Pure states

Let ${\displaystyle \{|{a_{i}}\rangle \}_{i=1}^{n}\subset H_{1}}$ and ${\displaystyle \{|{b_{j}}\rangle \}_{j=1}^{m}\subset H_{2}}$ be orthonormal bases for ${\displaystyle H_{1}}$ and ${\displaystyle H_{2}}$, respectively. A basis for ${\displaystyle H_{1}\otimes H_{2}}$ is then ${\displaystyle \{|{a_{i}}\rangle \otimes |{b_{j}}\rangle \}}$, or in more compact notation ${\displaystyle \{|a_{i}b_{j}\rangle \}}$. 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

${\displaystyle |\psi \rangle =\sum _{i,j}c_{i,j}(|a_{i}\rangle \otimes |b_{j}\rangle )=\sum _{i,j}c_{i,j}|a_{i}b_{j}\rangle ,}$

where ${\displaystyle c_{i,j}}$ is a constant. If ${\displaystyle |\psi \rangle }$ can be written as a simple tensor, that is, in the form ${\displaystyle |\psi \rangle =|\psi _{1}\rangle \otimes |\psi _{2}\rangle }$ with ${\displaystyle |\psi _{i}\rangle }$ a pure state in the i-th space, it is said to be a product state, and, in particular, separable. Otherwise it is called entangled. Note that, even though the notions of product and separable states coincide for pure states, they do not in the more general case of mixed states.

Pure states are entangled if and only if their partial states are not pure. To see this, write the Schmidt decomposition of ${\displaystyle |\psi \rangle }$ as

${\displaystyle |\psi \rangle =\sum _{k=1}^{r_{\psi }}{\sqrt {p_{k}}}(|u_{k}\rangle \otimes |v_{k}\rangle ),}$

where ${\displaystyle {\sqrt {p_{k}}}>0}$ are positive real numbers, ${\displaystyle r_{\psi }}$ is the Schmidt rank of ${\displaystyle |\psi \rangle }$, and ${\displaystyle \{|u_{k}\rangle \}_{k=1}^{r_{\psi }}\subset H_{1}}$ and ${\displaystyle \{|v_{k}\rangle \}_{k=1}^{r_{\psi }}\subset H_{2}}$ are sets of orthonormal states in ${\displaystyle H_{1}}$ and ${\displaystyle H_{2}}$, respectively. The state ${\displaystyle |\psi \rangle }$ is entangled if and only if ${\displaystyle r_{\psi }>1}$. At the same time, the partial state has the form

${\displaystyle \rho _{A}\equiv \operatorname {Tr} _{B}(|\psi \rangle \!\langle \psi |)=\sum _{k=1}^{r_{\psi }}p_{k}\,|u_{k}\rangle \!\langle u_{k}|.}$

It follows that ${\displaystyle \rho _{A}}$ is pure --- that is, is projection with unit-rank --- if and only if ${\displaystyle r_{\psi }=1}$, which is equivalent to ${\displaystyle |\psi \rangle }$ being separable.

Physically, this means that it is not possible to assign a definite (pure) state to the subsystems, which instead ought to be described as statistical ensembles of pure states, that is, as density matrices. A pure state ${\displaystyle \rho =|\psi \rangle \!\langle \psi |}$ is thus entangled if and only if the von Neumann entropy of the partial state ${\displaystyle \rho _{A}\equiv \operatorname {Tr} _{B}(\rho )}$ is nonzero.

Formally, the embedding of a product of states into the product space is given by the Segre embedding.^[1] That is, a quantum-mechanical pure state is separable if and only if it is in the image of the Segre embedding.

For example, in a two-qubit space, where ${\displaystyle H_{1}=H_{2}=\mathbb {C} ^{2}}$, the states ${\displaystyle |0\rangle \otimes |0\rangle }$, ${\displaystyle |0\rangle \otimes |1\rangle }$, ${\displaystyle |1\rangle \otimes |1\rangle }$, are all product (and thus separable) pure states, as is ${\displaystyle |0\rangle \otimes |\psi \rangle }$ with ${\displaystyle |\psi \rangle \equiv {\sqrt {1/3}}|0\rangle +{\sqrt {2/3}}|1\rangle }$. On the other hand, states like ${\displaystyle {\sqrt {1/2}}|00\rangle +{\sqrt {1/2}}|11\rangle }$ or ${\displaystyle {\sqrt {1/3}}|01\rangle +{\sqrt {2/3}}|10\rangle }$ are not separable.

Mixed states

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

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

where

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

Otherwise ${\displaystyle \rho }$ is called an entangled state. We can assume without loss of generality in the above expression that ${\displaystyle \{\rho _{1}^{k}\}}$ and ${\displaystyle \{\rho _{2}^{k}\}}$ 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 ${\displaystyle \{\rho _{1}^{k}\}}$ and ${\displaystyle \{\rho _{2}^{k}\}}$ are themselves states and ${\displaystyle \;\sum _{k}p_{k}=1.}$ 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 ${\displaystyle p_{k}}$, then the state can be expressed just as ${\textstyle \rho =\rho _{1}\otimes \rho _{2},}$ and is called simply separable or product state. One property of the product state is that in terms of entropy,

${\displaystyle S(\rho )=S(\rho _{1})+S(\rho _{2}).}$