The equivalent of vector addition for density operators?

Question

In quantum mechanics, pure states may be represented by (subspaces spanned by) vectors in a Hilbert space, which may be added. This is physically meaningful, and in wave mechanics leads to visible interference patterns.

But we can also discuss mixed states. They are usually introduced as a book-keeping device for classical uncertainty about the quantum state of a system, but it is also possible to have a formalism where they are foundational and kets and bras are the derived objects.

If we take such a perspective and think of density operators as our fundamental unit, then how would we "add" two pure state density operators in the sense of some operation that would map the density operators associated with two kets to the density operator of the sum of those kets?

Answer 1

First let me fix some notions: a density matrix (more accurately, density operator) is a hermitian operator $\rho = \rho^{\dagger}$ that is positive, $\langle \psi | \rho | \psi \rangle > 0$ for all $\psi \neq 0$, and has trace $1$, i. e. $\mathrm{Tr} \rho = \sum_j \langle \psi_j | \rho | \psi_j \rangle$ where $\{ \psi_j \}_j$ is any orthonormal basis of your Hilbert space. Moreover, you call $\rho$ pure state if it is a projection, $\rho^2 = \rho$; otherwise $\rho$ is a mixed state.

Now on to your question: you would take convex sums, namely if $\rho_0$ and $\rho_1$ are density matrices, then for any $s \in [0,1]$ the convex combination $\rho_s := (1-s) \, \rho_0 + s \, \rho_1$ is also a density operator. Those convex sums in general do not simplify to a pure state of the form $|\psi\rangle \langle\psi|$ for some $\psi$, though, so density matrices are generalizations of pure states.