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Separable Scattering

by llello
Tags: scattering, separable
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llello
#1
Nov22-10, 11:13 PM
P: 32
1. The problem statement, all variables and given/known data

Obtain the analytic expression for the N-channel T-matrix assuming a separable potential.
Hint: assume that T is proportional to V. Specialise your answer to N=1 and perform the required integral to get an explicit form for T, assume the given form for g(k).
[tex]$g=\frac{k_o^3}{k^2+k_o^2}$ [/tex]

2. Relevant equations

[tex] T(k',k) = V(k',k_o) + \frac{2}{\pi}\int_0^{\infty}\frac{q^2 V(k',q) T(k,q) - k_o^2 V(k',k_o)T(k',k_o)}
{\frac{k_o^2-q^2}{2\mu}}dq-2i\mu k_o V(k',k_o)T(k',k_o) [/tex]

3. The attempt at a solution

The T-matrix takes the form
[tex]
\begin{equation}
T(k',k) = V(k',k_o) + \frac{2}{\pi}\int_0^{\infty}\frac{q^2 V(k',q) T(k,q) - k_o^2 V(k',k_o)T(k',k_o)}
{\frac{k_o^2-q^2}{2\mu}}dq-2i\mu k_o V(k',k_o)T(k',k_o)
\end{equation}
[/tex]

Now we assume that T is proportional to V $(T=CV)$, which gives us

[tex]
\begin{equation}
T(k',k) = V(k',k_o) + \frac{2}{\pi}\int_0^{\infty}\frac{q^2 V(k',q) CV(k',q) - k_o^2 V(k',k_o)CV(k',k_o)}
{\frac{k_o^2-q^2}{2\mu}}dq-2i\mu k_o V(k',k_o)CV(k',k_o)
\end{equation}
[/tex]

Now we assume that $V(k,k') = \lambda g(k)g(k')$ which gives us

[tex]
\begin{equation}
T(k',k) = \lambda g(k')g(k_o) + C\frac{2}{\pi}\lambda^2\int_0^{\infty}\frac{q^2 g(k')^2 g(q)^2 - k_o^2 g(k')^2 g(k_o)^2}
{\frac{k_o^2-q^2}{2\mu}}dq-2i\mu k_o C \lambda^2 g(k')^2 g(k_o)^2
\end{equation}

\begin{equation}
T(k',k) = g(k')g(k_o) - \frac{4C\mu}{\pi}g(k')^2\int_0^{\infty}\frac{q^2 g(q)^2 - k_o^2 g(k_o)^2}
{q^2-k_o^2}dq-2i\mu k_o Cg(k')^2 g(k_o)^2
\end{equation}
[/tex]

Assume that [tex] $g(k)=\frac{k_o^3}{k^2+k_o^2}$ [/tex] then

(note, pretty sure from here on out is wrong and i just fudged it on the next line, i know, not kosher. will fix later)

[tex]
\begin{equation}
\int_0^{\infty}\frac{q^2 g(q)^2 - k_o^2 g(k_o)^2}{q^2-k_o^2}dq =
k_o^6\int_0^{\infty}\frac{q^2}{(q^2+k_o^2)^4} - \frac{k_o^2}{(q^2-k_o^2)^2(q^2+k_o^2)} dq
\end{equation}
[/tex]

Note that

[tex]
\begin{equation}
\frac{d^3}{da^3}\frac{1}{(x^2+a^2)} = \frac{24a}{(x^2+a^2)^3} - \frac{48a^3}{(x^2+a^2)^4}
\end{equation}
[/tex]

Ok.. So our prof isn't using a book and didn't give a reference for this stuff and the notes were crummy. I can't seem to find much online either. Basically, I'm not even sure if my starting equations are correct and I'm kinda feeling around in the dark. Any guidance whatsoever would be helpful. A link, a reference or anything would be great. Thanks alot.
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