On the quantum physical theory of scattering

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Void123
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I have some questions here regarding phase shift and scattering.

1). I am working out an exercise for my own benefit, for which I already have the solution, but there are parts of it I don't understand. I am given a scattering delta potential: b*delta(r - a), where b is a constant, and I am interested in finding the wave function for s-waves (i.e. angular momentum values, l = 0).

After imposing boundary conditions, the solution is:

u = Asin(kr), for r < a and u = Asin(kr + phase shift factor), for r > a.

So my question is, how does the post-scattering wave function (with phase shift contribution) come out to be this way, after imposing the boundary condition: u(0) = 0 and u(infinity) = finite number.

2). When we look at the whole wave function for large values of r, the first term in the expression (which is exponential) is the incident plane wave. But this is talking about the prescattering wave which is contained in the wave function. So, when we say for large values of r, does that mean the incident wave travels quite a distance to the target where it is to be scattered? In other words, the wave function contains the whole superpositioned history of the wave before it scatters and after, over a long distance. Correct?

3). When we are talking about an incident wave being reflected off some potential wall, like at z = 0, (where it then receives a phase shift factor), we get two exponential terms: exp(ikz) and exp(-ikz), and this is for z < -b after which we can call "the interaction region" until it hits the wall at z = 0. So, my question is, if the two exponential terms are defined over some range of values that can be anywhere less than (-b), then wouldn't -infinity force one of the exponential terms to blow up and hence making it physically unacceptable?

Sorry for the long inquiry, but I need to clarify some of this confusion which has been dragging me down.

Thank You.
 
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At least on the first one, can someone explain to me how for values r > a we get u(r) = Asin(kr + phase shift factor) when the sin of infinity produces a non-existent limit. How does this give u = 0 as r-> infinity? I also don't see how this leads to eliminating the cosine term but keeping the sin one instead.