# 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.

• Could you please give a reference where the Hamiltonian of a damped oscillator is written in this way? This is the first time I see anything like this... and I doubt that one gets from it the damped oscillator equations. – Vadim Jul 31 at 10:24
• physics.stackexchange.com/questions/311168/… , physics.stackexchange.com/questions/111017/… (see the answer by Valter Moretti), physics.stackexchange.com/questions/258395/… If you use the proper Lagrangian, you'll indeed get the damped oscillator EoM. As far as I know, this is one of the few dissipative systems for which a Hamiltonian can be written down. @Vadim – SRS Jul 31 at 11:18
• If we define "dissipative" to mean that the system loses kinetic energy as time passes, then the quantum system you described is clearly dissipative. This is consistent with the family of related definitions given in Wikipedia, but it's not the only justifiable definition. Do you have another specific definition in mind? For example, are you asking if this model can be derived from a larger model having a time-independent Hamiltonian and an explicit bath into which the energy can disperse? – Chiral Anomaly Jul 31 at 21:28
• No. In this post, I'm not asking for a "larger" model. I am asking in what sense the above quantum hamiltonian describes a dissipative system. The classical version is dissipative in the sense that, if you define the instantaneous total energy to be $E(t)=\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\omega^2x^2$ (because in this case, the Hamiltonian does not give the total energy), then it can be checked that $E(t)\sim e^{-\gamma t}$. Similarly, if the quantum version is dissipative too, which quantity should we look at to draw a conclusion? @ChiralAnomaly – SRS Aug 1 at 5:47