Ze Chen

## Scattering Theory

#### Asymptotic Solutions

The solution of scattering by a potential $$V(r)$$ has the form $\psi_k \xrightarrow[r\rightarrow\infty]{} e^{ikz} + f(\theta,\varphi)\frac{e^{ikr}}{r},$ where $$f$$ is called the scattering amplitude. The cross section is given by $\dv{\sigma}{\Omega} = \abs{f(\theta,\varphi)}^2.$

#### Approximations

##### Time-Dependent Approximation

Fermi's Golden Rule states that the transition rate is given by $\Gamma_{i\rightarrow f} = \frac{2\pi}{\hbar}\abs{\bra{f}V\ket{i}}^2\rho(E_f).$

The S Matrix is the operator defined by $S = \lim_{\substack{t_f \rightarrow \infty\\ t_i\rightarrow -\infty}} U(t_f,t_i).$

With the Fermi's golden rule the transition rate is found to be \begin{align*} R_{i\rightarrow \dd{\Omega}} &= \frac{2\pi}{\hbar}\abs{\bra{\vb*{p}_f}V\ket{\vb*{p}_i}}^2 \mu p_i \dd{\Omega}. \end{align*}

The incoming probability current is given by $j_i = \frac{\hbar k}{\mu}\qty(\frac{1}{2\pi\hbar})^3,$ from which we find that the differential cross section is given by $\dv{\sigma}{\Omega}\dd{\Omega} = \frac{R_{i\rightarrow \dd{\Omega}}}{j_i},$ i.e. $$\label{eq:diff_cross_sec} \dv{\sigma}{\Omega} = \abs{\frac{\mu}{2\pi\hbar^2}\int e^{-i\vb*{q}\cdot \vb*{r}'}V(\vb*{r}')\dd{^3 \vb*{r}'}}^2,$$ where $$\hbar \vb*{q} = \vb*{p}_f - \vb*{p}_i$$.

I am I little bit confused by Shankar's normalization. One may arrive at exactly the same result as in \eqref{eq:diff_cross_sec} with the conventional box normalization.

From \eqref{eq:diff_cross_sec} we infer that (up to a phase factor) $f(\theta,\varphi) = -\frac{\mu}{2\pi\hbar^2} \int e^{-i\vb*{q}\cdot\vb*{r}'}V(\vb*{r}')\dd{^3\vb*{r}'}.$

In particular, for a central potential $$V(r)$$ we have $f(\theta) = -\frac{2\mu}{\hbar^2}\int \frac{\sin qr'}{q}V(r')r'\dd{r'}.$

For the Yukawa potential $V(r) = \frac{g e^{-\mu_0 r}}{r}{}$ we find $\dv{\sigma}{\Omega} = \frac{4\mu^2 g^2}{\hbar^4\qty[\mu_0^2 + 4k^2\sin^2\qty(\theta/2)]^2}.$ Let $$g = Ze^2$$ and $$\mu_0 = 0$$, we find the differential cross section for Coulomb scattering $\dv{\sigma}{\Omega} = \frac{(Ze^2)^2}{16E^2\sin^4\qty(\theta/2)}.$

##### Born Approximation

We rewrite the scattering problem as $(\grad^2 + k^2)\psi_k = \frac{2\mu}{\hbar^2}V(r)\psi_k.$ The solution may be obtained using the Green's function by $\psi_{k}(r) = e^{i\vb*{k}\cdot\vb*{r}} + \frac{2\mu}{\hbar^2} \int G^0(\vb*{r},\vb*{r}')V(\vb*{r}')\psi_k(\vb*{r}')\dd{^3 \vb*{r}'},$ where $$G^0$$ is the solution to $(\grad^2 + k^2) G^0(\vb*{r},\vb*{r}') = \delta^3(\vb*{r} - \vb*{r}').$ By demanding $$G(\vb*{r},\vb*{r}')$$ be out-going we obtain $G(\vb*{r},\vb*{r}') = -\frac{e^{ik\abs{\vb*{r} - \vb*{r}'}}}{4\pi \abs{\vb*{r} - \vb*{r}'}}.$

Minutia in Obtaining the Green's Function

The operator $$\grad^2 + k^2$$ is not invertible since $$e^{ikz}$$ is in the null space thereof. Adding a $$i\epsilon$$ we obtain a operator $$\grad^2 + k^2 + i\epsilon$$ invertible in the physical Hibert space since elements in its null space diverges at infinity. A positive $$\epsilon$$ happened to yield the retarded Green's function, while a negative $$\epsilon$$ yields the advanced Green's functions.

To the first order we recover $f(\theta,\varphi) = -\frac{2\mu}{4\pi\hbar^2} \int e^{-i(\vb*{k}_f - \vb*{k}_i)\cdot \vb*{r}'}V(\vb*{r}')\dd{^3\vb*{r}'}.$

##### The Partial Wave Expansion

For the spherical Bessel function we have the asymptotic formula \begin{align*} j_l(kr) &\xrightarrow[r\rightarrow\infty]{}\frac{\sin(kr - l\pi/2)}{kr}; \\ n_l(kr) &\xrightarrow[r\rightarrow\infty]{}-\frac{\cos(kr - l\pi/2)}{kr}. \end{align*}

We expand $$f$$ in terms of Legendre Polynomials $f(\theta,k) = \sum_{l=0}^\infty (2l+1)a_l(k) P_l(\cos\theta)$ as well as the incident wave $e^{ikz} = \sum_{l=0}^\infty i^l (2l+1) j_l(kr) P_l(\cos\theta),$ which consists of an incoming and an outgoing spherical wave. By allowing a phase shift in $R_l(l) = \frac{U_l(r)}{r} \xrightarrow[r\rightarrow\infty]{} \frac{A_l \sin \qty[kr - l\pi/2 + \delta_l(k)]}{r}$ and demanding the amplitude of incoming spherical wave be the same as in $$e^{ikz}$$ we find $\psi_{k}(\vb*{r}) = e^{ikz} + \qty[\sum_{l=0}^\infty (2l+1)\qty(\frac{e^{2i\delta_l} - 1}{2ik})P_l(\cos\theta)]\frac{e^{ikr}}{r}.$ Note that from the ansatz of $$R_l$$ the S matrix in the basis of $$(k,l,m)$$ where $$m=0$$ is found to be $S_l(k) = e^{2i\delta_l(k)}.$ From the expression of $$\psi_k$$ we deduce $a_l(k) = \frac{e^{2i\delta_l} - 1}{2ik} = \frac{e^{i\delta_l}\sin\delta_l}{k},$ whence we obtain the total cross section $\sigma = \frac{4\pi}{k^2} \sum_{k=0}^\infty (2l+1)\sin^2\delta_l = \frac{4\pi}{k^2}\Im f(0),$ where the last equality gives the optical theorem.

###### The Hard Sphere

For the potential $V(r) = \begin{cases} \infty, & \text{if }r<r_0,\\ 0, & \text{if }r>r_0 \end{cases}$ we have the free particle solution $R_l(r) = A_l j_l(kr) + B_l n_l(kr)$ for $$r>r_0$$ subjected to the boundary condition $$R_l(r_0) = 0$$. The phase shift is found by using the asymptotic formulae of $$j_l$$ and $$n_l$$ to be $\delta_l = \arctan \frac{j_l(kr_0)}{n_l(kr_0)}.$ In particular, for $$k\ll 1$$ we have $$\delta_l \approx (kr_0)^{2l+1}$$.

2021/2/7 18:02:37

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#### 1 Comment

scattering theory is just so cool!

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Anonymous 2021/8/28 9:28:08

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