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--- adjoints_and_hermitian_operators [2021/02/24 07:20] – created admin
+++ adjoints_and_hermitian_operators [2022/10/06 00:45] (current) – [Projection Operators] admin
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 Note, in mathematics, the Hermitian adjoint is usually denoted $^*$.  This makes sense as, in many ways it is the generalization of complex conjugation to operators.  However, in physics we normally reserve $^*$ for taking the operator formed by taking the complex conjugates of each of its elements in a given basis.  This is not the same thing as the Hermitian adjoint.  We will discuss how to represent operators in a basis in section 2.iv.
-You should now do the first in class activity to determine some properties of Hermitian adjoints.
+===== Properties of Hermitian Adjoints =====
+In an in class activity, you will show that
+\[\left ( \hat{A}^{\dagger}\right )^{\dagger} = \hat{A},\]
+and
+\[\left ( \hat{A}\hat{B} \right )^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger},\]
+so taking adjoints reverses the order of products.
+===== Taking Adjoints of General Expressions =====
 It is convenient to define $\left ( \ket{\psi} \right )^{\dagger} = \bra{\psi}$ and $\left ( \bra{\psi} \right )^{\dagger} = \ket{\psi}$.  Then, we can write things like
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 \[\left ( \hat{A}\hat{B}\hat{C}\hat{D} \ket{\psi}\right )^{\dagger} = \bra{\psi}\hat{D}^{\dagger}\hat{C}^{\dagger}\hat{B}^{\dagger}\hat{A}^{\dagger}\]
 \[\left ( \ketbra{\psi}{\phi} \right )^{\dagger} = \ketbra{\phi}{\psi}\]
+===== Operators Acting Inside Bras and Kets =====
+It is convenient to define the action of operators inside bras and kets as
+\begin{align*}
+\ket{a\hat{A}\psi} & = a\hat{A}\ket{\psi}, & \bra{a\hat{A}\psi} = a^* \bra{\psi}\hat{A}^{\dagger}.
+\end{align*}
+In other words, you can think of the symbol $\bra{\text{expression}}$ as an instruction to take the Hermitian adjoint of $\text{expression}$.
+With this convention, we have
+\[\sand{\phi}{\hat{A}}{\psi} = \braket{\hat{A}^{\dagger}\phi}{\psi} = \braket{\phi}{\hat{A}\psi}.\]
+Personally, I tend to avoid writing operators inside of bras and kets as much as possible because it can get confusing to keep track of where the $^{\dagger}$'s have to go.  In rare cases, the notation is useful though.
+===== Hermitian and Skew-Hermitian Operators =====
+An operator $\hat{A}$ is //**Hermitian**// if
+\[\hat{A}^{\dagger} = \hat{A},\]
+or equivalently
+\[\sand{\phi}{\hat{A}}{\psi} = \sand{\psi}{\hat{A}}{\phi}^*,\]
+for all vectors $\ket{\psi}$, $\ket{\phi}$.
+It is //**skew-Hermitian**// or //**anti-Hermitian**// if
+\[\hat{A}^{\dagger} = -\hat{A},\]
+or equivalently
+\[\sand{\phi}{\hat{A}}{\psi} = -\sand{\psi}{\hat{A}}{\phi}^*,\]
+for all vectors $\ket{\psi}$, $\ket{\phi}$.
+Note that this implies that the expectation value of a Hermitian operator is always real, since
+\[\sand{\psi}{\hat{A}}{\psi} = \sand{\psi}{\hat{A}}{\psi}^*.\]
+It should come as no surprise that the operators representing physical quantities like position, momentum, and energy are all Hermitian, since their expectation values need to be real numbers.
+Similarly, the expectation value of an anti-Hermitian operator is always imaginary, since
+\[\sand{\psi}{\hat{A}}{\psi} = -\sand{\psi}{\hat{A}}{\psi}^*.\]
+For any operator, $\hat{A}$, we can define its //**Hermitian part**// as
+\[\hat{A}_h = \frac{\hat{A} + \hat{A}^{\dagger}}{2},\]
+which is a Hermitian operator, and its //**anti-Hermitian part**// as
+\[\hat{A}_a = \frac{\hat{A} - \hat{A}^{\dagger}}{2i},\]
+which is also a Hermitian operator.
+Then, the operator $\hat{A}$ can be written as
+\[\hat{A} = \hat{A}_h + i\hat{A}_a.\]
+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 =====
+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 $\mathcal{H}$.
+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 $V$.
+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 $\hat{P}_{\mathcal{H}} = \hat{I}$.
+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*}
+\ket{\psi_1} & = \ket{\phi_1} + \ket{\phi_1^{\perp}}, & \ket{\psi_2} & = \ket{\phi_2} + \ket{\phi_2^{\perp}},
+\end{align*}
+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.
+  * If two projectors $\hat{P}_1$ and $\hat{P}_2$ commute, then $\hat{P}_1\hat{P}_2$ is also a projector, since
+  \[\left ( \hat{P}_1 \hat{P}_2\right )^{\dagger} = \left ( \hat{P}_2 \hat{P}_1\right )^{\dagger} = \hat{P}_1^{\dagger} \hat{P}_2^{\dagger} = \hat{P}_1\hat{P}_2\]
+  \[\left ( \hat{P}_1 \hat{P}_2\right )^2 = \hat{P}_1\hat{P}_2\hat{P}_1\hat{P}_2 = \hat{P}_1\hat{P}_1\hat{P}_2\hat{P}_2 = \hat{P}_1^2 \hat{P}_2^2 = \hat{P}_1 \hat{P}_2.\]
+  In fact, $\hat{P}_1\hat{P}_2$ is the projector onto the intersection of the subspaces that $\hat{P}_1$ and $\hat{P}_2$ project onto.
+  * Two projectors are //**orthogonal**// if $\hat{P}_1\hat{P}_2 = 0$ (this is a special case of two commuting projectors).  It means that the subspaces that they project onto are orthogonal.
+  * A sum of two orthogonal projectors is a projector, since
+  \[\left ( \hat{P}_1 + \hat{P}_2 \right )^{\dagger} = \hat{P}_1^{\dagger} + \hat{P}_2^{\dagger} = \hat{P}_1 + \hat{P}_2\]
+  \[\left ( \hat{P}_1 + \hat{P}_2 \right )^2 = \hat{P}_1^2 + \hat{P}_2^2 + \hat{P}_1\hat{P}_2 + \hat{P}_2\hat{P}_1 = \hat{P}_1 + \hat{P}_2.\]
+  The projection operator $\hat{P}_1 + \hat{P}_2$ projects onto the linear span of the subspaces that $\hat{P}_1$ and $\hat{P}_2$ project onto.
+  * A sum of non-orthogonal projectors is not a projector.
+{{:question-mark.png?direct&50|}}
+====== In Class Activities ======
+  - **Properties of the Hermitian Adjoint**
+    - Prove that $\left ( \hat{A}^{\dagger} \right )^{\dagger} = \hat{A}$.
+    - Prove that $\left ( \hat{A}\hat{B} \right )^{\dagger} = \hat{B}^{\dagger}\hat{A}^{\dagger}$.\\
+    HINT: You will find it useful to do the Dirac notaty.  Insert an identity operator somewhere and decompose it as $\hat{I} = \sum_j \proj{e_j}$ for some orthonormal basis $\ket{e_j}$.