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# 物理代考| Model Problem 量子力学代写

## 物理代写

5.1 Model Problem
Consider two particles moving in one dimension along the $x$-axis. Particle one is free to move in a large circle as in Fig. 2.1, so it satisfies periodic boundary conditions. It has the eigenfunctions and eigenvalues of Eqs. (2.31) and (2.32)
\begin{aligned} \psi_{n}(x) &=\frac{1}{\sqrt{L}} e^{2 \pi i n x / L} & & ; n=0, \pm 1, \pm 2, \cdots \ E_{n} &=\frac{(2 \pi \hbar)^{2}}{2 m L^{2}} n^{2} & & ; \text { particle one } \end{aligned}
The eigenfunctions satisfy the orthonormality condition in Eq. (2.33).
The second particle is confined to a box of a much shorter length as in Fig. $3.1$ $(3.13)^{1}$
$\psi_{n}(x)=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right) \quad ; n=1,2,3, \cdots$
$$E_{n}=\frac{(\pi \hbar)^{2}}{2 m L^{2}} n^{2}$$
The starting hamiltonian and general solution for this two-particle them are then orthonormal. It is assumed here that the box is completely transparent the first particle, which passes right through it. $^{2}$ The $H_{0}=\frac{p_{1}^{2}}{2 m_{1}}+\frac{p_{2}^{2}}{2 m_{2}}+V_{\text {box }}\left(x_{2}\right)$ $\Psi_{0}\left(x_{1}, x_{2}, t\right)=\sum_{n_{1}, n_{2}} c_{n_{1}, n_{2}}^{0}(t) \psi_{n_{1}}\left(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left(x_{2}\right) e^{-i E_{n_{2}} t / \hbar}$ This wave function satisfies the Schrödinger equation $$i \hbar \frac{\partial \Psi_{0}\left(x_{1}, x_{2}, t\right)}{\partial t}=H_{0} \Psi_{0}\left(x_{1}, x_{2}, t\right)$$
We have left a time-dependence in the coefficients $c_{n_{1}, n_{2}}^{0}(t)$ for generality, but let us see what
\begin{aligned} &\sum_{n_{1}, n_{2}}\left[i \hbar \frac{d c_{n_{1}, n_{2}}^{0}(t)}{d t}+\left(E_{n_{1}}+E_{n_{2}}\right) c_{n_{1}, n_{2}}^{0}(t)\right] \psi_{n_{1}}\left(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left(x_{2}\right) e \ &=\sum_{n_{1}, n_{2}}\left(E_{n_{1}}+E_{n_{2}}\right) c_{n_{1}, n_{2}}^{0}(t) \psi_{n_{1}}\left(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left(x_{2}\right) e^{-i E_{n_{2}} t / \hbar} \end{aligned}
Upon cancellation of the common terms, and the use of the orthonormality of the eigenfunctions to extract a given coefficient, one obtains
$\frac{d c_{n_{1}, n_{2}}^{0}(t)}{d t}=0$
coef- (5.8) tiile
$c_{n_{1}, n_{2}}^{0}(t)=c_{n_{1}, n_{2}}^{0}(0) \quad ;$ time-independent
If we were to start in a given state $\psi_{n_{1}^{0}}\left(x_{1}\right) \psi_{n_{2}^{0}}\left(x_{2}\right)$, then all but one coefficient would vanish
$c_{n_{1}, n_{2}}^{0}=\delta_{n_{1}, n_{1}^{0}} \delta_{n_{2}, n_{2}^{0}} \quad ;$ given initial state
${ }^{2}$ The second particle might be bound to the target by a strong force that the projectile does not feel, for example.

Nransition Rate the two particles interact through some short-range potential, so that the hamiltonian becomes
\begin{aligned} H &=H_{0}+H^{\prime} \ H^{\prime} &=V\left(x_{1}, x_{2}\right) \end{aligned}

## 物理代考

5.1 模型问题

$$\开始{对齐} \psi_{n}(x) &=\frac{1}{\sqrt{L}} e^{2 \pi i n x / L} & & ; n=0, \pm 1, \pm 2, \cdots \ E_{n} &=\frac{(2 \pi \hbar)^{2}}{2 m L^{2}} n^{2} & & ; \text { 粒子一 } \end{对齐}$$

$\psi_{n}(x)=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right) \quad ; n=1,2,3, \cdots$
$$E_{n}=\frac{(\pi \hbar)^{2}}{2 m L^{2}} n^{2}$$

$$\开始{对齐} &\sum_{n_{1}, n_{2}}\left[i \hbar \frac{d c_{n_{1}, n_{2}}^{0}(t)}{dt}+\left (E_{n_{1}}+E_{n_{2}}\right) c_{n_{1}, n_{2}}^{0}(t)\right] \psi_{n_{1}}\左(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left(x_{2}\right) e \ &=\sum_{n_{1}, n_{2}}\left(E_{n_{1}}+E_{n_{2}}\right) c_{n_{1}, n_{2}}^{ 0}(t) \psi_{n_{1}}\left(x_{1}\right) e^{-i E_{n_{1}} t / \hbar} \psi_{n_{2}}\left (x_{2}\right) e^{-i E_{n_{2}} t / \hbar} \end{对齐}$$

$\frac{d c_{n_{1}, n_{2}}^{0}(t)}{d t}=0$
coef- (5.8) 平铺
$c_{n_{1}, n_{2}}^{0}(t)=c_{n_{1}, n_{2}}^{0}(0) \quad ;$ 与时间无关

$c_{n_{1}, n_{2}}^{0}=\delta_{n_{1}, n_{1}^{0}} \delta_{n_{2}, n_{2}^{0 }} \quad ;$ 给定初始状态
${ }^{2}$ 例如，第二个粒子可能会被射弹感觉不到的强大力绑定到目标。

Nransition Rate 两个粒子通过一些短程势能相互作用，因此哈密顿方程变为
$$\开始{对齐} H &=H_{0}+H^{\素数} \ H^{\prime} &=V\left(x_{1}, x_{2}\right) \end{对齐}$$

Matlab代写