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--- adjoints_and_hermitian_operators [2021/02/24 08:25] – admin
+++ adjoints_and_hermitian_operators [2022/10/06 00:45] (current) – [Projection Operators] admin
@@ Line 84: / Line 84: @@
 \[\hat{A}_h = \frac{\hat{A} + \hat{A}^{\dagger}}{2},\]
 which is a Hermitian operator, and its //**anti-Hermitian part**// as
-\[\hat{A}_h = \frac{\hat{A} - \hat{A}^{\dagger}}{2},\]
+\[\hat{A}_a = \frac{\hat{A} - \hat{A}^{\dagger}}{2i},\]
-which is an anti-Hermitian operator.
+which is also a Hermitian operator.
 Then, the operator $\hat{A}$ can be written as
-\[\hat{A} = \hat{A}_h + \hat{A}_a.\]
+\[\hat{A} = \hat{A}_h + i\hat{A}_a.\]
 You should think of this decomposition as analogous to writing a complex scalar as
 \[a = \text{Re}(a) + i\text{Im}(a).\]
+Note, the imaginary part of a complex number is a real number (not an imaginary number) and the anti-Hermitian part of an operator is Hermitian (not anti-Hermitian).  This terminology may be confusing, but at least it is consistent between complex numbers and operators.
 ===== Projection Operators =====
-A //**projection operator**// or //**projector**// $\hat{P}$ is a Hermitian operator that also satisfies $\hat{P}^2 = \hat{P}$.  Here are some examples.
+Let $V$ be a subspace of a Hilbert space $\mathcal{H}$.  The //**projection operator**// or //**projector**// $\hat{P}_V$ onto $V$ is defined via
+\begin{align*}
+\hat{P}_V \ket{\psi} & = \ket{\psi},\,\,\text{for all }\ket{\psi} \in V, \\
+\hat{P}_V \ket{\psi} & = \boldsymbol{0},\,\,\text{for all }\ket{\psi} \in V^{\perp},
+\end{align*}
+where $V^{\perp}$ is the orthogonal complement of $V$ in $\mathcal{H}$.
-  * The identity operator $\hat{I}$ is a projector, since it is Hermitian
+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
-  \[\sand{\phi}{\hat{I}}{\psi} = \braket{\phi}{\psi} = \braket{\psi}{\phi}^* = \sand{\psi}{\hat{I}}{\phi}^*,\]
+\[\hat{P}_V \ket{\psi} = \hat{P}_V \left ( a\ket{\phi} + b\ket{\phi^{\perp}} \right ) = a\hat{P}_V\ket{\phi} + b\hat{P}_V \ket{\phi^{\perp}} = a\ket{\phi}.\]
-  and it squares to itself
+In other words, the projection operator $\hat{P}_V$ simply returns the component of the vector $\ket{\psi}$ that lies in the subspace $V$.
-  \[\hat{I}^2 \ket{\psi} = \hat{I}\hat{I}\ket{\psi} = \hat{I}\ket{\psi}.\]
-  * For any normalized vector $\ket{\psi}$, the outer product $\proj{\psi}$ is a projector, since
-  \[\braket{\phi}{\psi}\braket{\psi}{\chi} = \braket{\psi}{\phi}^*\braket{\chi}{\psi}^* = \braket{\chi}{\psi}^*\braket{\psi}{\phi}^* = \left [ \braket{\chi}{\psi}\braket{\psi}{\phi}\right ]^*,\]
-  and
-  \[\left( \proj{\psi}\right )^2 = \ket{\psi}\braket{\psi}{\psi}\bra{\psi} = \ket{\psi}1\bra{\psi} = \proj{\psi}.\]
-In fact, the general form of a projection operator is as follows.  Consider a subspace of a Hilbert space and let $\ket{e_1}, \ket{e_2}, \cdots$ be an orthonormal basis for that subspace.  Then,
-\[\hat{P} = \sum_j \proj{e_j}\]
-is a projection operator and, conversely, every projection operator is of this form for some subspace.  We will prove this in section 2.iii.5.
-We can now understand why these operators are called projection operators.  Any vector $\ket{\psi}$ can be written as
+A trivial example of a projection operator is the projector $\hat{P}_{\mathcal{H}}$ onto the entire Hilbert space $\mathcal{H}$.  Since the orthogonal complement of $\mathcal{H}$ only consists of the zero vector $\boldsymbol{0}$, the projector $\hat{P}_{\mathcal{H}}$ maps every vector $\ket{\psi}$ to itself.  In other words, it is just the identity operator $\hat{P}_{\mathcal{H}} = \hat{I}$.
-\[\ket{\psi} = \sum_j b_j \ket{e_j} + c \ket{\psi^{\perp}},\]
-where $\ket{\psi^{\perp}}$ is orthogonal to the subspace spanned by $\ket{e_1}, \ket{e_2}, \codts$.  It follows that $\braket{e_j}{\psi^{\perp}} = 0$ for all $j$.  Now, if we act on $\ket{\psi}$ with the projector $\hat{P} = \sum_j \proj{e_j}$, we get
+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 $\boldsymbol{0}$.  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
+\[\hat{I} = \sum_j \proj{\phi_j}.\]
+This result can be extended to arbitrary projectors.  If $\ket{\phi_1},\ket{\phi_2},\cdots$ is an orthonormal basis for a //subspace// $V$ of $\mathcal{H}$ then
+\[\hat{P}_V = \sum_j \proj{\phi_j}.\]
+To see this, let note that any vector $\ket{\psi}$ can be written as $\ket{\psi} = \ket{\phi} + \ket{\phi^{\perp}}$ where $\ket{\phi} \in V$ and $\ket{\phi^{\perp}} \in V^{\perp}$.  Since all of the basis vectors $\ket{\phi_j}$ are in $V$, they must satisfy $\braket{\phi_j}{\phi^{\perp}} = 0$.  Also, since $V$ is itself a Hilbert space $\hat{P}_V = \sum_j \proj{\phi_j}$ is the identity operator on $V$, so $\hat{P}_V \ket{\phi} = \ket{\phi}$.
+All projection operators are Hermitian $\hat{P}_V^{\dagger} = \hat{P}_V$ and //**idempotent**//, which means $\hat{P}_V^2 = \hat{P}_V$.  In fact, the converse is also true: //all// operators that are Hermitian and idempotent are projectors, so it is common to //define// a projection operator to be an idempotent Hermitian operator.  We shall prove the converse in section 2.v, but for now we will show that projectors have these two properties.
+To show that $\hat{P}_V$ is Hermitian, let $\ket{\psi_1}$ and $\ket{\psi_2}$ be two vectors in $\mathcal{H}$, we can write them both as
 \begin{align*}
-\hat{P} \ket{\psi} & = \left (\sum_k \proj{e_k} \right ) \left ( \sum_j b_j \ket{e_j} + c \ket{\psi^{\perp}}\right ) \\
+\ket{\psi_1} & = \ket{\phi_1} + \ket{\phi_1^{\perp}}, & \ket{\psi_2} & = \ket{\phi_2} + \ket{\phi_2^{\perp}},
-& \sum_k \left ( \sum_j b_j \ket{e_k}\braket{e_k}{e_j} +  c \ket{e_k}\braket{e_k}{\psi^{\perp}}\right ) \\
-& = \sum_k \sum_j b_j \ket{e_k}\delta_{kj} + 0 \\
-& = \sum_j b_j \ket{e_j}.
 \end{align*}
-So, you can see that the projector $\hat{P}$ projects the vector onto the subspace spanned by $\ket{e_1}, \ket{e_2}, \cdots$, removing the orthogonal component.
+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 ( \bra{\phi_1} + \bra{\phi_1^{\perp}} \right ) \hat{P}_V \left (\ket{\phi_2} + \ket{\phi_2^{\perp}} \right ) \\
+   & = \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.