# How to compute the probability of finding a given state in the $|+\rangle$ state?

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 $$\left| + \right>$$?

• Base must contain 2 states. Jan 14 '20 at 11:04

## 1 Answer

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 $$\left|\frac{i + \sqrt{2}}{\sqrt{6}}\right|^2 = 0.5$$