# 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

$$\hat{G} = \frac{\hat{J_z}}{\hbar}$$. $$\hat{U}_{d\varphi} = \mathbb{I} + \frac{i}{\hbar}d\varphi \hat{J}_z$$

Created: 2024-05-30 Thu 21:19

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