Understanding Exponential Terms in Quantum Mechanics Problem

[NightHaWk]

Homework Statement

Consider an electric dipole consisting of two electric charges e and −e at a mutual distance 2a.
Consider also a particle of charge e and mass m with an incident wave vector k perpendicular to
the direction of the dipole.

1. Calculate the scattering amplitude in the Born approximation. Find the directions at which
the differential cross section is maximal.

Relevant Equations

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So I am trying to understand and solve the problem mentioned in the title.I found a solution online:

https://physics.bgu.ac.il/COURSES/QuantumMechCohen/ExercisesPool/EXERCISES/ex_9011_sol_Y09.pdf

The problem is, I can't understand this step :

I relly can't find out how the two expontential terms in the parenthesis show up. Any help greatly appreciated.

 

[MathematicalPhysicist]

Change of variables.
The first term he changed to $r \mapsto r+a$ and the second term you know what to do, don't you? ;-)

 

[NightHaWk]

If you could explain the second term as well i would be grateful because I still cannot get the same result as the given solution :)

 

[MathematicalPhysicist]

Ok, I'll explain.
If you change variables $r'=r+a$ you get that $r=r'-a$, so the exponential in the integral becomes
$e^{ir\cdot q} = e^{i(r'-a)\cdot q}$, and because you now integrate over $r'$ (the Jacobian of this change of variables is one), $|r+a|=|r'|$.

With the second term you use $r\mapsto r-a$, simple isn't it? :-)

 

[Delta2]

Ok, I'll explain.
If you change variables $r'=r+a$ you get that $r=r'-a$, so the exponential in the integral becomes
$e^{ir\cdot q} = e^{i(r'-a)\cdot q}$, and because you now integrate over $r'$ (the Jacobian of this change of variables is one), $|r+a|=|r'|$.

With the second term you use $r\mapsto r-a$, simple isn't it? :-)

So you basically mean we first break up the integral to a sum of two integrals, then do a separate change of variable to each integral, and then reunite the two integrals? Yes I think it is correct if the domain of integration is whole $\mathbb{R^3}$.

 

[MathematicalPhysicist]

@Delta2 , yes.

 

[NightHaWk]

That is what I was wondering about the second term, I did the separete change of variable,but I could not justify the reunion under one common integral.Thank you very much for the help!

 

[Karthik N]

Homework Statement:: Consider an electric dipole consisting of two electric charges e and −e at a mutual distance 2a.
Consider also a particle of charge e and mass m with an incident wave vector k perpendicular to
the direction of the dipole.

1. Calculate the scattering amplitude in the Born approximation. Find the directions at which
the differential cross section is maximal.
Relevant Equations:: -

So I am trying to understand and solve the problem mentioned in the title.I found a solution online:

https://physics.bgu.ac.il/COURSES/QuantumMechCohen/ExercisesPool/EXERCISES/ex_9011_sol_Y09.pdf

The problem is, I can't understand this step :

View attachment 250782

I relly can't find out how the two expontential terms in the parenthesis show up. Any help greatly appreciated.

In the same solution, can someone please explain how the integral that shows up later converges? After integrating over cos(theta) you get something oh the integral(sin(qr),r,0,inf). They somehow integrated this in the pdf

 

[vanhees71]

It's the Fourier transform of a Coulomb potential. The most simple solution is to solve for
$$\Delta \Phi=-\delta^{(3)}(\vec{x})$$
with the solution
$$\Phi(\vec{x})=\frac{1}{4 \pi r}.$$
In Fourier space, writing
$$\Phi(\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 q}{(2 \pi)^3} \tilde{\Phi}(\vec{q}).$$
Then from the equation you get
$$-\vec{q}^2 \tilde{\Phi}=-1 \; \Rightarrow \; \tilde{\Phi}(\vec{q})=\frac{1}{\vec{q}^2}.$$
To check we evaluate the Fourier integral
$$\Psi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 q \frac{1}{(2 \pi)^3 q^2} \exp(\mathrm{i} \vec{q} \cdot \vec{x}).$$
In spherical coordinates with $u=\cos \vartheta$ and $\vec{x} = r \vec{e}_z$ we have
$$\Psi(\vec{x}) = \frac{1}{4 \pi^2} \int_0^{\infty} \mathrm{d} q \int_{-1}^1 \mathrm{d} u \exp(\mathrm{i} q r u) = \frac{1}{2 \pi^2 r} \int_0^{\infty} \mathrm{d} q \frac{\sin(q r)}{q}=\frac{1}{4 \pi r}=\Phi(\vec{x}.$$
QED.