Would passing horizontally polarized light through a varying width vertical slit allow you to measure the positional (x) amplitude of light? I have found closely related questions on StackExchange, but (surprisingly) not this exact question.  Seems some answers say individual photons do not have amplitude, only when traveling with other photons, forming a wave.  But even then, can't the amplitude be measured?  If so, what are the range of amplitudes and how do they vary?  Based on the frequencies and energies of the light?  Does higher frequency correspond to lower amplitude?
Moderators: This question is not a duplicate to this question:
Amplitude of an electromagnetic wave containing a single photon
That prior question is more about the formula for calculating electromagnetic energy amplitude of a single photon. My question is about an experimental way to MEASURE the positional (x) amplitude of the wave passing through a specific apparatus.  Two very different things.  Thanks.
I have updated my question to a more specific thought experiment:
If you had a vertical slit with a horizontal width that could be varied, and you passed horizontally polarized light through this slit, wouldn't the slit block any photons from passing through it if the width of the slit was smaller than the amplitude of the light waves?
Wouldn't we at least get a predicted or average amplitude, even with uncertainty principle?
Wouldn't the energy/frequency/wavelength of the wave also have some correlation to the amplitude?
 A: Classically (since rob has done a thorough job on the quantum picture), the amplitude of a light wave is not related to any physical extent. It is not the size of the wave in space, it is the strength of the fields (electric and magnetic).
We often draw wavy lines, but if you look closely the transverse axes will be label differently for, say, waves on a string and electromagnetic waves. You should not take those lines to imply a displacement the way they do in ripple on a pond. They just mean differing values of the field.
Classically, you can not filter different amplitudes with slit width. You simple block more light and create more diffraction as the slit grows narrower.
A: If you twisted my arm and forced me to assign an amplitude to a single photon, I'd do it this way:


*

*The energy density of a classical electromagnetic field is 
\begin{align}
U &= \frac12 \left(
\epsilon_0 E^2 + \frac1{\mu_0} B^2
\right)
\\
&= \epsilon_0 E^2 &\text{(only for light in a vacuum)}
\end{align}
where $E,B$ are the amplitudes of the electric and magnetic fields

*The total energy carried by the photon is $T=cp=\hbar c k$, where the wave vector $k=2\pi/\lambda$. (I shouldn't use $E$ for field amplitude and total energy.)
We can connect these if we come up with a reasonable estimate for the volume of a single photon. Which we probably shouldn't. But if you continued to twist my arm, I would say that the photon "fills" a long, skinny cylinder.


*

*The length scale for a photon perpendicular to the propagation direction $\vec k$ is roughly the same as the wavelength, so the "area" of the cylinder will be roughly $\lambda^2$.

*The length scale parallel to $\vec k$ is given by the uncertainty principle:
\begin{align}
\Delta x \Delta p = \hbar \Delta x \Delta k \approx \frac \hbar2
\\
\Delta x \Delta k \approx \frac12
\end{align}
Relating the uncertainty in the wavevector to the uncertainty in the wavelength is a little tricky, because calculus,
\begin{align}
\Delta k = \Delta \left( \frac{2\pi}{\lambda} \right) = \Delta\lambda \frac{2\pi}{\lambda^2},
\end{align}
which gives a position uncertainty
$$
\Delta x = \frac1{2\Delta k} = \frac{\pi\lambda^2}{\Delta\lambda}.
$$
So we have total energy $T = hc/\lambda$, smeared out over a volume like $V \approx \pi\lambda^4/\Delta \lambda $, to compare to a classical energy density $T/V \approx U = \epsilon_0 E^2$. That gives us
$$
E^2 \approx \frac{hc \Delta\lambda}{\epsilon_0\pi\lambda^5}
$$
It's probably less incorrect to use the uncertainty in the direction of $\vec k$ to find the "area" of the cylinder, but I'll stop here.
