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Consider the state $$ \left| \varphi \right>=\frac{i}{\sqrt{3}} \left| 0 \right> + \sqrt{\frac{2}{3}} \left| 1\right>. $$ What is the probability of qubit system when measured in the state $\left| + \right>$?

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  • $\begingroup$ Base must contain 2 states. $\endgroup$
    – kludg
    Jan 14, 2020 at 11:04

1 Answer 1

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It is known that:

$$ \left| 0 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle + \left| - \right\rangle \right) \qquad \left| 1 \right\rangle = \frac{1}{\sqrt{2}} \left( \left| + \right\rangle - \left| - \right\rangle \right) $$

By substituting in the initial state:

$$ \left| \varphi \right\rangle = \frac{i}{\sqrt{3}}\left| 0 \right\rangle + \frac{\sqrt{2}}{\sqrt{3}}\left| 1 \right\rangle = \frac{i + \sqrt{2}}{\sqrt{6}} \left| + \right\rangle + \frac{i - \sqrt{2}}{\sqrt{6}} \left| - \right\rangle $$

So the probability to measure the qubit in the $\left| + \right\rangle$ state is equal to $\left|\frac{i + \sqrt{2}}{\sqrt{6}}\right|^2 = 0.5$

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