2
$\begingroup$

When looking at this question: How to apply a 1-qubit gate to a single qubit from an entangled pair?

And other questions, it makes it seem like when you have a quantum circuit involving $N$ qubits, that when you want to peform a gate operation on a subset of the qubits, that you must still make a matrix that applies to all qubits.

Is this always true, or is it only true in the case where qubits are entangled?

In other words... is it only entangled qubits that we have to process together, or must all qubits in a circuit be processed together?

|cite|improve this question
$\endgroup$
3
$\begingroup$

You can apply a quantum gate on any subset of the qubits, not affecting the rest.

This can be formalized as evolving certain qubits via $U$ while the other qubits evolve via $I$, thus the joint unitary is $U \otimes I$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ doesn't entanglement throw a wrench in this? When you change one entangled bit, doesn't the other then change as well? $\endgroup$ – Alan Wolfe Oct 16 '15 at 17:52
  • $\begingroup$ Nope, it works just the same with entangled qubits. $\endgroup$ – Ran G. Oct 16 '15 at 17:58
  • $\begingroup$ awesome... i guess i was overthinking it. Thank you so much!! $\endgroup$ – Alan Wolfe Oct 16 '15 at 18:15
  • $\begingroup$ @AlanWolfe note that what you ask is still true: you have to construct a matrix which applies to all qubits, but, as Ran G points out, you do that by first making your matrix U, and then taking the tensor with the identity matrix. More generally, if you want to do a 3-qubit gate $U$ on qubits 3,4 and 5 out of 6 qubits, you take the matrix $I \otimes I \otimes U \otimes I$. You can construct matrices for arbitrary subsets (e.g. $U$ for qubits 1,2 and 5), but it's a little more work, and it probably warrants a question of its own. $\endgroup$ – Lieuwe Vinkhuijzen Oct 16 '15 at 21:28
  • $\begingroup$ I ended up asking the question here in case you have the answer Liuwe: cs.stackexchange.com/questions/48397/… $\endgroup$ – Alan Wolfe Oct 18 '15 at 18:37
0
$\begingroup$

I think you will need that in the case that your operation is not using qubits in sequence. That is if you have 4 qubits (1, 2, 3 and 4) and you want to apply a 2-qubit quantum operation on qubit 1 and 4, e.g. a CNOT then you should construct a matrix of size $2^4 \times 2^4$.

|cite|improve this answer
$\endgroup$

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.