$\newcommand{ket}[1]{|#1\rangle} \newcommand{bbraket}[3]{\langle #1 | #2 | #3 \rangle}$ 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?
