Question

Consider the case of low-energy scattering from a spherical delta function shell is$\mathrm{V}\left(\mathrm{r}\right)=\mathrm{a\delta }\left(\mathrm{r}-\mathrm{a}\right)$.Where$\alpha$and$a$are constants. Calculate the scattering amplitude,$F\left(\theta \right)$, the differential cross-section,$D\left(\theta \right)$, and the total cross-section,$\mathbf{\sigma }$.

Short answer

For Scattering Amplitude,

$\mathrm{f}\left(\mathrm{\theta }\right)=-\frac{\mathrm{\beta a}}{1+\mathrm{\beta }}$

For Differential Cross Section,

$D\left(\theta \right)=\frac{{\beta }^2{a}^2}{{\left(1+\beta \right)}^2}$

For total Cross Section,

$\sigma =4\pi =\frac{{\beta }^2{a}^2}{{\left(1+\beta \right)}^2}$

Step-by-step solution

By using the Spherical delta function.

Given the spherical delta function shell,

$v\left(R\right)=A\delta \left(r-a\right)$

Where $\alpha$and$a$are constants. First we want to find the scattering amplitude$f\left(\theta \right)$, assume that the incident particle is a low energy particle, so that$ka\ll 1$, but $k=2\pi /\lambda$thus$\lambda \gg \alpha$, this means the wavelength of the particle is much greater than the size of the target.

The cross section can be expanded in powers of${\left(ka\right)}^l$, since$ka\ll 1$, then only the$l=0$term can be considered in equation 11.29, so we have:

${\psi }_{ext}=A\left[j_0\left(kr\right)+ik{a}_0{h}_0^{\left(1\right)}\left(kr\right)P_0\left(\cos \theta \right)\right]$

Where $a_0$and $h_0$are the Bessel and Henkel functions, and we take the small argument of them, that are:

$$\begin{gathered} j_0\left(kr\right)\approx \frac{\sin kr}{kr} \\ {h}_0^{\left(1\right)}\left(kr\right)\approx -i\frac{e^{ikr}}{kr} \end{gathered}$$

And also$P_0=1$, then the exterior solution is:

${\psi }_{ext}=\frac{a}{kr}\left[\sin kr+k{a}_0{e}^{ikr}\right]$……(1)

The radial equation is given by:

$-\frac{{\sout{h}}^2}{2m}\frac{d^2}{d{r}^2}+\left(V+\frac{{\sout{h}}^2}{2m}\frac{l\left(l+1\right)}{r^2}\right)u=Eu$

For the internal function the potential is zero (note that$l=1$), then the equation will be:

$-\frac{{\sout{h}}^2}{2m}\frac{d^{2}u}{d{r}^2}=Eu$

The solution of this equation is:

$u\left(r\right)=B\sin \left(kr\right)+D\cos \left(kr\right)$

But,

$R\left(r\right)=u\left(r/r\$\right)$

So,

$R\left(r\right)B\frac{\sin \left(kr\right)}{r}+D\frac{\cos \left(kr\right)}{r}$

The solution must be finite at$r=0$, to make it finite we set$D=0$, so the solution is:

${\psi }_{in}\left(r\right)=B\frac{\sin \left(kr\right)}{r}$…..(2)

Apply boundary condition to find scattering amplitude, Differential and total cross section.

Now we need to apply the boundary condition, which states that the two wave functions must equal at the boundary$r=a$, that is:

${\psi }_{ext}\left(a\right)={\psi }_{it}\left(a\right)$

$\frac{A}{ka}\left[\sin \left(ka\right)+k{a}_0{e}^{ika}\right]=B\frac{\sin \left(ka\right)}{a}$……(3)

Now we need to apply another boundary condition, but the delta function is infinite, so we cannot assume that the derivative of the wave function is continuous. However, we can follow the same method in section 2.52(with the delta function well), where the radial equation is the same as the one-dimensional Schrodinger equation in the delta function well, but in this case, we have a barrier rather than a well, so we just replace$-a$by$+a$, to get:

$-\frac{\sout{h}}{2m}\frac{d^{2}u}{d{x}^2}+a\delta \left(x\right)u=Eu$

Integrate this equation term by term across the boundary, so we get:

$$\begin{gathered} -\frac{{\sout{h}}^2}{2m}{\int }_{-\dot{o}}^{\dot{o}}\frac{d^{2}u}{d{r}^2}dr+a{\int }_{a-\dot{o}}^{a+\dot{o}}\delta \left(r-a\right)udr=E{\int }_{a-\dot{o}}^{a+\dot{o}}udr \\ -{\overline{)\frac{{\sout{h}}^2}{2m}\frac{d^{2}u}{d{r}^2}}}_{a-\dot{o}}^{a+\dot{o}}+au\left(a\right)=E{\int }_{a-\dot{o}}^{a+\dot{o}}udr \end{gathered}$$

Now if we take the limit$\dot{o}\to 0$, then the integral of the RHS tends to be zero, since the function in the integral is continuous, but this is not the case with the first integral on the LHS, so:

localid="1656061637439" $\underset{\dot{0}\to 0}{\lim }{\overline{)\frac{\mathrm{du}}{\mathrm{dr}}}}_{\mathrm{a}-\dot{\mathrm{o}}}^{\mathrm{a}+\dot{\mathrm{o}}}=\frac{2\mathrm{ma}}{\sout{\mathrm{h}}}\mathrm{u}\left(\mathrm{a}\right)$…..(4)

But,

$\underset{\dot{0}\to 0}{\lim }{\overline{)\frac{\mathrm{du}}{\mathrm{dr}}}}_{\mathrm{a}-\dot{\mathrm{o}}}^{\mathrm{a}+\dot{\mathrm{o}}}={\overline{)\frac{d{u}_{ext}}{dr}}}_{r=a}{\overline{)\frac{du}{dr}}}_{r=a}$

Where,

${\overline{)\frac{d{u}_{ext}}{dr}}}_{r=a}=A\left(\cos \left(ka\right)+ik{a}_0{e}^{ika}\right){\overline{)\frac{d{u}_{ext}}{dr}}}_{r=a}=Bk\cos \left(ka\right)$

Thus,

$\underset{\dot{0}\to 0}{\lim }{\overline{)\frac{\mathrm{du}}{\mathrm{dr}}}}_{\mathrm{a}-\dot{\mathrm{o}}}^{\mathrm{a}+\dot{\mathrm{o}}}=\left(A-Bk\right)\cos \left(ka\right)+Aik{a}_0{e}^{ika}$

Using equation (4) we get:

$\frac{2ma}{{\sout{h}}^2}u\left(a\right)=\left(A-Bk\right)\cos \left(ka\right)+Aik{a}_0{e}^{ika}$

Substitute with$u\left(r\right)$from (2) we get:

$B\frac{2ma\sin \left(ka\right)}{{\sout{h}}^2}=\left(A-Bk\right)\cos \left(ka\right)+Aik{a}_0{e}^{ika}$

$B\left[\frac{2ma\sin \left(ka\right)}{{\sout{h}}^2}+k\cos \left(ka\right)\right]=A\left[\cos \left(ka\right)+ik{a}_0{e}^{ika}\right]$

Substitute with$B$from equation (3) we get:

$\frac{1}{k\sin \left(ka\right)}\left[\sin \left(ka\right)+k{a}_0{e}^{ika}\right]\left[\frac{2ma}{{\sout{h}}^2}\frac{\sin \left(ka\right)}{a}+k\cos \left(ka\right)\right]=\left[\cos \left(ka\right)+ik{a}_0{e}^{ika}\right]$

Solve this equation for$a_0$by taking into account$\beta \equiv 2m\alpha a/{\sout{h}}^2$, we get:

$a_0=-\frac{\beta e^{-ika}{\sin }^2\left(ka\right)}{\left(\beta \sin \left(ka\right)+ka\cos \left(ka\right)-iak\sin \left(ka\right)\right)k}$

Since$ka\ll 1$, then we can use the approximations$\sin \left(ka\right)\approx ka,\cos \left(ka\right)\approx 1,\mathrm{and}{\mathrm{e}}^{-\mathrm{ika}}\approx 1$, so we get:

$a_0\approx -\frac{\beta {\left(ka\right)}^2}{\left(\beta ka+ka-i{\left(ak\right)}^2\right)k}$

Since$ka\ll 1$, so we can neglect${\left(ak\right)}^2$, thus:

$a_0\approx \frac{\mathbb{\beta }\mathrm{a}}{1+\mathrm{\beta }}$

The scattering amplitude is given by:

$f\left(\theta \right)\equiv \sum _{l=0}^{\infty }\left(2l+1\right)a_l{P}_l\left(\cos \theta \right)$

For$l=0,f\left(\theta \right)=a_0$, so:

$f\left(\theta \right)=-\frac{\beta a}{1+\beta }$

The differential cross section is given by:

$$\begin{gathered} D\left(\theta \right)={\left|f\left(\theta \right)\right|}^2 \\ D\left(\theta \right)==\frac{{\beta }^2{a}^2}{{\left(1+\beta \right)}^2} \end{gathered}$$

The total cross section is:

$$\begin{gathered} \sigma =4\pi {\left|a_0\right|}^2 \\ \sigma =4\pi \frac{{\beta }^2{a}^2}{{\left(1+\beta \right)}^2} \end{gathered}$$