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

SRS
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DanielSank
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  • I don't know about QM, but I recall from my gas dynamics class that the harmonic oscillator was a suitable model for the vibration of molecules. So it wasn't exactly how they behaved, but it was close enough. My guess is that's probably true here -- in reality, it's not a perfectly harmonic oscillator. But it's close, so harmonic oscillation is a good way to model it. In other words, it's the same because they are all using the same model. – tpg2114 Feb 04 '15 at 20:13
  • @tpg2114: That's not really my question. The formula I give for the decay rate for the circuit is true in general, even for highly anharmonic circuits. The question is why the harmonic limit of the quantum case exactly matches the harmonic limit of the classical case. – DanielSank Feb 04 '15 at 20:15
  • I guess I'm confused. I'm saying they match because the model is the same for both of them. Any system that can be modeled as harmonic will have functionally the same decay rate. Maybe I'm missing the subtlety here. – tpg2114 Feb 04 '15 at 20:16
  • In the anharmonic case I think the quantum and classical models do not agree. I could be wrong about that. – DanielSank Feb 04 '15 at 21:10
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    The quantum harmonic oscillator can be seen in many ways to "not contain quantum effects of higher order than $\mathcal{O}(\hbar)$". It seems to me that this is another way, but I do not know of a true "reason" for it. – ACuriousMind Feb 05 '15 at 00:24
  • @ACuriousMind I don't like $\mathcal{O}(\hbar)$. When we talk about "orders" we should be talking about expansions in terms of a dimensionless quantity. $\hbar$ is not dimensionless. – DanielSank Jul 01 '15 at 03:23
  • The answer is trivial enough since the quantity do not depend on $\hbar$ (I am supposing that $R$ and $C$ do not depend on $\hbar$) and hence converges uniformly in the limit. I could see no deep reason for that, it seems just a very favourable case where we have a quantum object that is $\hbar$-independent. The non-trivial fact is to interpret the left-hand side as converging to the corresponding classical expression... – yuggib Nov 12 '15 at 08:02
  • On a second thought, it is not so uncommon. Take the energy of a QHO, and normalize it dividing by $\hbar$. Then its expectation on quantum (eigen)states would be $\hbar$-independent. But how to justify/know what the left hand side converges to in the limit? For it has to be something classical as well (if interpreted in the correct way) – yuggib Nov 12 '15 at 08:06
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    I'm not sure if this counts as a "deep" reason, but I believe it is related to the linearity of the equations of motion (EoM) for the quadratures of the harmonic oscillator (HO). As is well known, the unitary dynamics of the quadratures are identical to the classical EoM. It turns out that even in the dissipative case of a single quantum HO coupled linearly to an infinite heat bath of quantum HOs, the quantum quadratures obey a Langevin equation that is identical in form to the classical one (within the Markov approximation). This is expounded in detail in Gardiner & Zoller's textbook. – Mark Mitchison Nov 13 '15 at 13:09
  • @Mark Mitchison : the $64,000 dollar question is why are the unitary dynamics of the quadratures identical to the classical EoM? – John Duffield Nov 13 '15 at 17:10
  • @JohnDuffield Yes, that is the $64,000 question. I wonder if Mark can expand on his comment. – DanielSank Nov 13 '15 at 17:56
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    @JohnDuffield If you mean how does this form arise, it is as simple as$[a^\dagger a,a]=-a$. Or perhaps the right way to think about it is that Gaussian systems are completely defined in terms of their first and second moments, both in the quantum and classical case. Or if you like, the quantum Gaussian path integral can be evaluated exactly and leads to merely the classical action, up to a fluctuation determinant due to non-commutativity. There are surely many ways of seeing how these equations arise (see ACuriousMind's comment). The question of why may be answerable, but not by me. Sorry. – Mark Mitchison Nov 13 '15 at 18:34
  • @MarkMitchison Didn't know we have an open quantum systems tag. Nice. I bet I can come up with some more questions for it :D – DanielSank Nov 14 '15 at 06:07
  • @DanielSank I created it :) Please do add more questions. As far as I can tell, it mainly exists for our own amusement anyway. – Mark Mitchison Nov 14 '15 at 17:35

1 Answers1

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Harmonic oscillators are—from a certain point of view—very boring quantum systems. Unless they're coupled to some nonlinear elements, their quantum evolution is exactly the same as the evolution of a corresponding classical oscillator would be. This has some important consequences: one of them states that Gaussian quantum systems (i.e., linearly coupled harmonic oscillators with linear decay and homodyne measurement; such systems can be described using Gaussian quasi-probability distribution in phase space) can be simulated classically and thus cannot be used for universal quantum computing. Another, trivial consequence is that the decay rates of a quantum and a classical harmonic oscillator have to be the same.

Note, however, that this is no longer the case if you add some sort of nonlinearity into the system. If you're able to measure the photon number in the real time, you will see photons entering and leaving the oscillator one by one, which is qualitatively very different from exponential decay. (You'd have to observe many such trajectories starting from the same initial state and average them to see the exponential decay.) And if you want to describe the decay of an atom, you have to use quantum mechanics.

Ondřej Černotík
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