Decoherence Timescale Estimation

A coherence time is meaningful only after specifying the measured envelope and pulse sequence. This calculation obtains Ramsey and Hahn-echo envelopes from the same noise spectrum, checks a frequency integral against exact formulas, and finds the time at which each envelope reaches $e^{-1}$. The two times differ even though the underlying noise is unchanged.

Helpful background. Noise spectra introduces spectral conventions, and dynamical decoupling explains the sign change produced by a refocusing pulse. The code uses Python 3 and NumPy.

Gaussian dephasing with declared units

Let $H(t)=\hbar[\omega_0+\xi(t)]Z/2$, where $\xi$ is stationary, zero-mean classical Gaussian noise in radians per second. Work in the rotating frame of $\omega_0$ and take instantaneous ideal control pulses. A sequence changes the accumulated phase to $\phi(T)=\int_0^T y(t)\xi(t)dt$.

For Ramsey, $y=1$. For Hahn echo, $y=1$ in the first half and $y=-1$ in the second. The normalized coherence is $W(T)=\langle e^{-i\phi}\rangle=e^{-\chi(T)}$, with

$$\chi(T)=\frac12\int_{-\infty}^{\infty}\frac{d\omega}{2\pi} S(\omega)|Y(\omega,T)|^2, \qquad Y(\omega,T)=\int_0^T y(t)e^{i\omega t}dt.$$

Our two-sided spectrum satisfies $C(t)=\int(d\omega/2\pi)S(\omega)e^{-i\omega t}$. It has units of inverse seconds, since $C=\langle\xi(t)\xi(0)\rangle$ has units of inverse seconds squared. The exponent $\chi$ is dimensionless. The Gaussian assumption makes this second-cumulant expression exact for this commuting classical-noise model; it does not automatically apply to non-Gaussian noise or arbitrary quantum baths.

The filter magnitudes are

$$|Y_R|^2=\frac{4\sin^2(\omega T/2)}{\omega^2}, \qquad |Y_E|^2=\frac{16\sin^4(\omega T/4)}{\omega^2}.$$

At zero frequency their limits are $T^2$ and $0$, respectively. Echo suppresses slowly varying frequency shifts.

A Lorentzian spectrum with two exact answers

Use an Ornstein–Uhlenbeck correlation $C(t)=\sigma^2e^{-|t|/\tau_c}$, so

$$S(\omega)=\frac{2\sigma^2\tau_c}{1+\omega^2\tau_c^2}.$$

Writing $x=T/\tau_c$, direct integration of the correlation function gives

$$\begin{aligned} \chi_R(T)&=\sigma^2\tau_c^2(x-1+e^{-x}),\\ \chi_E(T)&=\sigma^2\tau_c^2(x-3+4e^{-x/2}-e^{-x}). \end{aligned}$$

For $T\ll\tau_c$, Ramsey begins as $\sigma^2T^2/2$, whereas echo begins as $\sigma^2T^3/(12\tau_c)$. At long times both have slope $\sigma^2\tau_c$ but different offsets. A single exponential fitted across the short- and long-time regimes is therefore inappropriate.

Integrate the spectrum and locate the threshold

Take $\sigma=2\,\mathrm{s}^{-1}$ and $\tau_c=0.5\,\mathrm{s}$. The even spectrum lets us integrate only positive frequencies, with prefactor $1/(2\pi)$. NumPy’s sinc(x) means $\sin(\pi x)/(\pi x)$; the factors of $\pi$ below implement the zero-frequency limits without division by zero.

import numpy as np

sigma, tau = 2.0, 0.5

def chi_exact(T, sequence):
    x = T/tau
    if sequence == 'Ramsey':
        return sigma**2*tau**2*(x+np.expm1(-x))
    if sequence == 'Echo':
        return sigma**2*tau**2*(x+4*np.expm1(-x/2)-np.expm1(-x))
    raise ValueError('Unknown sequence')

def chi_spectral(T, sequence, cutoff, points):
    w = np.linspace(0, cutoff, points)
    S = 2*sigma**2*tau/(1+(w*tau)**2)
    if sequence == 'Ramsey':
        filter_squared = T**2*np.sinc(w*T/(2*np.pi))**2
    elif sequence == 'Echo':
        filter_squared = T**2*np.sin(w*T/4)**2*np.sinc(w*T/(4*np.pi))**2
    else:
        raise ValueError('Unknown sequence')
    return np.trapezoid(S*filter_squared, w)/(2*np.pi)

def threshold(sequence):
    lo, hi = 0.0, tau
    while chi_exact(hi, sequence) < 1:
        hi *= 2
    for _ in range(60):
        mid = (lo+hi)/2
        if chi_exact(mid, sequence) < 1:
            lo = mid
        else:
            hi = mid
    return (lo+hi)/2

for sequence in ['Ramsey', 'Echo']:
    for T in [0.2, 0.5, 1.0, 2.0]:
        exact = chi_exact(T, sequence)
        coarse = chi_spectral(T, sequence, 80/tau, 20001)
        fine = chi_spectral(T, sequence, 80/tau, 40001)
        wider = chi_spectral(T, sequence, 160/tau, 80001)
        assert abs(coarse-fine) < 1e-7
        assert abs(wider-exact) < 1e-6
        print(sequence, T, 'chi:', exact, 'integral error:', wider-exact)
    tstar = threshold(sequence)
    assert abs(chi_spectral(tstar, sequence, 160/tau, 80001)-1) < 1e-6
    print(sequence, '1/e time in seconds:', tstar)

Refining the grid at a fixed cutoff tests quadrature resolution. Raising the cutoff at fixed grid spacing tests the omitted high-frequency tail. They are different checks; a finely sampled but too-short frequency interval can give a stable wrong answer. Near extremely small $T/\tau_c$, evaluate the exact formulas by their series to avoid cancellation between nearly equal terms.

The resulting threshold times are about $0.921\,\mathrm{s}$ for Ramsey and $1.626\,\mathrm{s}$ for echo. Here these are explicitly defined $1/e$ times, not constant exponential decay times inferred from a fit. For envelopes with revivals, specify the first threshold crossing; a threshold might never be reached.

Compare white and slow noise

For ideal white noise, $S(\omega)=S_0$, Parseval’s identity gives $\chi=S_0T/2$ for both Ramsey and ideal echo because $y(t)^2=1$. Refocusing does not remove infinitely short-correlated noise. A finite frequency cutoff approximates white noise only over the relevant bandwidth.

For a quasistatic Gaussian frequency shift, Ramsey gives $W_R=e^{-\sigma^2T^2/2}$ and an ideal echo cancels the phase completely. A regularized $1/f$ spectrum lies between such limiting descriptions; its low-frequency cutoff represents part of the physical or measurement model. Report how the envelope changes when that cutoff changes.

If independent Markovian population relaxation also occurs, an exponential transverse envelope satisfies

$$\frac1{T_2}=\frac1{2T_1}+\frac1{T_\phi}.$$

For the Gaussian-noise envelopes above, one instead multiplies by $e^{-T/(2T_1)}$ under the corresponding independent-noise assumptions and solves the resulting threshold equation. Subtracting inverse threshold times generally does not isolate a physical pure-dephasing rate. In the constant-rate model, a negative inferred $1/T_\phi$ signals incompatible data or assumptions, not a negative dephasing mechanism.

Exercises

White-noise threshold. Find the $1/e$ time and state whether an ideal echo changes it.

Solution

$\chi=S_0T/2=1$ gives $T_*=2/S_0$. Ideal echo has the same result in this white-noise model.

Slow-noise expansion. Expand the echo exponent through its first nonzero power of $T$.

Solution

In $x-3+4e^{-x/2}-e^{-x}$, the constant, linear, and quadratic terms cancel; the cubic coefficient is $1/12$. Therefore $\chi_E=\sigma^2T^3/(12\tau_c)+O(T^4/\tau_c^2)$.

An exponential example. For genuinely exponential decay with $T_1=40\,\mu\mathrm{s}$ and $T_2=20\,\mu\mathrm{s}$, find $T_\phi$.

Solution

$1/T_\phi=1/20-1/80=3/80$ in inverse microseconds, so $T_\phi=80/3\,\mu\mathrm{s}\approx26.7\,\mu\mathrm{s}$.

A factor-of-two error. What happens if you integrate the even two-sided spectrum only over positive frequencies but retain the original prefactor $1/(4\pi)$?

Solution

You obtain half the correct $\chi$. In the white-noise case the inferred coherence time doubles. The missing negative-frequency contribution must be included by doubling the prefactor, unless the spectrum itself has already been converted to a one-sided convention.

References

• G. A. Álvarez and D. Suter, “Measuring the spectrum of colored noise by dynamical decoupling,” Physical Review Letters 107, 230501 (2011). doi:10.1103/PhysRevLett.107.230501.
• Ł. Cywiński, R. M. Lutchyn, C. P. Nave, and S. Das Sarma, “How to enhance dephasing time in superconducting qubits,” Physical Review B 77, 174509 (2008). doi:10.1103/PhysRevB.77.174509.
• C. L. Degen, F. Reinhard, and P. Cappellaro, “Quantum sensing,” Reviews of Modern Physics 89, 035002 (2017). doi:10.1103/RevModPhys.89.035002.
• G. Ithier et al., “Decoherence in a superconducting quantum bit circuit,” Physical Review B 72, 134519 (2005). doi:10.1103/PhysRevB.72.134519.