Thermal and Vacuum Noise

Thermal and vacuum noise are the two equilibrium limits that every open-system model must keep distinct. Vacuum noise is the zero-temperature quantum fluctuation structure of a field or bath. Thermal noise is the additional fluctuation caused by finite occupation of bath modes.

The difference is not cosmetic:

• vacuum can accept energy from an excited system, producing spontaneous emission;
• thermal occupation can also supply energy, producing excitation;
• symmetrized zero-point noise can be nonzero even when upward excitation is absent;
• the ratio of upward to downward rates is fixed by detailed balance at equilibrium.

For the general taxonomy of classical, quantum, vacuum, thermal, and technical noise, see Quantum Noise. This page focuses on the equilibrium bosonic formulas and their physical consequences.

Thermal Occupation

For a bosonic mode of angular frequency $\omega$, the thermal occupation is

$$\bar n(\omega,T) = \frac{1} {e^{\beta\hbar\omega}-1}, \qquad \beta=\frac{1}{k_BT}.$$

Two limits are especially important:

$$\bar n \approx \frac{k_BT}{\hbar\omega} \qquad \text{when } k_BT\gg\hbar\omega,$$

and

$$\bar n \approx e^{-\beta\hbar\omega} \qquad \text{when } k_BT\ll\hbar\omega.$$

Thus a reservoir can be thermally noisy for microwave or mechanical frequencies while being effectively in vacuum for optical frequencies at the same laboratory temperature.

In ordinary frequency $f=\omega/(2\pi)$,

$$\frac{\hbar\omega}{k_B} = \frac{hf}{k_B} \approx 0.048\,\mathrm K \left( \frac{f}{1\,\mathrm{GHz}} \right).$$

A 5 GHz microwave mode corresponds to about $0.24\,\mathrm K$. Its equilibrium occupation is approximately $6.2\times10^{-6}$ at $20\,\mathrm{mK}$ but $0.10$ at $100\,\mathrm{mK}$. These are mode temperatures, which need not equal the refrigerator thermometer reading if incoming radiation is poorly thermalized. A visible optical mode corresponds to tens of thousands of kelvin. This scale explains why room-temperature thermal photons are usually negligible in optical quantum optics but not in microwave engineering.

Vacuum Is Not No Noise

A harmonic oscillator or field mode in its ground state has

$$\langle a^\dagger a\rangle=0,$$

but it does not have zero quadrature fluctuations. For dimensionless quadratures

$$X = \frac{a+a^\dagger}{\sqrt2}, \qquad P = \frac{a-a^\dagger}{i\sqrt2},$$

the ground state has

$$\langle X^2\rangle = \langle P^2\rangle = \frac12.$$

For a thermal state,

$$\langle X^2\rangle = \langle P^2\rangle = \bar n+\frac12.$$

The $\frac12$ is the zero-point contribution. It is visible in symmetrized noise and uncertainty relations. It should not be confused with real thermal occupation.

Markov Input Correlations

For a broadband bosonic input field, the field commutator is

$$[b_{\mathrm{in}}(t),b_{\mathrm{in}}^\dagger(t')] = \delta(t-t').$$

In vacuum,

$$\langle b_{\mathrm{in}}(t)b_{\mathrm{in}}^\dagger(t')\rangle = \delta(t-t'), \qquad \langle b_{\mathrm{in}}^\dagger(t)b_{\mathrm{in}}(t')\rangle = 0.$$

Approximating the occupation by a constant $\bar n$ over the relevant band around a carrier gives

$$\langle b_{\mathrm{in}}^\dagger(t)b_{\mathrm{in}}(t')\rangle = \bar n\,\delta(t-t'),$$

and

$$\langle b_{\mathrm{in}}(t)b_{\mathrm{in}}^\dagger(t')\rangle = (\bar n+1)\delta(t-t').$$

The $+1$ is the vacuum contribution. It is why emission into an empty bath remains possible. A constant thermal occupation and delta correlations are a local-band Markov approximation; the exact Planck occupation varies with laboratory frequency.

Emission and Absorption

For a two-level system with transition frequency $\omega_0$, an equilibrium bosonic bath gives rates of the schematic form

$$\Gamma_\downarrow = \Gamma_0(\bar n+1), \qquad \Gamma_\uparrow = \Gamma_0\bar n.$$

The weak-coupling, rotating-wave model uses $\bar n=\bar n(\omega_0,T)$ and the vacuum rate $\Gamma_0$. Here $\Gamma_\downarrow$ is the rate for the system to emit energy $\hbar\omega_0$ into the bath, and $\Gamma_\uparrow$ is the rate for the bath to excite the system.

The ratio is

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = \frac{\bar n}{\bar n+1} = e^{-\beta\hbar\omega_0}.$$

This is detailed balance. At zero temperature, $\bar n=0$, so

$$\Gamma_\uparrow=0, \qquad \Gamma_\downarrow=\Gamma_0.$$

Vacuum does not thermally excite the atom, but it still permits spontaneous emission because the bath can absorb energy. The qubit population relaxes with total rate $\Gamma_0(2\bar n+1)$ toward $p_e^{\mathrm{eq}}=\bar n/(2\bar n+1)$. This differs from the harmonic oscillator below: the qubit saturates at population $1/2$, whereas an oscillator can hold arbitrarily many excitations.

Oscillator Thermalization

For a harmonic oscillator coupled weakly to a thermal bosonic reservoir, the standard Markovian thermal master equation is

$$\dot\rho = - \frac{i}{\hbar} [\hbar\omega a^\dagger a,\rho] + \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho.$$

The mean occupation obeys

$$\frac{d}{dt} \langle a^\dagger a\rangle = -\kappa \left( \langle a^\dagger a\rangle-\bar n \right).$$

Therefore

$$\langle a^\dagger a\rangle(t) = \bar n + \left[ \langle a^\dagger a\rangle(0)-\bar n \right] e^{-\kappa t}.$$

Vacuum damping is the special case $\bar n=0$. Finite temperature adds the $a^\dagger$ absorption term and changes the steady state.

Symmetrized Versus Ordered Noise

The symmetrized oscillator factor is

$$2\bar n+1 = \coth \left( \frac{\beta\hbar\omega}{2} \right).$$

It contains both thermal and zero-point contributions. Fluctuation–Dissipation Theorem derives this factor from KMS balance, while the Fluctuation–Dissipation Relation applies it to bath noise and damping.

Ordered spectra distinguish whether the bath absorbs or supplies energy. For $\omega>0$, in the convention used in Noise Spectra,

$$S(+\omega) \propto \bar n+1, \qquad S(-\omega) \propto \bar n.$$

The symmetrized spectrum averages these two directions. It is useful for detector noise power, but it cannot by itself tell whether an unexcited bath can drive a quantum transition.

Frequency and Temperature Scales

The relevant question is not whether the laboratory is “cold” in ordinary terms. The question is whether $k_BT$ is small or large compared with $\hbar\omega$ for the transition being modeled.

Examples:

System scale	Typical conclusion
optical transition	room-temperature thermal occupation is negligible
microwave cavity or qubit	thermal photons depend strongly on cryogenic filtering and temperature
mechanical resonator	thermal occupation can remain large unless deeply cooled
low-frequency dephasing noise	classical thermal or technical noise may dominate
high-frequency spontaneous emission	vacuum contribution can dominate

This is why the same laboratory can contain both nearly vacuum optical reservoirs and highly occupied low-frequency technical environments.

What Vacuum Noise Does Not Mean

Vacuum noise does not mean the bath contains hidden classical random photons. It means the quantum field has noncommuting operators and ground-state fluctuations.

Vacuum noise also does not mean every detector can extract energy from the vacuum. A ground-state bath cannot excite an ordinary detector at positive transition frequency in equilibrium. The distinction is encoded by ordered spectra and detailed balance.

Finally, vacuum fluctuations do not by themselves specify a measurement record. A record appears only after an output field is measured; otherwise the same field acts as an unobserved reservoir.

Exercises

Detailed-balance ratio

Using

$$\bar n = \frac{1} {e^{\beta\hbar\omega}-1},$$

show that $\bar n/(\bar n+1)=e^{-\beta\hbar\omega}$.

Solution

Let $x=e^{\beta\hbar\omega}$. Then

$$\bar n = \frac{1}{x-1}, \qquad \bar n+1 = \frac{x}{x-1}.$$

Therefore

$$\frac{\bar n}{\bar n+1} = \frac{1}{x} = e^{-\beta\hbar\omega}.$$

Vacuum damping limit

Take $\bar n=0$ in the thermal oscillator master equation. What remains?

Solution

The equation

$$\dot\rho = - \frac{i}{\hbar} [\hbar\omega a^\dagger a,\rho] + \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho$$

becomes

$$\dot\rho = - \frac{i}{\hbar} [\hbar\omega a^\dagger a,\rho] + \kappa\mathcal D[a]\rho.$$

Only loss remains; the oscillator relaxes toward vacuum.

Occupation scale

Estimate $\hbar\omega/k_B$ for a 5 GHz mode and explain why this matters for microwave experiments.

Solution

Using

$$\frac{\hbar\omega}{k_B} = \frac{hf}{k_B} \approx 0.048\,\mathrm K \left( \frac{f}{1\,\mathrm{GHz}} \right),$$

a 5 GHz mode has

$$\frac{\hbar\omega}{k_B} \approx 0.24\,\mathrm K.$$

Thermal occupation is negligible only when the effective mode temperature is well below this scale. This is why cryogenic attenuation, filtering, and thermalization matter in microwave quantum experiments.

Symmetrized zero-point noise

For a thermal oscillator, show that

$$\langle X^2\rangle = \bar n+\frac12$$

for $X=(a+a^\dagger)/\sqrt2$.

Solution

Use

$$X^2 = \frac12 \left( a^2+a^{\dagger 2}+aa^\dagger+a^\dagger a \right).$$

In a thermal state,

$$\langle a^2\rangle = \langle a^{\dagger 2}\rangle = 0, \qquad \langle a^\dagger a\rangle = \bar n,$$

and

$$\langle aa^\dagger\rangle = \bar n+1.$$

Therefore

$$\langle X^2\rangle = \frac12 \left[ (\bar n+1)+\bar n \right] = \bar n+\frac12.$$

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• H. J. Carmichael, An Open Systems Approach to Quantum Optics, Springer (1993).
• A. A. Clerk, 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.
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004).
• R. Kubo, “The fluctuation-dissipation theorem,” Reports on Progress in Physics 29, 255–284 (1966).
• D. F. Walls and G. J. Milburn, Quantum Optics, 2nd ed., Springer (2008).