## Green's function and scattering theory

I'm looking at scattering theory and eventually the Born approximation... In the notes I am reading it says we want to solve the Schrodinger equation written in the form:

##\left(\nabla ^2+k^2\right)\psi =V \psi##

Of which there are two solutions, the homogeneous solution which tends to the incident plane wave, and a particular solution which includes the scattering potential.

It then says that the general solution to this equation is

## \psi(r) = \phi_{\text{inc}} (r) + \frac{2 \mu}{\hbar^2} \int G(r-r') V(r') d^3 r ##

Then that the Green's function is obtained by solving the point source equation

## (\nabla^2 + k^2) G(r-r') = \delta (r-r') ##

I have some questions

1. Why does the homogeneous solutions tend to the incident plane wave? Does this just mean at large r the interaction potential is basically not felt?

2. How is the solution obtained with the Green's function? When will you know when to use a Green's function, why is this the time?

3. The solution to the point source equation is quite easy to get but I don't understand why we have asserted that that is the equation which we should solve

Mentor
 Quote by Gregg No reply again?
My servants! Attend me!

You might think about taking a different direction than the one you are on.

## Green's function and scattering theory

 Quote by Vanadium 50 My servants! Attend me! You might think about taking a different direction than the one you are on.
It's very common now that I post something, attempting to be as clear as possible, 100s look at the thread but no-one replies. It didn't used to be like that.

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 Quote by Gregg I'm looking at scattering theory and eventually the Born approximation... In the notes I am reading it says we want to solve the Schrodinger equation written in the form: ##\left(\nabla ^2+k^2\right)\psi =V \psi## Of which there are two solutions, the homogeneous solution which tends to the incident plane wave, and a particular solution which includes the scattering potential. It then says that the general solution to this equation is ## \psi(r) = \phi_{\text{inc}} (r) + \frac{2 \mu}{\hbar^2} \int G(r-r') V(r') d^3 r ## Then that the Green's function is obtained by solving the point source equation ## (\nabla^2 + k^2) G(r-r') = \delta (r-r') ## I have some questions 1. Why does the homogeneous solutions tend to the incident plane wave? Does this just mean at large r the interaction potential is basically not felt?
Basically, yes. The potential can modify the form of the plane waves somewhat - e.g., often these problems start with a plane wave of the form ##\exp(ikz)## incident on a localized potential which scatters the originally 'flat' plane wave into spherical plane waves. At least, at large distances from the potential the outgoing wave looks like a plane wave again. This is similar to how transients die out in ordinary differential equations. From a mathematical point of view, remember that far from the scattering potential there is basically no potential there, so if we just solved the equation in that region we would expect plane waves. The far and near solutions should match up somehow inbetween.

 2. How is the solution obtained with the Green's function? When will you know when to use a Green's function, why is this the time?
Green's functions are often used when you have an inhomogeneous differential equation to solve, as the Green's function enables you to write down the solution in terms of an integral over the inhomogeneous function times the Green's function.

In the case at hand, we are pretending the ##V(r)\psi(r)## term is the inhomogeneous term in our differential equation. Following what I just said about solutions to DE's being integrals over the inhomogeneous term times the Green's function, this allows us to recast the differential equation in terms of an integral equation for ##\psi(r)##:

$$\psi(\mathbf{r}) = \phi_0(\mathbf{r}) + \alpha \int_{\mathbb{R}^3} d\mathbf{r}~G(\mathbf{r}-\mathbf{r}')V(\mathbf{r'})\psi(\mathbf{r}').$$

Note that the equation you wrote down was missing the ##\psi(r)## in the integral. I also wrote ##\alpha## instead of all those constants separately.

The nice thing about doing this is that it's now relatively easy to develop a series expansion for ##\psi##. Given

$$\begin{eqnarray*} \psi(\mathbf{r}) & = & \phi_0(\mathbf{r}) + \alpha \int_{\mathbb{R}^3} d\mathbf{r}~ G(\mathbf{r}-\mathbf{r}')V(\mathbf{r'})\psi(\mathbf{r}') \\ & = & \phi_0(\mathbf{r}) + \alpha \int_{\mathbb{R}^3} d\mathbf{r}~G(\mathbf{r}-\mathbf{r}')V(\mathbf{r}')\phi_0(\mathbf{r}') + \alpha^2 \int_{\mathbb{R}^3\times \mathbb{R}^3}d\mathbf{r}'d\mathbf{r}''~G( \mathbf{r} -\mathbf{r}')V(\mathbf{r'})G( \mathbf{r}'-\mathbf{r}'')V(\mathbf{r}'')\psi(\mathbf{r}'') \\ & = & \phi_0(\mathbf{r}) + \alpha \int_{\mathbb{R}^3} d\mathbf{r}~G(\mathbf{r}-\mathbf{r}')V(\mathbf{r}')\phi_0(\mathbf{r}') + \alpha^2 \int_{\mathbb{R}^3 \times \mathbb{R}^3}d\mathbf{r}'d\mathbf{r}''~G( \mathbf{r}-\mathbf{r}')V(\mathbf{r'})G( \mathbf{r}'-\mathbf{r}'')V(\mathbf{r}'')\phi_0(\mathbf{r}'') + \dots \end{eqnarray*}$$

Typically we make some arguments that allow us to cut the series off at first order at this point. Note also that this sort of series expansion is similar to the kinds of series expansions done in Field Theories like Quantum Electrodynamics, where Feynman diagrams are employed to represent the terms in the series.

 3. The solution to the point source equation is quite easy to get but I don't understand why we have asserted that that is the equation which we should solve
I'm not sure what exactly you're asking for here, so let me ask, is the reason clearer with my discussion above, or is there something you're still unsure about?
 Mentor Well, we also used to have this thing called "civility" where people didn't complain that volunteers weren't responding fast enough - especially under a day. People also said "thank you" when they received help. If you absolutely need help and can't wait 17 hours, you should be hiring a tutor, not implicitly berating volunteers.

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 Quote by Gregg It's very common now that I post something, attempting to be as clear as possible, 100s look at the thread but no-one replies. It didn't used to be like that.
Have you been asking questions about more difficult material? Not everyone knows this subject, or other advanced subjects like it, off the top of their head. The only reason I knew it off the top of my head was that I nearly had to be substitute for a lecture about this very topic last semester.

 Quote by Vanadium 50 Well, we also used to have this thing called "civility" where people didn't complain that volunteers weren't responding fast enough - especially under a day. People also said "thank you" when they received help. If you absolutely need help and can't wait 17 hours, you should be hiring a tutor, not implicitly berating volunteers.
You have managed to extrapolate an awful lot from the three word post. Obviously it was a bad bump.. I'm sure the members will survive. I just wanted a reason for no reply after 100+ views (i.e. unclear etc.)

I have one now. There was no urgency. This used to be a place where I could discuss things at length with members and I occasionally I would help others - But I would not expect someone to essentially bump the thread with a "thank you"?

But to Mute, thank you, very helpful!!

Edit: Your area of interest: nuclear and particle physics!? What was wrong with the question about scattering? ha

Mentor
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 Quote by Gregg It's very common now that I post something, attempting to be as clear as possible, 100s look at the thread but no-one replies. It didn't used to be like that.
We have more visitors now. Did it occur to you that the people viewing your post are non-members just visiting the site? Being obnoxious will just cause members that could help you to ignore you instead. No one is obligated to help you.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Placing this technical question in the forum would have definitely drawn an immediate reaction.