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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Roger V.
    Jul 31, 2020 at 10:24
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    $\begingroup$ physics.stackexchange.com/q/311168 , physics.stackexchange.com/q/111017 (see the answer by Valter Moretti), physics.stackexchange.com/q/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 $\endgroup$
    – SRS
    Jul 31, 2020 at 11:18
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Aug 1, 2020 at 5:47
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    $\begingroup$ What about simply looking at the quantum version of your $E(t)$ in the comment above - i.e. the expectation value of the operator corresponding to it? $\endgroup$
    – ACuriousMind
    Jun 27, 2021 at 20:16
  • $\begingroup$ @ACuriousMind Expectation value in which state? It doesn't have a stationary state. $\endgroup$
    – SRS
    Jun 27, 2021 at 20:18

2 Answers 2


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.


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.


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$( \hat{a} \rho(t) )$


$<\dot{\hat{a}}(t)> = ( i \omega -\gamma/2) <\hat{a}(t)>$

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.


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.


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