Finite well scattering in the Born approximation

[THEODORE D SAUYET]

I'm preparing for an exam and I expect this or a similar question to be on it, but I'm running into problems with using the Born approximation and optical theorem for scattering off of a finite well.

1. Homework Statement
Calculate the cross sectional area σ for low energy scattering off of a finite well of depth V₀ and width a.

Homework Equations

Definition of f using the T-matrix:
$$f(k,k') = (\frac{2m}{\hbar^2})(\frac{-1}{4\pi})(2\pi)^3<k'|T|k>$$

Where the <k'| and |k> refer to incoming plane waves, I believe.

Optical theorem:
$$\sigma_{tot} = \frac{4\pi}{k}\textrm{Im}(f(k,k))$$

And in the Born approximation T $\approx$ V

The Attempt at a Solution

[/B]
To me, the matrix element $<k|V|k>$ should just be V₀a, because

$$\int_0^ae^{-ikx}V_0e^{ikx} = V_0a$$

Where the bounds are from 0 to a because the potential is zero outside of the range 0 to a. But then when you plug this into the optical theorem there is no imaginary component, so you get that the total cross section is zero.

Am I missing something about how to properly use the Born approximation/optical theorem? Is this what we expect? Am I doing something wrong in the math? I'm pretty skeptical of how I treated the <k'| and |k>, but it seems fairly consistent, because <k|$V_0$|k> would just be $V_0$, since the k's are orthonormal.

Any help/insights would be much appreciated!

 

[THEODORE D SAUYET]

So, according to the Wikipedia article on the Born approximation, "It is accurate if the scattered field is small compared to the incident field on the scatterer." Perhaps this assumption does not apply for arbitrary $V_0$? It's not immediately obvious to me why this would be the case though.

Although, if we take $V_0 \rightarrow +\infty$, then the scattered field would be just as large as the incoming field and this assumption would fail (completely and utterly)? This makes sense, but I don't see where we could assume $V_0$ small and change the above math. Perhaps you simply cannot.