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If I have the state $\frac{1}{\sqrt2}|0000\rangle + \frac{e^{i\phi}}{\sqrt2}|1111\rangle$ and need to get $\frac{1}{\sqrt2}|0\rangle + \frac{e^{i\phi}}{\sqrt2}|1\rangle$, does anyone have any ideas about how I might go about getting the state?

My first thought was to measure the last three qubits, but I don't think that's correct as I believe it would cause the first qubit to collapse to $|0\rangle$ or $|1\rangle$ due to entanglement.

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  • $\begingroup$ Why do you want to delete the question? I have rollbacked to the previous version of the question. If you still want to delete it, you should see a delete option at the bottom of your question below the tags. You can read more here. $\endgroup$
    – FDGod
    Apr 1 at 0:15

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Just apply $\text{CNOT}$ from the first qubit to every other qubit.

Let $|\psi\rangle$ be the initial state.

$$|\psi\rangle = \frac{1}{\sqrt{2}}|0000\rangle + \frac{e^{i\phi}}{\sqrt{2}}|1111\rangle\,.$$

Now, applying $\text{CNOT}$s gives us

\begin{align} \bigg(\text{CNOT}_{1\to 2} \cdot \text{CNOT}_{1\to 3} \cdot \text{CNOT}_{1\to 4}\bigg)|\psi\rangle &= \frac{1}{\sqrt{2}}|0000\rangle + \frac{e^{i\phi}}{\sqrt{2}}|1000\rangle\,,\\ &= \bigg(\frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\phi}}{\sqrt{2}}|1\rangle\bigg)|000\rangle \,. \end{align}

Now, you can just dump qubits 2,3 and 4 and keep qubit 1, which has the state you wanted.

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