# How does $H^2$ effect on the probability of photon detection?

Lets consider an electromagnetic wave, between two ideal conductive plates.

Maxwell's theory predicts appearance of a standing waves(of $$\textbf{H}$$ and $$\textbf{E}$$), at that nodes($$\textbf{E} = 0$$) of $$\textbf{E}$$-field are at the plate's surface(if it was not so, than the energy of the wave would be lost on the excitation of currents because of Ohm's law).However nodes of $$\textbf{H}$$ and $$\textbf{E}$$ are shifted by quarter wavelength and on the plate's surface located the anti-nodes of $$\textbf{H}$$-field where time-averaged $$ \neq 0$$. Which makes sense from the wave theory point of view, because the magnetic field does not work, therefore there is no energy loss.

But from the particle point of view it seems a bit strange, because vanishing of time-averaged $$\frac{ + }{2}$$ energy density on the plate's surface seems necessary, as a value proportional to the the probability of detecting a photon on the surface, because then electrons would not be able to catch any photon and there will be no current and no energy losses from the field. But from the Maxwell's theory it follows that: $$\frac{ + }{2} = 0 + \frac{}{2} \neq 0$$ And I do not see the way to conform this two points of view.

Where am I mistaken ?