Re: Why do light detectors always square the electric field ?

Hi there,
I have some experience in light detection so I'll explain from the point 
of view of a photomultiplier tube but this goes for all light detectors.
A photon of a given energy hits the front plate of the light detector, 
this knocks off an electron of an energy corresponding to hf - W (the 
energy of the photon - the work function of the material). Due to an  
electric field between the front and back plate of the detector, the 
electron is accelerated through the material. This causes further 
electron emissions as the initial electron hits electrons orbiting atoms 
in the material. 1 becomes 2, 2 to 4, 4 to 8 and so on. At the end plate 
the number of electrons which have been knocked off is proportional to 
the initial number of photons which hit the front plate of the detector 
in the first place. The number of photons hitting the front of the 
detector is called the intensity and from Maxwells equations this is 
directly proportional to the square of the electric field of the photon. 
So the reason the detectors take the square of the electric field is 
because their output signal is proportional to the intensity of the light 
inputted. 

With regard to the integration of the detector, the minimum amount of 
integrated (or summed = same thing) signal is a function of how quickly 
the material it is made of can transmit the charge carriers through the 
lattice. This is known as electron mobility and is tabulated for many 
useful materials used for detection. The more time the charge is left to 
build up in the detector tells you the integration time of the signal. 
Using electronics after the detector it is possible to increase this 
integration time so you can have as long a time as you want, but no 
shorter than the electron mobility.

For the electric energy, it is not defined as the square of the electric 
field, but this falls out as the relationship when you solve Gauss's law 
for the electric field and then integrating the electric field gives the 
electric potential. Its a straight forward proof and is in any university 
level electromagnetism text book such as Electromagnetism by Grant and 
Philips. 

I hope this answers most of your question.