In quantum theory,
$$E=\hbar \omega.$$
In classical theory, we have the Poynting vector: $$\vec{S}=\frac{1}{\mu_0}\vec{E}_0\times\vec{B}_0\cos^2(kr-\omega t).$$ Given that $S$ is the energy flux density (the rate of energy transfer per unit area), it seems to me that frequency cannot have an effect on energy larger than the factor of one.
Is this prediction of classical theory different from quantum theory?
No, because we can think of the classical wave as being made up of a large number of photons. If we have a low-frequency wave with the same energy as a high-frequency wave, it simply means that there are a larger number of low-energy photons.
You are looking at two different things assuming they should be the same.
In classical electrodynamics the Poynting vector is defined as $\mathbf{S} = \mathbf{E}\times \mathbf{H}$ and enters the variation of energy density as $$ \frac{\partial u}{\partial t} = -\textrm{div}\,\mathbf{S} - \mathbf{j}\cdot\mathbf{E}; $$ according to the functional form of $\mathbf{E}$ and $\mathbf{H}$ you will have different values. Usually fields can be expressed in Fourier transform and each single component will contribute with a factor proportional to a trigonometric function (with amplitudes in front); eventually, you still have to integrate over all contributions to get the actual value. Furthermore, the Poynting vector is still not the energy of the electromagnetic field, rather, as seen above, it just enters the equation for the energy flux variation.
In quantum field theory whenever you have integer spin field (and the electromagnetic field is so, the photon as its carrier having spin $1$) the Hamiltonian can be expressed as: $$ H = H_0 + \int \textrm{d}k\,\hbar \omega\,(a^{\dagger}(k)a(k)) $$ the latter contribution being interpreted as each oscillator adding $\hbar \omega$ pieces of energy.
Addendum: even if they were the same thing, yes, of course predictions from quantum theory are different from classical field theory: that is exactly why one introduces quantum mechanics. If they had the same values there would be no need to pass to the quantum description. Quantum field theory only reduces to classical field theory in the limit $\hbar\to 0$.
The quantum mechanical proportionality between energy and frequency of a photon originates in Planck's action quantum h, which has the dimension (energytime), i.e. the dimension of angular momentum. All physical interaction occurs in entire units of this elementary action quantum, also interaction of matter with electromagnetic radiation. Now, (energytime) is just the same as (energy / frequency), which means that a photon's energy must be proportional to its frequency.
The energy of a classical electromagnetic wave is given by the Poynting vector as you point out, and is proportional to E_0^2 or B_0^2 constants defined by the source of the radiation, and antenna, for example. The higher the voltages imposed on the circuits the higher the radiated energy, as Eoin points out. Therefore for calculating the energy input from the sun, for example , the frequency spectrum is necessary which is input from the measurements or from statistical calculations, as for example the Stefan Boltzman law.
Here is again from Panofski and Philips the frequency versus energy forexample , which is classcially constant , chapter 20 "radiation from circular orbits".
In my answer to your previous question I quoted how, classically, a four vector variable can be defined for the phase of the electric field. When this fourvector is identified with a quantum of energy, i.e. that the electromagnetic field is emerging from a huge ensemble of photons, the connection of frequency with the energy of the photon blends smoothly with the classical framework. You will need more photons for the same amount of energy the lower the frequency to build up the wave whose Poynting vector is fixed in frequency, as Edgar Mueller says.
So the prediction is not different, the two frameworks are different. The poynting vector is in the classical framework, the photons are in the underlying quantum mechanical framework .
Like Gremlin says....the classical electromagneticwave is made up of a large number of photons. If we have a low-frequency wave with the same energy as a high-frequency wave, it simply means that there are a larger number of low-energy photons.
Also we might add that photons are bosons which means they can occupy the same quantum state (unlike electrons and other fermions) and it is easy to have huge numbers of them which together form a coherent state which is easily measured observed on large human scales as the electric field. More photons gives more electric field and more energy. The electric field of one single photon would not be easily measurable by normal apparatus, especially at low frequencies in microwave región it is unheard of I believe. At higher frequencies single photons can be detected by photomultipliers (and perhaps by an animal eye if dark adapted with some luck as a single tiny speckle or flash at one point).
So the quantum theory gives the classical theory when you have superposition of lots of photons. Of course the quantum theory does differ from classical EM results in certain cases for example the usual problems which led to the development of QM such as thermal radiation, photoelectric effect, spectral lines, casimir effect, compton scattering. Also the quantum theory needs to be applied over electrons and nuclei to get a whole picture of whats going on which would then apply atomic structure and interactions and chemistry.
