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# 物理代写|量子力学代写Quantum mechanics代考|PX3511 Energy Shift and Decay Width

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## 物理代写|量子力学代写Quantum mechanics代考|Energy Shift and Decay Width

Our considerations so far have been restricted to the question of how states other than the initial state become populated. In other words, we have been concerned with the time development of the coefficient $c_n(t)$ with $n \neq i$. The question naturally arises, What happens to $c_i(t)$ itself?

To avoid the effect of a sudden change in the Hamiltonian, we propose to increase the perturbation very slowly. In the remote past $(t \rightarrow-\infty)$ the time-dependent potential is assumed to be zero. We then gradually turn on the perturbation to its full value; specifically,
$$V(t)=e^{\eta t} V$$
where $V$ is assumed to be constant and $\eta$ is small and positive. At the end of the calculation, we let $\eta \rightarrow 0$ (see Figure 5.14), and the potential then becomes constant at all times.

In the remote past, we take this time to be $-\infty$, so the state ket in the interaction picture is assumed to be $|i\rangle$. Our basic aim is to evaluate $c_i(t)$. However, before we do that, let us make sure that the old formula of the golden rule (see Section 5.7) can be reproduced using this slow-turn-on method. For $c_n(t)$ with $n \neq i$, we have [using (5.276)]

\begin{aligned} c_n^{(0)}(t) &=0 \ c_n^{(1)}(t) &=\frac{-i}{\hbar} V_{n i} \lim {t_0 \rightarrow-\infty} \int{t_0}^t e^{\eta t^{\prime}} e^{i \omega_{n i} t^{\prime}} d t^{\prime} \ &=\frac{-i}{\hbar} V_{n i} \frac{e^{\eta t+i \omega_{n i} t}}{\eta+i \omega_{n i}} . \end{aligned}
To lowest nonvanishing order, the transition probability is therefore given by
$$\left|c_n(t)\right|^2 \simeq \frac{\left|V_{n i}\right|^2}{\hbar^2} \frac{e^{2 \eta t}}{\eta^2+\omega_{n i}^2},$$
or
$$\frac{d}{d t}\left|c_n(t)\right|^2 \simeq \frac{2\left|V_{n i}\right|^2}{\hbar^2}\left(\frac{\eta e^{2 \eta t}}{\eta^2+\omega_{n i}^2}\right) .$$

## 物理代写|量子力学代写Quantum mechanics代考|Scattering as a Time-Dependent Perturbation

We assume that the Hamiltonian can be written as
$$H=H_0+V(\mathbf{x})$$
where
$$H_0=\frac{\mathbf{p}^2}{2 m}$$
stands for the kinetic-energy operator, with eigenvalues
$$E_{\mathbf{k}}=\frac{\hbar^2 \mathbf{k}^2}{2 m} .$$
We denote the plane wave eigenvectors of $H_0$ by $|\mathbf{k}\rangle$ and we assume that the scattering potential $V(\mathbf{r})$ is time independent.

Our treatment realizes that an incoming particle will “see” the scattering potential as a perturbation which is “turned on” only during the time that the particle is in the vicinity of the scatterer. Therefore, we can analyze the problem in terms of time-dependent perturbation theory in the interaction picture.

To review (see Section 5.7), the state $\left|\alpha, t_0 ; t_0\right\rangle_I$ evolves into the state $\left|\alpha, t_0 ; t\right\rangle_I$ according to
$$\left|\alpha, t_0 ; t\right\rangle_I=U_I\left(t, t_0\right)\left|\alpha, t_0 ; t_0\right\rangle_I$$
where $U_I\left(t, t_0\right)$ satisfies the equation
$$i \hbar \frac{\partial}{\partial t} U_I\left(t, t_0\right)=V_I(t) U_I\left(t, t_0\right)$$

with $U_I\left(t_0, t_0\right)=1$ and $V_I(t)=\exp \left(i H_0 t / \hbar\right) V \exp \left(-i H_0 t / \hbar\right)$. The solution of this equation can be formally written as
$$U_I\left(t, t_0\right)=1-\frac{i}{\hbar} \int_{t_0}^t V_I\left(t^{\prime}\right) U_I\left(t^{\prime}, t_0\right) d t^{\prime} .$$
Therefore, the “transition amplitude” for an initial state $|i\rangle$ to transform into a final state $|n\rangle$, where both are eigenstates of $H_0$, is given by
$$\left\langle n\left|U_I\left(t, t_0\right)\right| i\right\rangle=\delta_{n i}-\frac{i}{\hbar} \sum_m\langle n|V| m\rangle \int_{t_0}^t e^{i \omega_{n m} t^{\prime}}\left\langle m\left|U_I\left(t^{\prime}, t_0\right)\right| i\right\rangle d t^{\prime}$$
where $\langle n \mid i\rangle=\delta_{n i}$ and $\hbar \omega_{n m}=E_n-E_m$.

## 物理代写|量子力学代写Quantum mechanics代考|Energy Shift and Decay Width

$$V(t)=e^{\eta t} V$$

$$c_n^{(0)}(t)=0 c_n^{(1)}(t) \quad=\frac{-i}{\hbar} V_{n i} \lim t_0 \rightarrow-\infty \int t_0{ }^t e^{\eta t^{\prime}} e^{i \omega_{r i a} t^{\prime}} d t^{\prime}=\frac{-i}{\hbar} V_{n i} \frac{e^{\eta t+i \omega_{r i} t}}{\eta+i \omega_{n i}}$$

$$\left|c_n(t)\right|^2 \simeq \frac{\left|V_{n i}\right|^2}{\hbar^2} \frac{e^{2 \eta t}}{\eta^2+\omega_{n i}^2}$$

$$\frac{d}{d t}\left|c_n(t)\right|^2 \simeq \frac{2\left|V_{n i}\right|^2}{\hbar^2}\left(\frac{\eta e^{2 \eta t}}{\eta^2+\omega_{n i}^2}\right)$$

## 物理代写量子力学代写Quantum mechanics代考|Scattering as a TimeDependent Perturbation

$$H=H_0+V(\mathbf{x})$$

$$H_0=\frac{\mathbf{p}^2}{2 m}$$

$$E_{\mathbf{k}}=\frac{\hbar^2 \mathbf{k}^2}{2 m} .$$

$$\left|\alpha, t_0 ; t\right\rangle_I=U_I\left(t, t_0\right)\left|\alpha, t_0 ; t_0\right\rangle_I$$

$$i \hbar \frac{\partial}{\partial t} U_I\left(t, t_0\right)=V_I(t) U_I\left(t, t_0\right)$$

$$U_I\left(t, t_0\right)=1-\frac{i}{\hbar} \int_{t_0}^t V_I\left(t^{\prime}\right) U_I\left(t^{\prime}, t_0\right) d t^{\prime} .$$

$$\left\langle n\left|U_I\left(t, t_0\right)\right| i\right\rangle=\delta_{n i}-\frac{i}{\hbar} \sum_m\langle n|V| m\rangle \int_{t_0}^t e^{i \omega_{n n n t}}\left\langle m\left|U_I\left(t^{\prime}, t_0\right)\right| i\right\rangle d t^{\prime}$$

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