# Why would measuring a qubit determine the states of the other qubits?

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?

• 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 $\vert1\rangle$). Mar 24 at 20:34
• Thank you for pointing out that the text refers to Eq. 5. That straightens out my confusion. Mar 24 at 21:18

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 qbit is 0, then the first one will be 0 as well. You cannot be unsure about the first qbit and say “it might be zero or one” once you’ve measured the second one.