The dimension of Spectral Function

Question

For a free electron gas model with volume $V$, the spectral function is $$A(k,\epsilon)=2\pi\delta(\epsilon-\epsilon_k)$$ so the density of states, $$D(\epsilon)=\frac{1}{V}\sum_k\delta(\epsilon-\epsilon_k)=\frac{1}{2\pi}\int \frac{d^3k}{(2\pi)^3} A(k,\epsilon)$$ Because $D(\epsilon)\sim\frac{1}{V} \frac{dN}{d\epsilon}$, the spectral function is written as $A(k,\epsilon)\sim\frac{1}{V}\frac{dN}{d\mathbf{k}d\epsilon}\sim(eV)^{(-1)}$

Now my quenstion is, if we increase the volume $V$, the electron number $N$ increases and $\frac{1}{Vdk}$ does not change. Does the $A(k,\epsilon)$ keep increasing or even diverge with the increasing $V$?

Answer 1

If the Volume V is increasing in size, the number of primitive unit cells will increase as a result, which mean the resolution of the Brillouin zone will increase as it becomes quasi-continuous, since the distance between two adjacent $k$ points in the x-direction for example is $\Delta k_x=\frac{2\pi}{N_xa_x}$, so as $N_x$ increases, the step decreases.