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CyBeaR
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2nd Born term calculation problem (potential scattering)
I'm trying to get a 2nd Born term for scattering amplitude in momentum representation (potential scattering).
I take potential of the form:
[tex]V \left( r \right) = \gamma r^{-1} \exp \left( \frac{-r}{R} \right) \; , R >0[/tex]
and it's form in momentum representation is:
[tex]\left\langle \mathbf{p'} \vert V \vert \mathbf{p} \right\rangle = \frac{\gamma}{2 \pi^2} \frac{1}{R^{-2}+ \left(\mathbf{p'}-\mathbf{p}\right)^2}[/tex]
2nd Born term is of the form:
[tex]\left\langle \mathbf{p'} \vert VGV \vert \mathbf{p} \right\rangle = \int d^3 q \frac{\left\langle \mathbf{p'} \vert V \vert \mathbf{q} \right\rangle \left\langle \mathbf{q} \vert V \vert \mathbf{p} \right\rangle }{k^2 - q^2 + i \epsilon } [/tex]
where [tex]k[/tex] is the length of the momenta [tex] \mathbf{p'} [/tex] and [tex] \mathbf{p}[/tex] and [tex]0< k \leq 1[/tex]
After applying Feynman method:
[tex] \frac{1}{ab} = \frac{1}{2} \int_{-1}^1 d\alpha \left( a \frac{1+\alpha}{2}+b \frac{1-\alpha }{2} \right)^{-2} [/tex]
with:
[tex]a=R^{-2}+ \left(\mathbf{p'}-\mathbf{q}\right)^2 , \; b=R^{-2}+ \left(\mathbf{q}-\mathbf{p}\right)^2[/tex]
to:
[tex]\int d^3 \mathbf{q}\frac{1}{a b \left( k^2 - q^2 + i \epsilon \right)}[/tex]
i'm getting two integrals to evaluate. One (easly done by using partial fractions):
[tex]-2 i \pi^2 k \int_0^1 d\alpha (f^2 - k^2 \tau^2 \alpha^2)^{-1}=-\frac{i \pi^2}{\tau f} ln\left|\frac{f+k\tau}{f-k\tau}\right|[/tex]
and 2nd:
[tex]- 2 \pi^2 R^{-2} \int_0^1 d\alpha (4 R^{-2}+\tau^2 - \tau^2 \alpha^2)^{-1/2} (f^2 - k^2 \tau^2 \alpha^2)^{-1}[/tex]
which is the problematic one.
Variable info:
-> [tex]\tau = 2 k sin\left( \frac{\theta}{2} \right)[/tex]
-> [tex] \theta [/tex] is scattering angle and [tex]0\leq \theta \leq \pi[/tex]
-> [tex]f^2 = R^{-4} + 4R^{-2}k^2 + k^2 \tau^2[/tex]
I've tried substitution like:
[tex]\sqrt{4R^{-2}+\tau^2 -\tau^2 \alpha^2}=\alpha \lambda -\sqrt{4R^{-2}+\tau^2}[/tex]
or
[tex]\lambda = \frac{4R^{-2}+\tau^2-\tau^2 \alpha^2}{\sqrt{4R^{-2}+\tau^2}-\tau \alpha} [/tex]
- where [tex]\lambda[/tex] is new integration variable in both cases - but that leads to cumbersome results.
Maybe someone can explain to me how to calculate second integral or share some tricks which can be applied to such integrals.
Any advice will be greatly appreciated.
Cheers!
Homework Statement
I'm trying to get a 2nd Born term for scattering amplitude in momentum representation (potential scattering).
I take potential of the form:
[tex]V \left( r \right) = \gamma r^{-1} \exp \left( \frac{-r}{R} \right) \; , R >0[/tex]
and it's form in momentum representation is:
[tex]\left\langle \mathbf{p'} \vert V \vert \mathbf{p} \right\rangle = \frac{\gamma}{2 \pi^2} \frac{1}{R^{-2}+ \left(\mathbf{p'}-\mathbf{p}\right)^2}[/tex]
Homework Equations
2nd Born term is of the form:
[tex]\left\langle \mathbf{p'} \vert VGV \vert \mathbf{p} \right\rangle = \int d^3 q \frac{\left\langle \mathbf{p'} \vert V \vert \mathbf{q} \right\rangle \left\langle \mathbf{q} \vert V \vert \mathbf{p} \right\rangle }{k^2 - q^2 + i \epsilon } [/tex]
where [tex]k[/tex] is the length of the momenta [tex] \mathbf{p'} [/tex] and [tex] \mathbf{p}[/tex] and [tex]0< k \leq 1[/tex]
After applying Feynman method:
[tex] \frac{1}{ab} = \frac{1}{2} \int_{-1}^1 d\alpha \left( a \frac{1+\alpha}{2}+b \frac{1-\alpha }{2} \right)^{-2} [/tex]
with:
[tex]a=R^{-2}+ \left(\mathbf{p'}-\mathbf{q}\right)^2 , \; b=R^{-2}+ \left(\mathbf{q}-\mathbf{p}\right)^2[/tex]
to:
[tex]\int d^3 \mathbf{q}\frac{1}{a b \left( k^2 - q^2 + i \epsilon \right)}[/tex]
i'm getting two integrals to evaluate. One (easly done by using partial fractions):
[tex]-2 i \pi^2 k \int_0^1 d\alpha (f^2 - k^2 \tau^2 \alpha^2)^{-1}=-\frac{i \pi^2}{\tau f} ln\left|\frac{f+k\tau}{f-k\tau}\right|[/tex]
and 2nd:
[tex]- 2 \pi^2 R^{-2} \int_0^1 d\alpha (4 R^{-2}+\tau^2 - \tau^2 \alpha^2)^{-1/2} (f^2 - k^2 \tau^2 \alpha^2)^{-1}[/tex]
which is the problematic one.
Variable info:
-> [tex]\tau = 2 k sin\left( \frac{\theta}{2} \right)[/tex]
-> [tex] \theta [/tex] is scattering angle and [tex]0\leq \theta \leq \pi[/tex]
-> [tex]f^2 = R^{-4} + 4R^{-2}k^2 + k^2 \tau^2[/tex]
The Attempt at a Solution
I've tried substitution like:
[tex]\sqrt{4R^{-2}+\tau^2 -\tau^2 \alpha^2}=\alpha \lambda -\sqrt{4R^{-2}+\tau^2}[/tex]
or
[tex]\lambda = \frac{4R^{-2}+\tau^2-\tau^2 \alpha^2}{\sqrt{4R^{-2}+\tau^2}-\tau \alpha} [/tex]
- where [tex]\lambda[/tex] is new integration variable in both cases - but that leads to cumbersome results.
Maybe someone can explain to me how to calculate second integral or share some tricks which can be applied to such integrals.
Any advice will be greatly appreciated.
Cheers!
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