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In my quantum mechanics notes, in connection to the problem of representing a function $\psi \in L^2(V)$ (where V is a cubic box centered at the origin, $V=\{ \bar{x}:|x_1|,|x_2|,|x_3| \leq L/2\})$ in the basis of plane waves: $\{\phi_k(\bar x)=\frac{1}{\sqrt{L^3}}e^{i\bar{k}.\bar{x}},\bar{k}=\frac{2\pi}{L}(n_1,n_2,n_3),n_i \in \mathbb{Z}\}$, they argue that the following is a representation of the Dirac's delta function: $$ \delta^3(\bar{y}-\bar{x})=\sum_{\bar{k}} \frac{e^{i\bar{k}.(\bar{y}-\bar{x})}}{L^3}$$

Why is that true? Can someone give both a an intuitive/heuristic explanation and a rigorous one?

They call the space $V$ I just defined a space with a periodic boundary condition, since $\phi_{\bar{k}}(x_1,x_2,x_3)=\phi_{\bar{k}}(x_1+L,x_2,x_3)=\phi_{\bar{k}}(x_1,x_2+L,x_3)=\phi_{\bar{k}}(x_1,x_2,x_3+L)$, so L is the period.

I know that in $L^2(\mathbb{R}^3)$ the dirac delta function can be represented with: $$ \delta^3(\bar{y}-\bar{x})=\int_{\mathbb{R}^3} d^3k \frac{e^{i\bar{k}.(\bar{y}-\bar{x})}}{(2\pi)^3}$$ but I am having trouble convincing myself of the statement in the limited cubic domain case specialy because of that $\frac{1}{L^3}$ factor instead of the $\frac{1}{(2\pi)^3}$ which I fail to derive logically from a Fourier transformation as they do here Fourier Representation of Dirac's Delta Function for the $L^2(\mathbb{R}^3)$ case

Can someone clear this up?

some_math_guy
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    You can think of this as a product of three 1D cases, one for each direction. At that point, what you have is the Dirac comb. – Semiclassical Jan 12 '21 at 20:54
  • @Semiclassical I am confused, a Dirac comb is an impulse train, even in one dimension and what I should get is a single delta not a bunch of them – some_math_guy Jan 12 '21 at 21:49
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    A function with periodic boundary conditions over an interval is equivalent to a periodic function over the entire real line. As such, what you have is a train of Dirac deltas. It's just that you're only studying one interval. – Semiclassical Jan 12 '21 at 21:55
  • @Semiclassical Oh, I see, but since the interval is closed [-L/2, L/2], wouldn't I have 3 deltas at 0, L/2 and -L/2? How do I get rid of the last two? – some_math_guy Jan 12 '21 at 22:01
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    No. Your function is L-periodic, so there's also delta functions at $x=\pm L, \pm 2L$, etc. There's nothing strange going on at the $x=\pm L/2$ boundaries. Also, if you want to verify this for yourself, write down the Fourier expansion of your delta function as $\delta(x-y)=\sum_{n=-\infty}^\infty c_k e^{i 2\pi x n/L}$ where $k_n=2\pi n/L$. Then multiply both sides by $e^{-i 2\pi x m/L}$ and integrate both sides over the domain of periodicity. – Semiclassical Jan 12 '21 at 22:15
  • @Semiclassical So esentially the solution to my question is to 1) verify that the expresion for the Fourier series of the Dirac comb, which i found here https://en.wikipedia.org/wiki/Dirac_comb,2) then change variables to have the sum over k instead of n and 3)go to 3 dimensions. Right? – some_math_guy Jan 12 '21 at 23:43
  • Let us continue this discussion in chat. – Semiclassical Jan 12 '21 at 23:44

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