Malus law in quantum mechanics

[AlexHT]

Hi,

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?

 

[damgo]

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 $$c (\hat{x}+\hat{y}) ,\ c \in R$$ 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 $$| <\mbox{light}|\mbox{polarizer}>|^2$$ 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

$$|\mbox{rotated ket}> = \left( \begin{array}{cc} \cos(m\theta) & \sin(m\theta) \\ -\sin(m\theta) & \cos(m\theta) \end{array} \right) |\mbox{original ket}>$$

 

[denian]

Malus law in quantum mechanics states that the intensity of a polarized light beam passing through a polarizer is proportional to the square of the cosine of the angle between the polarization direction of the light and the transmission axis of the polarizer. In the example you provided, the polarizer is at a 45 degree angle from the horizontal, which means that the intensity of the light beam will be reduced by a factor of 1/2 (since cos(45)^2 = 1/2). This means that the beam will lose intensity, but the exact amount can only be calculated by knowing the initial intensity of the beam and the properties of the polarizer.

To calculate the intensity loss on your own, you can use the equation I = I_0 cos^2(theta), where I is the final intensity, I_0 is the initial intensity, and theta is the angle between the polarization direction and the transmission axis. In your example, theta is 45 degrees, so the intensity loss would be I = (1/2)I_0.

I hope this helps clarify the application of Malus law in quantum mechanics. It is an important principle in understanding the behavior of polarized light in quantum systems.