# Why does a damped quantum harmonic oscillator have the same decay rate as the equivalent classical system?

$\newcommand{ket}[1]{|#1\rangle} \newcommand{bbraket}[3]{\langle #1 | #2 | #3 \rangle}$ Why does the decay rate for a damped quantum harmonic oscillator exactly match the classical limit?

# Background

Consider a localized quantum system $S$ connected to a 1D wave-supporting continuum. If $S$ is excited, say into $\ket{1}$, then coupling between $S$ and the continuum means that there is a process wherein $S$ undergoes $\ket{1} \rightarrow \ket{0}$ and emits an excitation (e.g. a photon) into the continuum. The rate for this decay process is traditionally computed via Fermi's golden rule. Specifically, if the operator coupling $S$ to the continuum is $\hat{V}$, then the decay rate is

$$\Gamma_\downarrow = 2 \pi \frac{|\bbraket{0}{\hat{V}}{1}|^2}{\hbar} \rho$$

where $\rho$ is the density of states in the continuum at the energy $E_1 - E_0$.

# Example systems

This system is realized many ways in nature. A very canonical example is an atom in space. An excited electron state in the atom can decay with emission of a photon into the vacuum. In this case, $\hat{V}$ is typically the electric field operator multiplied by the dipole moment of the electron transition.

A superconducting circuit connected to a resistor $R$ also experiences decay; in this case, the resistor plays the role of the continuum. In fact, one can compute that the decay rate is

$$\Gamma_\downarrow = \frac{2 \omega_{10}}{\hbar R} \bbraket{0}{\hat{\Phi}}{1}^2$$

where $\omega_{10} \equiv E_{10}/\hbar = 1/\sqrt{LC}$ and $\hat{\Phi}$ is the operator for the flux in the circuit.

# Harmonic case

In the case that the circuit is an $LC$ oscillator, the matrix element is

$$\bbraket{0}{\hat{\Phi}}{1}^2 = \frac{\hbar}{2} \sqrt{\frac{L}{C}}$$

which when plugged into the decay rate gives

$$\Gamma_\downarrow = \frac{1}{RC} \, .$$

This is just exactly the classical result for the energy decay rate of an $LCR$ circuit! Is there a deep reason for this correspondence?

• I don't know about QM, but I recall from my gas dynamics class that the harmonic oscillator was a suitable model for the vibration of molecules. So it wasn't exactly how they behaved, but it was close enough. My guess is that's probably true here -- in reality, it's not a perfectly harmonic oscillator. But it's close, so harmonic oscillation is a good way to model it. In other words, it's the same because they are all using the same model. – tpg2114 Feb 4 '15 at 20:13
• @tpg2114: That's not really my question. The formula I give for the decay rate for the circuit is true in general, even for highly anharmonic circuits. The question is why the harmonic limit of the quantum case exactly matches the harmonic limit of the classical case. – DanielSank Feb 4 '15 at 20:15
• The quantum harmonic oscillator can be seen in many ways to "not contain quantum effects of higher order than $\mathcal{O}(\hbar)$". It seems to me that this is another way, but I do not know of a true "reason" for it. – ACuriousMind Feb 5 '15 at 0:24
• I'm not sure if this counts as a "deep" reason, but I believe it is related to the linearity of the equations of motion (EoM) for the quadratures of the harmonic oscillator (HO). As is well known, the unitary dynamics of the quadratures are identical to the classical EoM. It turns out that even in the dissipative case of a single quantum HO coupled linearly to an infinite heat bath of quantum HOs, the quantum quadratures obey a Langevin equation that is identical in form to the classical one (within the Markov approximation). This is expounded in detail in Gardiner & Zoller's textbook. – Mark Mitchison Nov 13 '15 at 13:09
• @JohnDuffield If you mean how does this form arise, it is as simple as$[a^\dagger a,a]=-a$. Or perhaps the right way to think about it is that Gaussian systems are completely defined in terms of their first and second moments, both in the quantum and classical case. Or if you like, the quantum Gaussian path integral can be evaluated exactly and leads to merely the classical action, up to a fluctuation determinant due to non-commutativity. There are surely many ways of seeing how these equations arise (see ACuriousMind's comment). The question of why may be answerable, but not by me. Sorry. – Mark Mitchison Nov 13 '15 at 18:34