# Quantum Mechanical Pictures

## Properties

### Expectation Value Equivalence

$$\langle A\rangle_S = \langle \alpha,t_0;t|A_S|\alpha,t_0;t\rangle = \langle \alpha,t_0|\hat{U}^\dagger A_S\hat{U}|\alpha,t_0\rangle = \langle A_H\rangle$$.

### Examples

#### Equations of Motion

$$\hat{H} = \frac{\hat{P}_x^2+\hat{P}_y^2 + \hat{P}_z^2}{2m}$$.

Then, $$\frac{d}{dt}P_i = -\frac{i}{\hbar}[\hat{P}_i,\hat{H}] = 0$$.

And, $$\frac{d}{dt}R_i = -\frac{i}{\hbar}[\hat{R}_i,\hat{H}] = \frac{P_i}{m}$$.

Since, momentum is a constant of the motion ($$P_i=C_i$$), $$\hat{R}_i(t) = \hat{R}_i(0) + \frac{\hat{P}_i}{m}t$$.

#### With a Potential - Ehrenfest Theorem Roots

$$\hat{H} = \frac{\hat{P}_x^2+\hat{P}_y^2 + \hat{P}_z^2}{2m} + \hat{V}(R)$$.

Then, $$\frac{d}{dt}P_i = -\frac{i}{\hbar}[\hat{P}_i,\hat{H}] = -\frac{\partial}{\partial R_i}\hat{V}(R)$$, by Virial Theorem.

And, $$\frac{d}{dt}R_i = -\frac{i}{\hbar}[\hat{R}_i,\hat{H}] = \frac{P_i}{m}$$.

$$\frac{d^2}{dt^2}R_i = \frac{1}{m}\frac{d}{dt}P_i = \frac{1}{m}\nabla_i\hat{V}(R)$$.

Thus, we have Ehrenfest Theorem, $$m\frac{d^2}{dt^2}\vec{R} = -\vec{\nabla}\hat{V}(\vec{R})$$.

Created: 2024-05-30 Thu 21:20

Validate