The Hermitian adjoint $\hat{A}^{\dagger}$ of an operator $\hat{A}$ is the unique operator such that \[\sand{\psi}{\hat{A}^{\dagger}}{\phi} = \sand{\phi}{\hat{A}}{\psi}^*,\] for all vectors $\ket{\psi}$ and $\ket{\psi}$.

Note that multiplication by a scalar $\hat{a}\ket{\psi} = a\ket{\psi}$ can be thought of as a linear operator. Its Hermitian adjoint can be found via \begin{align*} \sand{\psi}{\hat{a}^{\dagger}}{\phi} & = \sand{\phi}{\hat{a}}{\psi}^* \\ & = \left ( a \braket{\phi}{\psi}\right )^* \\ & = a^* \braket{\phi}{\psi}^* \\ & = a^* \braket{\psi}{\phi}, \end{align*} so $\hat{a}^{\dagger}$ is multiplication by $a^*$. For scalars, we use $a^{\dagger}$ and $a^*$ interchangeably.

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.

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.

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 \begin{align*} \left [ \sand{\phi}{\hat{A}\hat{B}\hat{C}\cdots}{\psi} \right ]^{\dagger} & = \left ( \ket{\psi} \right )^{\dagger} \cdots \hat{C}^{\dagger} \hat{B}^{\dagger} \hat{A}^{\dagger} \left ( \bra{\psi} \right )^{\dagger} \\ & = \sand{\psi}{\cdots \hat{C}^{\dagger} \hat{B}^{\dagger} \hat{A}^{\dagger}}{\phi}. \end{align*}

With these definitions, we can take the Hermitian adjoint of an arbitrary expression. The rules for taking ajoints of expressions are:

1. Reverse the order of any products.
2. Replace operators $\hat{A}$ by their adjoints $\hat{A}$.
3. Replace scalars $a$ by their complex conjugates $a^*$.
4. Replace kets $\ket{\psi}$ by bras $\bra{\psi}$ and vice versa.

Here are some examples of these rules in action. \[\left ( a\hat{A}\right )^{\dagger} = \hat{A}^{\dagger}a^* = a^*\hat{A}^{\dagger}\] \[\left ( \hat{A}^n\right )^{\dagger} = \left ( \hat{A}^{\dagger} \right )^{n}\] \[\left ( \hat{A} + \hat{B} + \hat{C} + \hat{D} \right )^{\dagger} = \hat{A}^{\dagger} + \hat{B}^{\dagger} + \hat{C}^{\dagger} + \hat{D}^{\dagger}\] \[\left ( \hat{A}\hat{B}\hat{C}\hat{D}\right )^{\dagger} = \hat{D}^{\dagger}\hat{C}^{\dagger}\hat{B}^{\dagger}\hat{A}^{\dagger}\] \[\left ( \hat{A}\hat{B}\hat{C}\hat{D} \ket{\psi}\right )^{\dagger} = \bra{\psi}\hat{D}^{\dagger}\hat{C}^{\dagger}\hat{B}^{\dagger}\hat{A}^{\dagger}\] \[\left ( \ketbra{\psi}{\phi} \right )^{\dagger} = \ketbra{\phi}{\psi}\]

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.

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.

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.

1. Properties of the Hermitian Adjoint

   1. Prove that $\left ( \hat{A}^{\dagger} \right )^{\dagger} = \hat{A}$.
   2. 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 $\ket{e_j}$.