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AccidentalTaylorExpansion
AccidentalTaylorExpansion
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Let's assume a fully polarized Stokes vectorStokes vector, i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrizedparametrized according to \begin{equation} \vec S=\pmatrix{ \cos 2\psi\cos2\chi\\ \sin 2\psi\cos2\chi\\ \sin 2\chi\tag{1} } \end{equation} In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by \begin{align} J=\pmatrix{ \cos\psi\cos\chi+i\sin\psi\sin\chi\\ \sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2} } \end{align} This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.

To go the other way around, start with an arbitrary Jones vector \begin{align} J=\pmatrix{ \cos\theta e^{i\alpha}\\ \sin\theta\tag{3} } \end{align} Then the corresponding Stokes vector is given by \begin{align} \vec S=\pmatrix{ \cos 2\theta\\ \sin2\theta \cos\alpha\\ \sin2\theta \cos\alpha }\tag{4} \end{align}

Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by \begin{align} \rho=|\psi\rangle\langle \psi |\tag{5} \end{align} while for a mixed state it is given by \begin{align} \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6} \end{align} where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.

For a general Jones vector, this matrix is given by \begin{align} \rho=\left( \begin{array}{cc} \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\ e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\ \end{array} \right)\tag{7} \end{align} To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using \begin{align} J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{ -\sin\psi\cos\chi-i\cos\psi\sin\chi\\ \cos\psi\cos\chi-i\sin\psi\sin\chi } \tag{8} \end{align} Finally, the density matrix of the Stokes vector is given by \begin{align} \rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp| \end{align}

\begin{align} \rho=\left( \begin{array}{cc} \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\ \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\ \end{array} \right) \tag{9} \end{align}

To reconstruct the Stokes vector, we can use$^1$: \begin{align} S_i=\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align}\begin{align} S_i=S_0\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align} where \begin{align} &\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\ &\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L| \end{align} Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.

Alternatively, we could construct a density matrix straight from the Stokes parameters using \begin{align} \rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i \end{align}

Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.

1: https://doi.org/10.48550/arXiv.quant-ph/0103121

Let's assume a fully polarized Stokes vector, i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrized according to \begin{equation} \vec S=\pmatrix{ \cos 2\psi\cos2\chi\\ \sin 2\psi\cos2\chi\\ \sin 2\chi\tag{1} } \end{equation} In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by \begin{align} J=\pmatrix{ \cos\psi\cos\chi+i\sin\psi\sin\chi\\ \sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2} } \end{align} This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.

To go the other way around, start with an arbitrary Jones vector \begin{align} J=\pmatrix{ \cos\theta e^{i\alpha}\\ \sin\theta\tag{3} } \end{align} Then the corresponding Stokes vector is given by \begin{align} \vec S=\pmatrix{ \cos 2\theta\\ \sin2\theta \cos\alpha\\ \sin2\theta \cos\alpha }\tag{4} \end{align}

Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by \begin{align} \rho=|\psi\rangle\langle \psi |\tag{5} \end{align} while for a mixed state it is given by \begin{align} \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6} \end{align} where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.

For a general Jones vector, this matrix is given by \begin{align} \rho=\left( \begin{array}{cc} \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\ e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\ \end{array} \right)\tag{7} \end{align} To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using \begin{align} J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{ -\sin\psi\cos\chi-i\cos\psi\sin\chi\\ \cos\psi\cos\chi-i\sin\psi\sin\chi } \tag{8} \end{align} Finally, the density matrix of the Stokes vector is given by \begin{align} \rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp| \end{align}

\begin{align} \rho=\left( \begin{array}{cc} \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\ \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\ \end{array} \right) \tag{9} \end{align}

To reconstruct the Stokes vector, we can use$^1$: \begin{align} S_i=\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align} where \begin{align} &\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\ &\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L| \end{align} Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.

Alternatively, we could construct a density matrix straight from the Stokes parameters using \begin{align} \rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i \end{align}

Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.

1: https://doi.org/10.48550/arXiv.quant-ph/0103121

Let's assume a fully polarized Stokes vector, i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrized according to \begin{equation} \vec S=\pmatrix{ \cos 2\psi\cos2\chi\\ \sin 2\psi\cos2\chi\\ \sin 2\chi\tag{1} } \end{equation} In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by \begin{align} J=\pmatrix{ \cos\psi\cos\chi+i\sin\psi\sin\chi\\ \sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2} } \end{align} This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.

To go the other way around, start with an arbitrary Jones vector \begin{align} J=\pmatrix{ \cos\theta e^{i\alpha}\\ \sin\theta\tag{3} } \end{align} Then the corresponding Stokes vector is given by \begin{align} \vec S=\pmatrix{ \cos 2\theta\\ \sin2\theta \cos\alpha\\ \sin2\theta \cos\alpha }\tag{4} \end{align}

Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by \begin{align} \rho=|\psi\rangle\langle \psi |\tag{5} \end{align} while for a mixed state it is given by \begin{align} \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6} \end{align} where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.

For a general Jones vector, this matrix is given by \begin{align} \rho=\left( \begin{array}{cc} \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\ e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\ \end{array} \right)\tag{7} \end{align} To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using \begin{align} J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{ -\sin\psi\cos\chi-i\cos\psi\sin\chi\\ \cos\psi\cos\chi-i\sin\psi\sin\chi } \tag{8} \end{align} Finally, the density matrix of the Stokes vector is given by \begin{align} \rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp| \end{align}

\begin{align} \rho=\left( \begin{array}{cc} \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\ \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\ \end{array} \right) \tag{9} \end{align}

To reconstruct the Stokes vector, we can use$^1$: \begin{align} S_i=S_0\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align} where \begin{align} &\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\ &\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L| \end{align} Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.

Alternatively, we could construct a density matrix straight from the Stokes parameters using \begin{align} \rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i \end{align}

Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.

1: https://doi.org/10.48550/arXiv.quant-ph/0103121

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AccidentalTaylorExpansion
AccidentalTaylorExpansion
  • 18.1k
  • 3
  • 33
  • 65

Let's assume a fully polarized Stokes vector, i.e. $I=p=1$. The last three components of the Stokes vector, $(S_1,S_2,S_3)$, are then parametrized according to \begin{equation} \vec S=\pmatrix{ \cos 2\psi\cos2\chi\\ \sin 2\psi\cos2\chi\\ \sin 2\chi\tag{1} } \end{equation} In the $HV$ basis where $H=(1,0)^T, D=\tfrac 1{\sqrt{2}}(1,1)^T$ and $R=\tfrac 1 {\sqrt 2}(1, -i)^T$, a Jones Vector that reproduces this Stokes vector is given by \begin{align} J=\pmatrix{ \cos\psi\cos\chi+i\sin\psi\sin\chi\\ \sin\psi\cos\chi-i\cos\psi\sin\chi\tag{2} } \end{align} This Jones vector is of course up to a global phase factor. It seems that this expression could be simplified further but I could not find a nicer expression.

To go the other way around, start with an arbitrary Jones vector \begin{align} J=\pmatrix{ \cos\theta e^{i\alpha}\\ \sin\theta\tag{3} } \end{align} Then the corresponding Stokes vector is given by \begin{align} \vec S=\pmatrix{ \cos 2\theta\\ \sin2\theta \cos\alpha\\ \sin2\theta \cos\alpha }\tag{4} \end{align}

Finally, we can express these quantities using the density operator, which allows us to connect the degree of polarization $p$ more closely to the Jones vector. The density matrix for a pure state is given by \begin{align} \rho=|\psi\rangle\langle \psi |\tag{5} \end{align} while for a mixed state it is given by \begin{align} \rho=\sum_i p_i|\psi_i\rangle\langle \psi_i |\tag{6} \end{align} where $\sum_i p_i=1$ and $\psi_i$ form an orthonormal basis.

For a general Jones vector, this matrix is given by \begin{align} \rho=\left( \begin{array}{cc} \cos ^2(\theta ) & e^{i \alpha } \sin (\theta ) \cos (\theta ) \\ e^{-i \alpha } \sin (\theta ) \cos (\theta ) & \sin ^2(\theta ) \\ \end{array} \right)\tag{7} \end{align} To find the density matrix for the Stokes vector, we have to create a mixed and we have to define an orthonormal basis first. We can easiliy construct a vector that is orthogonal to (2) using \begin{align} J_\perp=\pmatrix{-J_2^*\\J_1^*}=\pmatrix{ -\sin\psi\cos\chi-i\cos\psi\sin\chi\\ \cos\psi\cos\chi-i\sin\psi\sin\chi } \tag{8} \end{align} Finally, the density matrix of the Stokes vector is given by \begin{align} \rho=(\tfrac 1 2+\tfrac{p}{2})|J\rangle\langle J|+(\tfrac 1 2-\tfrac{p}{2})|J_\perp\rangle\langle J\perp| \end{align}

\begin{align} \rho=\left( \begin{array}{cc} \frac{1}{2} (1+p \cos (2 \xi ) \cos (2 \psi )) & \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )+i \sin (2 \xi )) \\ \frac{1}{2} p (\cos (2 \xi ) \sin (2 \psi )-i \sin (2 \xi )) & \frac{1}{2} (1-p \cos (2 \xi ) \cos (2 \psi )) \\ \end{array} \right) \tag{9} \end{align}

To reconstruct the Stokes vector, we can use$^1$: \begin{align} S_i=\mathrm{Tr}\left( \rho \,\hat\sigma_i \right) \end{align} where \begin{align} &\hat\sigma_0=1\!\!1, &&\hat\sigma_1=|R\rangle\langle L|+|L\rangle\langle R|, \\ &\hat\sigma_2=i(|R\rangle\langle L|-|L\rangle\langle R|), &&\hat\sigma_3=|R\rangle\langle R|-|L\rangle\langle L| \end{align} Note that $\hat \sigma_2$ has a minus sign compared to the source$^1$. This is because the authors use an alternative definition of $D$.

Alternatively, we could construct a density matrix straight from the Stokes parameters using \begin{align} \rho=\frac 1 2\sum_{i=0}^3 \frac{S_i}{S_0}\hat\sigma_i \end{align}

Disclaimer: I derived a lot of these formulas myself. Although I checked them with Mathematica, if you want to use these for anything serious you might want to check them yourself.

1: https://doi.org/10.48550/arXiv.quant-ph/0103121