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If we measure the first qubit and obtain $|0\rangle$, what does the second qubit collapses to?

$$ \left| \varphi \right>=\frac{1}{\sqrt{2}} \left| 00 \right> + {\frac{i}{2}} \left| 01\right> - {\frac{1}{2}} \left| 11\right> $$

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    $\begingroup$ Hint: note that $\left|\varphi\right\rangle =\left|0\right\rangle \otimes\left(\frac{1}{\sqrt{2}}\left|0\right\rangle +\frac{i}{2}\left|1\right\rangle \right)+\left|1\right\rangle \otimes\left(-\frac{1}{2}\left|1\right\rangle \right)$ $\endgroup$ – Shai Deshe Jan 4 at 13:27
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If first qubit is $|1\rangle$ there is no other possibility than second qubit to be $|1\rangle$ as well since probability of state $|10\rangle$ is zero. Hence probability of measuring $|1\rangle$ in second qubit is $1$.

In case first qubit is $|0\rangle$ there are two possibe results: $|00\rangle$ or $|01\rangle$. Since probability of state $|00\rangle$ is $\frac{1}{2}$ and probability of state $|01\rangle$ is $\frac{1}{4}$, conditional probabilities that second qubit is $|0\rangle$ is $\frac{2}{3}$ and that it is $|1\rangle$ is $\frac{1}{3}$.

This is about probabilities of measuring $|0\rangle$ or $|1\rangle$ in computational basis, regarding quantum state of second qubit before its measurement, please refer to Shai Dashe comment:

Hint: note that $|\varphi\rangle = |0\rangle \otimes \big(\frac{1}{\sqrt{2}}|0\rangle + \frac{i}{2}|1\rangle\big) + |1\rangle\otimes \big(-\frac{1}{2}|1\rangle\big)$

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    $\begingroup$ You're doing this question backwards. You should always start with probability amplitudes, and keep everything as probability amplitudes for as long as possible. You only convert to probabilities at the last possible moment. $\endgroup$ – DaftWullie Jan 7 at 7:41
  • $\begingroup$ @DaftWullie: However, the results are same in this case. Is it possible that my approach would lead to different results than keeping probabilities aplitudes until last step? $\endgroup$ – Martin Vesely Jan 7 at 8:02
  • $\begingroup$ The problem is that you lose all phase information when trying to talk about the state of the second qubit.. $\endgroup$ – DaftWullie Jan 7 at 8:06
  • $\begingroup$ @DaftWullie: I see, thanks for explanation. $\endgroup$ – Martin Vesely Jan 7 at 8:07

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