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

Question

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?

Answer 1

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 $\frac{1}{4}$.