Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Malus law in quantum mechanics

  1. Jun 11, 2007 #1

    For example, if I were to send |y> = 1/sqrt{2}( |H> + |V> ) photons into a polarizer with a 45 degre angle from horizontal, would the beam loose intensity? How can I calculate it on my own?

    EDIT: The picture below illustrate what is happening. I am changing the base from x-y to a diagonal base. Do I loose intensity?

    Attached Files:

    Last edited: Jun 11, 2007
  2. jcsd
  3. Jun 11, 2007 #2
    No. Or rather, it depends on which diagonal you're talking about. One will allow |H>+|V> through and completely block |H>-|V> ; the other will do the reverse. You can pretty much see what happens here just by looking at things: the lower-left to upper-right diagonal clearly corresponds to |H>+|V> -- that line is given by [tex]c (\hat{x}+\hat{y}) ,\ c \in R[/tex] after all -- and so a polarizer along that lines corresponds to a projection onto |H>+|V>. The intensity of light getting through is given by [tex] | <\mbox{light}|\mbox{polarizer}>|^2[/tex] where light is the normalized polarization vector of the light, and polarizer is the normalized vector of the polarizer.

    In a more general situation -- I mean when you're dealing with polarization of objects which are not spin-1 like light -- what you'd want to do is rotate the object's polarization vector into the basis of the polarizer. To do this you need to know how the object transforms under rotations; that can be found by exponentiating the angular momentum operators (which are the generators of rotations.) If this sounds like gobbedy-gook, you can study a good (grad-level) quantum mechanics textbooks. They will go into this in some depth. IIRC Sakurai's Modern Quantum Mechanics is particularly good on this topic. For a spin-m object polarized perpendicularly to the plane of the polarizer, this turns out to be pretty simple; you get roughly the same result as for light, except the rotation angle that shows up is proportional to m*theta, ie

    [tex]|\mbox{rotated ket}> = \left( \begin{array}{cc} \cos(m\theta) & \sin(m\theta) \\ -\sin(m\theta) & \cos(m\theta) \end{array} \right) |\mbox{original ket}>[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook