# SchrÃ¶dinger Picture

States have time dependence. $$|\alpha,t_0;t\rangle_S=|\alpha'\rangle = \hat{U}(t,t_0)|\alpha\rangle = \exp(-i\hat{H}(t-t_0)/\hbar)|\alpha\rangle$$. $$A_S(0) = A_S(t)$$.

$$\langle b'|A|a'\rangle = \langle Ub|A|Ua\rangle$$.

## Equation of Motion

$$\hat{H}|\Psi\rangle = i\hbar\frac{\partial}{\partial t}|\Psi\rangle$$.

## Position Space Hamiltonian

$$i\hbar \frac{\partial}{\partial t}\langle \vec{x}'|\alpha,t_0;t\rangle = \langle \vec{x}'|\hat{H}|\alpha,t_0;t\rangle$$, $$i\hbar\frac{\partial}{\partial t}\psi_\alpha(\vec{x}',t) = \langle \vec{x}'|p^2/2m|\alpha,t_0;t\rangle + V(\vec{x}')\psi_\alpha(\vec{x}',t) = \frac{-\hbar^2}{2m}\vec{\nabla}^2\psi_\alpha(\vec{x},t) + V(\vec{x}')\psi_\alpha(\vec{x}',t)$$.

$$i\hbar\frac{\partial}{\partial t}\psi_\alpha(\vec{x}',t) = \frac{-\hbar^2}{2m}\vec{\nabla}^2\psi_\alpha(\vec{x},t) + V(\vec{x}')\psi_\alpha(\vec{x}',t)$$.

$$-\frac{\hbar^2}{2m}\vec{\nabla}^2+\hat{V}(\vec{x}')=\hat{H}=i\hbar\frac{\partial}{\partial t}$$.

## Momentum Space Hamiltonian

$$i\hbar\frac{\partial}{\partial t}\Phi(\vec{p}',t) = \frac{\vec{p}'^2}{2m}\Phi(\vec{p}',t) + \langle \vec{p}'|V(i\hbar\vec{\nabla}_p)|\alpha,t_0;t\rangle$$.

### Example 1

$$\hat{H} = \frac{\hat{P}^2}{2m} + \hat{V}(x) = \frac{\hat{P}^2}{2m}+\frac{1}{\cosh^2\hat{X}}$$.

$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}u_e(x) + \frac{1}{\cosh^2x}u_E(x) = Eu_E(x)$$.

$$\frac{p^2}{2m}\Phi(p)+\frac{1}{\cosh^2(i\hbar\frac{d}{dp})}\Phi_E(p)=E\Phi_E(p)$$.

### Example 2 (Worksheet)

$$V(x) = fx$$, $$\left[\frac{p^2}{2m}+i\hbar f\frac{\partial}{\partial p}\right]\Phi_E(p)=E\Phi_E(p)$$.

### Example 3 Free Particle

$$\hat{H} = \frac{p^2}{2m}$$.

$$\Phi_E(p) = C_1\delta(p-\sqrt{2mE}) + C_2\delta(-p-\sqrt{2mE})$$

$$\Psi_E(x) = C_1\exp(-i(kx+\omega t))$$, $$\omega = \frac{E}{\hbar} = \frac{\hbar k^2}{2m}$$. $$|v_{ph}|=\frac{\omega(k)}{k}=\mathcal{O}(k)$$.

## Stationary State

$$|\alpha,t_0\rangle = |E_k\rangle$$, $$|\alpha,t_0;t\rangle = \exp(-iE_kt/\hbar)|\alpha,t_0\rangle$$. $$\langle \vec{x}'|E_k;t\rangle = \exp(-iE_kt/\hbar)\varphi_k(\vec{x}')$$. Plugging this into the Position State Hamiltonian, $$-\frac{\hbar^2}{2m}\vec{\nabla}^2\varphi_k(\vec{x}',t)+V(x')\varphi_k(\vec{x}',t) = i\hbar(-iE_k/\hbar)\varphi_k(\vec{x},t)$$. So, $$-\frac{\hbar^2}{2m}u_E(\vec{x}') + V(\vec{x}')u_E(\vec{x}') = E_k u_E(\vec{x}')$$.

$$H(\vec{x}')u_E(\vec{x}')=E_ku_E(\vec{x}')$$, $$u_E(\vec{x}') = \langle x'|\varphi_k\rangle$$.

Created: 2023-06-25 Sun 02:30

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