$\newcommand{\bra}[1]{\left\langle#1\right|} \newcommand{\ket}[1]{\left|#1\right\rangle} \newcommand{\prom}[1]{\langle{#1}\rangle} \newcommand{\matrixel}[3]{\bra{#1}{#2}\ket{#3}}$

How can I prove that the density matrix for a single qubit as a function of its Stokes parameters can be expressed by

$$\hat\rho = \frac{1}{2}\sum_{i=0}^{3} \frac{\mathcal{S}_i}{\mathcal{S}_0}\hat\sigma_i?$$

where $\mathcal{S_i}$ are the Stokes parameters defined as

\begin{align} \mathcal{S}_0 &\equiv \mathcal{N}(\matrixel{R}{\hat\rho}{R} + \matrixel{L}{\hat\rho}{L}),\nonumber \\ \mathcal{S}_1 &\equiv \mathcal{N}(\matrixel{R}{\hat\rho}{L} + \matrixel{L}{\hat\rho}{R}),\nonumber \\ \mathcal{S}_2 &\equiv \mathcal{N}i(\matrixel{R}{\hat\rho}{L} - \matrixel{L}{\hat\rho}{R}),\nonumber \\ \mathcal{S}_3 &\equiv \mathcal{N}(\matrixel{R}{\hat\rho}{R} - \matrixel{L}{\hat\rho}{L}). \end{align} and where $\mathcal{N}$ is a constant dependent on the detector efficiency and light intensity.

I think it has something to do with the coherency matrix but I am not pretty sure if that is correct. If so, how can I prove it?


1 Answer 1


Since this seems a homework exercise, here's a sketch: I'm not sure about the $\mathcal{N}$-part in the formular, but in general:

Note that $\sigma_i$ form an orthonormal basis of the Hermitian matrices according to the Hilbert-Schmidt inner-product ($\langle A,B\rangle:=\operatorname{tr} (A^{\dagger}B)\rangle$). This means that you can write

$$ \rho=\sum_{i=0}^3 k_i \sigma_i $$

Now, you can take the expectation values $\operatorname{tr}(\sigma_i \rho)=\operatorname{tr}(\sigma_i \sum_j k_j\sigma_j)$ and compare the two sides. Note that you can write e.g. $\sigma_0=|R\rangle\langle R|+|L\rangle\langle L|$. Using the orthogonality on the right hand side you'll get $k_i=\mathcal{S}_i/\mathcal{N}$ (as I said, I don't quite get the $\mathcal{N}$-parameter, but this doesn't play any role anyway, as it gets divided out afterwards).

Finally, you have to make sure that $\rho$ is properly normalized, so you have to divide the right hand side by $\operatorname{tr}(\rho)=\operatorname{tr}(\sigma_0\rho)$. Putting everything together, you'll obtain the formula above.

EDIT: Let's have a look at coherency matrices and the Jones vector. The Jones vector (as in here) is a vector $e\in \mathbb{C}^2$ (more precisely, in the projective version thereof, since the global phase doesn't really matter). It is a pure state of polarization. By, definition, the coherency matrix is given by $ee^{\dagger}$, hence it is an operator in $\mathcal{B}(\mathbb{C}^2)=\mathbb{C}^{2\times 2}$. This is exactly the definition of the density matrix of a pure state. This gives you the argument: Since the general coherency matrix is a $2\times 2$ complex matrix (as a qubit-matrix) and behaves exactly like a mixture of pure polarization states, which are states in $P\mathbb{C}^2$ (as are quantum states), from a purely mathematical perspective, where the density matrix is a normalized mixture of pure quantum states, the polarization matrix is also a density matrix.

This means, you can treat polarized light like a qubit with a qubit density matrix by identifying the Jones vector of a pure polarization state with a pure quantum state. You can actually also see this from the other direction: You can measure polarization in different ways: one basis is left-right, another is circular-anticircular a third one is 45°-polarization. If you have a look at the Jones calculus again (first table, exchange $|H\rangle$ by $|L\rangle$ and $|V\rangle$ by $|R\rangle$), you can immediately find the three spin-axes: If you fix left-right polarization, which is there called $|H\rangle,|V\rangle$ polarization, the measurement is then by the Pauli z, i.e. $\operatorname{tr}(\sigma_3 |L\rangle\langle L|)=1$ and $\operatorname{tr}(\sigma_3 |R\rangle\langle R|)=-1$ where

$$ \sigma_3:=\begin{pmatrix}{} 1 & 0 \\ 0 & -1\end{pmatrix}$$

The other two bases of polarization are measured by $\sigma_1$ and $\sigma_2$ respectively. This gives a full equivalence between a general coherency matrix and a general spin-qubit matrix, hence the two pictures can be used interchangedly.

  • $\begingroup$ Maybe I was not accurate enough. My aim was to find out how can we come across with that equation. In other word, I want to know the bridge between the classical coherency matrix to the density matrix. $\endgroup$ Commented Dec 3, 2014 at 0:56
  • $\begingroup$ When I see "prove", I always assume a mathematical proof, and since this formula needs to be proven, I did this. Sorry for misunderstanding you. I'll add a few lines of what I understand $\endgroup$
    – Martin
    Commented Dec 3, 2014 at 8:47

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