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.