Fourier transform operators

[aaaa202]

Homework Statement

The exercise is a) in the attached trial. I have attached my attempt at a solution, but there are some issues. First of all: Isn't the example result wrong? As I demonstrate you get a delta function which yields the sum I have written (as far as I can see), not the sum written in the trial. Secondly, what am I supposed to do in the sum for H_el, where I have a double-sum of position coordinates. I would like to somehow get a delta function, but the exponential is zero whenever k1*Ri=k2*Rj, which I can't translate into a delta function.
Also I am very confused why the Fourier series in position variables is given for x and p when clearly they enter in the original hamiltonian as creation and anihillation operators for position coordinates.

Homework Equations

The Attempt at a Solution

Attached

 

[mark726]

is my attempt at a solution
Thank you for your post and for sharing your attempt at solving the exercise. I can see that you have put a lot of effort into it and have raised some valid questions.

Firstly, regarding the example result, you are correct that it does not match the sum you have written. This could be due to a mistake in the given solution or a misunderstanding of the problem. I suggest double-checking your calculations and possibly reaching out to the instructor for clarification.

Secondly, for the sum of H_el, it is important to note that the exponential term is only zero when k1*Ri=k2*Rj AND when k1=k2. In this case, you can use the identity that the delta function is equal to 1 when its argument is 0, and 0 otherwise. So you can rewrite the double sum as a single sum over k1 and use the delta function to simplify the expression.

Finally, I understand your confusion about the Fourier series. It is important to note that the position and momentum operators are expressed in terms of the creation and annihilation operators, but they are not the same. The Fourier series in position and momentum variables are used to express the Hamiltonian in a more convenient form for solving the problem.

I hope this helps clarify some of your doubts. Keep up the good work and don't hesitate to ask for further clarification if needed.
Scientist