# Time-Dependent Potentials

## The Interaction Picture

Consider $$H = H_0 + V(t)$$ with $$H_0$$ being time-independent.

We then have a state: $$|\alpha,t_0;t\rangle = \exp\left(\frac{i}{\hbar}H_0(t-t_0)\right)|\alpha,t_0;t\rangle_S$$.

Operators are: $$A_I = \exp\left(\frac{i}{\hbar}H_0t\right)A_S\exp\left(-\frac{i}{\hbar}H_0t\right)$$.

Equation of state: $$\frac{dA_I}{dt} = \frac{1}{i\hbar}[A_I,H_0]$$.

Assuming $$t_0=0$$ for simplicity, $$i\hbar\frac{\partial}{\partial t}|\alpha,t_0;t\rangle_I = i\hbar\frac{\partial}{\partial t}\left(\exp\left(\frac{i}{\hbar}H_0t\right)|\alpha,t_0;t\rangle_S\right) = -H_0\exp\left(\frac{i}{\hbar}H_0t\right)|\alpha,t_0;t\rangle_S + \exp\left(\frac{i}{\hbar}H_0t\right)i\hbar\frac{\partial}{\partial t}|\alpha,t_0;t\rangle_S = -H_0\exp\left(\frac{i}{\hbar}H_0t\right)|\alpha,t_0;t\rangle_S + \exp\left(\frac{i}{\hbar}H_0t\right)(H_{0,S} + V_{(S)}(t))|\alpha,t_0;t\rangle_S = \exp\left(\frac{i}{\hbar}H_0t\right)V_{(S)}(t)|\alpha,t_0;t\rangle_S = V_I\exp\left(\frac{i}{\hbar}H_0t\right)|\alpha,t_0;t\rangle_S = V_I|\alpha,t_0;t\rangle_I$$.

Thus,

$$i\hbar\frac{\partial}{\partial t}|\alpha,t_0;t\rangle_I = V_I|\alpha,t_0;t\rangle_I.$$

Therefore, the time dependence is solely due to the time dependence of the potential.

Note: $$V_{(S)}(t) = \exp\left(-\frac{i}{\hbar}H_0t\right)V_I\exp\left(\frac{i}{\hbar}H_0t\right)$$.

### High-Level Example

$$H_0|n\rangle = E_n|n\rangle$$. Assume we have an initial state $$|i\rangle$$. Now we turn on the time-dependent potential, $$H = H_0 + V(t)$$. What is/are our final state(s), with what probabilities?

$$|\alpha,t_0;t\rangle = \sum_n c_n(t)|n\rangle$$. $$\mathcal{P}(t) = |c_n(t)|^2$$.

Ansatz: $$|\alpha,t_0;t\rangle = \sum_n c_n\exp\left(-\frac{i}{\hbar}V_It\right)|n\rangle$$.

More rigorously, letting $$|\alpha(t)\rangle = |\alpha,t_0;t\rangle$$,

\begin{align*} i\hbar\frac{\partial}{\partial t}|\alpha(t)\rangle_I &= V_I|\alpha(t)\rangle_I \\ i\hbar\frac{\partial}{\partial t}\langle n|\alpha(t)\rangle &= \langle n|V_I|\alpha(t)\rangle \\ i\hbar\frac{\partial}{\partial t}c_n(t) &= \sum_m\langle n|V_I|m\rangle\langle m|\alpha(t)\rangle \\ &= \sum_m\langle n|V_I|m\rangle c_m(t) \\ &= \sum_m\langle n|\exp\left(\frac{i}{\hbar}H_0t\right)V(t)\exp\left(-\frac{i}{\hbar}H_0t\right)|m\rangle c_m(t) \\ &= \sum_m\langle n|\exp\left(\frac{i}{\hbar}E_nt\right)V(t)\exp\left(-\frac{i}{\hbar}E_mt\right)|m\rangle c_m(t) \\ &= \sum_m\exp\left(\frac{i}{\hbar}(E_n-E_m)t\right)\langle n|V(t)|m\rangle c_m(t) \\ &= \sum_m\exp\left(i\omega_{nm}t\right) V_{nm}(t) c_m(t) \\ i\hbar\frac{dc_n(t)}{dt} &= \sum_m V_{nm}\exp(i\omega_{nm})c_m(t). \\ i\hbar \frac{d}{dt}\begin{pmatrix}\dot{c}_1\\\dot{c}_2\\\cdots\end{pmatrix} &= \begin{pmatrix}V_{11} & V_{12}\exp(i\omega_{12}t)&\cdots \\ V_{21}\exp(i\omega_{21}t)&V_{22}&\cdots\\\cdots&\cdots&\cdots\end{pmatrix}\begin{pmatrix}c_1\\c_2\\\cdots\end{pmatrix}. \end{align*}

This is useful for discrete system, i.e. non-infinite. Consider two-level systems: spin-1/2, laser or maser (NMR) with energies $$E_1,E_2$$.

#### Example Two-Level System

Let $$E_1,E_2$$, $$H_0=E_1|1\rangle\langle 1| + E_2|2\rangle\langle 2|$$. Apply a potential, $$V(t) = \gamma\exp(i\omega t)|1\rangle\langle 2| + \gamma\exp(-i\omega t)|2\rangle\langle 1|$$.

$$i\hbar \dot{c}_1 = V_{11}c_1 + c_2V_{12}\exp(i\omega_{12}t) = c_2V_{12}\exp(i\omega_{12}t)$$. $$i\hbar \dot{c}_2 = V_{22}c_2 + c_1V_{21}\exp(i\omega_{21}t) = c_1V_{21}\exp(-i\omega_{12}t)$$.

$$|c_2(t)|^2 = \frac{\frac{\gamma^2}{\hbar^2}}{\frac{\gamma^2}{\hbar^2}+\frac{(\omega+\omega_{12})^2}{4}}\sin\left(\sqrt{\frac{\gamma^2}{\hbar^2}+\frac{(\omega+\omega_{12})^2}{4}}t\right) = A\sin\Omega t$$. $$\Omega$$ is called the Rabi frequency. At resonance, $$\omega=\omega_21$$, the amplitude is maximized and the Rabi frequency is $$\frac{\gamma}{\hbar}$$ and $$|c_2(t)|$$ can reach 1. Whereas, off resonance, $$c_2(t)$$ has a maximum occupation probability and is never 1.

### SchrÃ¶dinger Picture

State evolves and operators are time-independent.

### Heisenberg Picture

State is time-independent and operators evolve.

$$\frac{dA_H}{dt} = \frac{1}{i\hbar}[A_H,H].$$

Created: 2024-05-30 Thu 21:17

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