Categories: Physics, Quantum information, Quantum mechanics.

Consider a composite quantum system which consists of two subsystems \(A\) and \(B\), respectively with basis states \(\ket{a_n}\) and \(\ket{b_n}\). All accessible states of the sytem \(\ket{\Psi}\) lie in the tensor product of the subsystemsâ€™ Hilbert spaces \(\mathbb{H}_A\) and \(\mathbb{H}_B\):

\[\begin{aligned} \ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B \end{aligned}\]

A subset of these states can be written as the tensor product (i.e.Â Kronecker product in a basis) of a state \(\ket{\alpha}\) in \(A\) and a state \(\ket{\beta}\) in \(B\), often abbreviated as \(\ket{\alpha} \ket{\beta}\):

\[\begin{aligned} \ket{\Psi} = \ket{\alpha} \ket{\beta} = \ket{\alpha} \otimes \ket{\beta} \end{aligned}\]

The states that can be written in this way are called **separable**, and states that cannot are called **entangled**. Therefore, we are dealing with **quantum entanglement** if the state of subsystem \(A\) cannot be fully described independently of the state of subsystem \(B\), and vice versa.

To detect and quantify entanglement, we can use the density operator \(\hat{\rho}\). For a pure ensemble in a given (possibly entangled) state \(\ket{\Psi}\), \(\hat{\rho}\) is given by:

\[\begin{aligned} \hat{\rho} = \ket{\Psi} \bra{\Psi} \end{aligned}\]

From this, we would like to extract the corresponding state of subsystem \(A\). For that purpose, we define the **reduced density operator** \(\hat{\rho}_A\) of subsystem \(A\) as follows:

\[\begin{aligned} \boxed{ \hat{\rho}_A = \Tr_B(\hat{\rho}) = \sum_m \bra{b_m} \Big( \hat{\rho} \Big) \ket{b_m} } \end{aligned}\]

Where \(\Tr_B(\hat{\rho})\) is called the **partial trace** of \(\hat{\rho}\), which basically eliminates subsystem \(B\) from \(\hat{\rho}\). For a pure composite state \(\ket{\Psi}\), the resulting \(\hat{\rho}_A\) describes a pure state in \(A\) if \(\ket{\Psi}\) is separable, else, if \(\ket{\Psi}\) is entangled, it describes a mixed state in \(A\). In the former case we simply find:

\[\begin{aligned} \boxed{ \ket{\Psi} = \ket{\alpha} \otimes \ket{\beta} \quad \implies \quad \hat{\rho}_A = \ket{\alpha} \bra{\alpha} } \end{aligned}\]

We call \(\ket{\Psi}\) **maximally entangled** if its reduced density operators are **maximally mixed**, where \(N\) is the dimension of \(\mathbb{H}_A\) and \(\hat{I}\) is the identity matrix:

\[\begin{aligned} \hat{\rho}_A = \frac{1}{N} \hat{I} \end{aligned}\]

Suppose that we are given an entangled pure state \(\ket{\Psi} \neq \ket{\alpha} \otimes \ket{\beta}\). Then the partial traces \(\hat{\rho}_A\) and \(\hat{\rho}_B\) of \(\hat{\rho} = \ket{\Psi} \bra{\Psi}\) are mixed states with the same probabilities \(p_n\) (assuming \(\mathbb{H}_A\) and \(\mathbb{H}_B\) have the same dimensions, which is usually the case):

\[\begin{aligned} \hat{\rho}_A = \Tr_B(\hat{\rho}) = \sum_n p_n \ket{a_n} \bra{a_n} \qquad \quad \hat{\rho}_B = \Tr_A(\hat{\rho}) = \sum_n p_n \ket{b_n} \bra{b_n} \end{aligned}\]

There exists an orthonormal choice of the subsystem basis states \(\ket{a_n}\) and \(\ket{b_n}\), such that \(\ket{\Psi}\) can be written as follows, where \(p_n\) are the probabilities in the reduced density operators:

\[\begin{aligned} \ket{\Psi} = \sum_n \sqrt{p_n} \Big( \ket{a_n} \otimes \ket{b_n} \Big) \end{aligned}\]

This is the **Schmidt decomposition**, and the **Schmidt number** is the number of nonzero terms in the summation, which can be used to determine if the state \(\ket{\Psi}\) is entangled (greater than one) or separable (equal to one).

By looking at the Schmidt decomposition, we can notice that, if \(\hat{O}_A\) and \(\hat{O}_B\) are the subsystem observables with basis eigenstates \(\ket{a_n}\) and \(\ket{b_n}\), then measurement results of these operators will be perfectly correlated across \(A\) and \(B\). This is a general property of entangled systems, but beware: correlation does not imply entanglement!

But what if the composite system is in a mixed state \(\hat{\rho}\)? The state is separable if and only if:

\[\begin{aligned} \boxed{ \hat{\rho} = \sum_m p_m \Big( \hat{\rho}_A \otimes \hat{\rho}_B \Big) } \end{aligned}\]

Where \(p_m\) are probabilities, and \(\hat{\rho}_A\) and \(\hat{\rho}_B\) can be any subsystem states. In reality, it is very hard to determine, using this criterium, whether an arbitrary given \(\hat{\rho}\) is separable or not.

As a final side note, the expectation value of an obervable \(\hat{O}_A\) acting only on \(A\) is given by:

\[\begin{aligned} \expval*{\hat{O}_A} = \Tr\big(\hat{\rho} \hat{O}_A\big) = \Tr_A\big(\Tr_B(\hat{\rho} \hat{O}_A)\big) = \Tr_A\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big) = \Tr_A\big(\hat{\rho}_A \hat{O}_A\big) \end{aligned}\]

- J.B. Brask,
*Quantum information: lecture notes*, 2021, unpublished.

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