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I have a question based on the principles described in "Quantum simulator of an open quantum system using superconducting qubits: exciton transport in photosynthetic complexes" by Mostame et al. In this paper they are interested in studying the remarkable efficiency of photosynthesis, and more specifically the transport of excitation in photosynthetic complexes such as FMO.

They model the complex as an 'electronic system' consisting of finite-dimensional system of two level systems, coupled to coupled to a bath of 'phonon bath' of harmonic oscillators. The reasoning behind this model is that the sites through which the excitations traverse either contain an excitation or not, making them a form of a two level system. In addition to that in real cells these sites interact with the vibtrational environment of the surrounding molecular structure, hence the phononic bath.

Based on these considerations one can write down a Hamiltonian of the form \begin{equation} H_\text{tot} = H_\text{el} + H_\text{ph} + H_\text{I} \tag{1} \end{equation} where $H_\text{el}$ describes the two-level systems, $H_\text{ph}$ describes the photons, and $H_\text{I}$ describes their interaction. The two-level system part has the form \begin{equation} H_\text{el} = \sum_{j=1}^N \epsilon_j \vert{j}\rangle\langle{j}\vert + \sum_{i < j}^{N} V_{ij}\left(\vert{j}\rangle\langle{i}\vert + \vert{i}\rangle\langle{j}\vert\right) \, . \end{equation} where the basis states $\vert j\rangle$ are defined by the electronic excitation residing on molecule (site) j and all other sites being in their electronic ground state. The photon part has the form \begin{equation} H_\text{ph} = \sum_{j=1}^{N} H_{\text{ph},j} \quad H_{\text{ph},j} = \sum_l \hbar \omega_l(a^{j\dagger}_l a^j_l + 1/2) \, . \end{equation} The interaction has the form \begin{equation} H_\text{I} = \sum_{j=1}^N \vert{j}\rangle\langle{j}\vert\left(\sum_l \chi_{jl}(a^{\dagger,j}_l + a_l^j)\right) \, . \end{equation}

In the paper they conveniently write this in terms of Pauli matrices \begin{align} H = & \frac{1}{2}\sum_{j=1}^N \epsilon_j \sigma_{z}^j + \frac{1}{2} \sum_{i<j}^N V_{ij} \left(\sigma_{x}^j\sigma_{x}^i + \sigma_{y}^j\sigma_{y}^i \right) \\ & + \sum_{j=1}^N\sum_l \hbar \omega_{l,j}(a^{j\dagger}_l a^j_l + 1/2) + \sum_{j=1}^N\sum_l \chi_{jl} \sigma_z^j\left(a^{j\dagger}_l + a^j_l\right) \tag{4} \end{align} where we see that the bath and the two level systems can exchange energy.

Note that in the text they write that the effect of this phononic environment on the two level systems is fully contained in the bath power spectral density \begin{equation} J_j(\omega) = \sum_l \vert\chi_{jl}\vert^2\delta(\omega-\omega_l) \end{equation}

This leads to my question, which relates to the section that follows the previous equations called the classical noise approximation. There they describe the so called Haken-Strobl-Reineke model, where one replaces the quantum mechanical bath of harmonic oscillators by a classical noise environment leading to time-dependent fluctuations of the transition energies $\epsilon$. Here one can rewrite the above Hamiltonian as \begin{equation} \frac{1}{2}\sum_{j=1}^N \left[\epsilon_j + \delta\epsilon_j(t)\right] \sigma_{z}^j + \frac{1}{2} \sum_{i<j}^N V_{ij} \left(\sigma_{x}^j\sigma_{x}^i + \sigma_{y}^j\sigma_{y}^i \right) \tag{5} \end{equation} where one assumes that $\delta \epsilon_j(t)$ is white and Gaussian distributed.

Now my question is, how does one derive or motivate \begin{equation}\sum_{j=1}^N\sum_l \hbar \omega_{l,j}(a^{j\dagger}_l a^j_l + 1/2) + \sum_{j=1}^N\sum_l \chi_{jl} \sigma_z^j\left(a^{j\dagger}_l + a^j_l\right) \rightarrow \frac{1}{2} \sum_{j=1}^N \delta\epsilon_j(t) \sigma_z^j \end{equation}

I already know that one has to assume something about the noise having a white power spectral density and it being Gaussian distributed. In addition to this I think one might have to assume the environment to have a high enough number of modes to be considered a continuum. But even with this it is not obvious to me at all. Does one substitute the above $J_j(\omega)$ for a white noise PSD to obtain the result?

I tried looking at the original Haken Strobl paper on this topic, but the derivations did not seem to start from the phononic bath, they just assumed the $\delta \epsilon_j(t)$ term from the start. I therefore understand that an exact derivation might not be there, but I would at least like to motivate the transition, perhaps with some high temperature limit or something of the sort.

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  • $\begingroup$ Also, I'm confused by what $\sum_j |j\rangle \langle j |$ means. If $j$ indexes the two-leve-systems, then what's $|j\rangle$? $\endgroup$ – DanielSank Oct 2 '16 at 0:25
  • $\begingroup$ @DanielSank I agree that this notation is confusing. It is what they use in the paper, but it is sloppy as it would normally refer to the different levels of a single site. What they seem to mean is just $\sigma_z^j$, as seen in their pauli matrix version of the hamiltonian. $\endgroup$ – user129412 Oct 2 '16 at 0:28
  • $\begingroup$ Although no, what I write above is not true. They are precise in their notation, I just missed the definition. The basis states $\vert j \rangle$ are defined by the electronic excitation residing on molecule (site) j and all other sites being in their electronic ground state. I will update the main post. $\endgroup$ – user129412 Oct 2 '16 at 0:30
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – DanielSank Oct 2 '16 at 0:31
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Looking at Eq. (4), we can see that the coupling is essentially between the $\sigma_z$ of the two level systems, and the position $x$ of the photon modes. Let's focus on a single site:

$$H_\text{I, one site} = \sum_{\text{mode }l} \chi_l \sigma_z x_l \, .$$

In English: the electron site interacts with a set of modes the photon field, indexed by $l$, with a coupling strength $\chi_l$, and that coupling is of the form $\sigma_z x$.

Let's think classically. Each photon mode is a harmonic oscillator. If those oscillators are in a thermal state, then the position $x$ of each oscillator is jiggling around randomly. Since the position is directly coupled to the $\sigma_z$ of each two level system, then classically we're just straight-up saying that the wiggling bath makes the frequencies of the two-level systems wiggle. In other words:

$$ \epsilon_j \sigma_z^j \stackrel{\text{classical}}{\longrightarrow} \left( \epsilon_j + \delta\epsilon_j(t)\right) \sigma_z \, .$$

Note that in the original post, the photon bare Hamiltonian is not present in Eq. (5). My guess is that whatever approximation the authors actually used implicitly uses the interaction picture for the photons, which makes the bare photon Hamiltonian go away at the cost of introducing time dependence into $H_\text{I}$.

Of course now we want to know

  1. What assumptions had to be made for this approximation to work?

  2. What are the statistics (i.e. spectral density) of $\delta \epsilon_j$?

For the first, I bet one approximation made in the paper is that the photons and site energies are largely detuned. If this weren't the case, the coupling might have $\sigma_{x,y}$ terms in it to begin with. Whether or not those terms exist depends on the details of the coupling. Anyway, that's one possible assumption.

The next assumption is probably something along the lines of the photon bath being "large" or "having short memory". This assumption directly leads to the white noise spectral density stated in the paper. To understand why, take a look at any text on the quantum master equation, for example in Exploring the Quantum by Serge Haroche.

In fact, another way to understand this is to realize that we have exactly the same situation as a qubit connected to a readout resonator. That system is known to experience dephasing from shot noise in the resonator, as described in the very well known 2004 paper by A. Blais.

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  • $\begingroup$ This answer contains the basic intuition for how one would proceed to derive the expression I asked about. As I asked for either a derivation or a motivation, I would say it answers the question. However, I am still interested in a more rigorous approach to the problem. I have discussed with Daniel that this is probably better done in a more focused question, one which I will write now. I'll link to it in an additional comment for those who are interested. $\endgroup$ – user129412 Oct 2 '16 at 1:24
  • $\begingroup$ @user129412 How about that link? :-) $\endgroup$ – DanielSank Sep 20 '17 at 21:57
  • $\begingroup$ About a year too late, but here it is! physics.stackexchange.com/questions/283574/… $\endgroup$ – user129412 Sep 21 '17 at 14:21

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