# 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? Sep 2, 2019 at 10:21
• @Ashish I am looking for both. Sep 2, 2019 at 11:51

## 2 Answers

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.

Regarding Q2, lets take the case of making a measurement on first qubit, then the probability of finding first qubit is 0.75 in |0> and 0.25 in |1> (explanation given in answer @czwang). Now say, the first qubit is found to be in |0> post measurement then the probability of second qubit to be in |0> is ONE and if the first qubit is found in |1> post measurement, then the probability of the second qubit to be in |0> is ZERO.

However, if you first perform the measurement on second qubit then the probability of second qubit to be in |0> is 0.75.