How to understand this Dirac delta function?

Question

I am reading this paper about quantization of the electromagnetic field, and there is a point where the author imposes the fundamental commutation relation between the vector potential and its canonical momentum: $$[A_i(\mathbf r,t), p_i(\mathbf r',t)] = \frac{i\hbar}{4V}\sum_{\mathbf k}{(2e^{i\mathbf k\cdot(\mathbf r-\mathbf r')} + 2e^{-i\mathbf k\cdot(\mathbf r-\mathbf r')})} = i\hbar \delta(\mathbf r-\mathbf r')$$ I could follow how the sum is derived. But not why this is a delta function.

The author makes a remark that this is a finite volume. I suppose that this delta function is then defined so that: $$\int_V\delta(\mathbf r-\mathbf r')d^3r' = 1.$$

Considering that this volume is a period, the meaning would be: $\int_V e^{i\mathbf k\cdot(\mathbf r-\mathbf r')} = 0$ for $\mathbf r\neq\mathbf r'$ for all $k$'s. But the problem is that for $\mathbf r = \mathbf r'$, $$\frac{i\hbar}{4V}\sum_{\mathbf k}{(2e^{i\mathbf k\cdot(\mathbf r-\mathbf r')} + 2e^{-i\mathbf k\cdot(\mathbf r-\mathbf r')})} = \frac{i\hbar}{V}\sum_{\mathbf k}cos(\mathbf k\cdot \mathbf 0)\implies $$

$$\int_V\sum_{\mathbf k}cos(\mathbf k\cdot \mathbf 0)d^3r' = V(1 + 1 + 1 +...) = \infty$$ instead of $V$ as required.

Answer 1

Consider Fourier series expansion of the periodic function $$\sum_{\ell,m,n}\delta(x-\ell L)\delta(y-m M)\delta(z-nN)=\sum_{\ell,m,n}c_{\ell mn}e^{i2\pi\ell x/L}e^{i2\pi m y/M}e^{i2\pi n z/N}$$ Multiply both sides by $e^{-i2\pi\ell' x/L}e^{-i2\pi m' y/M}e^{-i2\pi n' z/N}$, integrate within a single "unit cell" of volume $LMN$, and use the orthogonality of these functions to conclude $$c_{\ell'm'n'}=\frac{1}{LMN}$$ and thus $$\sum_{\ell,m,n}\delta(x-\ell L)\delta(y-m M)\delta(z-nN)=\frac{1}{LMN}\sum_{\ell,m,n}e^{i2\pi\ell x/L}e^{i2\pi m y/M}e^{i2\pi n z/N}$$ or in your language $$\sum_{\mathbf R}\delta(\mathbf r - \mathbf R) = \frac 1 V \sum_{\mathbf k}e^{i\mathbf k \cdot \mathbf r}$$ If we are only evaluating this function within a single "unit cell", e.g. the one containing $\mathbf r = 0$, then only one term of the sum on the left-hand side is non-zero and we are left with $$\delta(\mathbf r) = \frac 1 V \sum_{\mathbf k}e^{i\mathbf k \cdot \mathbf r}.$$

I've skipped some steps, and you can ask reasonable questions like "does a periodic impulse train even have a Fourier series", but this is one way to obtain this relation. Also see here.

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A more direct approach is to note that $$\sum_{\ell = -{\ell_0}}^{\ell_0} e^{i2\pi\ell x/L}=\frac{\sin[(l_0 + \frac 1 2)2\pi x/L]}{\sin(\pi x/L)}.$$ This follows from the partial sum of the geometric series and the fact that $\sin z = (e^{iz}-e^{-iz})/(2i)$. Lor large $\ell_0$, this is an oscillatory function bounded by the envelope provided by the denominator, with a very sharp peak at $x = 0$. It can be shown that its integral on any interval containing $x = 0$ converges to $1/L$ as $\ell_0\to\infty$, consistent with $\delta(x)/L$.

While the value of the sum does not strictly converge at $x\ne 0$, the sum gets increasingly oscillatory for large $\ell_0$ such that its integral on any interval not containing $x = 0$ converges to zero. As such, in a distribution sense we are justified in writing $$\frac 1 L\sum_{\ell = -\infty}^{\infty} e^{i2\pi\ell x/L}=\delta(x).$$

This can be extended in the straightforward way to 3D to arrive at the result obtained by the Fourier series approach.

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In your last line of equations, you are evaluating $\int_V \delta(0) d^3\mathbf r'$, so it is no surprise you get an infinity.