Infinitesimal and Finite Unitary Transformations
Recall
\(\hat{U}^\dagger = \hat{U}^{-1}, \hat{U}|\psi\rangle = |\psi\rangle, \hat{A}'=\hat{U}\hat{A}\hat{U}^\dagger, \hat{A}=\hat{U}^\dagger \hat{A}'\hat{U}\).
Definition
\(\hat{U}_\varepsilon(\hat{G})= \mathbb{I} + i\varepsilon \hat{G}\)
Check
\(\mathbb{I}=U_\varepsilon U_\varepsilon^\dagger = (\mathbb{I}+i\varepsilon\hat{G})(\mathbb{I}-i\varepsilon\hat{G}^\dagger) = \mathbb{I} + i\varepsilon (\hat{G}-\hat{G}^\dagger) + \mathcal{O}(\varepsilon^2) = \mathbb{I}\).
Thus, \(\hat{G}\) must be Hermitian.
Finite Transformation
\(\hat{U}_\alpha(\hat{G}) = \hat{U}_\varepsilon(\hat{G})^N = \mathbb{I} + Ni\varepsilon \hat{G} = \mathbb{I} + i\alpha \hat{G} = \exp(i\alpha \hat{G})\).
Applications
Time Translations
Let \(\hat{G}=\frac{\hat{H}}{\hbar}\). Then, \(\hat{U}_{\delta t}(\hat{H}) = \mathbb{I} - \frac{i}{\hbar} \hat{H}\delta t\). So,
- \(\hat{H}|\psi(t)\rangle = i\hbar \frac{1}{\partial t}|\psi(t)\rangle\)
- \(\hat{U}_{\delta t}|\psi(t)\rangle = (\mathbb{I} - i\hat{H}/\hbar \delta t)|\psi(t)\rangle = |\psi(t)\rangle - i^2\delta t \frac{\partial}{\partial t}|\psi(t)\rangle \approx |\psi(t+\delta t)\rangle\).
Spatial Translations
Let \(\hat{G}=\frac{1}{\hbar}\hat{P_x}\). Then,
- \(\hat{U}_{\delta x}|\psi(x)\rangle = |\psi(x)\rangle + \frac{i}{\hbar}(\delta x) \left(-i\hbar\frac{\partial}{\partial x}\right) = |\psi(x+\delta x)\rangle\).
- \(\hat{X}' = \hat{X}' + \frac{i}{\hbar}(\delta X)[\hat{P}_x,\hat{X}] \hat{X} + (\delta X)\)
- \(\hat{Y}' = \hat{Y}\)
Rotations
\(\hat{G} = \frac{\hat{J_z}}{\hbar}\). \(\hat{U}_{d\varphi} = \mathbb{I} + \frac{i}{\hbar}d\varphi \hat{J}_z\)