Pure and mixed states

Idea

In physics, recall that a state of a physical system is (in the Bayesian interpretation) a specification of the information that one might have about the system (typically relative to some fixed background information). States form (at least) a poset where $\psi \leq \phi$ means that $\phi$ contains all of the information of $\psi$ (and possibly more). A pure state is a maximal element of this poset: a state that specifies as much information as possible about the system. A mixed state is a state that is not pure.

Definitions

Fairly generally, a physical system has a complex $C^*$-algebra $A$ of observables and a state is a positive-semidefinite linear operator $\rho\colon A \to \mathbb{C}$ such that $\rho(1) = 1$. We might write $\rho$ as a convex-linear combination of two other states:

(1)$\rho = a \sigma + b \tau ,$

where necessarily $0 \leq a, b \leq 1$ and $a + b = 1$.

The state $\rho$ is pure if, whenever (1) holds for $\sigma \neq \tau$, then either $a = 0$ (hence $b = 1$ and $\rho = \tau$) or $a = 1$ (hence $b = 0$ and $\rho = \sigma$); conversely, $\rho$ is mixed if we ever have (1) for $\sigma \neq \tau$ and $0 \lt a \lt 1$ (hence $0 \lt b \lt 1$ and $\rho \neq \sigma, \tau$).

Really, this definition makes sense as long as the states form a convex space.

To define when one state gives at least as much information as another (the partial order from the Idea section), let $\rho \leq \sigma$ mean that the mutual information? $I(\rho,\sigma)$ equals the entropy $H(\rho)$, or equivalently that the conditional entropy $H(\rho|\sigma)$ is zero. (In the classical case, this partial order is attributed to Shannon (1953), which I have not read, by Li & Chong (2011), which I have only skimmed.) The maximal elements under this partial order should be precisely the pure states, but the direct definition of pure states is much simpler.

I need to check this claim. —Toby

Special cases

If $A$ is the algebra of continuous complex-valued functions on some compactum $X$, then the pure states on $A$ correspond precisely to the points in $X$; so pure states here are the states of classical mechanics (at least for a compact phase space). The mixed states, however, correspond more generally to Radon probability measures on $X$, with the pure states as the Dirac delta measure?s.

On the other hand, if $A$ is the algebra of all bounded operators on some Hilbert space $H$, then the pure states on $A$ correspond precisely to the rays in $H$, as is usual in quantum mechanics. The mixed states, however, correspond more generally to density matrices on $H$, with the pure states those matrices of the form ${|\psi\rangle}{\langle\psi|}$ for some unit vector ${|\psi\rangle}$.

Classical versus quantum

In any case, a pure state is a state of maximal information, while a mixed state is a state with less than maximal information. In the classical case, we may say that a pure state is a state of complete information, but this does not work in the quantum case; from the perspective of the information-theoretic or Bayesian interpretation of quantum physics, this inability to have complete information, even when having maximal information, is the key feature of quantum physics that distinguishes it from classical physics.

References

Not really references on this subject, but ones referred to in the text:

• Claude Shannon (1953): The lattice theory of information, IEEE Transactions on Information Theory 1, 105–107.

• Hua Li and Edwin Chong (2011): On a connection between information and group lattices, Entropy 13, 683–708; web.

Revised on July 14, 2012 13:08:14 by Toby Bartels (98.23.132.98)