1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.