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Equation 1.31 in Quantum Computation and Quantum Information a textbook by Isaac Chuang and Michael Nielsen is as follows,

$\left|\psi_2 \right> = \frac{1}{2}[\alpha(\left|0 \right>+\left|1 \right>)(\left|00\right>+\left|11\right>) + \beta(\left|0 \right>-\left|1 \right>)(\left|10\right>+\left|01\right>)]$

The book claims that after rearranging terms we obtain equation 1.32 as follows

$\left|\psi_2\right>=\frac{1}{2}[\left|00\right>(\alpha\left|0\right>+\beta\left|1\right>)+\left|01\right>(\alpha\left|1\right>+\beta\left|0\right>)+\left|10\right>(\alpha\left|0\right>-\beta\left|1\right>)+\left|11\right>(\alpha\left|1\right>-\beta\left|0\right>)]$

I don't know how to derive the second equation from the first. Can someone explain the steps please ?

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  • $\begingroup$ This is done by literally rearranging the terms. Open the brackets to get 8 terms, rearrange them so that terms with the same state of the first two qubits are next to each other, group them together. Could you please provide more details on what part of the derivation you're having trouble with? $\endgroup$ – Mariia Mykhailova Jul 23 at 6:59
  • $\begingroup$ Thanks for your comment. I did not know that $\left|001\right>$ was equal to $\left|0\right>\left|01\right>$, and that's why I could not follow along. Perhaps I should have mentioned this or shown my work. $\endgroup$ – rranjik Jul 23 at 15:13
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The tensor product is distributive, meaning that you can write $$ 2|\psi_2\rangle=\alpha|000\rangle+\alpha|100\rangle+\alpha|011\rangle+\alpha|111\rangle+\beta|010\rangle+\beta|001\rangle-\beta|110\rangle-\beta|101\rangle. $$ Next you reorder the terms in the sum, $$ 2|\psi_2\rangle=\alpha|000\rangle+\beta|001\rangle+\alpha|011\rangle+\beta|010\rangle+\alpha|100\rangle-\beta|101\rangle+\alpha|111\rangle-\beta|110\rangle. $$ Again, we can use distributivity to group these terms together based on the state of the first two qubits $$ 2|\psi_2\rangle=|00\rangle(\alpha|0\rangle+\beta|1\rangle)+|01\rangle(\alpha|1\rangle+\beta|0\rangle)+|10\rangle(\alpha|0\rangle-\beta|1\rangle)+|11\rangle(\alpha|1\rangle-\beta|0\rangle), $$ as required

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  • $\begingroup$ Thanks Daft! I did not know that $\left|000\right>$ was equal to $\left|0\right>\left|00\right>$. $\endgroup$ – rranjik Jul 23 at 15:11

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