I'm currently working with the cX gate but I have the question: This gate is a projection to sigmax axis on the bloch sphere if y partially trace the target qubit?

Or i mean, what is the relationship between this gate and the measurement?


No CX or CNOT gate is not a measurement in any sense. CNOT gate is still a unitary gate as all the rest, it just acts on two qubits. The action of CNOT is the following

$$|00\rangle\to|00\rangle$$ $$|01\rangle\to|01\rangle$$ $$|10\rangle\to|11\rangle$$ $$|11\rangle\to|10\rangle$$

where the first digit indicates the control qubit and the second digit the target qubit (the one that changes). But what happens when one of the qubits is in a superposition? Well, the rules above help to perform the calculation. Suppose the first qubit (first digit) is in a superposition of $|0\rangle$ and $|1\rangle$ as $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, then CNOT does the following:

$$|+ 0\rangle=(|00\rangle+|10\rangle)/\sqrt{2}\to (|00\rangle+|11\rangle)/\sqrt{2},$$ this resulting state is an entangled state (it cannot be factored into a product of two single states). If the target qubit (second digit) is in a superposition, we have $$|1+\rangle=(|10\rangle+|11\rangle)/\sqrt{2}\to (|11\rangle+|10\rangle)/\sqrt{2}=|1+\rangle,$$ which in this case the action of the CNOT returned the same initial state, even when the control qubit was set to 1.

Anyway, I do not know if this helps, but I hope it illustrates that a CNOT gate is not a measurement. If it was related to measurements you will not expect it to conserve superpositions and create entangled states.

Also, it can be written as $\mathrm{CNOT}=|00\rangle\langle 00|+|01\rangle\langle 01|+|11\rangle\langle 10|+|10\rangle\langle 11|$ which is not necessarily a projection, in the sense that it is a superposition that includes every basis state and does not project into any particular subspace.

  • $\begingroup$ Thank you! That was exactly what I was wondering about the gate. $\endgroup$
    – felipechoy
    Oct 4 at 23:28
  • $\begingroup$ @felipechoy no problem, you can approve/accept the answer or ask for more details. $\endgroup$
    – Mauricio
    Oct 4 at 23:34
  • $\begingroup$ Thanks, I have other questions. As you showed it creates an entangled state if one of the qubits is in superposition. If I partially trace one of those quits, the result will be I/2. More interesting is if I connect one of those qubits entangled o a third qubit and partially trace obtaining the reduced d.m. on the third qubit. For any initial q3 state, after tracing , the result state will be in the X axis. I was wondering... Why does that happen? I mean, It ends up projected in X, but i don't know why. $\endgroup$
    – felipechoy
    Oct 5 at 0:39
  • $\begingroup$ @felipechoy why do you think it would be in the x-axis I don’t get that, the result will depend on an additional measurement $\endgroup$
    – Mauricio
    Oct 5 at 10:36

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