# Getting rid of extra qubits without changing phase

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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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}