# Factoring some qubits with Kronecker product

I have this quantum state: $$|\phi \rangle =\frac{1}{4\sqrt{2}}|000\rangle +\frac{3-\sqrt{5}i}{8}|001\rangle +\frac{1}{4\sqrt{2}}|010\rangle +\frac{3-\sqrt{5}i}{8}|011\rangle \\+\frac{1}{4\sqrt{2}}|100\rangle +\frac{3-\sqrt{5}i}{8}|101\rangle +\frac{1}{4\sqrt{2}}|110\rangle +\frac{3-\sqrt{5}i}{8}|111\rangle$$ And I have to use Kronecker product to factorize that state into individual qubits (Image below). I'm not sure where to start, any hint would be appreciated

$$|\phi \rangle =\left| Q_2 \right. \rangle \left| Q_1 \right. \rangle \left| Q_0 \right. \rangle =\left| Q_2Q_1Q_0 \right. \rangle$$

• Partial trace your $|\phi\rangle$ for each qubit, if it can be written in the product state, then it's the answer. Aug 19, 2022 at 2:38
• Hopefully @narip can expand his suggestion a bit more, but consider playing around with products of $|+>$ with itself to see what it looks like. Aug 20, 2022 at 2:43
• With OP's notation, if $|\phi\rangle$ can be written as direct product form, we then can calculate it with $$Tr_{10}\left( |\phi \rangle \langle \phi | \right) =\left| Q_2 \right. \rangle \langle Q_2| \\ Tr_{20}\left( |\phi \rangle \langle \phi | \right) =|Q_1\rangle \langle Q_1| \\ Tr_{21}\left( |\phi \rangle \langle \phi | \right) =|Q_0\rangle \langle Q_0|.$$ Aug 20, 2022 at 3:34

Let $$a=\frac{1}{4 \sqrt{2}}$$ and $$b = \frac{3-\sqrt{5}i}{8}$$. Then we can factor out $$|0\rangle$$ from the first 4 terms, and we can factor out $$|1\rangle$$ from the last 4 terms. This gives
\begin{align*} |0\rangle \left(a|00\rangle + b|01\rangle + a|10\rangle + b|11\rangle \right ) + \\ |1\rangle \left(a|00\rangle + b|01\rangle + a|10\rangle + b|11\rangle \right). \end{align*} Now, factor out the common factor and you get $$\left(|0\rangle + |1\rangle \right) \left(a|00\rangle + b|01\rangle + a|10\rangle + b|11\rangle \right).$$ Repeat the same procedure on the right-most factor.
Hint: the first two qubits will be in the state $$|+\rangle |+\rangle$$.