Electromagnetic Field Intensity: Square law power = Poynting vector?

Question

Consider an arbitrary polychromatic wave in vacuum (an actual wave, not evanescent!), assuming real waves.

In scalar wave theory, optical intensity is defined as (Saleh & Teich, p.68):

$$I = \langle u^2(\mathbf{r}, t) \rangle_{\text{time}} $$

Where $u$ is the real wavefunction.

In electromagnetic wave theory,

$$I = \langle S(\mathbf{r}, t) \rangle_{\text{time}} = \langle \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B} \rangle_{\text{time}}$$

How does one prove that they are equivalent?

P.S. Also see wikipedia (link to highlight)

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It appears that both formulas(definitions) are NOT equivalent, as 'optical intensity' is operationally defined as what you would measure based on square-law detectors that only respond to the electric field! (check out any quantum optics text that deals with photoelectric theory of photon detection), right?

The units are not equivalent to begin with, differing by an uninteresting constant.

Answer 1

They are not equivalent. Scalar theory is like waves in simple elastic medium obeying Hooke's law, where we can thing of medium energy as sum of potential energies of springs, and kinetic energies of masses joining those springs.

Macroscopic EM theory is not like that, there is electric field and magnetic field vectors, and Poynting energy flux density is function of both these field vectors.

Scalar theory is easier to use in optics, and produces some useful results. But in general they are different results from the results of EM theory. From memory, scalar theory predicts somewhat different diffraction patterns.