Ex4 18

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.18}$

Show that if there is no measurement of one of the qubits that gives a single result with certainty, then the two qubits are entangled.

page revision: 0, last edited: 18 Nov 2012 18:29

Let's prove the contrapositive: If two qubits are

notentangled, then thereisa measurement of one of the qubits that gives a single result with certainty.If two qubits are not entangled, then we can write their state as a tensor product. In general, we can write an arbitrary unentangled two-qubit state as $\ket{\psi} = (a\ket{0}+b\ket{1}) \otimes (c\ket{0}+d\ket{1})$. Let $\ket{v} = a\ket{0}+b\ket{1}$ be the state of the first qubit. Then, if we measure the first qubit in the orthonormal basis $\{\ket{v}, \ket{v^\perp} \}$, the result will be $\ket{v}$ with certainty. So we have proven the desired result.

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