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What happens to multi-qubit quantum state after one of qubits is measured?
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Problem
If we have a quantum state consisting of (let's say) 3 entangled qubits and we read the value of one of them, what happens to the probabilities of the remaining possible states?
For example, if we have three entangled qubits and have an amplitude for every state from $|000\rangle$ to $|111\rangle$ and we measure the second (middle) qubit to be 1, the remaining states are $|010\rangle, |011\rangle, |110\rangle, |111\rangle$. If we know the amplitudes for all possible 9 states before the measurment, how do we get the amplitudes of the remaining states after measurment?
For example, if we have three entangled qubits and have an amplitude for every state from $|000\rangle$ to $|111\rangle$ and we measure the second (middle) qubit to be 1, the remaining states are $|010\rangle, |011\rangle, |110\rangle, |111\rangle$. If we know the amplitudes for all possible 9 states before the measurment, how do we get the amplitudes of the remaining states after measurment?
Solution
The amplitudes are scaled so that their norm is 1. In your case, if the original amplitudes are $\alpha_{000},\ldots,\alpha_{111}$, then after measuring the second bit to be 1, which happens with probability $p = \alpha_{010}^2 + \alpha_{011}^2 + \alpha_{110}^2 + \alpha_{111}^2$, the quantum state is
$$
\frac{\alpha_{010} \left|010\right\rangle + \alpha_{011} \left|011\right\rangle + \alpha_{110} \left|110\right\rangle + \alpha_{111} \left|111\right\rangle}{\sqrt{\alpha_{010}^2 + \alpha_{011}^2 + \alpha_{110}^2 + \alpha_{111}^2}}.
$$
$$
\frac{\alpha_{010} \left|010\right\rangle + \alpha_{011} \left|011\right\rangle + \alpha_{110} \left|110\right\rangle + \alpha_{111} \left|111\right\rangle}{\sqrt{\alpha_{010}^2 + \alpha_{011}^2 + \alpha_{110}^2 + \alpha_{111}^2}}.
$$
Context
StackExchange Computer Science Q#37388, answer score: 8
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