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?


1 Answer 1


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.

  • $\begingroup$ 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
    Commented Sep 10, 2020 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
    Commented Sep 10, 2020 at 20:28
  • $\begingroup$ I have updated my question to better illustrate my point. $\endgroup$
    – user171780
    Commented Sep 11, 2020 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.