Find the differential and total cross sections in the first Born approximation for the elastic scattering of a particle of mass m, which is initially traveling along the z-axis, from a nonspherical, double-delta potential V (r) = V0δ(r - ak) + V0 δ(r + ak), where k is the unit vector along the

Question

Find the differential and total cross sections in the first Born approximation for the elastic scattering of a particle of mass m, which is initially traveling along the z-axis, from a nonspherical, double-delta potential [latex]V(\vec{r})=V_{0} \delta(\vec{r}-a \vec{k})+V_{0} \delta(\vec{r}+a \vec{k})[/latex], where [latex]\vec{k}[/latex] is the unit vector along the z-axis.

Accepted Answer

Since [latex]V(\vec{r})[/latex] is not spherically symmetric, the differential cross section can be obtained from (11.66):

[latex]\frac{d \sigma}{d \Omega}=|f( heta, \varphi)|^{2}=\frac{\mu^{2}}{4 \pi^{2} \hbar^{4}}\left|\int e^{i \vec{q} \cdot \vec{r}^{\prime}} V\left(\vec{r}^{\prime} ight) d^{3} r^{\prime} ight|^{2}[/latex], (11.66)

[latex]\frac{d \sigma}{d \Omega}=\frac{m^{2}}{4 \pi^{2} \hbar^{4}}\left|\int V_{0}[\delta(\vec{r}-a \vec{k})+\delta(\vec{r}+a \vec{k})] e^{i \vec{q} \cdot \vec{r}} d^{3} r ight|^{2}=\frac{m^{2} V_{0}}{4 \pi^{2} \hbar^{4}}|I|^{2}[/latex]. (11.173)

Since [latex]\delta(\vec{r} \pm a \vec{k})=\delta(x) \delta(y) \delta(z \pm a)[/latex] we can write the integral I as

[latex]I=\int \delta(x) e^{i x q_{x}} d x \int \delta(y) e^{i y q_{y}} d y \int[\delta(z-a)+\delta(z+a)] e^{i z q_{z}} d z[/latex]

[latex]=e^{i a q_{z}}+e^{-i a q_{z}}=2 \cos \left(a q_{z} ight)[/latex]. (11.174)

The calculation of qz is somewhat different from that shown in (11.67).

[latex]q=\left|\overrightarrow{k_{0}}-\vec{k} ight|=\sqrt{k_{0}^{2}+k^{2}-2 k k_{0} \cos heta}=k \sqrt{2\left(1-\cos ^{2} heta ight)}=2 k \sin \left(\frac{ heta}{2} ight)[/latex]. (11.67)

Since the incident particle is initially traveling along the z-axis, and since it scatters elastically from the potential [latex]V(\vec{r})[/latex], the magnitudes of its momenta before and after collision are equal. So, as shown in Figure 11.8, we have [latex]q_{z}=q \sin ( heta / 2)=2 k \sin ^{2}( heta / 2)[/latex], since [latex]q=\left|\vec{k}_{0}-\vec{k} ight|=2 k \sin ( heta / 2)[/latex]. Thus, inserting [latex]I=2 \cos \left(a q_{z} ight)=2 \cos \left[2 a k \sin ^{2}( heta / 2) ight][/latex] into (11.173), we obtain

[latex]\frac{d \sigma}{d \Omega}=\frac{m^{2} V_{0}}{\pi^{2} \hbar^{4}} \cos ^{2}\left(2 a k \sin ^{2} \frac{ heta}{2} ight)[/latex]. (11.175)

The total cross section can be obtained at once from (11.175):

[latex]\sigma=\int \frac{d \sigma}{d \Omega} \sin heta d heta d \varphi=2 \pi \int_{0}^{\pi} \frac{d \sigma}{d \Omega} \sin heta d heta[/latex]

[latex]=2 \pi \frac{m^{2} V_{0}}{\pi^{2} \hbar^{4}} \int_{0}^{\pi} \sin heta \cos ^{2}\left(2 a k \sin ^{2} \frac{ heta}{2} ight) d heta[/latex], (11.176)

which, when using the change of variable [latex]x=2 a k \sin ^{2}( heta / 2)[/latex] with [latex]d x=2 a k \sin ( heta / 2) \cos ( heta / 2) d heta[/latex], leads to

[latex]\sigma=\frac{2 m^{2} V_{0}}{\pi \hbar^{4}} \int_{0}^{\pi} 2 \sin \left(\frac{ heta}{2} ight) \cos \left(\frac{ heta}{2} ight) \cos ^{2}\left(2 a k \sin ^{2} \frac{ heta}{2} ight) d heta[/latex]

[latex]=\frac{2 m^{2} V_{0}}{\pi a k \hbar^{4}} \int_{0}^{1} \cos ^{2}(x) d x[/latex]

[latex]=\frac{m^{2} V_{0}}{\pi a k \hbar^{4}} \int_{0}^{1}[1+\cos (2 x)] d x[/latex]

[latex]=\frac{m^{2} V_{0}}{\pi a k \hbar^{4}}[/latex]. (11.177)

Question 11.P.5: Find the differential and total cross sections in the first ...

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