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?
 A: Harmonic oscillators are—from a certain point of view—very boring quantum systems. Unless they're coupled to some nonlinear elements, their quantum evolution is exactly the same as the evolution of a corresponding classical oscillator would be. This has some important consequences: one of them states that Gaussian quantum systems (i.e., linearly coupled harmonic oscillators with linear decay and homodyne measurement; such systems can be described using Gaussian quasi-probability distribution in phase space) can be simulated classically and thus cannot be used for universal quantum computing. Another, trivial consequence is that the decay rates of a quantum and a classical harmonic oscillator have to be the same.
Note, however, that this is no longer the case if you add some sort of nonlinearity into the system. If you're able to measure the photon number in the real time, you will see photons entering and leaving the oscillator one by one, which is qualitatively very different from exponential decay. (You'd have to observe many such trajectories starting from the same initial state and average them to see the exponential decay.) And if you want to describe the decay of an atom, you have to use quantum mechanics.
