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 \[\hat{A}_h = \frac{\hat{A} - \hat{A}^{\dagger}}{2},\] which is an anti-Hermitian operator.

Then, the operator $\hat{A}$ can be written as \[\hat{A} = \hat{A}_h + \hat{A}_a.\] You should think of this decomposition as analogous to writing a complex scalar as \[a = \text{Re}(a) + i\text{Im}(a).\]

A projection operator or projector $\hat{P}$ is a Hermitian operator that also satisfies $\hat{P}^2 = \hat{P}$. Here are some examples.

• The identity operator $\hat{I}$ is a projector, since it is Hermitian \[\sand{\phi}{\hat{I}}{\psi} = \braket{\phi}{\psi} = \braket{\psi}{\phi}^* = \sand{\psi}{\hat{I}}{\phi}^*,\] and it squares to itself \[\hat{I}^2 \ket{\psi} = \hat{I}\hat{I}\ket{\psi} = \hat{I}\ket{\psi}.\]
• 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 \[\ket{\psi} = \sum_j b_j \ket{e_j} + c \ket{\psi^{\perp}},\] 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 \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 ) \\ & \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.

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 $\ket{e_j}$.