19th Ave New York, NY 95822, USA

# 物理代考| S-Wave Scattering 量子力学代写

## 物理代写

4.2 S-Wave Scattering
The separated solutions in Eq. (4.4) satisfy the Schrödinger equation in spherical coordinates. Let us focus on the $l=0$ term, which is the dominant term at low energy where $k r \rightarrow 0$,
$$\left(\nabla^{2}+k^{2}\right) j_{0}(k r)=\left(\nabla^{2}+k^{2}\right) \frac{\sin (k r)}{k r}=0$$
Evidently the radial part of the laplacian in spherical coordinates is
${ }^{1}$ These relations actually hold to all orders; see also Prob. $4.2 .$
for then the above becomes $^{2}$
$$\left(-k^{2}+k^{2}\right) \frac{\sin (k r)}{k r}=0$$
27
Let us now include a potential $V(r)$, and work at very separated $l=0$ Schrödinger equation, or $s$-wave equation, $$\left[\frac{1}{r} \frac{\partial^{2}}{\partial r^{2}} r-v(r)+k^{2}\right] \psi(r)=0 \quad ; v(r) \equiv \frac{2}{\hbar^{2}}$$ Let us define $$\psi(r) \equiv \frac{u(r)}{r} \quad ; s \text {-wave }$$
The $s$-wave Schrödinger equation for $u(r)$ then becomes $$\left[\frac{d^{2}}{d r^{2}}-v(r)+k^{2}\right] u(r)=0 \quad ; s \text {-wave eqn }$$
4.3 Spherical Square Well
Let us solve the $s$-wave Schrödinger equation for an attractive square-well potential of the form $$v(r)=-v_{0}$$ cosine, which we can write
Scattering4.3 Spherical Square Well
Let us solve the $s$-wave Schrödinger equation for an attractive square-well potential of the form
$$v(r)=-v_{0} \quad ; rd$$
where $\delta_{0}$ is the $s$-wave phase shift. Inside the potential, if we assume there is no bound-state and keep just the solution that is non-singular at the origin, we have
\begin{aligned} u_{\text {in }}(r) &=B \sin (\kappa r) & & ; r<d \ \kappa^{2} & \equiv k^{2}+v_{0} & & \end{aligned}
${ }^{2}$ The laplacian in spherical coordinates is actually
$$\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}$$
The first term is the same as in Eq. (4.10).
28
Introduction to Quantum Mechanics
Upon equating the logarithmic derivative at the potential boundary, we obtain an equation for the phase shift $\delta_{0}(k)$
$$k \cot \left(k d+\delta_{0}\right)=\kappa \cot (\kappa d)$$

## 物理代考

4.2 S 波散射

$$\left(\nabla^{2}+k^{2}\right) j_{0}(kr)=\left(\nabla^{2}+k^{2}\right) \frac{\sin ( kr)}{kr}=0$$

${ }^{1}$ 这些关系实际上适用于所有订单；另见概率。 $4.2 .$

$$\left(-k^{2}+k^{2}\right) \frac{\sin (k r)}{k r}=0$$
27

$u(r)$ 的 $s$-wave Schrödinger 方程变为 $$\left[\frac{d^{2}}{dr^{2}}-v(r)+k^{2}\对] u(r)=0 \quad ; s \text {-wave eqn }$$
4.3 球形方形井

Scattering4.3 球形方形井

$$v(r)=-v_{0} \quad ; rd$$

$$\开始{对齐} u_{\text {in }}(r) &=B \sin (\kappa r) & & ; r<d \ \kappa^{2} & \equiv k^{2}+v_{0} & & \end{对齐}$$
${ }^{2}$ 球坐标中的拉普拉斯算子实际上是
$$\nabla^{2}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial}{\partial r}\右)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta} \right)+\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}$$

28

$$k \cot \left(k d+\delta_{0}\right)=\kappa \cot (\kappa d)$$

Matlab代写