Unitary Operators

Definition

A Unitary operator satisfies $A^{†} = A^{- 1}$ .

Thus $A^{†} A = A A^{†} = I$ .

Theorem - Product

Let $A, B, C, D, \dots$ be Unitary. Then their product is Unitary.

Proof. Then $(A B C D \dots) (A B C D \dots)^{†} = (A B C D \dots) (\dots D^{†} C^{†} B^{†} A^{†}) = I^{\dots} = I$ . 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. $U | φ_{n, m} ⟩ = a_{n, m} | φ_{n, m} ⟩$ . Then $⟨ φ_{m} | U^{†} U | φ_{n} ⟩ = a_{m}^{*} a_{n} ⟨ φ_{m} | φ_{m} ⟩$ . Or, $⟨ φ_{m} | I | v a r p h i_{n} ⟩ = ⟨ φ_{m} | φ_{n} ⟩$ .

For $m = n$ , $a_{n}^{*} a_{n} = 1$ . For $a_{m} \neq a_{n}$ , the inner product ensures orthogonality.

Identity Operator - Completeness Relation

$I = \sum_{i} | φ_{i} ⟩ ⟨ φ_{i} |$ .

Examples

$| φ_{1} ⟩ ≐ (\begin{matrix} 1 \\ 0 \end{matrix})$ $| φ_{2} ⟩ ≐ (\begin{matrix} 0 \\ 1 \end{matrix})$

Then, $I ≐ (\begin{matrix} 1 \\ 0 \end{matrix}) (10) + (\begin{matrix} 0 \\ 1 \end{matrix}) (01) = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) + (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ .