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# 物理代考| Photoionization 量子力学代写

## 物理代写

#### Photoionization

We are now in a position to make a more realistic calculation of photoionization by the radiation field. We work in three dimensions with initial and
remember that now $\vec{E}=-\partial \vec{A} / \partial t$, and the magnitude of the time-average Poynting vector for the electromagnetic field is $S_{\text {inc }}=\left\langle\varepsilon_{0} \vec{E}^{2}\right\rangle c$.
$50 \quad$ Introduction to Quantum Mechanics
final particle wave functions and energies
\begin{aligned} \psi_{i}(\vec{x}) &=\psi_{0}(\vec{x}) & & ; E_{i}=E_{0} \ \psi_{f}(\vec{x}) &=\frac{1}{\sqrt{L^{3}}} e^{i \vec{k}{f} \cdot \vec{x}} & & ; E{f}=\frac{\left(\hbar k_{f}\right)^{2}}{2 m} \end{aligned}
The transition rate multiplied by the number of final states, and divided by the incident flux, is then
\begin{aligned} \frac{1}{S_{\text {inc }}} R_{f i} d n_{f}=&\left(\frac{e^{2}}{2 \varepsilon_{0} \omega^{2} c}\right)\left(\frac{1}{m^{2} L^{3}}\right) \times\left(\frac{2 \pi}{\hbar}\right)\left|\vec{e}{\vec{k} s} \cdot \int d^{3} x e^{i\left(\vec{k}-\vec{k}{f}\right) \cdot \vec{x}} \vec{p} \psi_{0}(\vec{x})\right|^{2} \ & \times \delta\left(E_{f}-E_{0}-\hbar \omega\right)\left[\frac{L^{3}}{(2 \pi)^{3}} d^{3} k_{f}\right] \end{aligned}
Write $d^{3} k_{f}=k_{f}^{2} d \Omega_{f} d k_{f}$, and use
$$\frac{d E_{f}}{d k_{f}}=\frac{\hbar^{2} k_{f}}{m}$$
This yields
\begin{aligned} \omega_{f i} & \equiv \frac{1}{S_{\mathrm{inc}}} R_{f i} d n_{f} \ &=\frac{\alpha}{2 \pi c^{2}}\left(\frac{k_{f}}{2 E}\right)\left|\vec{e}{\vec{k} s} \cdot \int d^{3} x e^{i\left(\vec{k}-\vec{k}{f}\right) \cdot \vec{x}}\left(\frac{\vec{p}}{m}\right) \psi_{0}(\vec{x})\right|^{2} d \Omega_{f} \end{aligned}
where the energy $E$ is defined by
$$E \equiv \frac{(\hbar k)^{2}}{2 m}$$
This is a nice result. It is the general expression for photoionization by the classical radiation field to lowest order in $\alpha$. Note that the factors of $L^{3}$ have again cancelled. One can again check that this has the correct dimensions.

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## 物理代考

$6.5$ 光电离

${ }^{5}$ 见概率。 6.3.
${ }^{6}$ 见 [Walecka (2018)];请记住，现在 $\vec{E}=-\partial \vec{A} / \partial t$，电磁场的时间平均坡印廷矢量的大小为 $S_{\text {inc }}=\左\langle\varepsilon_{0} \vec{E}^{2}\right\rangle c$。
$50 \quad$ 量子力学导论

\begin{aligned} \frac{1}{S_{\text {inc }}} R_{f i} d n_{f}=&\left(\frac{e^{2}}{2 \varepsilon_{0} \omega^{2} c}\right)\left(\frac{1}{m^{2} L^{3}}\right) \times\left(\frac{2 \pi}{\hbar}\right)\left|\vec{e}{\vec{k} s} \cdot \int d^{3} x e^{i\left(\vec{k}-\vec{k}{f}\right) \cdot \vec{x}} \vec{p} \psi_{0}(\vec{x})\right|^{2} \ & \times \delta\left(E_{f}-E_{0}-\hbar \omega\right)\left[\frac{L^{3}}{(2 \pi)^{3}} d^{3} k_{f}\right] \end{aligned}

$$\frac{d E_{f}}{d k_{f}}=\frac{\hbar^{2} k_{f}}{m}$$

\begin{aligned} \omega_{f i} & \equiv \frac{1}{S_{\mathrm{inc}}} R_{f i} d n_{f} \ &=\frac{\alpha}{2 \pi c^{2}}\left(\frac{k_{f}}{2 E}\right)\left|\vec{e}{\vec{k} s} \cdot \int d^{3} x e^{i\left(\vec{k}-\vec{k}{f}\right) \cdot \vec{x}}\left(\frac{\vec{p}}{m}\right) \psi_{0}(\vec{x})\right|^{2} d \Omega_{f} \end{aligned}

$$E \equiv \frac{(\hbar k)^{2}}{2 m}$$

Matlab代写