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I'm confused about the last complete sentence in the following paragraphs. If ab=0, that means either a or b equals to 0. As a result, doesn't $|\psi\rangle|\psi\rangle$ equal to either $b^2|11\rangle$ or $a^2|00\rangle$? If so, how is this related to $a|00\rangle + b|11\rangle$? Also, how do $|\psi\rangle|\psi\rangle = b^2|11\rangle$ or $=a^2|00\rangle$ show that the quantum state input is copied?

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Claire
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1 Answers1

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Note that if $|\psi \rangle = |0\rangle$ or $|\psi\rangle = |1\rangle$ then

$$CNOT|\psi \rangle|0\rangle = CNOT|0 \rangle|0\rangle = CNOT |0 0\rangle = |00\rangle = |0\rangle_{\textrm{original qubit}}|0\rangle_{\textrm{copied qubit}} $$

and

$$CNOT|\psi \rangle|0\rangle = CNOT|1 \rangle|0\rangle = CNOT|10\rangle = |11\rangle = |1\rangle_{\textrm{original qubit}}|1\rangle_{\textrm{copied qubit}}$$

This says that if the qubit is in a definite state $|0\rangle$ or $|1\rangle$ then we can use the CNOT gate to copy the state of the qubit. This is not very surprising since the qubit in this case behave just like a classical bit...

KAJ226
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  • Thank you for the reply! I'm just a bit confused. We know that |ψ⟩|ψ⟩ equal to either b^2|11⟩ or a^2|00⟩ and the output from our circuit is either a|00⟩ or b|11⟩ (since either a or b is 0). The amplitudes are apparently different: one is the square of the other. How does this still count as successfully copied the state? – Claire Aug 05 '21 at 15:35
  • Note that in this case, either $a= 1$ or $b =1$. This is because the state must be normalized in the first place. Thus, $a^2 = a = 1$ or $b^2 = b = 1$. – KAJ226 Aug 05 '21 at 16:55