---
title: "Bell state"
firstLetter: "B"
publishDate: 2021-03-09
categories:
- Quantum mechanics
- Quantum information

date: 2021-03-09T17:31:29+01:00
draft: false
markup: pandoc
---

# Bell state

In quantum information, the **Bell states** are a set of four two-qubit states
which are simple and useful examples of [quantum entanglement](/know/concept/quantum-entanglement/).
They are given by:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \ket*{\Phi^{\pm}}
            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{0}_B \pm \ket{1}_A \ket{1}_B \Big)
            \\
            \ket*{\Psi^{\pm}}
            &= \frac{1}{\sqrt{2}} \Big( \ket{0}_A \ket{1}_B \pm \ket{1}_A \ket{0}_B \Big)
        \end{aligned}
    }
\end{aligned}$$

Where e.g. $\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B$
is the tensor product of qubit $A$ in state $\ket{0}$ and $B$ in $\ket{1}$.
These states form an orthonormal basis for the two-qubit
[Hilbert space](/know/concept/hilbert-space/).

More importantly, however,
is that the Bell states are maximally entangled,
which we prove here for $\ket*{\Phi^{+}}$.
Consider the following pure [density operator](/know/concept/density-operator/):

$$\begin{aligned}
    \hat{\rho}
    = \ket*{\Phi^{+}} \bra*{\Phi^{+}}
    &= \frac{1}{2} \Big( \ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big) \Big( \bra{0}_A \bra{0}_B + \bra{1}_A \bra{1}_B \Big)
\end{aligned}$$

The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows:

$$\begin{aligned}
    \hat{\rho}_A
    &= \Tr_B(\hat{\rho})
    = \sum_{b = 0, 1} \bra{b}_B \Big( \ket*{\Phi^{+}} \bra*{\Phi^{+}} \Big) \ket{b}_B
    \\
    &= \sum_{b = 0, 1} \Big( \ket{0}_A \braket{b}{0}_B + \ket{1}_A \braket{b}{1}_B \Big)
    \Big( \bra{0}_A \braket{0}{b}_B + \bra{1}_A \braket{1}{b}_B \Big)
    \\
    &= \frac{1}{2} \Big( \ket{0}_A \bra{0}_A + \ket{1}_A \bra{1}_A \Big)
    = \frac{1}{2} \hat{I}
\end{aligned}$$

This result is maximally mixed, therefore $\ket*{\Phi^{+}}$ is maximally entangled.
The same holds for the other three Bell states,
and is equally true for qubit $B$.

This means that a measurement of qubit $A$
has a 50-50 chance to yield $\ket{0}$ or $\ket{1}$.
However, due to the entanglement,
measuring $A$ also has consequences for qubit $B$:

$$\begin{aligned}
    \big| \bra{0}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{0}{0}_B + \braket{0}{1}_A \braket{0}{1}_B \Big)^2
    = \frac{1}{2}
    \\
    \big| \bra{0}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \braket{0}{0}_A \braket{1}{0}_B + \braket{0}{1}_A \braket{1}{1}_B \Big)^2
    = 0
    \\
    \big| \bra{1}_A \bra{0}_B \: \ket*{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{0}{0}_B + \braket{1}{1}_A \braket{0}{1}_B \Big)^2
    = 0
    \\
    \big| \bra{1}_A \bra{1}_B \: \ket*{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \braket{1}{0}_A \braket{1}{0}_B + \braket{1}{1}_A \braket{1}{1}_B \Big)^2
    = \frac{1}{2}
\end{aligned}$$

As an example, if $A$ collapses into $\ket{0}$ due to a measurement,
then $B$ instantly also collapses into $\ket{0}$, never $\ket{1}$,
even if it was not measured.
This was a specific example for $\ket*{\Phi^{+}}$,
but analogous results can be found for the other Bell states.