Density operators are the mathematical objects used to describe the preparation of quantum systems. A density operator, or state on the system's Hilbert space H is any (bounded) linear operators \rho on H which is positive and has trace one (in symbols: \rho >=0 and tr\rho =1). The set of such operators is called the state space over H, and is denoted by S(H).
Let us recall some of the notions appearing in this definition. A (bounded) linear operator A on a Hilbert space H is called positive, if any one (and hence all) of the following three equivalent conditions hold: (i) for all \psi \in H, we have <\psi ,A\psi >>=0; (ii) A can be written as A=B*B, for some other bounded operator B. (iii) A=A* is hermitian, and the spectrum (in finite dimension this is just the set of eigenvalues) is contained in the positive real line. All three of these characterizations should be memorized, because they are all useful at times.
The trace of an operator on a finite dimensional space is the
sum over its diagonal matrix elements with respect to an orthonormal
basis, which is a maximal set of vectors \phi i such that
<\phi i,\phi j>=\delta ij. It is easy to see that this
does not depend on the basis chosen.
[5]
Example: Let us compute the simplest case: the state space S(C2) of a two-dimensional Hilbert space, or a "two-level" system, or one "qubit" [6] . With respect to any orthonormal basis, a density operator is represented by a matrix \rho ={1\over 2}(\matrix{x0+x3&x1-ix2\cr x1+ix2&x0-x3}) \hskip 1em\relax , where x0,x1,x2,x3 are complex parameters. This is just a funny way of parametrizing an arbitrary 2×2-matrix, and in fact, just the expansion of \rho in the basis consisting of the Pauli spin matrices, and the identity operator as the 0th element. What are the conditions on the parameters xi for \rho to be a density operator? First of all, tr\rho =x0=1, so only three parameters remain. By the third criterion for positivity we have that \rho must be hermitian, which is to say that all xi are real. The determinant of \rho is the product of the eigenvalues, and since these are positive, we must also have det\rho =(1/4)(1-x12-x22-x32)>=0. Hence (x1,x2,x3), as a vector in euclidean 3-space must have length <=1. This condition is also sufficient: if the product of the eigenvalues is positive, they might also be both negative. But since their sum, namely tr\rho =1 is positive, this cannot happen. To summarize, in the parameters xi, S(C2) is the unit ball in Euclidean 3-space. [7]
Let us compare this with the classical case, where the state space is the space of probability distributions pi over the finite set X. When X has just two elements, i.e., for a classical bit, the condition p1+p2=1 allows us to eliminate p2, and the states are uniquely parametrized by 0<=p1<=1. For |X|=3 (here the "absolute value" sign means (cardinal) number elements in a set), we get a triangle, for |X|=4 a tetrahedron, and so on through the whole range of simplices.
The difference in the geometrical structure of classical and quantum state spaces is crucial. It is best expressed by analyzing their structure as convex sets. By definition, a subset K of a vector space is called convex, if together with any two of its points the whole straight line segment connecting the two is also contained in the set. In formulas: x,y\in K and \lambda \in R, 0<=\lambda <=1 imply that \lambda x+(1-\lambda )y\in K. [8] Clearly, the two types of state spaces are convex, and this is no coincidence, because the convex combination has an immediate physical interpretation as a mixture of preparing procedures. Mixing is a fairly trivial operation. Preparing the density operator (1/2)\rho 1+(1/2)\rho 2 requires flipping a coin, and using the procedure for \rho 1 for heads and the procedure for \rho 2 for tails. (Other values of \lambda correspond to biased random generators making the choice.) It is clear from this description of mixtures that the same combination must also apply to all expectation values obtained with the preparing procedures involved. Indeed, the formalisms of quantum mechanics and classical probability guarantee this, because the formula for expectation values is linear in \rho (resp. p).
Of course, no experimentalist in her right mind would prepare a mixture in the way described: usually the aim is to make the preparation as sharp as possible, and mixing clearly goes in the opposite direction. The "best" preparations from this point of view are those which cannot be obtained as non-trivial mixtures at all. Substances which are not mixed from others are called "pure", and in this spirit we call the unmixed states pure states. In the language of convex geometry such points are called extreme points. We make the phrase "no non-trivial decompositions" precise in the following way: a point z\in K is an extreme point of a convex set K, if z=\lambda x+(1-\lambda )y\in K, x,y\in K and 0<\lambda <1 imply x=y=z.
For a single qubit the pure states are just the extreme points of the unit ball, i.e., the points on the surface. [9] From our determination of the state space it is clear that these are precisely the matrices, for which the product det\rho of the eigenvalues vanishes, i.e., those which have one zero eigenvalue. Since the trace is 1, this means that the other eigenvalue is 1. Hence the pure states are precisely the one-dimensional projections, i.e., the operators of the form \rho =|\psi ><\psi | with ||\psi ||=1. This statement holds also for Hilbert spaces of arbitrary dimension.
In the classical case the pure states are exactly those for which the probability is concentrated in a single configuration x\in X. Hence in contrast to the quantum case, the finite systems have only finitely many extreme points. We can now state the condition which singles out precisely the classical state spaces from all finite dimensional convex sets: in these convex sets every point has a unique convex decomposition into extreme points. This is the definition of a simplex. [10]
It is obvious that the qubit state space does not have this property: any point in the interior of the sphere has many decompositions into two pure states, and still more into larger numbers of extreme points. Since the density operator contains the complete description of a preparing procedure with respect to all statistical measurements, there is no experiment which could distinguish between preparations in which the same density operator is obtained in different ways as a mixture. Such a mixed state analyzer is yet another impossible machine of quantum mechanics. [11]
Please send comments and corrections to the author,
R.F. Werner.
Document created: Oct. 26, 1996;
last updated: Nov. 10, 1998