What percentage of photons pass through a polarization filter? Assuming all photons have random states, what percentage of photons make it through a polarization filter with preferred axis $\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}
\ket{v} = 1\ket{\uparrow} + 0\ket{\rightarrow}$ ?
Does 100% of the light pass through and get polarized vertically? Or does 50% of the light pass through and get polarized vertically?
My knowledge of the subject suggests that 100% of the light passes and gets polarized vertically, but the answers I've found for my textbook say 50% get through.
 A: (See edit below)You can’t get a half photon. If a photon passes through a polarizer, It will be 100% and it will come out polarized parallel with the polarizer, no matter what polarization it had before it went in.
EDIT: Because you changed your question, the answer would be 50% if the light is unpolarized and $\cos^2(\theta)$ if the light is polarized.
A: The percentage of photons of a wavelength depends on the design of the polarizer. If the slits are too wide for the specific wavelength, more than 50% of the photons will pass through. If the slits are too narrow, less than 50% or even no EM radiation will pass through.
Background. Each photon has exactly two possibilities when interacting with a polarizer. As you probably know, electromagnetic radiation in free space has an oscillating electric field component, whose oscillation direction is perpendicular to the direction of motion (and additionally a magnetic field component, perpendicular to the direction of motion and to the electric field component).
When interacting with the surface electrons of the slit, the photons are polarized with an orientation between -45° and 45° and 135° and 215° (i.e., all oriented at 0° and 180°, respectively). The other 50% of the EM radiation is absorbed by the polarizer. The specific angles depend, of course, on the orientation of the polarizer, but the principle should be clear.
