# 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, $$\mathbb{I} \doteq \begin{pmatrix}1\\0\end{pmatrix}(1 0) + \begin{pmatrix}0\\1\end{pmatrix}(0 1) = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} + \begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$$.

Created: 2024-05-30 Thu 21:16

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