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Hi, How to convert $|100\rangle$ 3-qubit quantum state into $\frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$ state using only Hadamard and CNOT gates? Also, is output state an entangled one?

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If the first qubit is in state $|1\rangle$, i.e. the input state $|100\rangle$ then resulting GHZ state is $\frac{1}{\sqrt{2}}(|000\rangle - |111\rangle)$, i.e. the phase is $\pi$. To have phase $0$, $Z$ has to be applied but this gate is not allowed. But you can use controlled $Z$ which is composed only with $H$ and $CNOT$. The circuit is this

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A part before orange line produces the state $\frac{1}{\sqrt{2}}(|000\rangle - |111\rangle)$ from $|100\rangle$, a part after the line is controlled $Z$.

Here is a resulting state from IBM Q:

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As you can see, the phase is correct.

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The upper figure shows the quantum circuit to do so, the first $X$ gate is to let the first qubit be in the $|1\rangle$ state coherently.

The state $\frac{|000\rangle+|111\rangle}{\sqrt2}$ is the GHZ state, it is one of the most famous tri-qubit entangled state (another one is the W state), you may see Wikipedia for detail.

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  • $\begingroup$ Hi Yitian, the circuit above produces |000>-|111>, but, I need |000>+|111> state. Phase is incorrect. I understand the confusion, considering you're viewing only probabilities. So, how to generate |000>+|111> ?? $\endgroup$ – Sscr Nov 15 '20 at 7:28
  • $\begingroup$ Note that $X$ gate is forbidden, only H and CNOT are allowed. $\endgroup$ – Martin Vesely Nov 15 '20 at 7:39
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    $\begingroup$ Notice that if the X gate were allowed then it is very simple.... all you have to do is to bit flip the qubit to get it back to the state $|00\cdots 0\rangle$ and generate the GHZ states from there.. $\endgroup$ – KAJ226 Nov 15 '20 at 7:53
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    $\begingroup$ Also if the Z gate is available then you don't have to do controlled Z either... you just need to apply the $Z$ gate to the top qubit to fix the phase. But without the Z gate, @MartinVesely is right! +1 $\endgroup$ – KAJ226 Nov 15 '20 at 8:00
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    $\begingroup$ @MartinVesely You are absolutely correct. I just made my comments in a general setting... nothing more. :) $\endgroup$ – KAJ226 Nov 16 '20 at 16:19

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