classical-electrodynamicselectromagnetic-radiationphotonsquantum mechanics
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?
Ok, so since your reasoning seems to be ok, the real question is : "How can one measure experimentally a laser spot width?".
In your case, since you are using a gaussian beam, an equivalent is "How to measure the waist of my laser?"
If you are working in a wealthy lab, the simpliest way seems to buy and use a CCD camera. Otherwise, an other "easy peasy lemon squeezy" (and cheaper) method is possible.
For this, you will need :
Step 1 : Align the photodiode with the laser.
At this point, be sure to get the maximum power from your laser.
Step 2 : Attach the piece of cardboard on the translation stage.
This blocks the laser light so that it can't reach the photodiode.
Step 3 : Place the all thing between your fiber opening and the photodiode.
Step 4 : Move the translation stage with the spanner and measure the power collected by the photodiode in respect of the displacement (using the graduation).
Since your beam is gaussian, what you expect is an error function because the power that you are measuring is simply the integral of a gaussian... $$\mathcal{P}\sim\int \exp \left[ - \left(\frac{2(x-x_0)}{w}\right)^2 \right] \mathrm{d}x\;\sim \mathrm{erf}\left(\frac{x-x_0}{w/\sqrt 2}\right)$$
where $w$ is the waist.
Step 5 : Use your favorite software and do a fit. Then you have $w$.
Note that this method is not that precise. But it gives a good order of the waist, thought.
The difference isn't electromagnetic versus mechanical. All classical waves behave the same way, as do all quanta. The difference arises because you're treating the light as quantum and the mechanical wave as classical.
Combining these two results shows that the density of quanta making up a classical ideal plane wave of fixed amplitude and frequency $\omega$ is proportional to $\omega$.
Best Answer
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.