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. \begin{equation} \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, \end{equation} 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!




  Expand

Anonymous 2021/8/28 9:28:08

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