# Units of $\hat{a}$ and $\hat{a}^\dagger$ in discrete vs continuous $k$ and normalization

Consider the quantization of the electromagnetic field. In the discrete case, given in Wikipedia, the operators $$\hat{a}$$ and $$\hat{a}^\dagger$$ are dimensionless $$[\hat{a}]=[\hat{a}^\dagger] = 1,$$ which is consistent with the commutation relation $$[\hat{a},\hat{a}^\dagger]=\delta_{\vec{k},\vec{k}'}$$. This is also consistent with $$\vec{E} = i\sum_{\vec{k},s}\sqrt{\frac{\hbar\omega}{2V\epsilon_0}}[\vec{e}_s\hat{a}_{\vec{k},s}e^{i\vec{k}\cdot\vec{r}} + \vec{e}_s\hat{a}_{\vec{k},s}^\dagger e^{-i\vec{k}\cdot\vec{r}}]$$ meaning that $$[\vec{E}] = \text{V}/\text{m}$$ in SI units.

Now consider the continuous case in which $$[\hat{a},\hat{a}^\dagger]=\delta^3(\vec{k}-\vec{k}')$$. This commutation relation implies that the units of the creation and annihilation operators are $$[\hat{a}]=[\hat{a}^\dagger] = \text{distance}^{3/2}.$$ However if I take the limit to $$k$$ continuous using $$\lim_{V\to\infty}\frac{1}{V} \sum_\vec{k} \to \int d^3k$$ in the expression of $$E$$ given in Wikipedia (in this page) now the units don't match. Also, the volume $$V$$ does not cancel out so in this limit $$E=\infty$$. I mean, taking the previous expression for $$\vec{E}$$ and applying this limit gives $$\vec{E} = i\sum_{s} \int d^3k \sqrt{\frac{\hbar\omega V}{2\epsilon_0}}[\vec{e}_s\hat{a}_{\vec{k},s}e^{i\vec{k}\cdot\vec{r}} + \vec{e}_s\hat{a}_{\vec{k},s}^\dagger e^{-i\vec{k}\cdot\vec{r}}]$$ which, obviously, has the same units as before thus being non consistent with the "new units" of the operators.

How are these "units problem" and "infinity problem" solved?

If you write $$H = \int \frac{d^3 k}{(2\pi)^3}E({\bf k}) \hat a^\dagger({\bf k})\hat a({\bf k})e^{i{\bf k}\cdot {\bf x}}$$ then the dimensions $$[a]= {\rm L}^{3/2}$$ work out as $$[k]= {\rm L}^{-1}$$
$$k= \frac{2\pi n} {L}$$ and for quantities that vary smoothly with $$k$$ we can write $$\sum_n \to \int dn = L \int \frac {dk}{2\pi}.$$
Rather more heuristically we have $$\delta(k-k')= \delta\left(\frac {2\pi}{ L}(n-n')\right)= \frac L{2\pi} \delta(n-n')\equiv \frac L{2\pi}\delta_{nn'}$$ which is consistent with $$2\pi \delta(k=0) = L$$, which also follows from setting $$k=0$$ in $$2\pi \delta(k)= \int_{-\infty}^{\infty} e^{ikx}dx.$$ Thus the the factors of the length $$L$$'s are also consistent.
In the expression for $$H$$ we have absorbed the factor $$L^3$$ in front of the integral into the $$a$$'s.
• Thanks for your answer. Do you know which would be the "normalization factor" or whatever that multiplies $E$ in order to have the correct units? Sep 10 '20 at 20:25
• It should already be correct: If $E(k)$ is in Joules then the eigenvalue of the eigenstate $|k\rangle =\hat a^\dagger(k) |0\rangle$ of $H$ is $E(k)$ and is in Joules. Sep 10 '20 at 20:28