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---
title: "Bell state"
sort_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.
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