I learned that power intensity in EM (electromagnetic) radiation is $$I=\frac12c\varepsilon_0E_0^2$$ This equation implies that the energy in EM radiation is frequency-independent

I also learned that photon energy (quantized energy) in EM waves is $E=hf$; this equation implies that EM energy is amplitude-independent

The first equation describes the wave nature of light, while the second one describes the particle nature of light, can we combine the two equations like this

$$\frac{hfN}{At}=\frac12c\epsilon_0E_0^2$$

where $N$ is the number of photons that flow through the surface area $A$ during $t$ interval of time?

• That seems valid. It's basically saying that for a given field intensity $E_0$ in free space, there is a certain volumetric number density of photons in that space. I.e., $\rho=\frac{N}{c A t}=\frac{1}{2hf}c\epsilon_0E_0^2$. – DumpsterDoofus May 9 '14 at 17:16

For classical light intensities, the manipulation you did is perfectly valid (I have no idea about whether it makes sense in the low-intensity quantum optic regime).

In short, you just equate the classical intensity $I=\frac{1}{2}c\epsilon_0E_0^2$ with the flux of photons per units area $I=\frac{Nhf}{A \Delta t}$.

Further intuition can be gained by noting that during a time interval $\Delta t$, the photons passing through an area $A$ would, if you somehow were able to freeze time, fill up a volume of $V=A c \Delta t$. As a result, you can rearrange $\frac{hfN}{At}=\frac12c\epsilon_0E_0^2$ to get an expression for the photon density, giving $$\rho=\frac{N}{V}=\frac{c\epsilon_0E_0^2}{2hf}$$ which has units of photons per volume.

(I have no idea about whether it makes sense in the low-intensity quantum optic regime)

Indeed one has to be careful. In classical physics, what appears as purely mathematical objects, the complex representation of an electric field and its complex conjugate have different meaning. In the classical calculation of the intensity one takes the modulus squared of the complex representation for instance. However, in QED, the corresponding complex representation of the electric field is associated to the annihilation of a photon in the field whereas its conjugate is associated with the creation of a photon, these two do not commute.

The first equation describes the wave nature of light, while the second one describes the particle nature of light, can we combine

The wave nature of light is described via the canonical momentum. The "particle" nature of light via the mechanical (linear) momentum. As long as you are in free space, both equal to $$p = \hbar k$$ so yes you can combine, and get an hypothetic equivalent photon flux which has ultimately no real meaning in quantum optics, (see < http://link.springer.com/article/10.1007%2FBF01135846> )

good luck,