Q:

Scattering from a potential, matrix elements of momentum eigenstates, and the Fourier transform

I am working on my last quantum homework and don't know where to begin with part (i) in this question 4.  Do I need to use a product rule in the FT and use convolution?  Not sure how to go about the FT in bra-ket notation.  I think if I get help with the first question I'll be able to do the rest.  If there are any good examples out there I'd be interested in seeing them.

A:

What's your problem? How do you pass from a certain vector |Q> expressed according to some base of vectors   {|$B_i$>} to the same vector |Q> expressed according to another base of vectors {|$C_j$>}? 
Let's see it together. Take for example
(1) |Q> = $∑_i$ $a_i$ |$B_i$>,
where $a_i$ are the amplitudes. For passing to the new base you project |Q> on each new vector, and get by reorganizing the product, 
(2)  = $∑_i$ $a_i$ <$C_j$|$B_i$>,
So, the vector |Q> in the new base is
(3) |Q> = $∑_j$|$C_j$> $∑_i$ $a_i$ <$C_j$|$B_i$>,,
This is what you do with the function v(k). It is similar to the vector |Q> i.e. instead of ($a_1$, $a_2$, $a_3$, ...), we should have (v($k_1$), v($k_2$), v($k_3$), ...). But k is CONTINUOUS, not discrete, and so is r in whose representation we want to pass.
Now, for us is important the relation (2). The matrix element <$C_j$|$B_i$> is the expression exp(ikr), and the summation over i becomes in your case is an integral over k.
Similarly with v(k').
Are the things clear? 
You don't need convolution here, v(k) and v(k') are expressed in DIFFERENT variables, k and k'. You look when do you need convolution, it is when you have a product of two functions of the same variable.
Good luck !

