New answers tagged

5

I think you'll find that most hardware, at the hardware level, gives you arbitrary single qubit rotations. So, in that sense, it is true that Solovay-Kitaev is not directly applicable to current systems. However, obsolete is certainly not the right word. It's rather the opposite - it's ahead of its time. The importance of the Solovay-Kitaev algorithm is ...


1

The idea is to realise that classical physics is a limiting case of quantum physics. The limits (usually) are: Number of particles, $N \to \infty$ Degrees of Freedom (DoF), $\omega \to \infty$ There are various other ways to think about it, but these are sufficient. Now, the case of a transmon circuit being a quantum system comes from the low temperatures ...


3

The transmon is a slightly nonlinear $LC$ resonator, with a resonance frequency of a few GHz. The image below, from Bardin et al., IEEE Microwave Magazine, 2020 (of which I am an author), shows in the main panel three wavefunctions corresponding to the first three eigenstates of harmonic $LC$ oscillator (not a transmon, but close enough for the purpose of ...


2

I recommend checking out this reference if you haven't already, specifically page 28. Basically, starting from the classical Hamiltonian of the circuit described in figure 12, we see that the classical external voltage drive $V_d(t)$ ends up scaling the circuit charge variable $Q$, which, when quantized, can be expressed as the operator $Q \rightarrow \hat{...


3

Basically, the current and voltage oscillate - they have uncertain values that would correspond to oscillation in the classical limit. In the ground state, the uncertainty is at or near the minimum allowed by the uncertainty principle. In the first excited state, the uncertainty is greater. That is, if you were to perform a measurement you would get a larger ...


2

Basis dependence First note that superposition is not a distinct type of quantum state. In quantum mechanics, every state may be cast as a superposition by appropriate choice of basis$^1$. For example, even though $|0\rangle$ is not a superposition in the computational basis, it is a superposition in the $|+\rangle$, $|-\rangle$ basis since $|0\rangle = \...


Top 50 recent answers are included