Let's say two qubits are both in $|+\rangle$ state. We need to find $a_1$, $a_2$, $a_3$, and $a_4$ in $|\phi\rangle = a_1|00\rangle + a_2|01\rangle + a_3|10\rangle + a_4|11\rangle$, how do we find these amplitudes? How do we do it in general case, when each of the qubits are not necessarily in $|+\rangle$ state, but in some $|?\rangle$ state?
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asked Jun 8, 2018 at 10:15
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Start by writing out what the $|+\rangle$ state actually is: $$ |+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) $$ So, two qubits in the state $|+\rangle$ are in the state $$ |+\rangle|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)\otimes \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) $$ You then need to expand this out, making use of the distributivity of the tensor product, and match up with the specified form of $|\phi\rangle$. Since this sounds a bit like a homework problem, I'm not going to do that for you explicitly. (If you've tried something and got stuck, show us what you tried!)
The general case is absolutely equivalent, you just replace $|+\rangle|+\rangle$ with something like $$ (b_0|0\rangle+b_1|1\rangle)\otimes (b_2|0\rangle+b_3|1\rangle) $$
answered Jun 8, 2018 at 10:21
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$\begingroup$ I tried multiplying each of the amplitudes, but for |00⟩ I got the amplitude 1/2 (obviously, the rest of the amplitudes would be the same), so the sum of the squares of the modules of the amplitudes would not equal 1, so I knew I did something wrong. I tried normalizing, but I didn't know if I was doing it right. $\endgroup$ Commented Jun 8, 2018 at 10:32
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2$\begingroup$ @ArchilZhvania $|\frac{1}{2}|^2+|\frac{1}{2}|^2+|\frac{1}{2}|^2+|\frac{1}{2}|^2=1$. $\endgroup$ Commented Jun 8, 2018 at 10:35
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$\begingroup$ Does the generalization question want $\mid ? \rangle \mid ? \rangle$ with the same ? so $b_2=b_0$ and $b_3=b_1$? $\endgroup$– AHusainCommented Jun 9, 2018 at 0:18