In what sense is a quantum damped harmonic oscillator dissipative? The classical Hamiltonian of a damped harmonic oscillator $$H=\frac{p^2}{2m}e^{-\gamma t}+\frac{1}{2}m\omega^2e^{\gamma t}x^2,~(\gamma>0)\tag{1}$$ when promoted to quantum version, remains hermitian. Therefore, the time evolution of the system is unitary and probability conserving. The Heisenberg equation of motion, for the operators $x$ and $p$ derived from this Hamiltonian matches perfectly with the classical Hamilton's EoM, in appearance: $$\dot{x}=\frac{p}{m}e^{-\gamma t}, ~\dot{p}=m\omega^2xe^{\gamma t}.\tag{2}$$
Question Quantum mechanically, how to show that this is a dissipative system? Note that this system has no stationary states.
 A: I'll use the convention of writing the exponent as $\gamma t / m$ rather than $\gamma t$.
The actual energy of the HO is
$$ E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2x^2 = \frac{1}{2m}p^2\mathrm{e}^{-2\gamma t/m} + \frac{1}{2}m\omega^2x^2 = \mathrm{e}^{-\gamma t/m} H$$
since $p = \partial_{\dot{x}}L = \mathrm{e}^{\gamma t/m}m\dot{x}$.
Ehrenfest's theorem means that
$$ \frac{\mathrm{d}}{\mathrm{d}t}\langle E\rangle  = -\mathrm{i}\langle [E,H]\rangle + \langle \partial_t E\rangle = -\frac{\gamma}{m}\mathrm{e}^{-\gamma t/m}\langle H\rangle  + \mathrm{e}^{-\gamma t / m}\langle \partial_t H\rangle= -\frac{\gamma}{m}\langle E\rangle + \mathrm{e}^{-\gamma t / m}\langle \partial_t H\rangle,$$
so as $t\to \infty$ (meaning we can ignore the second term, not a literal limit) we have that $\langle E\rangle(t) \to \mathrm{e}^{-\gamma t / m }\langle E\rangle (0)$, same as in the classical case.
A: In quantum mechanics a Hamiltonian can only capture the coherent (non-disipative) dynamics of a closed quantum system. Where closed means that is does not interact with its environment.
Interactions with the outside environment can introduce friction, which cannot be modeled using only a Hamiltonian.
The most common way of dealing with these problems is using a Lindbladian (Lindblad Master equation). This models the evolution of the density operator, $\rho$. For the problem you mention the most natural representation might be (at zero temperature):
$\dot {\rho } =-{i \over \hbar }[H,\rho ]+ (\gamma/2) ( 2 \hat{a} \rho \hat{a}^{\dagger}  - \hat{a}^{\dagger} \hat{a}\rho - \rho \hat{a}^{\dagger} \hat{a} ) $
Where $H = \hbar \omega \hat{a}^{\dagger} \hat{a}$,  the ordinary (unchanged) Hamiltonian for a simple Harmonic oscilator at frequency $\omega$ ($\hat{a}$ is the anhilation operator,  $\hat{a} = (\hat{x} + \hat{p}) / \sqrt{2}$ in dimensionless units.)
Ref: https://en.wikipedia.org/wiki/Lindbladian
Their are complex-valued Hamiltonians that people sometimes use to approximate dissipative systems, but those are approximations. A dissipative Harmonic oscillator prepared in a pure state can arrive in a mixed state, something no Hamiltonian alone can do.
Example
Now we try and calcualte how the expected position of the particle changes with time. We will first do $\hat{a}$:
$<\dot{\hat{a}}(t)> = $ Trace$( \hat{a} \dot{\rho}(t) )$
Some re-arrangement (commutators and the cyclic property of the trace) gives:
$<\dot{\hat{a}}(t)> = ( i \omega -\gamma/2) $ Trace$( \hat{a} \rho(t) )$
IE:
$<\dot{\hat{a}}(t)> = ( i \omega -\gamma/2) <\hat{a}(t)>$
Re-arranging and using $<x> = (<\hat{a}> + <\hat{a}^\dagger>) / \sqrt{2}$ and similarly $<p> = (<\hat{a}> - <\hat{a}^\dagger>) / i \sqrt{2}$ you find:
$<\dot{\hat{x}}> = -(\gamma/2)<\hat{x}> + \omega <\hat{p}>$
With something similar for $p$. The solutions to the combined equations are exponentially decaying at a rate set by $\gamma$ and oscillating at frequency $\omega$ as expected.
EDIT:
I just re-read your question, and either you edited it for clarity or I just read it badly the first time. I hope what I have written helps but I am not sure it really answers your question at all.
As an aside, I am worried about your equation. If I prepare a damped harmonic oscillator in the state $(|0> + |N> )/ \sqrt{2}$ where N is some colossal number of photons, then I expect to find that I very rapidly evolve into a statistically mixed state, $\rho(t) \sim (|0><0| + |N><N|)/2$. Unitary (Hamiltonain) evolution appears unable to achieve that.
