Partial Measurement for 2 and 3 Qubit Systems

When dealing with a multi-qubit system, we might only want to measure one specific qubit, not the entire system.

This is called partial measurement. Consider a general 2-qubit (pure) state:

$$|\Psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle$$

where $|00\rangle = |0\rangle \otimes |0\rangle$, etc., and the state is normalized such that $|\alpha_{00}|^2 + |\alpha_{01}|^2 + |\alpha_{10}|^2 + |\alpha_{11}|^2 = 1$.

Suppose we measure only the first qubit in the computational basis $\{|0\rangle, |1\rangle\}$. To find the probability of measuring the first qubit as $|0\rangle$, we sum the squared magnitudes of the amplitudes of all basis states where the first qubit is $|0\rangle$:

$$P(0_1) = |\alpha_{00}|^2 + |\alpha_{01}|^2$$

where we have used the notation $0_{1}$ to denote the fact that it's the probability of measuring the first qubit in the state $0$.

If this outcome occurs, we must perform state update.

The new state of the system, which we'll denote by $|\Psi'\rangle_{0_1}$, will have the first qubit as $|0\rangle_1$.

The second qubit's state will be a renormalized superposition of the states where the first qubit was $|0\rangle$:

$$|\Psi'\rangle_{0_1} = \frac{\alpha_{00}|00\rangle + \alpha_{01}|01\rangle}{\sqrt{P(0_1)}} = |0\rangle_1 \otimes \left( \frac{\alpha_{00}|0\rangle_2 + \alpha_{01}|1\rangle_2}{\sqrt{P(0_1)}} \right)$$

Similarly, the probability of measuring the first qubit as $|1\rangle_1$ is:

$$P(1_1) = |\alpha_{10}|^2 + |\alpha_{11}|^2$$

And the post-measurement state if this outcome occurs is:

$$|\Psi'\rangle_{1_1} = \frac{\alpha_{10}|10\rangle + \alpha_{11}|11\rangle}{\sqrt{P(1_1)}} = |1\rangle_1 \otimes \left( \frac{\alpha_{10}|0\rangle_2 + \alpha_{11}|1\rangle_2}{\sqrt{P(1_1)}} \right)$$

Notice that $P(0_1) + P(1_1) = 1$.

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