Momentum space representation of a wavefunction.

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The normalization is indeed sqrt(a/2). The absolute value squared of the wavefunction is |C|^2/Cosh^2(ax), the antiderivative if this is |C|^2/a tanh(ax) and it then follows that the integral from minus infinity to infinity is 2/a |C|^2, equating this to 1 gives the result that we can take C = sqrt(a/2).

a(p) can be easily computed using contour integration. In case you are not familiar with this, let me explain the basic idea (this is a very important technique, so you'll have to master it). Just like you can integrate real functions over some segment of the real axis, you can integrate complex functions over some contour in the complex plane. The integral of f(z) dz over a contour parametrized by z(t) = x(t) + iy(t) where x(t) and y(t) are real functions is defined as

integral of f(z(t)) (dx(t) + i dy(t)) = integral of f(z(t)) (dx/dt + i dy/dt) dt

Now, let's skip some 50 pages of complex function theory and move to the basic results on contour integration. Suppose that f(z) is given by a convergent series expansion (such functions are called analytic). Then a contour integration along a contour from z = z1 to z = z2 is path independent. It is then simply given by the difference of the antiderivative of f at z2 and z1, completely analogous to the case of real integration. Now I didn't explain how derivatives are defined, but this is formally similar to real functions (you need to take the limit [f(z+h)-f(z)]/h, in the complex plane you can let h approach zero from different directions, so complex differentiability is a more restrictive condition...). E.g. the derivative of sin(z) is cos(z), the antiderivative of cos(z) is sin(z).

Now, if a function f(z) is analytic, and has an antiderivative F(z), then any integral around a closed contour ...