Categories: Physics, Quantum information, Quantum mechanics.

# Quantum entanglement

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. 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:

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:

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:

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:

$\begin{aligned} \expval{\hat{O}_A} = \Tr\!\big(\hat{\rho} \hat{O}_A\big) = \Tr_A\!\big(\Tr_B(\hat{\rho} \hat{O}_A)\big) = \Tr_A\!\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big) = \Tr_A\!\big(\hat{\rho}_A \hat{O}_A\big) \end{aligned}$## References

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*Quantum information: lecture notes*, 2021, unpublished.