Completely Entangled Random Subspaces Jonathan Walgate and Andrew

Completely Entangled Random Subspaces Jonathan Walgate and Andrew Scott ar. Xiv: 0709. 4238

Completely entangled subspaces … Consider a multipartite Hilbert space, When does a subspace, . , contain only entangled states? The answer has diverse implications: • Almost all mixed states of limited rank are entangled. • Almost all many-qubit subspaces are wholly locally indistinguishable. • Necessary and sufficient conditions for the locally unambiguously distinguishing almost all sets of pure quantum states.

Completely entangled subspaces … If is bipartite, and is relatively small, then with high probability it will contain only near-maximally entangled states. [Hayden, Leung, Winter, 2004]

Completely entangled subspaces … If the dimension of exceeds a high bound: then it must contain product states. There are completely entangled subspaces at the bound. [Wallach, 2002] [Parthasarathy, 2004] In fact, almost all subspaces at or below this bound are completely entangled.

Hilbert space… and …projective space … d-dimensional Hilbert space maps to d-1 dimensional projective space. State vectors map to points. n dimensional subspaces map to (n-1)-planes. The set of all product states maps to the Segre variety.

Zariski topology Projective space has two key features: The general plane If the general (n-1)-plane has property P, then the set of n dimensional subspaces with property P has full Haar measure. Dimension If then the two sets must intersect. If one is the general plane, then when the two sets do not intersect.

Segre variety We know the dimension of the Segre variety: So the general n-plane does not intersect with the Segre variety iff Thus almost all n-dimensional subspaces are completely entangled iff:

Example: 2 x 2 This result applies to all multipartite situations, but the simplest example is provided by a two qubit system. The system has a 3 -dimensional projective space, and its Segre variety has dimension 2. Thus… Since is entangled. , a random pure state Since , a two-dimensional subspace contains product states. (Generically, it contains exactly two. )

Schmidt rank We needn’t just look at the product states. We can consider any algebraically well-defined set of states, and exactly the same reasoning will hold. For example, “states of Schmidt rank less than r. ” We have constructions for subspaces containing only high-Schmidt rank states. [Cubitt, Montanaro, Winter, 2004] These constructions are generic.

Mixed states on a dimension d system have ranks ranging from 1 (pure) all the way to d. A rank-n mixed state on a Hilbert space, , can be thought of as the reduced density matrix derived from a larger pure entangled state, on a Hilbert space, . … The support of a rank n mixed state is an n-dimensional subspace. If : almost all rank-n mixed states are entangled, and their span supports no separable mixed states at all. If : all subspaces contain separable mixed states of all possible ranks from 1 to n.

Local state discrimination An unknown state is given to local parties, who must identify it. Unambiguous discrimination When the parties hazard a guess, they must be correct. A D B C [Chefles, 2004]: A member state of a set only if there is some product state orthogonal to all the others. is conclusively locally identifiable if and that is nonorthogonal to that state and A set of states is unambiguously locally distinguishable only if every subset’s complementary subspace contains a product state.

Local state discrimination A set of states is unambiguously locally distinguishable only if every subset’s complementary subspace contains a product state. What if is a set of randomly chosen states? All subsets of a set of random states are sets of random states. The span of a set of random states is a random subspace, as is its complement. , etc… Almost all sets of n pure quantum states can be locally unambiguously distinguished iff:

Entangled vs. separable states Simple parameter counting shows that sets of n randomly chosen product states obey the exact same bound: In almost all cases, product states are neither harder nor easier to unambiguously distinguish than entangled states. From a local perspective, ‘entangled’ information is just as easy to unambiguously obtain as ‘separable’ information.