# Dirac delta identity, and intuition on normalization

I'm working through Quantum Field Theory for the Gifted Amateur by Lancaster and Blundell, and in Chapter 3 one of the problems is written like this:

For boson operators satisfying $$[\hat{a}_{\mathbf{p}}, \hat{a}^\dagger_{\mathbf{q}}] = \delta_{\mathbf{p}\mathbf{q}}$$, show that $$\frac{1}{\mathcal{V}} \sum_{\mathbf{pq}} e^{i(\mathbf{p}\cdot\mathbf{x} - \mathbf{q}\cdot\mathbf{y})} [\hat{a}_{\mathbf{p}}, \hat{a}^\dagger_{\mathbf{q}}] = \delta^{(3)}(\mathbf{x}-\mathbf{y}).$$

(The system isn't yet taken to the infinite-size limit here, so the possible momenta are discrete.)

I think the idea here is straightforward: the Kronecker delta $$\delta_{\mathbf{pq}}$$ from the commutator reduces the pair of sums to one, and the resulting sum kind of looks like the Fourier transform of the Dirac delta, just discretized.

But I have two questions, one about formalism and one about a small detail that I'd like to get intuition on. The answers are likely related.

1. Can I take the intermediate step $$\frac{1}{\mathcal{V}} \sum_{\mathbf{p}} e^{i\mathbf{p}\cdot (\mathbf{x} - \mathbf{y})} = \delta^{(3)}(\mathbf{x}-\mathbf{y})$$ as an identity for discrete $$\mathbf{p}$$? When is that identity applicable (i.e. do we need infinitely many $$\mathbf{p}$$ in the sum, etc.)?
2. Is there any intuition for why a sum over exponentials like $$\sum_{\mathbf{p}} e^{i\mathbf{p}\cdot (\mathbf{x} - \mathbf{y})}$$ should yield a factor of the system volume $$\mathcal{V}$$ ? I've tried thinking about taking an integral over one of the positions and getting that factor of volume, but I don't really see it.
• Just interprete $\sum_p$ as $\int \frac{d^3p}{(2\pi)^3}$. When you take continuum limit, the volume $V$ appears Dec 23, 2021 at 14:44
• Why does volume appear? We're integrating over momenta so I'm not quite sure. Dec 23, 2021 at 14:49
• Related: this and this. Dec 23, 2021 at 14:52
• In answer to $1$, that follows from the commutator being a Kronecker delta.
– J.G.
Dec 23, 2021 at 17:33

While ekardnam's answer is completely correct, I think it's worth seeing a slightly more direct proof as well. I'll look at the one-dimensional version, since your three-dimensional version is just a product of three one-dimensional sums of the form $$\frac{1}{L} \sum_p e^{i p(x-y)}, \quad p = \frac{2\pi}{L}\mathbb{Z}$$ First, note that since the desired sum is invariant under $$x-y \rightarrow x-y + L$$, the result of the sum must be periodic in $$L$$. If you allow $$x$$ and $$y$$ to take any real values (rather than be constrained to $$[0,L]$$), then the correct result for the sum is actually $$\frac{1}{L} \sum_p e^{ip(x-y)} = \sum_n \delta(x-y-nL), \quad n \in \mathbb{Z}$$ From this expression, you can see that the sum is essentially just the Poisson summation formula. (To get the version on Wikipedia, just integrate both sides against an arbitrary function.)
For functions defined in a volume $$V$$ the functions $$\phi_{\vec{p}}(\vec{x}) = \frac{1}{\sqrt{V}}\exp(-i \vec{p}\cdot\vec{x})$$ is a complete orthonormal basis hence $$\sum_{\vec{p}}\phi_{\vec{p}}(\vec{x})\phi^*_{\vec{p}}(\vec{y}) = \delta^3(\vec{x} - \vec{y})$$ is basically the resolution to the identity and it is the equation of your point 1. The point 2 I guess is also covered in this explanation in order to have the $$\phi$$s be properly normalized.