photon polarization, uncertainty in Energy A beam of red light is sent along the $z$ axis through a polaroid filter that passes only $x$ polarized light. The beam is initially polarized at $30$°, and the total energy is $10$ Joules. Estimate the fluctuations in the energy of the beam. 
After it passes the polaroid, the quantum state will collapse into $\vert\psi\rangle =\vert H\rangle$. And the energy will be $5\sqrt{3}J$, but how can I find the fluctuation? I have tried to use the uncertainty principle without success. 
Based on the comments, how is the Poisson distribution for the process?
Thanks 
 A: Let us first assume a (fictitious) beam carrying precisely $n$ photons at wavelength $\lambda$. The energy $E=nhc/\lambda$, where $h$ is the Planck constant and $c$ is the velocity of light, of such a beam would be fixed. In other words, such a beam would not exhibit any energy fluctuations. 
Apart from the fact it would make the answer to this question trivial ( $\Delta E = 0$), 
such light beams are anyway not commonplace as producing them experimentally is quite challenging. (Quantum-mechanically, the state of the beam is a $n$-photon Fock state, represented by $|n\rangle$).
Therefore, it is likely that in the question, the photon number of the beam is described by some kind of probability distribution. Typical laser beams exhibit Poisson statistics (http://en.wikipedia.org/wiki/Poissonian) described by parameter $\mu$. For such beams, the variance of the photon number $Var(n)$ is equal to the mean photon number $\mu$. This implies the absolute fluctuations in the energy $\Delta E$ would scale as $\sqrt{\mu}$ 
Since in the question the beam is also polarized at an angle of $30^{\circ}$  w.r.t. to the polaroid filter, the relative energy standard deviation after the filter can be calculated as $\Delta E/\langle E \rangle = \cos^2(\pi/6) \times \sqrt{hc/\langle E \rangle \lambda}$ where $\langle E \rangle = 10$J is the given average energy of the beam. 
Note that the question is incomplete because $\lambda$ is not precisely stated; it only says a beam of red light. 
