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

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

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

• Hint: note that $\left|\varphi\right\rangle =\left|0\right\rangle \otimes\left(\frac{1}{\sqrt{2}}\left|0\right\rangle +\frac{i}{2}\left|1\right\rangle \right)+\left|1\right\rangle \otimes\left(-\frac{1}{2}\left|1\right\rangle \right)$ Jan 4 '20 at 13:27

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