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 \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.