# How to find measurement probabilities of a single qbit in a tensored state

Given a tensored state of qbits such as $$\frac{1}{\sqrt{3}}|0\rangle_1|1\rangle_2 + \sqrt{\frac{2}{3}}|1\rangle_1|0\rangle_2$$ or $$\frac{1}{\sqrt{2}}(|0\rangle_1|+\rangle_2 + |+\rangle_1|-\rangle_2)$$ Then how do you calculate the probabilities of getting $$|0\rangle$$ or $$|1\rangle$$ if you measure qbit one in the above states?

• Hi and welcome to Quantum Computing SE. Just note that the first state represent entangled one, so it cannot be expressed as a tensor product. Feb 10, 2020 at 7:36

Only consider first qubit.

First case:

$$\frac{1}{\sqrt 3}|0\rangle _1|1\rangle_2 + \sqrt{\frac{2}{3}}|1\rangle_1|0\rangle_2$$

$$P(|0\rangle_1) = \frac{1}{3}$$ and $$P(|1\rangle_1) = \frac{2}{3}$$

Second case:

$$\frac{1}{\sqrt 2}(|0\rangle_1|+\rangle_2 + |+\rangle_1|-\rangle_2) = \frac{1}{\sqrt 2}(|0\rangle_1|+\rangle_2 + \frac{1}{\sqrt 2}|0\rangle_1|-\rangle_2 + \frac{1}{\sqrt 2}|1\rangle_1|-\rangle_2 )=\frac{1}{\sqrt 2}|0\rangle_1|+\rangle_2 + \frac{1}{2}|0\rangle_1|-\rangle_2 + \frac{1}{2}|1\rangle_1|-\rangle_2$$

Therefore, probabilities are:

$$P(|0\rangle_1|+\rangle_2) = \frac{1}{2}$$ and $$P(|0\rangle_1|-\rangle_2) = P(|1\rangle_1|+\rangle_2) = \frac{1}{4}$$

So, probability of measuring $$|0\rangle$$ in the first qubit is $$\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$$ and probability of measuring $$|1\rangle$$ is $$\frac{1}{4}$$.