# Quantum Mechanical Unitary Transformations

$$\hat{U}|\psi\rangle = |\psi'\rangle$$ and $$\langle\psi'|=\langle|\hat{U}^\dagger$$

## Definition

$$\hat{U}^\dagger\hat{U}=\mathbb{I}$$.

### Consider the Unitary Transformation

$$A|\psi\rangle = |\varphi\rangle \Rightarrow^{\text{Unitary Transformation}} \hat{A}'|\psi'\rangle = |\varphi'\rangle$$. $$\hat{A}'\hat{U}|\psi\rangle=\hat{U}|\varphi\rangle$$. Then $$\hat{U}^\dagger \hat{A}'\hat{U}|\psi\rangle = |\varphi\rangle$$. Thus, $$\hat{A}' = \hat{U}A\hat{U}^\dagger$$. $$\hat{A}'$$ and $$\hat{A}$$ are unitarily equivalent observables.

## Properties

• Let $$\hat{A}$$ be Hermitian. Then, $$\hat{A}'=\hat{U}\hat{A}\hat{U}^\dagger$$, $$\hat{A}'^\dagger = \hat{U}\hat{A}^\dagger\hat{U}^\dagger = \hat{A}'$$.
• Let $$\hat{A}|\psi\rangle=a|\psi\rangle$$. Then, $$\hat{A}'|\psi'\rangle = \hat{U}\hat{A}\hat{U}^\dagger\hat{U}|\psi\rangle = a|\psi'\rangle$$.
• Consider $$\langle\psi|\hat{A}|\chi\rangle$$.Then, $$\langle\psi'|\hat{A}'|\chi'\rangle = \langle\psi|\hat{U}^\dagger \hat{U} \hat{A} \hat{U}^\dagger \hat{U}|\chi = \langle\psi|\hat{A}|\chi\rangle$$.
• So, in general, $$\langle\psi'|\chi'\rangle = \langle\psi|\chi\rangle$$. Then, the norm and expectation values are invariant under unitary transformations.

Created: 2024-05-30 Thu 21:21

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