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.

You should now do the first in class activity to determine some properties of Hermitian adjoints.

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. \[\left ( a\hat{A}\right )^{\dagger} = \hat{A}^{\dagger}a^* = a^*\hat{A}^{\dagger}\] \[\left ( \hat{A}^n\right )^{\dagger} = \left ( \hat{A}^{\dagger} \right )^{n}\] \[\left ( \hat{A} + \hat{B} + \hat{C} + \hat{D} \right )^{\dagger} = \hat{A}^{\dagger} + \hat{B}^{\dagger} + \hat{C}^{\dagger} + \hat{D}^{\dagger}\] \[\left ( \hat{A}\hat{B}\hat{C}\hat{D}\right )^{\dagger} = \hat{D}^{\dagger}\hat{C}^{\dagger}\hat{B}^{\dagger}\hat{A}^{\dagger}\] \[\left ( \hat{A}\hat{B}\hat{C}\hat{D} \ket{\psi}\right )^{\dagger} = \bra{\psi}\hat{D}^{\dagger}\hat{C}^{\dagger}\hat{B}^{\dagger}\hat{A}^{\dagger}\] \[\left ( \ketbra{\psi}{\phi} \right )^{\dagger} = \ketbra{\phi}{\psi}\]