Converting momentum sums to integrals in curved spacetime

[KariK]

TL;DR Summary

Is Parker and Toms book wrong calculating number density of particle creation?

I am studying particle pair production using Parker and Toms book: Quantum Field Theory in Curved Spacetime. On page 48 they talk about converting the sum over momentum (k) into an integral. You assume boundary conditions so that k = 2*Pi*n/L, where n is an integer and L is the coordinate length of the cube, the volume of which is L^3. Then on page 60 he derives a formula for the number density of the created particles. There he has the volume as (L*a)^3, with a being the cosmological scale factor. But then in the next line (Equation 2.103) he uses the integral and uses (2*Pi^2*a)^-3 as the factor? (I understand that you get the 4*Pi from the angles integration.) In my mind if the side of the cube is L*a, then k = 2*Pi*n/(L*a), so we have a factor [L*a*dk/(2*Pi)]^3. So the integral in Equation 2.103 should be just (2*Pi)^-3 times the integral. Why do they have the additional a in there? Am I missing something or is the book wrong?

 

[KariK]

Well, nobody responded to this question and some time ago I understood the issue, so here is the answer for posterity.
If you solve the Klein-Gordon equation for a scalar field, the answer depends on the phase:
Phi = ...*e^(i*k.x)*..., where k is a constant and x is the comoving coordinate. We demand that the phase is the same at opposite edges of the cube, so we indeed get k*L = 2*Pi*n. For density quantities (number or energy) we have to divide by the physical volume (a*L)^3. So indeed we get the factor (2*Pi*a)^-3.