# Disentanglement of qubits from output of CNOT gate

Suppose we have a control qubit with initial state $$\binom{a}{b}$$ and a target qubit with initial state $$\binom{c}{d}$$.What a CNOT does is transform the state of the target qubit into:

$$\begin{pmatrix} ac\\ ad\\ bd\\ bc \end{pmatrix}$$

If we measure the qubit we destroy the entanglement and target qubit is again in a pure state.However which state would that be?

• The target qubit is not of dimension 4. It is not a 1x4 matrix; it is a 1x2 matrix. Dec 17, 2023 at 3:25
• I want to object.What a quantum gate does is entangle the control qubit with the target qubit and since both the pure states of the control qubit and the target qubit are 2x1 vectors then the result will be a 4x1 vector. Dec 17, 2023 at 3:29
• You are not using the word qubit in the way I understand it. A qubit is a two dimensional vector. You are correct about what an entangling 2 qubit gate does, and yes two entangled qubits are of dimension 4, but that doesn’t mean that the target qubit itself is of dimension 4. Dec 17, 2023 at 3:45
• Can you elaborate why are you saying the following- "If we measure the qubit we destroy the entanglement and target qubit is again in a pure state." I believe there are some mistakes in your understanding of a few terminologies. Dec 17, 2023 at 9:22
• related by op quantumcomputing.stackexchange.com/q/35218/55
– glS
Dec 21, 2023 at 10:37

The CNOT gate does not only transform the target qubit, but instead transforms the state of the whole 2-qubit system to $$|\psi\rangle = \begin{pmatrix} ac \\ ad \\ bd \\ bc \end{pmatrix} = ac |00\rangle + ad |01\rangle + bd |10\rangle + bc |11\rangle,$$ where here we are assuming that the target qubit is written on the right.
When analyzing the post-measurement state of the 2-qubit system when the target qubit alone is measured, it is helpful to rewrite $$|\psi\rangle$$ as $$|\psi\rangle = (ac | 0\rangle + bd|1\rangle)|0\rangle + (ad|0\rangle + bc|1\rangle)|1\rangle.$$ If the target qubit is measured in state $$|0\rangle$$, then the post-measurement state of the 2-qubit system is $$\frac{(ac | 0\rangle + bd|1\rangle)|0\rangle}{\sqrt{(ac)^2 + (bd)^2}}.$$ If the target qubit is measured in state $$|1\rangle$$, then the post-measurement state of the 2-qubit system is $$\frac{(ad|0\rangle + bc|1\rangle)|1\rangle}{\sqrt{(ad)^2 + (bc)^2}}.$$ The square-roots in the denominators of these post-measurement states are to make sure that the post-measurement states are normalized (have norm 1).
• If the target qubit is measured in state |1⟩ , then the post-measurement state of the 2-qubit system is $(ad|0⟩+bc|1⟩)|1⟩\sqrt{(ad)2+(bc)2}$.How does this make any sense? you have altered the control qubit! Dec 19, 2023 at 14:35