1
$\begingroup$

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

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Pablo Palacios Dec 3 '14 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 Dec 3 '14 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.