4
$\begingroup$

I have been studying open quantum systems for some time now. I have learnt about something known as spectral density that confers information about the physical structure and are found in the definition of the correlation functions. Now, the definition of spectral density is:

$$J(ω) = \sum_i ω^2_i λ^2_i δ(ω − ω_i)$$

where $ω_i$ are frequencies of the harmonic-oscillator bath modes and $λ_i$ are dimensionless couplings to the respective modes.

However, sometimes I have seen the spectral density defined as

$$J(\omega)=2\hbar\gamma\lambda\omega/(\omega^2+\gamma^2)$$

where gamma is the reorganization energy and lambda is the cutoff frequency. I hear this is called Ohmic spectral density with Lorentz-Drude cutoff function, correct? How is this related to the formal definition of spectral density?

Also, I have seen that the spectral density is also written as

$$J(\omega)=\frac{\gamma}{\hbar\lambda}\omega e^{-\omega/\lambda}$$

I hear this is also called Ohmic spectral density. My main question is, how is the second equation and third equation, both called Ohmic density, equal to each other or related to each other?

$\endgroup$
  • $\begingroup$ Note that the best starting point for spectral density is the definition $S_{uu}(\omega) = \int_{-\infty}^\infty dt \, \langle u(t) u(0) \rangle e^{i \omega t}$. The thing you called $J$ is related. See this other question. $\endgroup$ – DanielSank Aug 7 '15 at 21:09
4
$\begingroup$

When studying Markovian quantum systems, the low-frequency ($\omega \ll \lambda$) behaviour of the spectral density is most important. This is because the most relevant modes of the bath, which control the open system dynamics, have frequencies commensurate with the frequencies of the open system. Meanwhile, the open system frequency scales must all be much less than $\lambda$: this implies that the relevant bath correlation functions decay on a time scale $\lambda^{-1}$ much less than system time scales, ensuring that the Markov assumption holds. However, in order to avoid certain divergent integrals, a high-frequency cutoff function, e.g. the exponential cutoff $$ f_\lambda(\omega) = e^{-\omega/\lambda}$$ or the Drude-Lorentz cutoff $$f_\lambda(\omega) = \frac{\lambda^2}{\omega^2 + \lambda^2}$$ must be introduced. As long as the cutoff function satisfies the conditions $$ \lim_{\omega\to 0}f_\lambda(\omega) = 1 $$ and $$ \lim_{\omega\to \infty}f_\lambda(\omega) = 0, $$its exact form is usually unimportant for the qualitative physics at frequencies $\omega\ll \lambda$. For this reason, spectral densities of the form $$ J(\omega) = \frac{\gamma}{\lambda} \omega f_\lambda(\omega)$$ are generally called Ohmic (up to factors of, say, $2$, and I have set $\hbar = 1$ in order to avoid a discussion about the different possible definitions/units for $J(\omega)$). Here, Ohmic simply means that the low-frequency behaviour of $J(\omega)$ is linear in $\omega$. This may be contrasted with super-Ohmic or sub-Ohmic spectral densities, whose low-frequency behaviour is $J(\omega) \sim \omega^s$, with $s > 1$ or $s<1$, respectively.

When one writes down an effective bath model, usually the cutoff function $f(\omega)$ can be chosen to make the theoretical manipulations easier. However, if you actually have a microscopic model for the system-bath interaction then the form of the spectral density, and in particular the cutoff, is usually dictated by the microscopic physics. Furthermore, outside of the Markovian regime, when the relevant frequencies may be on the order of $\lambda$, of course the form of the cutoff is extremely important.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great answer! To confirm the values close to 0 (or values less than gamma, for that matter) are more important for spectral density? $\endgroup$ – TanMath Aug 14 '15 at 5:41
  • $\begingroup$ @TanMath For open quantum systems whose dynamics are (approximately) Markovian, yes. $\endgroup$ – Mark Mitchison Aug 15 '15 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.