---
title: "Quantum entanglement"
date: 2021-03-07
categories:
- Physics
- Quantum mechanics
- Quantum information
layout: "concept"
---

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 subsystems'
[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ 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](/know/concept/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:

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


## References
1.  J.B. Brask,
    *Quantum information: lecture notes*,
    2021, unpublished.