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While going through the IBM qiskit textbook online, I came across the following question in section 2.2:

Write the state: $ |\Psi⟩ =\frac{1}{\sqrt{2}}|00⟩+\frac{i}{\sqrt{2}}|01⟩$ as two separate qubits.

I understand tensor products with qubits, but I don't know how to even begin this problem. Does anyone have some advice on how to separate a state into its constituent qubits?

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As is the case with ordinary multiplication, tensor product distributes over addition, so we can pull $|0\rangle$ on the first qubit out in front

$$ \begin{align} |\Psi⟩ &= \frac{1}{\sqrt{2}}|\color{red}{0}0\rangle+\frac{i}{\sqrt{2}}|\color{red}{0}1\rangle \\ &= \frac{1}{\sqrt{2}}\color{red}{|0\rangle}\otimes|0\rangle+\frac{i}{\sqrt{2}}\color{red}{|0\rangle}\otimes|1\rangle \\ &= \color{red}{|0\rangle}\otimes\left(\frac{1}{\sqrt{2}}|0\rangle+\frac{i}{\sqrt{2}}|1\rangle\right) \\ &= \color{red}{|0\rangle}\left(\frac{1}{\sqrt{2}}|0\rangle+\frac{i}{\sqrt{2}}|1\rangle\right) \end{align} $$

and what remains in parenthesis is the state of the second qubit.

Note that people generally tend to make tensor product signs $\otimes$ implicit. I marked them explicitly to highlight the distributive law familiar from ordinary multiplication.

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Giving $|\psi \rangle = \dfrac{1}{\sqrt{2}}|00\rangle + \dfrac{i}{\sqrt{2}}|01\rangle$ we can see that the first qubit is in the state $|0\rangle$ so we can rewrite the state $|\psi\rangle$ as a tensor product:

$$ |\psi \rangle = |0\rangle \otimes \bigg( \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}}\bigg)$$

So the first qubit is in the state $|0\rangle$ and the second qubit is in the state $\dfrac{|0\rangle + i|1\rangle}{\sqrt{2}}$.

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  • $\begingroup$ That makes a lot of sense, actually. Thanks! $\endgroup$
    – aikky
    Jan 11, 2021 at 16:19
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Other important methods to check if a state is a separable or entangled are the Peres-Horodecki criterion and Schmidt decomposition.

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  • $\begingroup$ Could you please the derivation for this decomposition. $\endgroup$
    – RSW
    Aug 16, 2022 at 7:52

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