# 物理代写|量子力学代写quantum mechanics代考|PHYS3034

## 物理代写|量子力学代写quantum mechanics代考|Quantum Electrodynamics

In order to do more physics, we need to get more realistic. An essential part of modern physics is the interaction of a charged particle with an electromagnetic field. We certainly cannot do all of quantum electrodynamics here, and we will be content to include the electromagnetic field through vector and scalar potentials $(\vec{A}, \Phi)$. This will allow us to describe
(1) A static Coulomb field, as in an atom,
$$\Phi(\vec{x}, t)=\Phi_{\text {Coulomb }}(r) \quad ; \vec{A}=0$$
(2) A static magnetic field ${ }^1$
$$\vec{B}(\vec{x})=\vec{\nabla} \times \vec{A}(\vec{x}) \quad ; \Phi=0$$
(3) A transverse radiation field ${ }^2$ with
\begin{aligned} \vec{B}(\vec{x}, t) &=\vec{\nabla} \times \vec{A}(\vec{x}, t) & ; \Phi=0 \ \vec{E}(\vec{x}, t) &=-\frac{\partial \vec{A}(\vec{x}, t)}{\partial t} & \ \vec{A}(\vec{x}, t) &=\operatorname{Re}\left[\vec{e}{\vec{k} s} e^{i(\vec{k} \cdot \vec{x}-\omega t)}\right] & & ; \omega=k c \end{aligned} where $\vec{e}{\vec{k} s}$ with $s=(1,2)$ are transverse unit vectors.
To proceed, we need to construct the hamiltonian for a charged particle in such an electromagnetic field.

## 物理代写|量子力学代写quantum mechanics代考|Ionization in Oscillating Electric Field

We start the discussion of electromagnetic interactions with a very simple example, where we explicitly have all the wave functions. Suppose a charged particle is moving in the ground-state of the one-dimensional box, and it is boosted into the continuum by an electric field that is oscillating along the $x$-axis according to
$$\mathcal{E}x=\mathcal{E}_0 \cos \omega_0 t=\mathcal{E}_0 \frac{1}{2}\left(e^{i \omega_0 t}+e^{-i \omega_0 t}\right)$$ The interaction hamiltonian is then $$H^{\prime}=-e \mathcal{E}_0 x \cos \left(\omega_0 t\right) \doteq-\left(\frac{e \mathcal{E}_0}{2}\right) x e^{-i \omega_0 t}$$ where it is only the final term that will increase the energy of the bound particle in Eqs. (5.17) and (5.18). The initial and final wave functions and energies are $^4$ \begin{aligned} &\psi_i(x)=\psi{n_0}(x) \quad ; E_i=\frac{\hbar^2 \pi^2}{2 m d^2} n_0^2 \ &\psi_f(x)=\frac{1}{\sqrt{L}} e^{i k_f x} \quad ; E_f=\frac{\left(\hbar k_f\right)^2}{2 m} \quad \text {; p.b.c. } \ & \end{aligned}
The transition rate times the number of final states is then
$$R_{f i} d n_f=\left(\frac{e \mathcal{E}_0}{2}\right)^2 \frac{2 \pi}{\hbar}|\langle f|x| i\rangle|^2 \delta\left(E_f-E_i-\hbar \omega_0\right)\left[\frac{L}{(2 \pi)} d k_f\right]$$
We can now carry out some familiar manipulations and use
$$\frac{d E_f}{d k_f}=\frac{\hbar^2 k_f}{m}$$
The maximum energy density of the electric field is
$$U_0=\frac{\varepsilon_0}{2} \mathcal{E}_0^2 \quad ; \text { field energy density }$$

## 物理代写|量子力学代写quantum mechanics代考|Quantum Electrodynamics

（1）静态库仑场，如在原子中，
$$\Phi(\vec{x}, t)=\Phi_{\text {Coulomb }}(r) \quad ; \vec{A}=0$$
(2)静硑场 ${ }^1$
$$\vec{B}(\vec{x})=\vec{\nabla} \times \vec{A}(\vec{x}) \quad ; \Phi=0$$
(3) 横向辐射野 ${ }^2$ 和
$$\vec{B}(\vec{x}, t)=\vec{\nabla} \times \vec{A}(\vec{x}, t) \quad ; \Phi=0 \vec{E}(\vec{x}, t)=-\frac{\partial \vec{A}(\vec{x}, t)}{\partial t} \quad \vec{A}(\vec{x}, t)=\operatorname{Re}\left[\vec{e} \vec{k} s e^{i(\vec{k} \cdot \vec{x}-\omega t)}\right] \quad ; \omega=k c$$

## 物理代写|量子力学代写quantum mechanics代考|Ionization in Oscillating Electric Field

$$\mathcal{E} x=\mathcal{E}0 \cos \omega_0 t=\mathcal{E}_0 \frac{1}{2}\left(e^{i \omega_0 t}+e^{-i \omega_0 t}\right)$$ 那么相互作用的哈密顿量是 $$H^{\prime}=-e \mathcal{E}_0 x \cos \left(\omega_0 t\right) \doteq-\left(\frac{e \mathcal{E}_0}{2}\right) x e^{-i \omega_0 t}$$ 其中只有最后一项会增加方程式中束缚粒子的能量。(5.17) 和 (5.18)。初始和最終波函数和能量是 4 $$\psi_i(x)=\psi n_0(x) \quad ; E_i=\frac{\hbar^2 \pi^2}{2 m d^2} n_0^2 \quad \psi_f(x)=\frac{1}{\sqrt{L}} e^{i k_f x} \quad ; E_f=\frac{\left(\hbar k_f\right)^2}{2 m} \quad ; \text { p.b.c. }$$ 转换率乘以最終状态的数量是 $$R{f i} d n_f=\left(\frac{e \mathcal{E}_0}{2}\right)^2 \frac{2 \pi}{\hbar}|\langle f|x| i\rangle|^2 \delta\left(E_f-E_i-\hbar \omega_0\right)\left[\frac{L}{(2 \pi)} d k_f\right]$$

$$\frac{d E_f}{d k_f}=\frac{\hbar^2 k_f}{m}$$

$U_0=\frac{\varepsilon_0}{2} \mathcal{E}_0^2 \quad ;$ field energy density

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