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In a semiclassical treatment, the noise Hamiltonian of a quantum system can be treated as a summation over its noise channels in the following way: $$\hat{H}_{\text{noise}}=\sum_{k}\beta_{k}\left(t\right)\hat{N}_{k}$$ where $\hat{N}_k$ is an operator indicating the noise axis in Hilbert space, and $\beta_k(t)$ is a classical noise field. Considering the case of a single qubit that is not subjected to control fields, the trivial noise channel is $N_z = \sigma_z$ - fluctuations in the magnetic field in the environment of the qubit results in Zeeman splitting and therefore to an undesirable Z rotation. However, I would like to know what physical process can lead to a noise channel in an orthogonal direction to the quantization axis (XY rotations)? To make it clear, all I am asking for is an example.

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    $\begingroup$ Given any physical channel for causing intentional rotation about X or Y, noise on that channel will cause undesired rotation about X or Y. What exactly are you asking for? $\endgroup$ – DanielSank Mar 9 at 4:56
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    $\begingroup$ @DanielSank That's true, but I meant in the absence of a control channel. Actually, I read the paragraph about "single qubit noise sources" in the documentation in the website of q-ctrl, and in their analysis they are taking into account 2 different noise channels: 1) Z rotation due to interaction of the qubit with the environment 2) X and Y rotations due to a noise in the control itself. I wondered why they are not taking into account XY rotation due to the interaction of the qubit with the environment? $\endgroup$ – Qexp Mar 9 at 7:33
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    $\begingroup$ Well if a qubit experiences X/Y rotations when subject to field $\beta$, then we usually introduce $\beta$ intentionally to do logic gates. If $\beta$ exists when we're not trying to do gates, then it's noise. So, I guess you're asking for a few examples, is that right? $\endgroup$ – DanielSank Mar 9 at 23:03
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    $\begingroup$ @DanielSank Yes! Few examples will be great! I guess there are no massive such noise channels otherwise q-ctrl were taking it into consideration. $\endgroup$ – Qexp Mar 10 at 2:03

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