# Tag Info

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The spectral density, or spectral function, describes the coupling between a small quantum system that is coupled to a larger environment. In many cases, this environment can be modelled effectively as a system of free bosonic or fermionic modes, with Hamiltonian (working in units with $\hbar = 1$) $$H_B = \sum_k \omega_k b_k^{\dagger}b_k.$$ The mode ...

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General form, properties A Lindblad form $$\dot \rho = -i[\eta, \rho] + A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A$$ has three important properties: It is still linear dynamics, in terms of $\rho$. It is trace-free regardless of the trace of $\rho$. This means that the total sum of the eigenvalues, which starts out as 1, does ...

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I actually don't think that this view of light being in a quantum superposition is anything new: what Discover magazine is describing (I believe) is the stock standard picture of how one would describe a system of cells, molecules, chloroplasts, fluorophores, whatever interacting with the quantised electomagnetic field. My simplified account here (answer to ...

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The role of coherence in biological electron transport, e.g. within chromophores, is an open and actively researched problem in quantum optics/quantum chemistry. The two classic theoretical treatments which kick-started the field are by Plenio & Huelga and Mohseni et al.. Since then an enormous literature has emerged on the topic. A basic, generic model ...

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I like to think of the spectral density as a filter for the bosonic field frequencies, it tells you "how much" of each frequency there is. In this case, if $S=1$ you have a linear increase for low frequencies ($\omega<\omega_C$) and an exponential decay for $\omega>\omega_C$. If you put $S=2$ the increase for low $\omega$ is parabolic instead of linear....

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The examples you give are of two particle systems. The open quantum system presupposes a many body state. The underlying nature of reality is quantum mechanical , thus all particles in the universe, in theory, should belong to one universal wavefunction solution, this would be the equivalent closed solution to the two body problems of your example. As ...

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First of all, let me point out that there are theories that propose nonlinear extensions to quantum mechanics (for instance Weinberg's nonlinear quantum mechanics). But there are very strong arguments against those approaches. Here are my favourites: Experiments. If we could have nonlinear quantum evolution, this should be visible in experiments. However, ...

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Indeed, for a product operator $\hat{C} = \hat{A}\hat{B}$, it is not true that $\hat{C}(t) = \hat{A}(t) \hat{B}(t)$ for a general (i.e. non-unitary) evolution in the Heisenberg picture. It is instructive to consider the simple example of a harmonic oscillator equilibrating with a thermal bath. This is described by a Lindblad equation $$\dot{\rho} = -i[\omega ... 6 Roughly speaking, it is the Schrödinger-picture density operator which has rapidly oscillating phase factors. Transforming to the interaction picture removes these phase factors. The residual time dependence of \rho_I(t) is generated only by coupling to the reservoir and is therefore slow, assuming weak coupling. Indeed, you have already shown that {\rm ... 5 Excitation loss can be modelled using a Lindblad equation (with a Hermitian Hamiltonian) which is guaranteed to be trace-preserving. However, one must work in a slightly larger Hilbert space. In particular, we need to introduce one more state, the vacuum state \lvert \mathrm{vac} \rangle, representing the situation where there is no excitation in the ... 5 This is an active field of research with very many open problems, a summary of which would constitute a partially subjective list and thus be inappropriate for this site. You might find the following review useful, which provides a nice survey of the current status of the field and extensive references: Goold et al., J. Phys. A: Math. Theor. 49 143001 (2016).... 5 Before tackling the OP's questions, let us first quickly establish some conventions and notation. A stochastic master equation (SME) describing quantum-jump trajectories can be written in the form$$ \mathrm{d}\rho = \mathrm{d}t \mathcal{L}_0\rho - {\rm d}t\sum_m \mathcal{H}[ \tfrac{1}{2}M^\dagger_m M_m]\rho + \sum_m {\rm d}\mu_m \mathcal{G}[M_m]\rho. \...

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Let me start by asking the question: What do you think the definition of complete positivity should be instead? You want it to "ensure that I will never find a non positive global transformation", but that can not be possible in that generality. What I am trying to say here is: if complete positivity is supposed to be a condition on $\mathcal L_A$, then $\... 5 The system + environment will not go into its ground state if it is isolated. In fact, it will not even go to an equilibrium state but keep evolving under the unitary dynamics govered by the total Hamiltonian. However, if the environment is much larger than the system, then effectively, the reduced density matrix of the system will be in a steady state (up ... 4 When studying Markovian quantum systems, the low-frequency ($\omega \ll \lambda$) behaviour of the spectral density is most important. This is because the most relevant modes of the bath, which control the open system dynamics, have frequencies commensurate with the frequencies of the open system. Meanwhile, the open system frequency scales must all be much ... 4 The analogy is based on writing the Lindblad equation in a Liouville form$\dot{\rho} = \cal{L} \rho$with the Liouville superoperator$\cal{L}$being the generator of the (semigroup) evolution for the density matrix$\rho$. The solution is then formally$\rho(t) = e^{\cal{L} t} \rho(0)$. 4 One detail First thing I would like to note is that the operator you are talking about is called the Lindblad superoperator. A superoperator is like an operator that acts on other linear operators (in this case, the density matrix). Lindblad Equation What you have written is known as the Lindblad Equation. The Lindblad Equation is one example of the many ... 4 The distinction between an open and a closed quantum system is mostly about whether information about the system is copied into the outside world, not about interaction per se. Suppose, for example, that a photon is reflected from a mirror. That is an interaction that changes the photon's momentum, but the interaction is not a measurement because the ... 4 Introduction With some assumptions and approximations, the time-evolution of an open quantum system can often be adequately modelled using completely positive (CP) maps, as reviewed in [3], [4], and [5], specifically using CP trace-preserving (CPTP) maps, which are also called quantum channels. (The definitions are given below.) For example, page 1 in [6] ... 3 The statement is almost true, although one has to redefine both$H_S$and$H_{SB}$to make it work. You need to also make use of the standard assumption of factorising initial conditions, $$\rho(0) = \rho_S(0)\otimes \rho_B,$$ where$\rho_B$is the bath reference state, which must commute with the bath Hamiltonian, i.e.$[\rho_B,H_B]=0$. Often (but not ... 3 Here's a general idea for a proof: Suppose a linear CPTP$\Lambda$gives a reversible but not unitary evolution. Then$\Lambda$must map some pure state$|\psi\rangle$into a mixed state$\rho = \Lambda(|\psi\rangle\langle \psi|)$. But if$\Lambda$is reversible, its inverse$\Lambda^{-1}$must map the mixed state$\rho$back into$|\psi\rangle$,$\Lambda^{...

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In all generality, this is an open question. The "Lindblad-like" Master equation you are referring to is $$\dot \rho_t = \hat L_t\, \rho_t = -\mathrm i\, [H_t, \rho_t] + \sum_\mu \gamma^\mu_t \left( L^\mu_t \rho_t L^{\mu\dagger}_t - \frac 1 2 \left\{ L^{\mu\dagger}_t L^\mu_t, \rho_t \right\} \right)$$ with a system Hamiltonian $H$, Lindblad operators $L^\... 3 A well-known reference for this is the book by Breuer and Petruccione: The theory of Open Quantum Systems (Oxford University Press; 1 edition (August 29, 2002)). It seems to largely overlap with what you want. It is well written and modern, covers the basic reasonably well if you have suitable background in statistics and in quantum mechanics, but does ... 3 I have done more reading and see an important point I missed above. In my last line for$\dot{\rho_A(t)}$I made the step $$\text{Tr}_B(\mathcal{L}[\rho(t)]) = \mathcal{L}_A[\rho_A(t)]$$ This step is not justified in general. Namely, the problem is as follows. I think it is true that$\text{Tr}_B(\mathcal{L}[\rho(t)])$can be written as a function of$\...

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