# Measuring a single qubit in 2 bit entangled state

consider a 2 qubit entangled state ($$\sqrt 3/2)\left|00\right> + (1/2)\left|11\right>$$

Q1) Is the above entangled state a valid one? q2) If valid then What is the probability of measuring the second qubit as $$\left|0\right>$$?

• Q1) Yes, the 2 qubit entangled state (√3/2)|00⟩+(1/2)|11⟩ is a valid state. Q2) Do you wish to say that you will perform measurement on first qubit and then on second qubit or you will be performing measurement on second qubit alone? Are you looking for a mathematical formula or an explanation? – Ashish Sep 2 '19 at 10:21
• @Ashish I am looking for both. – Chaitanya Reddy Sep 2 '19 at 11:51

It is valid, as $$|00\rangle$$ and $$|11\rangle$$ are the two computational basis states and the amplitudes are normalized $$(\frac{\sqrt{3}}{2})^2+(\frac{1}{2})^2=1$$.
The probability of measuring the second qubit as $$|0\rangle$$ is $$\frac{3}{4}$$. BTW, as the result of the measurement, the first qubit will be also in $$|0\rangle$$ due to the entanglement.