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In Quantum Algorithm Implementations for Beginners, page 7 it is stated,

Suppose we have a three qubit state, $\vert\psi\rangle$, but we only measure the first qubit and leave the other two qubits undisturbed. What is the probability of observing a $\vert 0\rangle$ in the first qubit? This probability will be given by,

$$\sum_{(x_2,x_3)\in\{0,1\}^2}\vert\langle0x_2x_3\vert\phi\rangle\vert^2.$$

The state of the system after this measurement will be obtained by normalizing the state

$$\sum_{(x_2,x_3)\in\{0,1\}^2}\langle0x_2x_3\vert\phi\rangle\vert0x_2x_3\rangle.$$

Applying this paradigm to the state in Eq. (5) we see that the probability of getting $\vert0\rangle$ in the first qubit will be 0.5, and if this result is obtained, the final state of the system would change to $\vert 000\rangle$. On the other hand, if we were to measure $\vert 1\rangle$ in the first qubit we would end up with a state $\vert 111\rangle$.

This feels wrong to me. Why would measuring the first qubit determine the states of the other two qubits? Is this an error in the paper?

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    $\begingroup$ There might be some confusion between $\phi$ and $\psi$, but the statement otherwise is correct. Equation (5) of the paper is a GHZ state; the three qubits are entangled and upon measuring one of the qubits to be $\vert0\rangle$, you "collapse" the other two to be $\vert00\rangle$ (and vice-versa for $\vert1\rangle$). $\endgroup$
    – Mark Spinelli
    Mar 24, 2021 at 20:34
  • $\begingroup$ Thank you for pointing out that the text refers to Eq. 5. That straightens out my confusion. $\endgroup$
    – S. McGrew
    Mar 24, 2021 at 21:18
  • $\begingroup$ This is one of the key points of what entanglement is. It seems unintuitive, but that's what quantum does to your mind! (Also, while not quite as strong, there are classical situations which are partially equivalent. You might look up "Bertleman's socks".) $\endgroup$
    – DaftWullie
    Nov 3, 2021 at 7:32

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Simplest example is the bell state: if you know that the bell state is $$\dfrac{1}{\sqrt{2}}|00\rangle+\dfrac{1}{\sqrt{2}}|11\rangle$$ Then you know that the possible read outs are $00,11$. This trivially implies that if the outcome of a measurement on the second qubit is 0, then the first one will be 0 as well. You cannot be unsure about the first qubit and say “it might be zero or one” once you’ve measured the second one.

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