-2

The full Hilbert space of a quantum system will generally contain entangled states, and thus when entanglement is lost through decoherence, parts of Hilbert space become inaccessible. Is there a simple way to quantify how many pure states become inaccessible when all entanglement is lost? In other words, what fraction of pure states in a Hilbert space are entangled?

A concrete example is the two-qubit system spanned by the product states $\{|00\rangle,|01\rangle,|10\rangle,|11\rangle\}$. This space contains entangled states, e.g. the Bell state $(|01\rangle + |10\rangle)/\sqrt{2}$, so the fraction of entangled states is greater than zero. But ignoring entangled states does not simply correspond to projecting the 4-dimensional Hilbert space onto something of lower dimension, as the product state basis still spans a 4D space.

So is there some way to quantify, or at least estimate, how much of Hilbert space contains entangled states?

Note that my question does not simply ask for an argument as to whether there are more entangled than un-entangled states, rather it is about how those two amounts can be quantified and compared.

Codename 47
  • 2,785

1 Answers1

0

(Almost) All the states in a Hilbert space are entangled. To see this, we consider the reduced density matrix of a given state (obtained by tracing out a subsystem), $\rho$. The state is pure iff. $$ \text{Tr} [- \rho \ln \rho] = - \sum_i \lambda_i \ln \lambda_i = 0 . $$ where $\lambda_i \in [0,1]$ and $\sum_i \lambda_i = 1$ are eigenvalues of $\rho$. The only way that the equation above can be true is if any one of the eigenvalues is 1 and the rest are zero. In every other case, the state is entangled.

Prahar
  • 29,157