Separable state

In quantum mechanics, separable quantum states are states without quantum entanglement. For simplicity, the following assumes all relevant state spaces are finite-dimensional. First, consider separability for pure states. Let H 1 {displaystyle H_{1}} and H 2 {displaystyle H_{2}} be quantum mechanical state spaces, that is, finite-dimensional Hilbert spaces with basis states { | a i ⟩ } i = 1 n {displaystyle {|{a_{i}} angle }_{i=1}^{n}} and { | b j ⟩ } j = 1 m {displaystyle {|{b_{j}} angle }_{j=1}^{m}} , respectively. By a postulate of quantum mechanics, the state space of the composite system is given by the tensor product with base states { | a i ⟩ ⊗ | b j ⟩ } {displaystyle {|{a_{i}} angle otimes |{b_{j}} angle }} , or in more compact notation { | a i b j ⟩ } {displaystyle {|a_{i}b_{j} angle }} . 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 where c i , j {displaystyle c_{i,j}} is a constant. If a pure state | ψ ⟩ ∈ H 1 ⊗ H 2 {displaystyle |psi angle in H_{1}otimes H_{2}} can be written in the form | ψ ⟩ = | ψ 1 ⟩ ⊗ | ψ 2 ⟩ {displaystyle |psi angle =|psi _{1} angle otimes |psi _{2} angle } where | ψ i ⟩ {displaystyle |psi _{i} angle } 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.