# 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. Jul 31, 2020 at 10:24
• 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
– SRS
Jul 31, 2020 at 11:18
• 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, 2020 at 5:47
• 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? Jun 27, 2021 at 20:16
• @ACuriousMind Expectation value in which state? It doesn't have a stationary state.
– SRS
Jun 27, 2021 at 20:18

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.)

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 $$ = (<\hat{a}> + <\hat{a}^\dagger>) / \sqrt{2}$$ and similarly $$

= (<\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>. Unitary (Hamiltonain) evolution appears unable to achieve that.