Question

Find the scattering amplitude for low-energy soft-sphere scattering in the second Born approximation.

Short answer

The scattering amplitude for low energy is$f\left(\theta ,\varphi \right)\approx -\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}\left(1-\frac{4m{V}_0{a}^2}{5{\sout{h}}^2}\right)$

Step-by-step solution

To find the scattering amplitude.

The first order Born approximation in scattering theory is given by:

$\mathrm{\psi }\left(\mathrm{r}\right)={\mathrm{\psi }}_0\left(\mathrm{r}\right)+\int \mathrm{g}\left(\mathrm{r}-{\mathrm{r}}_0\right)\mathrm{V}\left({\mathrm{r}}_0\right)\mathrm{\psi }\left({\mathrm{r}}_0\right){\mathrm{d}}^3{\mathrm{r}}_0$…….(1)

Here,

$g\left(r-r_0\right)\equiv \frac{m}{2\pi {\sout{h}}^2}\frac{e^{ik\left|r-r_0\right|}}{\left|r-r_0\right|}$

And, the second order approximations is given by:

$\mathrm{\psi }\left(\mathrm{r}\right)={\mathrm{\psi }}_0\left(\mathrm{r}\right)+\int \mathrm{g}\left(\mathrm{r}-{\mathrm{r}}_1\right)\mathrm{V}\left({\mathrm{r}}_1\right)\mathrm{\psi }\left({\mathrm{r}}_1\right){\mathrm{d}}^3{\mathrm{r}}_1$……..(2)

$\psi \left(r\right)\approx \left[\begin{array}{c}{\mathrm{\psi }}_0\left(\mathrm{r}\right)+\int \mathrm{g}\left(\mathrm{r}-{\mathrm{r}}_1\right)\mathrm{V}\left({\mathrm{r}}_1\right){\mathrm{\psi }}_0\left({\mathrm{r}}_1\right){\mathrm{d}}^3{\mathrm{r}}_1\\ +\int \mathrm{g}\left(\mathrm{r}-{\mathrm{r}}_1\right)\mathrm{V}\left({\mathrm{r}}_1\right)\int \mathrm{g}\left({\mathrm{r}}_1-{\mathrm{r}}_0\right)\mathrm{V}\left({\mathrm{r}}_0\right){\mathrm{\psi }}_0\left({\mathrm{r}}_0\right){\mathrm{d}}^3{\mathrm{r}}_0{\mathrm{d}}^3{\mathrm{r}}_1\end{array}\right]$……..(3)

Note that we generate this equation by substituting from equation (1) in itself, but we relabeling $r_0$as$r_1$. Compare (1) with (2) we can see that the first two terms in (2) are the same as in (1), so only the third term indicates the second approximations. That is:

$l_2={\left(\frac{m}{2\pi {\sout{h}}^2}\right)}^2\iint \frac{e^{ik\left|r-r_1\right|}}{\left|r-r_1\right|}\frac{e^{ik\left|r_1-r_0\right|}}{\left|r_1-r_0\right|}V\left(r_1\right)V\left(r_0\right){\psi }_0\left(r_0\right)d^3{r}_0{d}^3{r}_1$…..(4)

Taketo be a far point from the scattering region, we can approximate the Green's function as:

$g\left(r-r_0\right)\approx -\frac{m}{2\pi {\sout{h}}^2}\frac{e^{ikr}}{r}{e}^{ik-r_0}$

Thus:

$l_2\approx {\left(\frac{m}{2\pi {\sout{h}}^2}\right)}^2\frac{e^{ikr}}{r}\iint e^{-ik-r_1}\frac{e^{ik\left|r_1-r_0\right|}}{\left|r_1-r_0\right|}V\left(r_1\right)V\left(r_0\right){\psi }_0\left(r_0\right)d^3{r}_0{d}^3{r}_1$…….(5)

Now consider a soft-sphere scattering, where the soft sphere is defined by the potential

$V\left(r\right)=\left\{\begin{array}{l}Vr\le a\\ 0r\ge a\end{array}\right.$……(6)

Where$V_0>0$is a constant. Substitute into(5), so we get:

$l_2\approx {\left(\frac{m}{2\pi {\sout{h}}^2}\right)}^2\frac{e^{ikr}}{r}{V}_0^2\iint e^{-ik-r_1}\frac{e^{ik\left|r_1-r_0\right|}}{\left|r_1-r_0\right|}{\psi }_0\left(r_0\right)d^3{r}_0{d}^3{r}_1$

To find the integral.

To find the integral we take the incident plane wave to be:

${\psi }_0\left(r_0\right)=A{e}^{ik-r0}$

$k\equiv k\hat{z}$

Take the polar axis to be along$r_1$, then we can write:

$\left|r_1-r_0\right|=\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(\theta \right)}$

Also consider only the case of low energy scattering, which is defined by the condition$ka\ll 1$, so all the exponential are approximately 1 since all includes$k$, so we get:

$l_2\approx {\left(\frac{m}{2\pi {\sout{h}}^2}\right)}^2\frac{A{e}^{ikr}}{r}{V}_0^2\iint \frac{d^3{r}_0{d}^3{r}_1}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(\theta \right)}}$

First, we need to do the integral for$r_0$as;

$$\begin{gathered} \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}={\int }_0^a{\int }_0^{\pi }{\int }_0^{2\pi }\frac{r_1^2\sin \left(0\right)d\varphi d0d{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \theta }} \\ \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}=2\pi {\int }_0^a{\int }_0^{\pi }\frac{r_{1)}^2\sin \theta dd\theta d{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \theta }} \\ \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}=\frac{2\pi }{r_1}{\int }_0^a{r}_0\left(r_0+r_1-\left|r_0-r_1\right|\right)d{r}_0 \end{gathered}$$

On further calculating;

$$\begin{gathered} \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}=\frac{2\pi }{r_1}{\int }_0^{r_1}{r}_0\left(r_0+r_1-\left(r_0-r_1\right)\right)d{r}_0+{\int }_{r_1}^a{r}_0\left(r_0+r_1-\left(r_0-r_1\right)\right)d{r}_0 \\ \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}=\frac{4}{3}\pi r_1^2+2\pi a^2-2\pi r_1^2 \\ \int \frac{d^3{r}_0}{\sqrt{r_1^2+r_0^2-2r_0{r}_1\cos \left(0\right)}}=2\pi a^2-\frac{2}{3}\pi r_1^2 \end{gathered}$$

Now the integral is easy to solve for, so we get:

$l_2\approx {\left(\frac{m}{2\pi {\sout{h}}^2}\right)}^2\frac{A{e}^{ikr}}{r}{V}_0^2{\int }_0^a{\int }_0^{\pi }{\int }_0^{2\pi }\left(2\pi a^2-\frac{2}{3}\pi r_1^2\right)r_1^2\sin \left(\theta \right)d\varphi d\theta d{r}_1$

The integral over $\theta$and $\varphi$gives$4\pi$, and the integral over $r_1$is easy to solve so we get:

$l_2\approx \frac{A{e}^{ikr}}{r}\frac{8m^2{V}_0^2{a}^5}{15{\sout{h}}^4}$

The second order correction to the scattering is a coefficient for,

$\frac{A{e}^{ikr}}{r}$

Then the correction to the scattering amplitude is:

$f(\theta ,\varphi )=-\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}$

The amplitude for the second order correction is:

role="math" localid="1656042606830" $$\begin{gathered} f(\theta ,\varphi )\approx -\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}+\frac{8m{a}^3{V}_0}{15{\sout{h}}^4} \\ f(\theta ,\varphi )=-\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}+\left(1-\frac{4m{V}_0{a}^2}{15{\sout{h}}^2}\right) \\ f(\theta ,\varphi )\approx -\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}+\left(1-\frac{4m{V}_0{a}^2}{15{\sout{h}}^2}\right) \end{gathered}$$

Thus, the amplitude for the second order correction is$f(\theta ,\varphi )\approx -\frac{2m{a}^3{V}_0}{3{\sout{h}}^2}+\left(1-\frac{4m{V}_0{a}^2}{15{\sout{h}}^2}\right)$