
 \end{align*}
+Now, the terms $\sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}}, \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2}$ and $\sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}}$ are all zero because $\ket{\phi_1^{\perp}}$ and $\ket{\phi_2^{\perp}}$ are in a subspace orthogonal to $V$, so $\hat{P}_V\ket{\phi_1^{\perp}} = \hat{P}_V \ket{\phi_2^{\perp}} = 0$.  In particular, since the complex conjugate of zero is zero, this means that
+\begin{align*}
+  \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^*, &
+  \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} & = \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^*, &
+   \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}}^*.
+\end{align*}
+In addition
+\begin{align*}
+   \sand{\phi_1}{\hat{P}_V}{\phi_2} & = \braket{\phi_1}{\phi_2} \\
+   & = \braket{\phi_2}{\phi_1}^* \\
+   & = \sand{\phi_2}{\hat{P}_V}{\phi_1}^*.
+\end{align*}
+This is because $\ket{\phi_1},\ket{\phi_2}\in V$, so $\hat{P}_V \ket{\phi_1} = \ket{\phi_1}$ and $\hat{P}_V \ket{\phi_2} = \ket{\phi_2}$.
+All together then, we have
+\begin{align*}
+  \sand{\psi_1}{\hat{P}_V}{\psi_2} & =\sand{\phi_1}{\hat{P}_V}{\phi_2} + \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} \\
+  & = \sand{\phi_2}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}}^* \\
+  & = \left [ \left (  \bra{\phi_2} + \bra{\phi_2^{\perp}}\right ) \hat{P}_V \left ( \ket{\phi_1} + \ket{\phi_1^{\perp}}\right ) \right ]^* \\
+  & = \sand{\psi_2}{\hat{P}_V}{\psi_1}^*,
+\end{align*}
+so $\hat{P}_V^{\dagger} = \hat{P}_V$.
+To prove that projection operators are idempotent, write an arbitrary vector $\ket{\psi}$ as
+\[\ket{\psi} = \ket{\phi} + \ket{\phi^{\perp}},\]
+with $\ket{\phi} \in V$ and $\ket{\phi^{\perp}} \in V^{\perp}$.  Then we have
+\[\hat{P}_V \ket{\psi} = \ket{\phi},\]
+but also
+\[\hat{P}_V^2 \ket{\psi} = \hat{P}_V \hat{P}_V \ket{\psi} = \hat{P_V} \ket{\phi} = \ket{\phi},\]
+since $\ket{\phi} \in V$.  Since this is true for an arbitrary vector $\ket{\psi}$, we have $\hat{P}_V^2 = \hat{P}_V$.
 We conclude with some more basic properties of projectors.