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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?

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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 $[k]= {\rm L}^{-1}$

In one dimension with periodic boundary conditions
$$ k= \frac{2\pi n} {L} $$ 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.

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  • $\begingroup$ 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? $\endgroup$
    – user171780
    Sep 10 '20 at 20:25
  • $\begingroup$ 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. $\endgroup$
    – mike stone
    Sep 10 '20 at 20:28
  • $\begingroup$ I have updated my question to better illustrate my point. $\endgroup$
    – user171780
    Sep 11 '20 at 8:55

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