Unitary Operators
Definition
A Unitary operator satisfies \(A^\dagger=A^{-1}\).
Thus \(A^\dagger A=A A^\dagger = \mathbb{I}\).
Theorem - Product
Let \(A,B,C,D,\cdots\) be Unitary. Then their product is Unitary.
Proof. Then \((ABCD\cdots)(ABCD\cdots)^\dagger = (ABCD\cdots)(\cdots D^\dagger C^\dagger B^\dagger A^\dagger) = \mathbb{I}^{\cdots} = \mathbb{I}\). Therefore, the product of unitary operators is Unitary.
Theorem - Eigenvalues
The eigenvalues of Unitary operators are complex with moduli 1 and the eigenvectors (no degenerate eigenvalues) are mutually orthogonal.
Proof. \(U|\varphi_{n,m}\rangle = a_{n,m}|\varphi_{n,m}\rangle\). Then \(\langle \varphi_m|U^\dagger U|\varphi_n\rangle = a_m^*a_n\langle\varphi_m|\varphi_m\rangle\). Or, \(\langle\varphi_m|\mathbb{I}|varphi_n\rangle = \langle\varphi_m|\varphi_n\rangle\).
For \(m=n\), \(a_n^*a_n = 1\). For \(a_m\neq a_n\), the inner product ensures orthogonality.
Identity Operator - Completeness Relation
\(\mathbb{I}=\sum_i |\varphi_i\rangle\langle\varphi_i|\).
Examples
\(|\varphi_1\rangle \doteq \begin{pmatrix}1\\0\end{pmatrix}\) \(|\varphi_2\rangle \doteq \begin{pmatrix}0\\1\end{pmatrix}\)
Then, \(\mathbb{I} \doteq \begin{pmatrix}1\\0\end{pmatrix}(1 0) + \begin{pmatrix}0\\1\end{pmatrix}(0 1) = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\).