Qubit measurement of the state $\frac{1}{\sqrt2}|00\rangle+\frac{i}{2}|01\rangle-\frac{1}{2}|11\rangle$

Question

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

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

Answer 1

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