Common Noise Spectra

A noise spectrum describes correlations resolved by frequency. A plotted line shape becomes physically useful only after its normalization, frequency units, and operator ordering are specified. This reference gives normalized examples and connects them to qubit transition rates.

Required background. Noise Spectra defines stationary correlations and distinguishes ordered from symmetrized quantum noise.

Correlations, Fourier factors, and units

Subtract the mean of the fluctuating quantity $B$. For a stationary Hermitian bath observable or a real classical process, use

$$\begin{gathered} C(t)=\langle B(t)B(0)\rangle,\\ S(\omega)=\int_{-\infty}^{\infty}e^{i\omega t}C(t)\,dt,\\ C(t)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi} e^{-i\omega t}S(\omega). \end{gathered}$$

$S$ has units $[B]^2\times\text{time}$. The variance is $C(0)=\int S(\omega)d\omega/(2\pi)$ when finite. A nonzero mean would add $2\pi\langle B\rangle^2\delta(\omega)$, which should not be confused with fluctuating low-frequency noise.

Centering does not remove every zero-frequency line. A conserved quantity can fluctuate between preparations while remaining constant during each run, leaving $2\pi\operatorname{Var}(B)\delta(\omega)$ and zero commutator response. The centered-energy example illustrates why such elastic weight must be treated separately from the regular fluctuation–dissipation formula.

A real stationary classical scalar process has an even spectrum. The ordered quantum spectrum generally is not even. Its symmetrized version is $S_{\rm sym}(\omega)=[S(\omega)+S(-\omega)]/2$.

If $S_f(f)=\int e^{i2\pi ft}C(t)dt$, then $S_f(f)=S(2\pi f)$ and the variance is $\int S_f(f)df$. For real classical noise the usual one-sided spectrum at $f>0$ is $S_f^{(1)}(f)=2S_f(f)$. Stating the integral for variance avoids ambiguous “per hertz” versus “per angular frequency” conventions.

Classical white, Lorentzian, and 1/f spectra

Model	Two-sided spectrum	Meaning
Ideal white noise	$S(\omega)=S_0$	Generalized correlation $C(t)=S_0\delta(t)$
Exponential correlation	$2\sigma^2\tau/[1+(\omega\tau)^2]$	$C(t)=\sigma^2e^{-\lvert t\rvert/\tau}$
Band-limited 1/f noise	$A/\lvert\omega\rvert$ between two cutoffs	Many slow timescales; no single decay time

White noise is a useful approximation over the frequencies sampled by the system. Extending it to arbitrarily large frequency gives divergent instantaneous variance. Unlike a $1/|\omega|$ spectrum, a constant spectrum has no infrared divergence and does not require a low-frequency cutoff for that reason. With the Fourier convention above, the delta-correlation coefficient is exactly $S_0$, without an unspecified $2\pi$ factor.

Lorentzian noise has half-width $1/\tau$ and variance $\sigma^2$. A narrow line means long correlation time, but whether it invalidates a Markov approximation also depends on coupling strength and the timescale of the reduced dynamics. A real classical oscillating correlation,

$$C(t)=\sigma^2e^{-|t|/\tau}\cos(\omega_c t),$$

has two peaks:

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

A single displaced Lorentzian is not the full spectrum of a real stationary classical scalar process. It can instead describe a complex amplitude or one part of a quantum ordered spectrum. See Memory Kernels and Pseudomode Methods.

1/f noise needs both infrared and ultraviolet limits for a finite variance. For $S(\omega)=A/|\omega|$ on $\omega_{\rm ir}\le|\omega|\le\omega_{\rm uv}$ and zero outside,

$$\sigma^2=\frac A\pi\ln\frac{\omega_{\rm uv}}{\omega_{\rm ir}}.$$

The same slow fluctuations can be almost constant during one run and drift between runs. State the measurement duration and averaging protocol when connecting this spectrum to dephasing. See Pure Dephasing Model and Decoherence Timescales.

Oscillator spectral densities and thermal ordering

A bath spectral density $J$ is not automatically a noise spectrum. To fix factors, take an energy-valued operator

$$\begin{gathered} B=\sum_k(g_k b_k+g_k^*b_k^\dagger),\\ J(\omega)=\sum_k|g_k|^2\delta(\omega-\omega_k), \qquad\omega>0. \end{gathered}$$

$J$ has units energy squared times time. For independent thermal modes,

$$\begin{gathered} S(+\omega)=2\pi J(\omega)[n_B(\omega)+1],\\ S(-\omega)=2\pi J(\omega)n_B(\omega),\\ n_B(\omega)=\frac1{e^{\beta\hbar\omega}-1}. \end{gathered}$$

An Ohmic density has $J(\omega)\propto\omega$ at low positive frequency, with a specified ultraviolet cutoff. For example, $J(\omega)=\alpha\hbar^2\omega e^{-\omega/\omega_c}$ uses dimensionless $\alpha$. In this convention its displacement energy scale is $\int_0^\infty J(\omega)d\omega/(\hbar\omega)=\alpha\hbar\omega_c$; alternative coupling normalizations change this coefficient. Coordinate-coupling spectral densities use different units and factors. See Spectral Densities and Caldeira–Leggett Model before translating them.

For $H_I=A\otimes B$ with dimensionless $A$, a transition of gap $\hbar\omega_0>0$ has weak-coupling rates

$$\begin{gathered} \Gamma_\downarrow=\frac{|A_{ge}|^2}{\hbar^2}S(+\omega_0),\\ \Gamma_\uparrow=\frac{|A_{ge}|^2}{\hbar^2}S(-\omega_0). \end{gathered}$$

Thus $S(-\omega)=e^{-\beta\hbar\omega}S(+\omega)$ expresses equilibrium detailed balance. In this sign convention, positive frequency describes the bath absorbing energy from the system. See Thermal Master Equations and Fluctuation–Dissipation Relation.

At zero temperature, $S(-\omega)=0$ for $\omega>0$: the vacuum can accept energy but does not thermally excite a positive-gap system in this stationary weak-coupling setting. The symmetrized vacuum spectrum is still nonzero on both sides. Substituting it for the ordered spectrum in a transition-rate formula would incorrectly add upward transitions.

Exercises

A normalized noise peak. For $C(t)=\sigma^2e^{-|t|/\tau}$, double $\tau$ at fixed variance. What changes in the spectrum?

Solution

The zero-frequency height $2\sigma^2\tau$ doubles and the half-width $1/\tau$ halves. The integrated area remains $2\pi\sigma^2$. Narrower does not mean less total noise.

Temperature from asymmetry. At a transition frequency, $S(-\omega_0)/S(+\omega_0)=1/3$. Find the temperature and stationary qubit excitation probability for a single equilibrium reservoir.

Solution

$k_BT=\hbar\omega_0/\ln3$ and $p_e^*=\Gamma_\uparrow/(\Gamma_\uparrow+\Gamma_\downarrow)=1/4$. The temperature inference requires equilibrium; a driven reservoir can produce an asymmetric spectrum without a single bath temperature.

References

• Breuer, H.-P., and F. Petruccione. The Theory of Open Quantum Systems. Oxford University Press, 2002. Thermal correlations and transition rates.
• Clerk, A. A., M. H. Devoret, S. M. Girvin, F. Marquardt, and R. J. Schoelkopf. “Introduction to quantum noise, measurement, and amplification.” Reviews of Modern Physics 82, 1155–1208 (2010). doi:10.1103/RevModPhys.82.1155.
• Gardiner, C. W., and P. Zoller. Quantum Noise. 3rd edition, Springer, 2004. Stochastic and operator-valued noise.
• Paladino, E., Y. M. Galperin, G. Falci, and B. L. Altshuler. “1/f noise: Implications for solid-state quantum information.” Reviews of Modern Physics 86, 361–418 (2014). doi:10.1103/RevModPhys.86.361.