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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.
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  • $\begingroup$ Just interprete $\sum_p$ as $\int \frac{d^3p}{(2\pi)^3}$. When you take continuum limit, the volume $V$ appears $\endgroup$ Commented Dec 23, 2021 at 14:44
  • $\begingroup$ Why does volume appear? We're integrating over momenta so I'm not quite sure. $\endgroup$ Commented Dec 23, 2021 at 14:49
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    $\begingroup$ Related: this and this. $\endgroup$ Commented Dec 23, 2021 at 14:52
  • $\begingroup$ In answer to $1$, that follows from the commutator being a Kronecker delta. $\endgroup$
    – J.G.
    Commented Dec 23, 2021 at 17:33

2 Answers 2

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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.)

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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.

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  • $\begingroup$ Interesting, I somehow managed not to make this connection but it's really useful. Thanks for the short and clear answer. $\endgroup$ Commented Dec 23, 2021 at 15:24

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