The Bell states · Quantum Mechanics

Intuition

Four entangled states of two qubits make a basis, the Bell basis: $\ket{\Phi^{\pm}}=(\ket{00}\pm\ket{11})/\sqrt{2}$ and $\ket{\Psi^{\pm}}=(\ket{01}\pm\ket{10})/\sqrt{2}$ . Each is maximally entangled — each qubit alone is completely mixed — and $\ket{\Psi^{-}}$ is the singlet. They turn into one another by acting on one qubit alone, with $\sigma_{x}$ , $\sigma_{z}$ or both, so either holder of a shared pair can choose among all four. Measuring in this basis, a Bell measurement, is the step at the heart of teleportation and dense coding.

Four agreements two players can make: always answer the same, or always answer differently, each in two versions that only a joint test can tell apart. The Bell states are these four agreements, and between them they cover every joint state.

The Bell basis

Writing $\ket{00}$ for $\ket{0}\ket{0}$ , and so on:

\begin{aligned}\ket{\Phi^{\pm}}&=\tfrac{1}{\sqrt2}\bigl(\ket{00}\pm\ket{11}\bigr)\\ \ket{\Psi^{\pm}}&=\tfrac{1}{\sqrt2}\bigl(\ket{01}\pm\ket{10}\bigr)\end{aligned}

Properties

The four are orthonormal and span the four-dimensional space of two qubits.
Each has both reduced states equal to $\tfrac12\hat{1}$ : maximally entangled.
$(σ_{x} \otimes \hat{1}) ∣ Φ^{+} ⟩ =$ , and .

One qubit reaches all four

\begin{aligned}&(\sigma_{z}\otimes\hat{1})\ket{\Phi^{+}}=\ket{\Phi^{-}},\quad (\sigma_{x}\otimes\hat{1})\ket{\Phi^{+}}=\ket{\Psi^{+}}\\ &(\sigma_{x}\sigma_{z}\otimes\hat{1})\ket{\Phi^{+}}=-\ket{\Psi^{-}}\end{aligned}

Act on the first qubit of $\ket{\Phi^{+}}$ : $\sigma_{z}$ flips the sign of the term with that qubit 1, $\sigma_{x}$ flips the qubit itself, and doing both does both. The results are the other three Bell states, up to a sign, and a direct check of their overlaps shows the four are orthonormal.

Proof steps

(\sigma_{z}\otimes\hat{1})\ket{\Phi^{+}}=\tfrac{1}{\sqrt2}\bigl(\ket{00}-\ket{11}\bigr)=\ket{\Phi^{-}}

$\sigma_{z}\ket{1}=-\ket{1}$ .

(\sigma_{x}\otimes\hat{1})\ket{\Phi^{+}}=\tfrac{1}{\sqrt2}\bigl(\ket{10}+\ket{01}\bigr)=\ket{\Psi^{+}}

$\sigma_{x}$ swaps $\ket{0}$ and $\ket{1}$ .

(\sigma_{x}\sigma_{z}\otimes\hat{1})\ket{\Phi^{+}}=\tfrac{1}{\sqrt2}\bigl(\ket{10}-\ket{01}\bigr)=-\ket{\Psi^{-}}

First $\sigma_{z}$ , then $\sigma_{x}$ .

\braket{\Phi^{+}|\Phi^{-}}=\tfrac12(1-1)=0,\qquad \braket{\Phi^{\pm}|\Psi^{\pm}}=0

$\Phi$ and $\Psi$ share no basis term; the two $\Phi$ cancel.

Applications

Practice

Four Maximally Entangled States

The Bell states pair the two qubits either equally $(\Phi)$ or oppositely $(\Psi)$ , with a plus or a minus sign between the two terms.

\begin{aligned}\ket{\Phi^{\pm}}&=\tfrac{1}{\sqrt2}\bigl(\ket{00}\pm\ket{11}\bigr)\\ \ket{\Psi^{\pm}}&=\tfrac{1}{\sqrt2}\bigl(\ket{01}\pm\ket{10}\bigr)\end{aligned}

Try it

Which Bell state is the singlet?