From 540d23bff03bedbc8f68287d71c8b5e7dc54b054 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Mon, 8 Mar 2021 15:04:06 +0100
Subject: Expand knowledge base

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+---
+title: "Quantum entanglement"
+firstLetter: "Q"
+publishDate: 2021-03-07
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-07T15:36:15+01:00
+draft: false
+markup: pandoc
+---
+
+# 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 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:
+
+$$\begin{aligned}
+    \expval*{\hat{O}_A}
+    = \Tr\big(\hat{\rho} \hat{O}_A\big)
+    = \Tr_A\big(\Tr_B(\hat{\rho} \hat{O}_A)\big)
+    = \Tr_A\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big)
+    = \Tr_A\big(\hat{\rho}_A \hat{O}_A\big)
+\end{aligned}$$
+
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