# Can the Stokes Parameters Make Photon Polarization Probability Zero?

• pere
This corresponds to the photon having horizontal polarization ($|\leftrightarrow>$) and the angle $\theta$ being a multiple of $\pi/2$. In summary, the density matrix and Stokes parameters can be used to calculate the probability of a photon having linear polarization at a certain angle, and there is a specific combination of Stokes parameters that results in a zero probability.
pere
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](http://en.wikipedia.org/wiki/Stokes_parameters) for which this probability is zero?

Yes, there is a value of the Stokes parameters for which the probability is zero. Specifically, if $\xi_1 = -1$ and $\xi_2 = \sin(2\theta)$, then $P_\theta = 0$.

## 1. What is a density operator in quantum mechanics?

A density operator, also known as a density matrix, is a mathematical representation of the state of a quantum system. It describes the statistical probabilities of the possible states that the system can be found in. It takes into account both the quantum state and the classical probabilities of the system.

## 2. How is the density operator different from a wavefunction?

A wavefunction is a mathematical function that describes the quantum state of a single particle or system, while a density operator describes the quantum state of a larger, more complex system with multiple particles. The density operator takes into account the statistical probabilities of the wavefunctions of all the particles in the system.

## 3. What is the significance of the trace of the density operator?

The trace of the density operator is equal to the probability of finding the system in any state. It is a measure of the overall probability distribution of the system's states. Additionally, the trace of the density operator is conserved under unitary transformations, making it a useful tool in quantum mechanics calculations and analyses.

## 4. How is the density operator used in quantum mechanics calculations?

The density operator is used to calculate the expectation values of physical quantities, such as energy or momentum, in a quantum system. It is also used in calculating the time evolution of a system and in describing the behavior of mixed states, which are systems with both classical and quantum probabilities.

## 5. Can the density operator be used to describe entangled quantum states?

Yes, the density operator can be used to describe entangled states, which are states where the quantum states of two or more particles are correlated with each other. The density operator takes into account the correlations between the particles and allows for the calculation of expectation values and probabilities for entangled states.

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