# Unitary

$$U^\dagger = U^{-1}$$.

They are often called isometries (symmetries of the system).

## Theorem

An operator $$U$$ is unitary iff it preserves the inner product of all vectors. I.e. if we define $$|V\rangle=U|v\rangle$$ and $$|W\rangle=U|w\rangle$$ then $$(|v\rangle,|w\rangle)=(|V\rangle,|W\rangle)$$.

In short, it maps orthonormal sets to orthonormal sets.

Proof. Assume $$U$$ is unitary. Then $$\langle V|W\rangle = \langle Uv|Uw\rangle = \langle v|U^\dagger U|w\rangle = \langle v|\mathbb{I}|w\rangle = \langle v|w|rangle$$.

Assume $$(|V\rangle,|W\rangle) = (|v\rangle,|w\rangle)$$. Then $$\langle Uv|Uw\rangle = \langle v|U^\dagger U|w\rangle = \langle v|w\rangle$$. So, $$U^\dagger U = \mathbb{I}$$. Then, $$U^\dagger U=\mathbb{I}$$. Also, $$\langle j|i\rangle = \langle Uj|Ui\rangle = \langle j|U^\dagger U|i\rangle = \langle j|i\rangle = \delta^i_j = \delta^j_i = \langle i|j\rangle = \langle j|i\rangle^* = \langle j|(U^\dagger U)^\dagger|i\rangle^*$$. Thus, $$U^\dagger = U^{-1}$$. Hence, $$U$$ is unitary.

## Examples

### Explicit example

$$U=\frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix}, UU^\dagger=U^\dagger U=\mathbb{I}$$.

### Rotation operators

Created: 2024-05-30 Thu 21:15

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