Categories: Quantum information, Quantum mechanics.

# 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. 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.

More importantly, however, is that the Bell states are maximally entangled, which we prove here for $\ket{\Phi^{+}}$. Consider the following pure 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.