---
title: "Bell state"
date: 2021-03-09
categories:
- Quantum mechanics
- Quantum information
layout: "concept"
---

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:

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

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 $B$:

$$\begin{aligned}
    \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2
    = \frac{1}{2}
    \\
    \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2
    = 0
    \\
    \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2
    = 0
    \\
    \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
    &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2
    = \frac{1}{2}
\end{aligned}$$

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.


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    2021, unpublished.