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The k-state is given by: $$ |𝐾⟩ = \dfrac{\sqrt{3}|100⟩ − 𝑒^{𝑖π/4}|010⟩ + \sqrt{2}|001⟩}{ \sqrt{6}}$$ I am fairly new to quantum computing and do not have much background in the field. I understand the amplitudes of the state and the phases, however, how would one go about making a circuit for the same using the different quantum gates. I did refer to entangled states such as the W state, but I am unable to understand the underlying principle behind the combination of the gates used to create the circuit. In what direction should I go about constructing the k-state circuit.

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I did the following: let $U$ be the single-qubit unitary $$ U=\frac{1}{\sqrt{3}}\left(\begin{array}{cc} \sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right) $$ This is a partial rotation about the y axis. We use this definition to construct a circuitenter image description here

Here's the thinking: after the Hadamard, I've got $(|000\rangle+|100\rangle)/\sqrt{2}$, so the $|100\rangle$ term is what I need it to be. So, I need to concentrate on making the other two terms without disturbing the first one. If everything else we do is controlled off the first qubit being $|0\rangle$, we achieve that lack of disturbance. Thus, really what we have to concentrate on is a circuit that converts $|00\rangle$ on the last two qubits into $(\sqrt{2}|01\rangle-e^{i\pi/4}|10\rangle)/\sqrt{3}$. We'll be able to get the phase right with a $T$ gate at the end, so you just needed to convert $|00\rangle$ into $(\sqrt{2}|01\rangle-e^{i\pi/4}|10\rangle)/\sqrt{3}$. The gate $U$ is a large part of doing that.

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