Fit a chip to a measurement

The mathematics

The starting point, in three steps. From a measured anharmonicity α (negative for a transmon), f₀₁ and the readout shift χ at a detuning Δ = f₀₁ − ω_r from the resonator:

The first two are the inversion of the transmon formulas of topic 29 (Koch et al. 2007); the third is the inversion of the shift of topic 31. In the example chip of the previous topic (E_J = 14.901 GHz, E_C = 0.2421 GHz) the exact f₀₁ is 5.1178 GHz and α is −273.5 MHz, so the first line gives E_C = 0.2735 GHz, 13% too big, and the second E_J = 13.28 GHz, 11% too small. That is the size of the correction the exact model removes: starting from the closed forms (E_C = 0.2735 GHz, E_J = 13.28 GHz) the exact fit lands on E_J = 14.900 GHz and E_C = 0.2421 GHz.

The exact fit. Let p = (E_J, E_C) and let F(p) be the model’s (f₀₁, α), computed by diagonalising the Hamiltonian. The fit minimises the sum of squared differences, S(p) = Σ_k (F_k(p) − m_k)², over the measurements m_k, by the Gauss–Newton step p ← p − (J^TJ)⁻¹J^Tr, where r = F(p) − m and J is the table of derivatives ∂F_k/∂p_j. Each step is a few diagonalisations; the routine below takes a handful of steps.

How well the numbers pin the knobs down. The table J says how much each measurement moves when a knob moves. For the example chip, from full diagonalisations: raising E_C by 1% raises f₀₁ by 0.47% (24 MHz) and makes α 1.08% larger in size (2.9 MHz); raising E_J by 1% raises f₀₁ by 0.53% (27 MHz) and changes α by only 0.07% (0.2 MHz). Both knobs move f₀₁ by about the same amount per percent, so f₀₁ alone cannot tell them apart; α is moved by E_C about fourteen times more than by E_J, so α separates them. That is the reason the closed forms are chained in the order of the figure, and the reason two measurements are enough for two knobs. If a measurement has an error of σ, the error in a knob is about σ divided by that column’s sensitivity, which is the quickest way to ask whether a parameter is really determined by the data.

Try it in Python. The routine below does the exact fit with SciPy’s least_squares and scqubits as the forward model, starting from the closed forms, for the f₀₁ and α printed in the previous topic. Run on scqubits 4.3.1 with QuTiP 5.2.0 and SciPy 1.14.1, 25 October 2026:

import numpy as np
import scqubits as scq
from scipy.optimize import least_squares

scq.settings.PROGRESSBAR_DISABLED = True

def transmon_numbers(EJ, EC):
    t = scq.Transmon(EJ=EJ, EC=EC, ng=0.0, ncut=30)
    E = t.eigenvals(evals_count=3)
    return E[1] - E[0], (E[2] - E[1]) - (E[1] - E[0])        # f01 and the anharmonicity, GHz

target = (5.1178, -0.2735)                                     # "measured" f01 and alpha, GHz

EC0 = -target[1]                                               # closed-form guess: alpha ~ -EC
EJ0 = (target[0] + EC0) ** 2 / (8 * EC0)                       # then f01 ~ sqrt(8 EJ EC) - EC
print("starting guess: EJ = %.3f, EC = %.4f GHz" % (EJ0, EC0))

fit = least_squares(lambda p: np.subtract(transmon_numbers(*p), target) * 1e3, x0=[EJ0, EC0], x_scale=[1.0, 0.01])
print("fitted:         EJ = %.3f, EC = %.4f GHz" % tuple(fit.x))
print("largest residual: %.3f MHz" % np.max(np.abs(fit.fun)))

starting guess: EJ = 13.284, EC = 0.2735 GHz
fitted:         EJ = 14.900, EC = 0.2421 GHz
largest residual: 0.000 MHz