Question

Find theS-wave(l=0) partial wave phase shift${\mathbf{\delta }}_{\mathbf{0}}\mathbf{\left(}\mathbf{k}\mathbf{\right)}$ for scattering from a delta-function shell (Problem 11.4). Assume that the radial wave functionu(r)goes to 0 as$\mathbf{r}\mathbf{\to }\mathbf{\infty }$.

Short answer

Forphase shift,

${\delta }_0=co{t}^{-1}\left(cot\left(ka\right)+\frac{ka}{\beta {\sin }^2\left(ka\right)}\right)$

Step-by-step solution

By using spherical delta function shell. 

Given the spherical delta function shell,

$V\left(r\right)=\alpha \delta \left(r-a\right)$

Where $\alpha$and are constants. Assume that the incident particle is a low energy particle, so that $ka\ll 1$, but $k=2\pi /\lambda$thus $\lambda \gg a$, 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 l = 0 only the 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:

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

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 (r) =u (r) /r, 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, to make it finite we set, so the solution is:

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

To apply boundary conditions and find phase shifts.

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

$$\begin{gathered} {\dot{\Psi }}_{ext}\left(a\right)={\dot{\Psi }}_{it}\left(a\right) \\ \frac{A}{ka}\left[\sin \left(ka\right)+k{a}_0{e}^{ika}\right]=B\frac{\sin \left(ka\right)}{a} \end{gathered}$$ ………. (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 2.5.2 (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 $\mathit{-}\alpha \mathit{}by\mathit{}\mathit{+}\alpha$by , to get:

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

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

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

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:

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

But.

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

Where,

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

Thus,

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

Using (4) we get:

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

Substitute with u (r) from (2) we get:

$$\begin{gathered} B\frac{2m\alpha \sin \left(ka\right)}{{\sout{h}}^2}\frac{\sin }{a}=\left(A-Bk\right)\cos \left(ka\right)+Aik{a}_0{e}^{ika} \\ B\left[\frac{2m\alpha }{{\sout{h}}^2}\frac{\sin \left(ka\right)}{a}+k\cos \left(ka\right)\right]=A\left[\cos \left(ka\right)+ik{a}_0{e}^{ika}\right] \end{gathered}$$

Substitute with B from equation (3) we get:

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

Solve this equation for $a_0$by taking into account localid="1658309795102" $\beta =2m\alpha a/{\sout{h}}^2$, we get:

$a_0=-\frac{\mathrm{\beta }{e}^{-ika}{\sin }^2\left(ka\right)}{\left(\mathrm{\beta sin}\mathit{\left(}ka\mathit{\right)}\mathit{+}ka\mathit{}cos\mathit{\left(}ka\mathit{\right)}\mathit{-}iak\mathit{}\sin \mathit{\left(}ka\mathit{\right)}\right)k}$ ..........(5)

The relation between ${\delta }_1$and $a_l$is given by equation 11.46 as:

$a_l=\frac{1}{k}{e}^{i{\delta }_l}\sin \left({\delta }_l\right)$ ………… (6)

Dividing the first equation by the second one, we get:

$$\begin{gathered} \cot \left({\delta }_0\right)=\frac{Re\left(k{a}_0\right)}{im\left(k{a}_0\right)} \\ {\delta }_0={\cot }^{-1}\left(\frac{\mathrm{le}\left(k{a}_{\mathit{0}}\right)}{\mathrm{lm}\left(k{a}_{\mathit{1}}\right)}\right)\dots \dots \left(7\right) \end{gathered}$$

To find real and imaginary parts.

We need to find the real and the imaginary parts of equation (5), by multiplying
(5) top and bottom by the complex conjugate of the denominator:

$k{a}_0=-\frac{\beta e^{ika}{\sin }^2\left(ka\right)\beta \sin \left(ka\right)+ka\cos \left(ka\right)-iak\sin \left(ka\right))}{{\left[\mathrm{\beta sin}\left(ka\mathit{\right)}+ka\mathit{}\cos \mathit{\left(}ka\mathit{\right)}\right]}^2+a^2{k}^2{\sin }^2\left(ka\right)}$

The real and the imaginary parts are:

$Re\left(k{a}_0\right)=-\frac{{\mathrm{\beta sin}}^2\left(ka\right)ka+\mathrm{\beta sin}\left(ka\right)\cos \left(ka\right))}{{\left(ka\mathit{\right)}}^{\mathit{2}}+{\mathrm{\beta }}^2+2ka\mathrm{\beta }\sin \mathit{\left(}ka\mathit{\right)}\mathit{}\cos \left(ka\right)\mathit{-}{\mathrm{\beta }}^{\mathit{2}}{\cos }^2\left(ka\right)}$

$\mathrm{Im}\left(k{a}_0\right)=\frac{{\mathrm{\beta }}^2{\sin }^4\left(ka\right)}{{\left(ka\right)}^2+{\beta }^2+2ka\beta \sin \left(ka\right)\cos \left(ka\right)-{\mathrm{\beta }}^2{\cos }^2\mathit{\left(}ka\mathit{\right)}}$

Therefore, the phase shift is,

$$\begin{gathered} {\delta }_0={\cot }^{-1}\left(\frac{-{\mathrm{\beta sin}}^2\left(ka\right)ka+\mathrm{\beta sin}\left(ka\right)\cos \left(ka\right))}{{\mathrm{\beta }}^2{\sin }^4\left(ka\right)}\right) \\ {\mathrm{\delta }}_0={\cot }^{-1}\left(\cot (ka\right)+\frac{ka}{{\mathrm{\beta sin}}^2\left(ka\right)} \end{gathered}$$