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 $0.24\,\mathrm K$. 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,

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

In a thermal state with occupation $\bar n$ over the relevant bandwidth,

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

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.

Emission and Absorption

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

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

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.

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.

Common Mistakes

• Treating vacuum as zero noise.
• Treating symmetrized zero-point noise as if it automatically excites systems.
• Forgetting the $+1$ spontaneous-emission term in bosonic damping.
• Ignoring thermal occupation at microwave or mechanical frequencies.
• Using room-temperature intuition without comparing $k_BT$ to $\hbar\omega$.
• Applying finite-temperature upward rates to an effectively zero-temperature optical bath.
• Calling technical noise thermal without checking equilibrium detailed balance.
• Mixing ordered spectra, symmetrized spectra, one-sided spectra, and two-sided spectra.

Cross-Links

• Thermal Light for mode-resolved thermal photon statistics, bunching, and optical coherence measurements.
• Quantum Noise for the broader noise taxonomy.
• Noise Spectra for ordered and symmetrized spectral conventions.
• Correlation Functions for time-domain thermal correlations.
• Fluctuation–Dissipation Relation for equilibrium response constraints.
• Fluctuation–Dissipation Theorem for the many-body KMS derivation and quantum-to-classical limits.
• Input–Output Theory for Markov traveling-field normalization.
• Quantum Langevin Equations for damping with input noise operators.
• Thermal Master Equations for Gibbs-relaxing Markov generators.
• Quantum Optical Master Equation for optical and cavity examples.
• Transition Rates in Light–Matter Interaction for the mode-resolved origin of the $N$ and $N+1$ factors.
• Approximation Checklist for thermal, Markov, and detailed-balance checks.

References

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

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

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