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I have the following situation: About the polarization of the photon, I introduce the basis:

Horizontal polarization $|\leftrightarrow>=\binom{1}{0}$ Vertical polarization $|\updownarrow>=\binom{0}{1}$

The density matrix in this problem is:

$$\rho =\frac{1}{2}\begin{pmatrix} 1+\xi _{1} & \xi_{2}-i\xi _{3}\\ \xi_{2}+i\xi _{3} & 1-\xi _{1} \end{pmatrix}$$

The Stokes parameters are: $\xi _{1}, \xi _{2}, \xi _{3}$

The probability that if the photon has got lineal polarization whose axis forms an angle $\theta$ with de horizontal is:

$$|w>=cos\theta |\leftrightarrow>+sin\theta|\updownarrow>$$

$$ P_{\theta}=<w|\rho|w>=\frac{1}{2}\left ( 1+\xi_{1}cos(2\theta)+\xi_{2}sin(2\theta) \right )$$

Is there any value of the Stokes parameters for which this probability is zero?

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What about $\xi_1 = \xi_2 = 0$? – user1504 Jan 3 '13 at 19:02
$\xi_1 = \xi_2 = 0$ won't do the trick because then the probability is 1/2. But you have three Stokes parameters that have to satisfy two conditions (first, the probability has to be zero, and second, $\xi_1^2+\xi_2^2+\xi_3^2 = 1$) so there sure is a solution. – Ondřej Černotík Jan 3 '13 at 19:32
Sorry, yes, total idiocy on my part. – user1504 Jan 3 '13 at 19:44
What about $\xi_{1}cos(2\theta)+\xi_{2}sin(2\theta=-1$? – user15940 Jan 3 '13 at 23:07
@user15940 Yup, that will do the trick. – Ondřej Černotík Jan 4 '13 at 10:34

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