# Pure states

A state is called '''pure''' if it cannot be represented as a mixture (convex combination) of other states, i.e., if it is an extreme point of the convex set of states. This is equivalent to the density matrix being a one dimensional projector, i.e., of the form $\rho=\vert\psi\rangle\langle\psi\vert$ for some unit vector $\vert\psi\rangle$. In this case the formula for the expectation value of an operator $A$ in the state simplifies as $tr\left(\rho A\right)=\langle\psi,A\psi\rangle$. Equivalently, a state $\rho$ is pure if $tr\left(\rho^2\right)=1$. In the Bloch sphere (the state space of a qubit), the pure states exactly form the surface of the sphere. For Hilbert space dimension larger than 2, however, the topological boundary (consisting of all density operators with at least one zero eigenvalue) is much larger than the set of extreme points (which have all but one eigenvalue zero). Caratheodory's Theorem guarantees that every point in a compact convex set of dimension $D$ can be represented as a mixture of at most $D+1$ points. The state space of a a system with $d$-dimensional Hilbert space is $d^2-1$ dimensional, so from this we would expect $d^2$ pure states to be necessary to represent a general mixed state. However, as the spectral theorem shows, the geometry of state spaces is such that $d$ pure states always suffice. Category:Handbook of Quantum Information Category:Quantum States

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Monday, October 26, 2015 - 17:56