Exercise 3.12 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.12}$

Give an example of a two-qubit state which is a superposition with respect to the standard basis but which is not entangled.

page revision: 0, last edited: 11 Dec 2011 23:24

Consider a state

$\ket{\psi} = \frac{1}{2}(\ket{00} + \ket{01} + \ket{10} + \ket{11} )$

By definition, this state is a superposition WRT the standard basis. It's easy to see that

$\ket{\psi} = (\frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1}) \otimes (\frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1})$

Therefore, $\ket{\psi}$ is not entangled.

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