Separable state

From Wikipedia, the free encyclopedia

In quantum mechanics, separable quantum states are states without quantum entanglement.

Separable pure states

For simplicity, the following assumes all relevant state spaces are finite dimensional. First, consider separability for pure states.

Let $H_{1}$ and $H_{2}$ be quantum mechanical state spaces, that is, finite dimensional Hilbert spaces with basis states $\{|{a_{i}}\rangle \}_{{i=1}}^{n}$ and $\{|{b_{j}}\rangle \}_{{j=1}}^{m}$ , respectively. By a postulate of quantum mechanics, the state space of the composite system is given by the tensor product

$H_{1}\otimes H_{2}$

with base states $\{|{a_{i}}\rangle \otimes |{b_{j}}\rangle \}$ , or in more compact notation $\{|a_{i}b_{j}\rangle \}$ . From the very definition of the tensor product, any vector of norm 1, i.e. a pure state of the composite system, can be written as

$|\psi \rangle =\Sigma _{{i,j}}c_{{i,j}}|a_{i}\rangle \otimes |b_{j}\rangle =\Sigma _{{i,j}}c_{{i,j}}|a_{i}b_{j}\rangle$

If a pure state $|\psi \rangle \in H_{1}\otimes H_{2}$ can be written in the form $|\psi \rangle =|\psi _{1}\rangle \otimes |\psi _{2}\rangle$ where $|\psi _{i}\rangle$ is a pure state of the i-th subsystem, it is said to be separable. Otherwise it is called entangled. When a system is in an entangled pure state, it is not possible to assign states to its subsystems. This will be true, in the appropriate sense, for the mixed state case as well.

Formally, the embedding of a product of states into the product space is given by the Segre embedding. That is, a quantum-mechanical pure state is separable if and only if it is in the image of the Segre embedding.

The above discussion can be extended to the case of when the state space is infinite dimensional with virtually nothing changed.

Separability for mixed states

Consider the mixed state case. A mixed state of the composite system is described by a density matrix $\rho$ acting on $H_{1}\otimes H_{2}$ . ρ is separable if there exist $p_{k}\geq 0$ , $\{\rho _{1}^{k}\}$ and $\{\rho _{2}^{k}\}$ which are mixed states of the respective subsystems such that

$\rho =\sum _{k}p_{k}\rho _{1}^{k}\otimes \rho _{2}^{k}$

where

$\;\sum _{k}p_{k}=1.$

Otherwise $\rho$ is called an entangled state. We can assume without loss of generality in the above expression that $\{\rho _{1}^{k}\}$ and $\{\rho _{2}^{k}\}$ are all rank-1 projections, that is, they represent pure ensembles of the appropriate subsystems. It is clear from the definition that the family of separable states is a convex set.

Notice that, again from the definition of the tensor product, any density matrix, indeed any matrix acting on the composite state space, can be trivially written in the desired form, if we drop the requirement that $\{\rho _{1}^{k}\}$ and $\{\rho _{2}^{k}\}$ are themselves states and $\;\sum _{k}p_{k}=1.$ If these requirements are satisfied, then we can interpret the total state as a probability distribution over uncorrelated product states.

In terms of quantum channels, a separable state can be created from any other state using local actions and classical communication while an entangled state cannot.

When the state spaces are infinite dimensional, density matrices are replaced by positive trace class operators with trace 1, and a state is separable if it can be approximated, in trace norm, by states of the above form.

If there is only a single non-zero $p_{k}$ , then the state is called simply separable (or it is called a "product state").

Extending to the multipartite case

The above discussion generalizes easily to the case of a quantum system consisting of more than two subsystems. Let a system have n subsystems and have state space $H=H_{1}\otimes \cdots \otimes H_{n}$ . A pure state $|\psi \rangle \in H$ is separable if it takes the form

$|\psi \rangle =|\psi _{1}\rangle \otimes \cdots \otimes |\psi _{n}\rangle .$

Similarly, a mixed state ρ acting on H is separable if it is a convex sum

$\rho =\sum _{k}p_{k}\rho _{1}^{k}\otimes \cdots \otimes \rho _{n}^{k}.$

Or, in the infinite dimensional case, ρ is separable if it can be approximated in the trace norm by states of the above form.

Separability criterion

The problem of deciding whether a state is separable in general is sometimes called the separability problem in quantum information theory. It is considered to be a difficult problem. It has been shown to be NP-hard.^[1]^[2] Some appreciation for this difficulty can be obtained if one attempts to solve the problem by employing the direct brute force approach, for a fixed dimension. We see that the problem quickly becomes intractable, even for low dimensions. Thus more sophisticated formulations are required. The separability problem is a subject of current research.

A separability criterion is a necessary condition a state must satisfy to be separable. In the low dimensional (2 X 2 and 2 X 3) cases, the Peres-Horodecki criterion is actually a necessary and sufficient condition for separability. Other separability criteria include the range criterion and reduction criterion.

Characterization via algebraic geometry

Quantum mechanics may be modelled on a projective Hilbert space, and the categorical product of two such spaces is the Segre embedding. In the bipartite case, a quantum state is separable if and only if it lies in the image of the Segre embedding.

References

↑ Gurvits, L., Classical deterministic complexity of Edmonds’ problem and quantum entanglement, in Proceedings of the 35th ACM Symposium on Theory of Computing, ACM Press, New York, 2003.
↑ Sevag Gharibian, Strong NP-Hardness of the Quantum Separability Problem, Quantum Information and Computation, Vol. 10, No. 3&4, pp. 343-360, 2010. arXiv:0810.4507.