What is the probability of finding the second qubit as $0$ in the state $|\psi\rangle=\frac1{\sqrt2}|00\rangle+\frac12|10\rangle-\frac12|11\rangle$?

Assuming two qubits start in the state: $$|\psi\rangle = \frac{1}{\sqrt 2}|00\rangle + \frac{1}{2}|10\rangle- \frac{1}{2}|11\rangle$$

What is the probability of measuring the second qubit as 0? And what is the new state of the system after measuring the first qubit as 1?

I know that for a single qubit state that the probability amplitudes are the coefficient squared. In a two qubit system are the probabilities distributed to the individual states? I.e. from this example does the each zero state in the state: $$|00⟩$$ have a 50% chance? And I don't really understand the second question, any suggestion on where to review or study?

• State update rules are usually given in books alongside the Born rule (rule to calculate probabilities of measurement outcomes). Nov 21 '20 at 18:07

If we have the state $$|\psi \rangle = \dfrac{1}{\sqrt{2}}|00\rangle + \dfrac{1}{2}|10\rangle - \dfrac{1}{2}|11\rangle$$ then the probability of the second qubit being in the state $$|0\rangle$$ is the probability of the state $$|\psi \rangle$$ having $$|0\rangle$$ on the second qubit. In this case, it is from the states $$|00\rangle$$ and $$|10\rangle$$. So The probability of measuring the second qubit in the state $$|0\rangle$$ is $$\bigg| \dfrac{1}{\sqrt{2}} \bigg|^2 + \bigg| \dfrac{1}{2} \bigg|^2 = \dfrac{3}{4}$$.

You can also work this out more explicitly as well. That is, we have

$$|\psi \rangle = \begin{pmatrix} 1/\sqrt{2} \ \ \\ 0 \\ 1/2 \\ -1/2 \end{pmatrix}$$

We are looking for the probability that the second qubit is in the state $$|0\rangle$$ so the projective measurement $$M$$ is

$$M = I \otimes |0\rangle \langle 0 | = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$ and so according to Born's rule we have that the probability to measure the second qubit in the state $$|0\rangle$$ is

$$\langle \psi | M | \psi \rangle = \begin{bmatrix} 1/\sqrt{2} & 0 & 1/2 &-1/2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1/\sqrt{2} \ \ \\ 0 \\ 1/2 \\ -1/2 \end{bmatrix} = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}$$

Also, the state post measurement is $$|\psi_{post} \rangle = \dfrac{M|\psi\rangle}{\sqrt{3/4}}$$.

You can extend this to the case where the first qubit is mesured in the state $$|1\rangle$$ too. In this case, the projective measurement $$M = |1\rangle \langle 1| \otimes I$$