1
$\begingroup$

I am trying to find out about the mechanism of measurement in IBM Q devices. To be specific, if I apply two Hadamard gates on the first qubit and identity on the second then is it possible to immediately measure so that the second qubit is not affected by the second Hadamard on the first qubit? Also, are measurements on qubits simultaneous? Does it differ with devices?

$\endgroup$
1
  • $\begingroup$ The time sequence of your quantum operation is quite confusing. Besides, the answer to the first question is apparently no, quantum operation is non-local(but this property does not provide any FTL effect). Then yes, once you measured the second qubit, the first qubit is also collapsed to a classical state. $\endgroup$ Dec 3 '20 at 0:59
2
$\begingroup$

I don't know if this is what you are looking for but I will try to write out what I see.


If you are thinking of a circuit like this:

enter image description here

Then the gates on the first qubit, $q_0$, will not be affected the second qubit at all as they not interacting with one another. The entire system is in a separable state. So it doesn't really matter when you make the measurement on the second qubit, the result will be the same.

Now, if you actually run this on the quantum computer then I don't think the measurement is actually being executed until after the second Hadamard gate on the first qubit has been executed. You can check the pulse schedule for the circuit above:

enter image description here

Now, if you looking at a circuit that have entangled gate between the two qubits,

enter image description here

then the measurement on the second qubit $q_1$ won't be executed until the second Hadamard gate is being executed as well. But again, the Hadamard gate on the first qubit won't affect the second qubit as it only acts on the first qubit. Hence, it is fine that the measurement is being executed after the second Hadamard gate have been applied to $q_0$. The result won't change.


Also, if you have a circuit like:

enter image description here

Then the Hadamard gate on the second qubit, $q_1$, won't be executed until when the 4th Hadamard gate is being executed on the first qubit. This is because if we apply the Hadamard gate to the second qubit too early, it will idling for a long time and more decoherence effects will take place.

$\endgroup$
3
  • $\begingroup$ Hi, thank you for your informative reply. Can you please tell me how did you get the pulse diagram? $\endgroup$
    – Gem
    Dec 3 '20 at 5:24
  • $\begingroup$ Could you also explain the difference between the blue sinusoidal function with the red step functions? I am guessing the latter represent the pulses (i.e. application of gates). $\endgroup$
    – Gem
    Dec 3 '20 at 5:43
  • $\begingroup$ @Gem I just realized that not all the public open machines actually have pulse control access... The only machine that is available is ibmq_armonk which is only a 1 qubit machine. The sinuisoidal function is the microwave pulse that drives the qubit to execute the gates, which in this case is the Hadamard gate. However, if you want to know more about qiskit pulse, take a look at this documentation: arxiv.org/abs/2004.06755 I think it will be very helpful. $\endgroup$
    – KAJ226
    Dec 3 '20 at 6:21

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.