How Does the Derivative Term Arise in the Commutator of Electromagnetic Fields?

[naima]

I start from these formulas(transverse electric and magnetic fields)
$E_\perp(r) = \Sigma_i i \mathscr E_{\omega_i}\epsilon_i [\alpha_i e^{i k_i . r}- \alpha^\dagger_i e^{-i k_i . r}]$
and
$B(r) = \Sigma_i i(1/c) \mathscr E_{\omega_i}(\kappa_i \times \epsilon_i) [\alpha_i e^{i k_i . r}- \alpha^\dagger_i e^{-i k_i . r}]$
where epsilon is a unit vector orthogonal to $\kappa_i = k_i/|k_i|$.
The authors compute their commutator and write it as
$[E_{x \perp}(r),B_y(r')] = -i (\hbar/\epsilon_0) \partial_z \delta(r - r')$

I do not see where this $\partial_z$ comes from.
Have you an idea?

Is it related to FT(f'(z)) = i k FT(f(k))

 

[naima]

I think that the solution is closer.
As $\mathscr(E)_{\omega_i} = \sqrt{\frac{\hbar \omega_i}{2 \epsilon_0 L}}$ and $h \omega_i = |k_i| c$
I get in the commutator a $-1/ \epsilon_0$ and the $|k_i|$ disappears in $\kappa_i$ and i come closer to a "i z FT(f(z)" term.
Help will be appreciated!