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:

Φ±=12(0A0B±1A1B)Ψ±=12(0A1B±1A0B)\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. 0A1B=0A1B\ket{0}_A \ket{1}_B = \ket{0}_A \otimes \ket{1}_B is the tensor product of qubit AA in state 0\ket{0} and BB in 1\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 Φ+\ket{\Phi^{+}}. Consider the following pure density operator:

ρ^=Φ+Φ+=12(0A0B+1A1B)(0A0B+1A1B)\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 ρ^A\hat{\rho}_A of qubit AA is then calculated as follows:

ρ^A=TrB(ρ^)=b=0,1bB(Φ+Φ+)bB=b=0,1(0Ab0B+1Ab1B)(0A0bB+1A1bB)=12(0A0A+1A1A)=12I^\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 Φ+\ket{\Phi^{+}} is maximally entangled. The same holds for the other three Bell states, and is equally true for qubit BB.

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

0A ⁣0BΦ+2=12(00A00B+01A01B)2=120A ⁣1BΦ+2=12(00A10B+01A11B)2=01A ⁣0BΦ+2=12(10A00B+11A01B)2=01A ⁣1BΦ+2=12(10A10B+11A11B)2=12\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 AA collapses into 0\ket{0} due to a measurement, then BB instantly also collapses into 0\ket{0}, never 1\ket{1}, even if it was not measured. This was a specific example for Φ+\ket{\Phi^{+}}, but analogous results can be found for the other Bell states.


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 J.B. Brask, Quantum information: lecture notes, 2021, unpublished.