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  Can the density matrix be measured?

+ 3 like - 0 dislike

An interesting aspect of partial polarization is that the Stokes parameters can in principle be measured. A simple analogy to the density matrix in quantum-mechanics is the coherency matrix description of is partial polarization. It can be computed in terms of the Stokes parameters

\(J=\begin{bmatrix} E(u_{x}u_{x}^{*})&E(u_{x}u_{y}^{\ast})\\ E(u_{y}u_{x}^{*})&E(u_{y}u_{y}^{\ast}) \end{bmatrix}=\frac12\begin{bmatrix} S_0+S_1&S_2+iS_3\\ S_2-iS_3&S_0-S_1 \end{bmatrix}\)

and hence can in principle be measured. But can the density matrix in quantum-mechanics in principle be measured?

Well, the measurement process of the Stokes parameters can be described by the following Hermitian matrices \(\hat{S}_0=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), \(\hat{S}_1=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\), \(\hat{S}_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}\) and \(\hat{S}_3=\begin{bmatrix}0&i\\-i&0\end{bmatrix}\). It's easy to see that we get back \(S_i\) if we compute \(\operatorname{Tr}J\hat{S}_i\). Only \(\hat{S}_0\) commutes with all other Hermitian matrices, which somehow means that each individual (quantum analog of a) Stokes parameter can be measured in isolation, but we can't measure them all simultaneously. However, we do not measure all (classical) Stokes parameters simultaneous either, or at least that is not what we mean when we say that the Stokes parameters can be measured in principle.

A more intuitive set of Hermitian matrices for measuring the density matrix might be \(e_x e_x^T=\begin{bmatrix}1&0\\0&0\end{bmatrix}\), \(e_y e_y^T=\begin{bmatrix}0&0\\0&1\end{bmatrix}\), \(\frac12(e_x+e_y)(e_x+e_y)^T=\frac12\begin{bmatrix}1&1\\1&1\end{bmatrix}\) and \(\frac12(e_x-ie_y)(e_x+ie_y)^T=\frac12\begin{bmatrix}1&i\\-i&1\end{bmatrix}\). Here it is straightforward to understand the statistic outcome of a single measurement. So the individual measurements would just ask whether the system is in a certain pure state. But for the density matrix itself, we don't really want to measure whether the system has a certain property or not, but just the probabilities. So can we say that the density matrix can be measured (or observed), but that the meaning of this statement is slightly different from the meaning of the statement that the position or the velocity of a particle can be measured?

asked Jul 17, 2014 in Theoretical Physics by Thomas Klimpel (280 points) [ revision history ]

The density matrix of a single photon spin in quantum-mechanics is in fact identical with the coherency matrix of its partial polarization. See Optical models for quantum mechanics.

1 Answer

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Yes, the density matrix $\rho$ can be measured, and this is done routinely in many quantum optics experiments. But one usually calls this state estimation or density matrix estimation.

The principle is the following: One measures $N$ observables corresponding to projection $P_k$ of the state to certain pure states $\psi_k$. Since $P_k=\psi_k\psi_k^*$, we have $\psi_k^*\rho \psi_k=\langle P_k\rangle$, and the latter is accumulated through statistics on a corresponding experiment. If the system is an $n$-level system (i.e., if all other levels can be neglected) then $N=n^2$ linearly independent projectors suffice to determine $\rho$ since $n\times n$ density matrices belong to the $n^2$-dimensional real vector space of Hermitian matrices.  (Indeed one less suffices since all density matrices have trace 1). 

answered Jul 19, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
edited Jul 19, 2014 by Arnold Neumaier

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