Entangled states

Intuition

A pure state of two parts that is not a product is entangled: the whole has a definite state and neither part has one of its own. The singlet is the example to keep in mind. Each spin alone is completely mixed, yet measured along any common axis the two always disagree. Entanglement is more than correlation: a separable mixture can make two spins opposite along one chosen axis, but only an entangled state makes them opposite along every axis at once. Measuring one part changes what can be predicted about the other — once $A$ is found up along $z$ , $B$ is certainly down along $z$ — though, as this chapter shows, it changes nothing $B$ can notice on his own.

Two people who give opposite answers to any yes-or-no question, even ones they could not have prepared for. Answers agreed in advance could cover a list of questions; covering every question at once is the mark of entanglement, and the chapter turns that difference into an experiment.

Entangled pure states

A pure state $\ket{\psi}$ of two parts $A$ and $B$ is entangled when no states $\ket{\alpha}$ of $A$ and of make it a product; with :

\begin{aligned}\ket{\psi}\ \text{entangled}&\iff\ket{\psi}\neq\ket{\alpha}\ket{\beta}\\ &\iff\operatorname{tr}\hat{\rho}_{A}^{2}<1\end{aligned}

Properties

For a pure state, not a product, Schmidt rank above one and a mixed reduced state are the same condition.
The singlet $\tfrac{1}{\sqrt2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})$ is unchanged when both spins are rotated together, so its spins come out opposite along every common axis.

Measuring one spin of the singlet

\bigl(\ket{\uparrow}\bra{\uparrow}\otimes\hat{1}\bigr)\ket{\Psi^{-}}=\tfrac{1}{\sqrt2}\ket{\uparrow\downarrow}

Spin $A$ alone is completely mixed, so up and down along $z$ each come with probability one half. The projector on up for $A$ keeps the one term of the singlet with $A$ up, and that term has $B$ down: after the result, $B$ is down along for certain, and half up, half down along .

Proof steps

\ket{\Psi^{-}}=\tfrac{1}{\sqrt2}\bigl(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\bigr)

The singlet.

P(A{=}\uparrow)=\bra{\uparrow}\hat{\rho}_{A}\ket{\uparrow}=\tfrac12

$A$ alone is $\tfrac12\hat{1}$ .

\bigl(\ket{\uparrow}\bra{\uparrow}\otimes\hat{1}\bigr)\ket{\Psi^{-}}=\tfrac{1}{\sqrt2}\ket{\uparrow\downarrow}

The projector keeps the term with $A$ up.

\ket{\uparrow}\ket{\downarrow}

Normalised: after the result the pair is a product, with $B$ down along $z$ .

P(B{=}\uparrow_{z})=0,\qquad P(B{=}\uparrow_{x})=\bigl|\braket{+x|\downarrow}\bigr|^{2}=\tfrac12

What can now be predicted about $B$ .

Applications

Practice

A pure state of two parts is entangled when it cannot be written as a state of one part times a state of the other. For two qubits the test is $c_{00}c_{11}=c_{01}c_{10}$ : entangled when it fails.

Try it

Which of these states of two qubits is entangled?