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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?

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    $\begingroup$ State update rules are usually given in books alongside the Born rule (rule to calculate probabilities of measurement outcomes). $\endgroup$ – Rammus Nov 21 at 18:07
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If we have the state $|\psi \rangle = \dfrac{1}{\sqrt{2}}|00\rangle + \dfrac{1}{2}|10\rangle - \dfrac{1}{2}|11\rangle$ which can be written as

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 $M = |1\rangle \langle 1| \otimes I$

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