Light with circular polarization passes through a polarimeter

[Albfey]

Homework Statement

Consider light of circular polarazed, that can be either right or left. If this light passes through a polarimeter in the direction ##\textbf{a} = (cos \theta, sin \theta), what is the probability of find a photon with polarization in this direction?

Relevant Equations

$$|R \rangle = \frac{1}{\sqrt{2}}\binom{1}{i}$$

$$|L \rangle = \frac{1}{\sqrt{2}}\binom{1}{-i}$$

I think I can write the density matrix as $$\rho = \frac{1}{2} ( |R \rangle \langle R | + |L \rangle \langle L | ).$$ The state of a linear polarized light in the direction ##\textbf{a}## can be write as $$|\theta \rangle = \frac{1}{\sqrt{2}} ( e^{-i \theta} |R\rangle + e^{i \theta} |L\rangle ).$$ The propability of find a system described by ##\rho## in a state ##|\theta \rangle## is given by $$Tr(\rho |\theta \rangle \langle \theta |)$$ Is this correct? I haven't calculated this yet because I don't know if it's right.

 

[_coyote_]

Yes, that is correct. To calculate the probability, you just need to compute the trace:$$Tr(\rho |\theta \rangle \langle \theta |) = \frac{1}{2} \left[ \langle R | \theta \rangle \langle \theta | R \rangle + \langle L | \theta \rangle \langle \theta | L \rangle \right] $$Using the definition of $|\theta \rangle$, you can simplify this to$$Tr(\rho |\theta \rangle \langle \theta |) = \frac{1}{2} \left[ | \langle R | \theta \rangle |^2 + | \langle L | \theta \rangle |^2 \right]$$which is equal to 1.