Editing Density matrix (section)

== Pure and mixed states ==
A pure quantum state is a state that can not be written as a probabilistic mixture, or [[convex combination]], of other quantum states.<ref name=Hall2013pp419-440 /> There are several equivalent characterizations of pure states in the language of density operators.<ref name=":0">{{cite book|last=Peres |first=Asher |author-link=Asher Peres |title=[[Quantum Theory: Concepts and Methods]] |year=1995 |publisher=Kluwer |isbn=978-0-7923-3632-7 |oclc=901395752}}</ref>{{rp|73}} A density operator represents a pure state if and only if:
* it can be written as an [[outer product]] of a state vector <math>|\psi\rangle</math> with itself, that is, <math display="block"> \rho = |\psi \rangle \langle \psi|.</math>
* it is a [[projection (linear algebra)|projection]], in particular of [[Rank (linear algebra)|rank]] one.
* it is [[idempotent]], that is <math display="block">\rho = \rho^2.</math>
* it has [[Purity (quantum mechanics)|purity]] one, that is, <math display="block">\operatorname{tr}(\rho^2) = 1.</math>

It is important to emphasize the difference between a probabilistic mixture (i.e. an ensemble) of quantum states and the [[quantum superposition|superposition]] of two states. If an ensemble is prepared to have half of its systems in state <math>| \psi_1 \rangle</math> and the other half in <math>| \psi_2 \rangle</math>, it can be described by the density matrix: 
: <math>\rho = \frac12\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, </math>
where <math>| \psi_1 \rangle</math> and <math>| \psi_2 \rangle</math> are assumed orthogonal and of dimension 2, for simplicity. On the other hand, a '''quantum superposition''' of these two states with equal [[probability amplitude]]s results in the pure state <math>| \psi \rangle = (| \psi_1 \rangle + | \psi_2 \rangle)/\sqrt{2},</math> with density matrix 
: <math>|\psi\rangle\langle\psi| = \frac12\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}.</math>
Unlike the probabilistic mixture, this superposition can display [[quantum interference]].<ref name="mikeandike" />{{rp|81}}

[[File:Bloch sphere.svg|thumb|In the [[Bloch sphere]] representation of a [[qubit]], each point on the unit sphere stands for a pure state. All other density matrices correspond to points in the interior.]] Geometrically, the set of density operators is a [[convex set]], and the pure states are the [[extremal point]]s of that set. The simplest case is that of a two-dimensional Hilbert space, known as a [[qubit]]. An arbitrary mixed state for a qubit can be written as a [[linear combination]] of the [[Pauli matrices]], which together with the identity matrix provide a basis for <math>2 \times 2</math> [[Hermitian matrix|self-adjoint matrices]]:<ref name=":2" />{{Rp|126}}
: <math>\rho = \frac{1}{2}\left(I + r_x \sigma_x + r_y \sigma_y + r_z \sigma_z\right),</math>
where the real numbers <math>(r_x, r_y, r_z)</math> are the coordinates of a point within the [[unit sphere|unit ball]] and
: <math>
  \sigma_x =
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix}, \quad
  \sigma_y =
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix}, \quad
  \sigma_z =
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} .</math>
Points with <math>r_x^2 + r_y^2 + r_z^2 = 1</math> represent pure states, while mixed states are represented by points in the interior. This is known as the [[Bloch sphere]] picture of qubit state space.

=== Example: light polarization ===
[[File:vertical polarization.svg|right|thumb|200px|The incandescent light bulb{{nbsp}}(1) emits completely random polarized photons{{nbsp}}(2) with mixed state density matrix:<br />
<div class="center"><math>\begin{bmatrix}
 0.5 & 0  \\
 0 & 0.5
\end{bmatrix}</math><span style="vertical-align:bottom">.</span></div>{{paragraph}}
After passing through vertical plane polarizer{{nbsp}}(3), the remaining photons are all vertically polarized{{nbsp}}(4) and have pure state density matrix:<br />
<div class="center"><math>\begin{bmatrix}
 1 & 0  \\
 0 & 0
\end{bmatrix}
</math><span style="vertical-align:bottom">.</span></div>]]
An example of pure and mixed states is [[Photon polarization|light polarization]]. An individual [[photon]]
can be described as having right or left [[circular polarization]], described by the orthogonal quantum states <math>|\mathrm{R}\rangle</math> and <math>|\mathrm{L}\rangle</math> or a [[Quantum superposition|superposition]] of the two: it can be in any state <math>\alpha|\mathrm{R}\rangle+\beta|\mathrm{L}\rangle</math> (with <math>|\alpha|^2+|\beta|^2=1</math>), corresponding to [[linear polarization|linear]], [[circular polarization|circular]], or [[elliptical polarization]]. Consider now a vertically polarized photon, described by the state <math>|\mathrm{V}\rangle = (|\mathrm{R}\rangle+|\mathrm{L}\rangle)/\sqrt{2}</math>. If we pass it through a [[circular polarizer]] that allows either only <math>|\mathrm{R}\rangle</math> polarized light, or only <math>|\mathrm{L}\rangle</math> polarized light, half of the photons are absorbed in both cases. This may make it ''seem'' like half of the photons are in state <math>|\mathrm{R}\rangle</math> and the other half in state <math>|\mathrm{L}\rangle</math>, but this is not correct: if we pass <math>(|\mathrm{R}\rangle+|\mathrm{L}\rangle)/\sqrt{2}</math> through a [[linear polarizer]] there is no absorption whatsoever, but if we pass either state <math>|\mathrm{R}\rangle</math> or <math>|\mathrm{L}\rangle</math> half of the photons are absorbed.

[[Unpolarized light]] (such as the light from an [[incandescent light bulb]]) cannot be described as ''any'' state of the form <math>\alpha|\mathrm{R}\rangle+\beta|\mathrm{L}\rangle</math> (linear, circular, or elliptical polarization). Unlike polarized light, it passes through a polarizer with 50% intensity loss whatever the orientation of the polarizer; and it cannot be made polarized by passing it through any [[wave plate]]. However, unpolarized light ''can'' be described as a statistical ensemble, e. g. as each photon having either <math>|\mathrm{R}\rangle</math> polarization or <math>|\mathrm{L}\rangle</math> polarization with probability 1/2. The same behavior would occur if each photon had either vertical polarization <math>| \mathrm{V}\rangle </math> or horizontal polarization <math>| \mathrm{H} \rangle </math> with probability 1/2. These two ensembles are completely indistinguishable experimentally, and therefore they are considered the same mixed state. For this example of unpolarized light, the density operator equals<ref name=":0" />{{Rp|75}}
: <math>\rho = \frac{1}{2} |\mathrm{R}\rangle \langle \mathrm{R}| + \frac{1}{2}|\mathrm{L}\rangle \langle \mathrm{L}| = \frac{1}{2} |\mathrm{H}\rangle \langle \mathrm{H}| + \frac{1}{2}|\mathrm{V}\rangle \langle \mathrm{V}| = \frac12\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}.</math>

There are also other ways to generate unpolarized light: one possibility is to introduce uncertainty in the preparation of the photon, for example, passing it through a [[birefringent crystal]] with a rough surface, so that slightly different parts of the light beam acquire different polarizations. Another possibility is using entangled states: a radioactive decay can emit two photons traveling in opposite directions, in the quantum state <math>(|\mathrm{R},\mathrm{L}\rangle+|\mathrm{L},\mathrm{R}\rangle)/\sqrt{2}</math>. The joint state of the two photons ''together'' is pure, but the density matrix for each photon individually, found by taking the partial trace of the joint density matrix, is completely mixed.<ref name="mikeandike" />{{Rp|106}}