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# The quantum Zeno and anti-Zeno effects with strong system-environment coupling

### Subjects

To date, studies of the quantum Zeno and anti-Zeno effects focus on quantum systems that are weakly interacting with their environment. In this paper, we investigate what happens to a quantum system under the action of repeated measurements if the quantum system is strongly interacting with its environment. We consider as the quantum system a single two-level system coupled strongly to a collection of harmonic oscillators. A so-called polaron transformation is then used to make the problem in the strong system-environment coupling regime tractable. We find that the strong coupling case exhibits quantitative and qualitative differences as compared with the weak coupling case. In particular, the effective decay rate does not depend linearly on the spectral density of the environment. This then means that, in the strong coupling regime that we investigate, increasing the system-environment coupling strength can actually decrease the effective decay rate. We also consider a collection of two-level atoms coupled strongly with a common environment. In this case, we find that there are further differences between the weak and strong coupling cases since the two-level atoms can now indirectly interact with one another due to the common environment.

## Introduction

By repeatedly measuring a quantum system very frequently, the evolution of the quantum system can be slowed down, an effect that has been dubbed as the Quantum Zeno effect (QZE)1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22. On the other hand, if the quantum system is measured repeatedly not very rapidly, the measurements can actually speed up the temporal evolution. This effect, the opposite of the QZE, is known as the Quantum anti-Zeno effect (QAZE)23,24,25,26,27,28. Both the QZE and the QAZE have attracted tremendous theoretical and experimental interest due to their great importance for emerging quantum technologies as well as their fundamental theoretical interest. However, it is worth noting that the emphasis in studies performed on the QZE and the QAZE to date has been on the population decay of quantum systems. In these studies, the quantum system is prepared in an excited state, and then the system is repeatedly checked to see if the system is still in the excited state or not23,24,25,26,27,28,29,30,31,32,33,34. It is well-known then that the decay rate of the quantum system depends on the overlap of the spectral density of the environment and a measurement-induced level width23. Depending on this overlap, decreasing the measurement interval can lead to a decrease (the QZE) or an increase (the QAZE) of the decay rate.

While studies of the QZE and the QAZE performed to date by and large focus on the population decay model where only decay takes place, we also know from the study of open quantum systems that, in general, quantum systems interacting with their environment also undergo dephasing. To this end, the QZE and the QAZE were studied for the exactly solvable pure dephasing model in ref. 35 where it was shown that the QZE and the QAZE are significantly different for the pure dephasing case as compared with the population decay case. This study was then extended to arbitrary system-environment models in ref. 36 where a general framework for calculating the effective decay rate of the system for an arbitrary system-environment model was presented. It was found that the effective decay rate can be written as an overlap integral of the spectral density of the environment and an effective ‘filter function’ that depends on the system-environment model at hand, the measurement interval, and the measurement being repeatedly performed. This general formalism was then used to study the QZE and the QAZE when both dephasing and population decay are present. For example, repeated measurements for the paradigmatic spin-boson model37 were considered and it was shown that the presence of both population decay and dephasing make the results differ considerably both quantitatively and qualitatively as compared to the pure population decay case.

It should be pointed out, however, that the results presented in ref. 36 were derived under the assumption that the system-environment coupling is weak. This is consistent with studies performed for the population decay models, where the effective decay rate can be derived using time-dependent perturbation theory24. On the other hand, the behavior of a quantum system, subjected to repeated measurements, that is interacting strongly with its environment is not well understood38. For instance, one could ask whether or not the effective decay rate is still an overlap integral of the spectral density function and a ‘filter’ function. This paper intends to answer precisely such questions by looking at what happens to the spin-boson model under the action of repeated measurements if the central two-level system is interacting strongly with a surrounding environment of harmonic oscillators. Since the system-environment coupling is strong, the system-environment interaction cannot be treated perturbatively, and thus the treatment given in ref. 36 is no longer applicable. Our strategy then is to perform a unitary transformation, known as the polaron transformation, on the system-environment Hamiltonian39, 40, 42,43,44, 46. One then finds that the system and the environment can end up interacting weakly in this new ‘polaron’ frame. Perturbation theory can then be applied and the effect of repeated measurements is analyzed. We find that the analysis of the QZE and QAZE are in general very different compared to the population decay case. For example, it is clear that for the usual population decay case, increasing the system-environment strength increases the effective decay rate. However, for the strong system-environment regime that we investigate, we find that increasing the system-environment coupling regime can actually decrease the effective decay rate. We also study the QZE and the QAZE for more than one two-level system interacting with a common environment. For the weak coupling regime, the effective decay rate is directly proportional to the number of two-level systems coupled to the common environment36. On the other hand, for the strong system-environment coupling regime, we find that the effective decay rate for more than one two-level system is very different compared to the single two-level system case. The indirect interaction between the two-level systems due to their interaction with a common environment now plays a very important role, and the effective decay rate is no longer simply proportional to the number of two-level systems coupled to the common environment.

## Results

### Spin-boson model with strong system-environment coupling

We start with the paradigmatic spin-boson model Hamiltonian37, 47, 48 which we write as (we set ħ = 1 throughout)

$${H}_{L}=\frac{\varepsilon }{2}{\sigma }_{z}+\frac{{\rm{\Delta }}}{2}{\sigma }_{x}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}+{\sigma }_{z}\sum _{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger }),$$
(1)

where the system Hamiltonian is $${H}_{S,L}=\frac{\varepsilon }{2}{\sigma }_{z}+\frac{{\rm{\Delta }}}{2}{\sigma }_{x}$$, the environment Hamiltonian is $${H}_{B}={\sum }_{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}$$, and the system-environment coupling is $${V}_{L}={\sigma }_{z}{\sum }_{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger })$$. ε is the energy level difference of the two-level system, Δ is the tunneling amplitude, ω k are the frequencies of the harmonic oscillators, b k and $${b}_{k}^{\dagger }$$ are the annihilation and creation operators for the harmonic oscillators, and σ x and σ z are the standard Pauli operators. The ‘L’ denotes the ‘lab’ frame. If the system-environment coupling is strong, we cannot treat the system-environment coupling perturbatively. Furthermore, the system-environment correlation effects are significant as well in general. To motivate our basic approach in this strong coupling regime, we note that if the system tunneling amplitude is negligible and the initial system state is an eigenstate of σ z , then, even though the system and the environment are strongly interacting, the evolution of the system state is negligible. This then means that we should look to unitarily transform H L such that the effective system-environment coupling contains the tunneling amplitude Δ. This unitary transformation is provided by the ‘polaron’ transformation, whereby the system-environment Hamiltonian in this new ‘polaron’ frame becomes $$H={e}^{\chi {\sigma }_{z}\mathrm{/2}}{H}_{L}{e}^{-\chi {\sigma }_{z}/2}$$, where $$\chi ={\sum }_{k}[\frac{2{g}_{k}}{{\omega }_{k}}{b}_{k}^{\dagger }-\frac{2{g}_{k}^{\ast }}{{\omega }_{k}}{b}_{k}]$$ 39,40,41,42,43,44,45,46. The system-environment Hamiltonian in the polaron frame is then H = H S  + H B  + V, where $${H}_{S}=\frac{\varepsilon }{2}{\sigma }_{z}$$, $${H}_{B}={\sum }_{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}$$, and $$V=\frac{{\rm{\Delta }}}{2}[{\sigma }_{+}X+{\sigma }_{-}{X}^{\dagger }]$$, with $$X={e}^{\chi }$$ (see Methods for details). Now, if the tunneling amplitude is small, we can use time-dependent perturbation theory, treating V as the perturbation. This is the key idea to deal with the strong system-environment coupling regime. Although the system and the environment are strongly interacting, in the polaron frame, they are effectively interacting weakly. Let us now use this fact in order to calculate the survival probability, and thereby the effective decay rate. For concreteness, we assume that the initial state prepared is |↑〉, where σ z |↑〉 = |↑〉. In other words, we consider the same initial state as that considered in the analysis of the usual population decay model23, 24. At time t = 0, we prepare the system state |↑〉, and we subsequently perform measurements with time interval τ to check if the system state is still |↑〉 or not. The survival probability after time interval τ is then s(τ) = Tr s,B [|↑〉〈↑|ρ L(τ)], where $${\rho }_{L}(\tau )$$ is the combined density matrix of the system and the environment at time τ just before the projective measurement. Then,

$$s(\tau )={{\rm{Tr}}}_{S,B}[|\uparrow \rangle \langle \uparrow |{e}^{-i{H}_{L}\tau }{\rho }_{L}\mathrm{(0)}{e}^{i{H}_{L}\tau }]\mathrm{.}$$
(2)

It is important to note that the initial state that we have prepared cannot simply be taken as the usual product state |↑〉〈↑|$$\otimes {e}^{-\beta {H}_{B}}/{Z}_{B}$$, with $${Z}_{B}={{\rm{Tr}}}_{B}[{e}^{-\beta {H}_{B}}]$$ since the system and the environment are strongly interacting and consquently there will be significant initial system-environment correlations49, 50. Rather, the initial state that we should consider is $${\rho }_{L}\mathrm{(0)}={P}_{\uparrow }{e}^{-\beta {H}_{L}}{P}_{\uparrow }/Z$$, where P  = |↑〉〈↑|, and $$Z={{\rm{Tr}}}_{S,B}[{P}_{\uparrow }{e}^{-\beta {H}_{L}}]$$. Keeping this in mind, we use the polaron transformation to cast the expression for the survival probability $$s(\tau )$$ after the measurement at time τ in terms of quantities in the polaron frame. Doing so leads us to

$$s(\tau )={{\rm{Tr}}}_{S,B}[|\,\uparrow \,\rangle \langle \,\uparrow \,|{e}^{-iH\tau }{P}_{\uparrow }\frac{{e}^{-\beta H}}{Z}{P}_{\uparrow }{e}^{iH\tau }],$$
(3)

with the Hamiltonian H now in the polaron frame. Now, for small Δ, to a first approximation, the initial state in the polaron frame can be written as $${P}_{\uparrow }\otimes {e}^{-\beta {H}_{B}}/{Z}_{B}$$. This is a similar approximation as the usual assumption that the initial system-environment state is $${\rho }_{S}\mathrm{(0)}\otimes {\rho }_{B}$$ since, in the polaron frame, the system and the environment are weakly interacting. We thus get

$$s(\tau )={{\rm{Tr}}}_{S,B}[|\uparrow \rangle \langle \uparrow |{e}^{-iH\tau }(|\uparrow \rangle \langle \uparrow |\otimes {e}^{-\beta {H}_{B}}/{Z}_{B}){e}^{iH\tau }]={{\rm{Tr}}}_{S}[|\uparrow \rangle \langle \uparrow |{\rho }_{S}(\tau )],$$
(4)

where $${\rho }_{S}(\tau )={{\rm{Tr}}}_{B}[{e}^{-iH\tau }(|\uparrow \rangle \langle \uparrow |\otimes {e}^{-\beta {H}_{B}}/{Z}_{B}){e}^{iH\tau }]$$. Our objective then is to find $${\rho }_{S}(\tau )$$, given the initial system-environment state ρ(0) = |↑〉〈↑| $$\otimes \,{e}^{-\beta {H}_{B}}/{Z}_{B}$$. We find that (see the Methods section)

$$\begin{array}{rcl}{\rho }_{S}(\tau ) & = & {U}_{S}(\tau )({\rho }_{S}\mathrm{(0)}+i\sum _{\mu }{\int }_{0}^{\tau }d{t}_{1}[{\rho }_{S}\mathrm{(0),}{\tilde{F}}_{\mu }({t}_{1})]{\langle {\tilde{B}}_{\mu }({t}_{1})\rangle }_{B}\\ & & +\,\sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}\{{C}_{\mu \nu }({t}_{1},{t}_{2})[{\tilde{F}}_{\nu }({t}_{2}){\rho }_{S}\mathrm{(0),}{\tilde{F}}_{\mu }({t}_{1})]+{\rm{h}}{\rm{.c}}{\rm{.}}\}){U}_{S}^{\dagger }(\tau \mathrm{).}\end{array}$$
(5)

Here $${U}_{S}(\tau )={e}^{-i{H}_{S}\tau }$$, $${F}_{1}=\frac{{\rm{\Delta }}}{2}{\sigma }_{+}$$, $${B}_{1}=X$$, $${F}_{2}=\frac{{\rm{\Delta }}}{2}{\sigma }_{-}$$, $${B}_{2}={X}^{\dagger }$$, $${\tilde{F}}_{\mu }(t)={U}_{S}^{\dagger }(t){F}_{\mu }{U}_{S}(t)$$, $${\tilde{B}}_{\mu }(t)={U}_{B}^{\dagger }(t){B}_{\mu }{U}_{B}(t)$$ with $${U}_{B}(t)={e}^{-i{H}_{B}t}$$, $${\langle \ldots \rangle }_{B}={{\rm{Tr}}}_{B}[{\rho }_{B}(\ldots )]$$ where $${{\rm{Tr}}}_{B}$$ denotes taking trace over the environment, the environment correlation functions are defined as $${C}_{\mu \nu }({t}_{1},{t}_{2})={\langle {\tilde{B}}_{\mu }({t}_{1}){\tilde{B}}_{\nu }({t}_{2})\rangle }_{B}$$, and h.c. denotes the hermitian conjugate. Now, since the system-environment coupling in the polaron frame is weak, we can neglect the build up of correlations between the system and the environment. Thus, we can write the survival probability after time $$t=N\tau$$, where N is the number of measurements performed after time t = 0, as $$S(t=N\tau )={[s(\tau )]}^{N}\equiv {e}^{-{\rm{\Gamma }}(\tau )N\tau }$$, thereby defining the effective decay rate $${\rm{\Gamma }}(\tau )$$. It then follows that $${\rm{\Gamma }}(\tau )=-\frac{1}{\tau }\,\mathrm{ln}\,s(\tau )$$. Since we have the system density matrix in the polaron frame, we can work out the survival probability $$s(\tau )$$ and hence the effective decay rate $${\rm{\Gamma }}(\tau )$$. The result is that (see the Methods section for details)

$${\rm{\Gamma }}(\tau )=\frac{{{\rm{\Delta }}}^{2}}{2\tau }{\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} {e}^{-{{\rm{\Phi }}}_{R}(t^{\prime} )}\,\cos [\varepsilon t^{\prime} -{{\rm{\Phi }}}_{I}(t^{\prime} )],$$
(6)

where

$${{\rm{\Phi }}}_{R}(t)={\int }_{0}^{\infty }\,d\omega \,J(\omega )\frac{1\,-\,\cos (\omega t)}{{\omega }^{2}}\,\coth (\frac{\beta \omega }{2}),$$
(7)
$${{\rm{\Phi }}}_{I}(t)={\int }_{0}^{\infty }\,d\omega \,J(\omega )\frac{\sin (\omega t)}{{\omega }^{2}},$$
(8)

and the spectral density of the environment has been introduced as $${\sum }_{k}{|{g}_{k}|}^{2}(\ldots )\to {\int }_{0}^{\infty }d\omega J(\omega )(\ldots )$$. At this point, it is useful to compare this expression for the effective decay rate for the case of strong system-environment coupling with the case of the usual population decay model where the effective decay rate is $${\rm{\Gamma }}(\tau )=\tau {\int }_{0}^{\infty }d\omega \,J(\omega ){{\rm{sinc}}}^{2}[\frac{(\varepsilon -\omega )\tau }{2}]$$ 23, 24. It should be clear that for the strong system-environment coupling case, the effective decay rate given by Eq. (6) has a very different qualitative behavior. In particular, the effective decay rate can no longer be regarded as simply an overlap integral of the spectral density of the environment with a sinc-squared function. Rather, the effective decay rate now has a very prominent non-linear dependence on the spectral density, leading to very different behavior as compared with the population decay case. For example, as the system-environment coupling strength increases, $${{\rm{\Phi }}}_{R}(t)$$ increases, and thus we expect Γ(τ) to decrease. To make this claim concrete, let us model the spectral density as $$J(\omega )=G{\omega }^{s}{\omega }_{c}^{1-s}{e}^{-\omega /{\omega }_{c}}$$, where G is a dimensionless parameter characterizing the system-environment coupling strength, ω c is the cutoff frequency, and s is the Ohmicity parameter48. For concreteness, we look at the Ohmic case (s = 1). In this case, $${{\rm{\Phi }}}_{R}(t)=\frac{G}{2}\,\mathrm{ln}(1+{\omega }_{c}^{2}{t}^{2})$$, while $${{\rm{\Phi }}}_{I}(t)=G{\tan }^{-1}({\omega }_{c}t)$$, leading to

$${\rm{\Gamma }}(\tau )=\frac{{{\rm{\Delta }}}^{2}}{2\tau }{\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} \frac{\cos [\varepsilon t^{\prime} -G{\tan }^{-1}({\omega }_{c}t^{\prime} )]}{{\mathrm{(1}+{\omega }_{c}^{2}t{^{\prime} }^{2})}^{G\mathrm{/2}}}.$$
(9)

The double integral can be worked out numerically. Results are shown in Fig. 1(a) for different system-environment coupling strengths G. For the strong system-environment regime that we are dealing with, it is clear that increasing the system-environment coupling strength G actually decreases the effective decay rate. This is in contrast with what happens in the weak system-environment regime for the paradigmatic population decay model [see Fig. 1(b)]. Here it is clear that increasing the system-environment coupling strength increases the effective decay rate as expected. It should also be noted that the behaviour of Γ(τ) as a function of τ allows us to identify the Zeno and anti-Zeno regimes. One approach is to simply say that if Γ(τ) decreases when τ decreases, we are in the Zeno regime, while if Γ(τ) increases if τ decreases, then we are in the anti-Zeno regime23, 30, 33, 35. From Fig. 1(b), it should also be noted that increasing the coupling strength does not change the qualitative behavior of the Zeno to anti-Zeno transition, but for the strong coupling regime [see Fig. 1(a)], while we only observe the Zeno effect for G = 1, both the Zeno and anti-Zeno effects are observed for G = 2.5. Similarly, as shown in Fig. 2(a), increasing the cutoff frequency for the strong coupling case decreases the effective decay rate, but the opposite behaviour is observed for the weak coupling case [see Fig. 2(b)].

In our treatment until now, we have considered the change in the system state due to the tunneling term. This tunneling term, due to its presence in $${H}_{S,L}=\frac{\varepsilon }{2}{\sigma }_{z}+\frac{{\rm{\Delta }}}{2}{\sigma }_{x}$$, leads to the system state changing even if the system and the environment are not coupled to each other. Thus, an alternative way to quantify the effective decay rate would be to remove the evolution due to the system Hamiltonian (in the ‘lab’ frame) $${H}_{S,L}$$ before performing each measurement since what we are really interested in is the change in the system state due to the system-environment interaction. A similar approach has been followed in refs 35, 36 and 51 Therefore, we now derive an expression for the effective decay rate of the system state when, just before each measurement, we remove the system evolution due to $${H}_{S,L}$$. The survival probability, after one measurement, is now (starting from the state |↑〉)

$$s(\tau )={{\rm{Tr}}}_{S,B}[(|\uparrow \rangle \langle \uparrow |){e}^{i{H}_{S,L}\tau }{e}^{-i{H}_{L}\tau }{\rho }_{{\rm{L}}}\mathrm{(0)}{e}^{i{H}_{L}\tau }{e}^{-i{H}_{S,L}\tau }].$$
(10)

Notice now the presence of $${e}^{i{H}_{S,L}\tau }$$ and $${e}^{-i{H}_{S,L}\tau }$$ which remove the evolution of the system due to the system Hamiltonian before performing the measurement. Once again transforming to the polaron frame, we obtain

$$s(\tau )=1-{{\rm{Tr}}}_{S,B}[(|\downarrow \rangle \langle \downarrow |){e}^{i{H}_{S,P}\tau }{e}^{-iH\tau }(|\uparrow \rangle \langle \uparrow |\otimes {\rho }_{B}){e}^{iH\tau }{e}^{-i{H}_{S,P}\tau }],$$
(11)

where $${H}_{S,P}=\frac{\varepsilon }{2}{\sigma }_{z}+\frac{{\rm{\Delta }}}{2}({\sigma }_{+}X+{\sigma }_{-}{X}^{\dagger })$$ and $${\rho }_{B}={e}^{-\beta {H}_{B}}/{Z}_{B}$$. Since we are assuming that the tunneling amplitude is small, the unitary operator $${e}^{-i{H}_{S,P}\tau }$$ can be expanded as a perturbation series. At the same time, $${e}^{-iH\tau }$$ can also expanded as a perturbation series. Keeping terms to second order in the tunneling amplitude (see the Methods section), we find that now the modified decay rate Γ n (τ) is

$${{\rm{\Gamma }}}_{n}(\tau )={\rm{\Gamma }}(\tau )+{{\rm{\Gamma }}}_{{\rm{mod}}}(\tau ),$$
(12)

where the modification to the previous decay rate is

$$\begin{array}{rcl}{{\rm{\Gamma }}}_{{\rm{mod}}}(\tau ) & = & \frac{{{\rm{\Delta }}}^{2}}{\tau }\{\frac{1}{{\varepsilon }^{2}}{\sin }^{2}(\frac{\varepsilon \tau }{2}){e}^{-{{\rm{\Phi }}}_{R}\mathrm{(0)}}{e}^{-i{{\rm{\Phi }}}_{I}\mathrm{(0)}}\\ & & -\,\frac{1}{\varepsilon }\,\sin (\frac{\varepsilon \tau }{2}){\int }_{0}^{\tau }dt{e}^{-{{\rm{\Phi }}}_{R}(t)}\,\cos \,[\varepsilon (t-\frac{\tau }{2})-{{\rm{\Phi }}}_{I}(t)]\}.\end{array}$$
(13)

Using these expressions, we have plotted the behavior of Γ n (τ) for the strong system-environment coupling regime in Fig. 3(a). It should be clear that once again increasing the system-environment coupling strength generally decreases the effective decay rate Γ n (τ). This is in sharp contrast with what happens in the weak coupling regime. For the weak coupling case, it is known that36

$${{\rm{\Gamma }}}_{n}(\tau )={\int }_{0}^{\infty }d\omega J(\omega )Q(\omega ,\tau ),$$
(14)

where the filter function $$Q(\omega ,\tau )$$ is

$$Q(\omega ,\tau )=\frac{2}{\tau }\{\coth (\frac{\beta \omega }{2}){D}_{1}(\omega ,\tau )+{D}_{2}(\omega ,\tau )\},$$
(15)

with

$${D}_{1}(\omega ,\tau )={\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} \,\cos (\omega t^{\prime} )[{a}_{x}(t-t^{\prime} ){a}_{x}(t)+{a}_{y}(t-t^{\prime} ){a}_{y}(t)],$$
(16)

and

$${D}_{2}(\omega ,\tau )={\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} \,\sin (\omega t^{\prime} )[-{a}_{x}(t-t^{\prime} ){a}_{y}(t)+{a}_{x}(t){a}_{y}(t-t^{\prime} )]\mathrm{.}$$
(17)

Here $${a}_{x}(t)=\frac{2\varepsilon {\rm{\Delta }}}{{{\rm{\Omega }}}^{2}}{\sin }^{2}(\frac{{\rm{\Omega }}t}{2})$$ and $${a}_{y}(t)=\frac{{\rm{\Delta }}}{{\rm{\Omega }}}\,\sin ({\rm{\Omega }}t)$$ with $${{\rm{\Omega }}}^{2}={\varepsilon }^{2}+{{\rm{\Delta }}}^{2}$$. Using these expressions, we can investigate how the decay rate varies as the measurement interval changes for different system-environment coupling strengths in the weak coupling regime. Typical results are illustrated in Fig. 3(b) from which it should be clear that increasing the coupling strength in the weak coupling regime increases the effective decay rate. Furthermore, changing the coupling strength has no effect on the measurement time interval at which the Zeno to anti-Zeno transition takes place for the weak coupling regime as the three curves in Fig. 3(b) achieve their maximum value for the same value of τ. This is not the case for the strong coupling regime [see Fig. 3(a)].

At this point, it is worth pausing to consider where the qualitative difference in the behavior of the effective decay rate in the weak and the strong coupling regime comes from. The effective decay rate is derived from the survival probability after one measurement $$s(\tau )$$. For both the weak and the strong coupling regimes, the survival probability after one measurement is given by Eq. (10). For both cases, the Hamiltonian $${H}_{S,L}$$ and $${H}_{L}$$ are the same. The only difference is in the choice of the system-environment state $${\rho }_{L}\mathrm{(0)}$$. For the weak coupling case, this state is simply the product state |↑〉〈↑|$$\otimes {e}^{-\beta {H}_{B}}/{Z}_{B}$$. This is not the case for the strong coupling due to the significant system-environment correlations. Thus, we can say that the qualitative difference in the behavior of the effective decay rate is because of the presence of the system-environment correlations. It seems that these correlations can protect the quantum state of the system - as the coupling strength increases, these correlations become more and more significant, and at the same time, the effective decay rate goes down.

### Large spin-boson model with strong system-environment coupling

Let us now generalize the usual spin-boson model to deal with N S two-level systems interacting with a common environment. In this case, the system-environment Hamiltonian (in the ‘lab’ frame) is given by36, 40, 50

$${H}_{L}=\varepsilon {J}_{z}+{\rm{\Delta }}{J}_{x}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}+2{J}_{z}\sum _{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger }),$$
(18)

where $${J}_{x,y,z}$$ are the usual angular momentum operators obeying the commutation relations $$[{J}_{k},{J}_{l}]=i{\varepsilon }_{klm}{J}_{m}$$. We now start from the spin coherent state |j〉 such that J z |j〉 = j|j〉 with $$j={N}_{S}\mathrm{/2}$$. Other eigenstates of $${J}_{z}$$ can be considered as the initial state in a similar manner. Our objective is to again perform repeated projective measurements, described by the projector |j〉〈 j|, with time interval τ and thereby investigate what happens to the effective decay rate. As before, the survival probability after one measurement is $$s(\tau )={{\rm{Tr}}}_{S,B}[|j\rangle \langle \,j|{e}^{-i{H}_{L}\tau }{\rho }_{L}\mathrm{(0)}{e}^{i{H}_{L}\tau }]$$. Since we consider the system and the environment to be strongly interacting, we once again perform the polaron tranformation given by $$H={e}^{\chi {J}_{z}}{H}_{L}{e}^{-\chi {J}_{z}}$$, with $$\chi$$ the same as before. Then, we find that

$$H=\varepsilon {J}_{z}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}-\kappa {J}_{z}^{2}+\frac{{\rm{\Delta }}}{2}({J}_{+}X+{J}_{-}{X}^{\dagger }),$$
(19)

where $$\kappa =4{\sum }_{k}\frac{{|{g}_{k}|}^{2}}{{\omega }_{k}}$$, and $${J}_{\pm }={J}_{x}\pm i{J}_{y}$$ are the standard raising and lowering operators. Interestingly, the transformed Hamiltonian now contains a term proportional to $${J}_{z}^{2}$$. This term arises because the collection of two-level systems interacting with the collective environment are indirectly interacting with each other. This term is obviously proportional to the identity operator for a single two-level system, and thus has no influence for a single two-level system. If the tunneling amplitude is small, then we again use perturbation theory and assume that, in the polaron frame, the system-environment correlations can be neglected. We find that now the effective decay rate is (see the Methods section)

$${\rm{\Gamma }}(\tau )=\frac{{{\rm{\Delta }}}^{2}j}{\tau }{\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} {e}^{-{{\rm{\Phi }}}_{R}(t^{\prime} )}\,\cos [\varepsilon t^{\prime} +\kappa \mathrm{(1}-2j)t^{\prime} -{{\rm{\Phi }}}_{I}(t^{\prime} )]\mathrm{.}$$
(20)

Here $${{\rm{\Phi }}}_{R}(t)$$ and $${{\rm{\Phi }}}_{I}(t)$$ are the same as defined before. This result obviously agrees with the result that we obtained for a single two-level system. Moreover, it is clear from Eq. (20) that increasing the system-environment coupling strength G should reduce the effective decay rate due to the $${e}^{-{{\rm{\Phi }}}_{R}(t^{\prime} )}$$ factor in the integrand. This is precisely what we observein Fig. 4(a). Furthermore, it may be thought that increasing j (or, equivalently, N S ) increases the effective decay rate. On the other hand, the dependence on j is not so clear because of the presence of the indirect interaction. Namely, increasing j increases the oscillatory behavior of the integrand due to the dependence of the integrand on $$\cos [\varepsilon t^{\prime} +\kappa \mathrm{(1}-2j)t^{\prime} -{{\rm{\Phi }}}_{I}(t^{\prime} )]$$. Thus, once the integral over this rapidly oscillating integrand is taken, we can again get a small number. Such a prediction is borne out by Fig. 4(b) where the effective decay rate has been plotted for different values of j. It is obvious that there is a big difference between the single two-level system case and the more than one two-level system case. Furthermore, it seems that increasing j can largely reduce the value of the effective decay rate, meaning that in the strong coupling regime, the indirect interaction helps in keeping the quantum state alive.

Let us now consider the situation where the evolution to the system Hamiltonian $${H}_{S,L}=\varepsilon {J}_{z}+{\rm{\Delta }}{J}_{x}$$ is removed before each measurement. In the polaron frame $${H}_{S,L}$$ becomes $${H}_{S,P}=\varepsilon {J}_{z}+\frac{{\rm{\Delta }}}{2}({J}_{+}X+{J}_{-}{X}^{\dagger })$$. The major difference now compared to the previous single two-level system case is that the total system-environment Hamiltonian in the polaron frame $$H={H}_{S,P}+{H}_{B}-\kappa {J}_{z}^{2}$$ contains a term (namely, $$-\kappa {J}_{z}^{2}$$) that is not part of the system Hamiltonian in the polaron frame. As a result, when the system evolution is removed just before performing each measurement, the evolution induced by this extra term survives. Keeping this fact in mind, the effective decay rate $${{\rm{\Gamma }}}_{n}(\tau )$$ is now

$${{\rm{\Gamma }}}_{n}(\tau )={\rm{\Gamma }}(\tau )+{{\rm{\Gamma }}}_{{\rm{mod}}}(\tau ),$$
(21)

where $${\rm{\Gamma }}(\tau )$$ is given by Eq. (20) and

$$\begin{array}{rcl}{{\rm{\Gamma }}}_{{\rm{mod}}}(\tau ) & = & \frac{{{\rm{\Delta }}}^{2}}{\tau }\mathrm{(2}j)\{\frac{1}{{\varepsilon }^{2}}{\sin }^{2}(\frac{\varepsilon \tau }{2}){e}^{-{{\rm{\Phi }}}_{R}\mathrm{(0)}-i{{\rm{\Phi }}}_{I}\mathrm{(0)}}\\ & & -\,\frac{1}{\varepsilon }\,\sin (\frac{\varepsilon \tau }{2}){\int }_{0}^{\tau }dt{e}^{-{{\rm{\Phi }}}_{R}(t)}\,\cos \,[\kappa \mathrm{(2}j-\mathrm{1)(2}\tau -t)+\varepsilon (t-\frac{\tau }{2})-{{\rm{\Phi }}}_{I}(t)]\}\mathrm{.}\end{array}$$
(22)

In Fig. 5(a), we have shown the behavior of $${{\rm{\Gamma }}}_{n}(\tau )$$ when the system-environment coupling strength is increased for $${N}_{S}=2$$. It should be obvious that we observe multiple Zeno-anti Zeno regimes. Also, increasing the coupling strength does not generally increase the effective decay rate $${{\rm{\Gamma }}}_{n}(\tau )$$. This behavior should be contrasted with the weak coupling scenario. For weak coupling, it has been found that the effective decay rate is still given by Eq. (14), but now the filter function is N S times the filter function given by Eq. (15)36. Thus, increasing the coupling strength should now increase the effective decay rate. This is precisely what is observed in Fig. 5(b). Consequently, the weak coupling and the strong coupling regimes are very different for the strong and the weak coupling regimes. The difference is again due to the system-environment correlations.

## Discussion

We have investigated the quantum Zeno and anti-Zeno effects for a single two-level system interacting strongly with an environment of harmonic oscillators. Although it seems that perturbation theory cannot be applied, we have applied a polaron transformation that can make the coupling strength effectively small in the transformed frame and thereby validate the use of perturbation theory. We have obtained general expressions for the effective decay rate, independent of any particular form of the spectral density of the environment. Thereafter, we have shown that the strong coupling regime shows both qualitative and quantitative differences in the behavior of the effective decay rate as a function of the measurement interval and the QZE to QAZE transitions as compared with the weak system-environment coupling scenario. The effective decay rate is no longer an overlap integral of the spectral density of the environment and some other function. Rather, there is a very pronounced non-linear dependence on the spectral density of the environment. Most importantly, increasing the coupling strength in the strong coupling regime can actually reduce the effective decay rate. These differences can be understood in terms of the significant role played by the system-environment correlations. Moreover, we have extended our results to many two-level systems interacting with a common environment. Once again, we obtained expressions for the effective decay rate that are independent of the spectral density of the environment. We illustrated that in this case as well the behavior of the effective decay rate is very different from the commonly considered weak coupling regime. Our results should be important for understanding the role of repeated measurements in quantum systems that are interacting strongly with their environment.

## Methods

### The polaron transformation

For completeness, let us sketch how to transform the spin-boson Hamiltonian to the polaron frame39, 40, 42,43,44, 46. We need to find $$H={e}^{\chi {\sigma }_{z}\mathrm{/2}}{H}_{L}{e}^{-\chi {\sigma }_{z}\mathrm{/2}}$$. We use the identity

$${e}^{\theta A}B{e}^{-\theta A}=B+\theta [A,B]+\frac{{\theta }^{2}}{\mathrm{2!}}[A,[A,B]]+\cdots$$
(23)

Now, it is clear that $$[\chi {\sigma }_{z}\mathrm{/2,}{\sigma }_{z}\mathrm{/2}]=0$$. Also, $$[{\sum }_{k}(\frac{{g}_{k}}{{\omega }_{k}}{b}_{k}^{\dagger }-\frac{{g}_{k}^{\ast }}{{\omega }_{k}}{b}_{k}),{\sum }_{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}]=-\,{\sigma }_{z}{\sum }_{k}({g}_{k}{b}_{k}^{\dagger }+{g}_{k}^{\ast }{b}_{k}).$$ Carrying on, we find that $$[{\sigma }_{z}\chi \mathrm{/2,}{\sigma }_{z}{\sum }_{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger })]=-\,2{\sum }_{k}\frac{{|{g}_{k}|}^{2}}{{\omega }_{k}}\mathrm{.}$$ This is simply a c-number, so the higher-order commutators are zero. Furthermore, this c-number leads to a constant shift in the transformed Hamiltonian, and can thus be dropped. Putting all the commutators together, we find that

$${e}^{\chi {\sigma }_{z}\mathrm{/2}}[\frac{\varepsilon }{2}{\sigma }_{z}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}+{\sigma }_{z}\sum _{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger })]{e}^{-\chi {\sigma }_{z}\mathrm{/2}}=\frac{\varepsilon }{2}{\sigma }_{z}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}.$$
(24)

Next, we observe that $${e}^{\chi {\sigma }_{z}\mathrm{/2}}\frac{{\rm{\Delta }}}{2}{\sigma }_{x}{e}^{-\chi {\sigma }_{z}\mathrm{/2}}={e}^{\chi {\sigma }_{z}\mathrm{/2}}\frac{{\rm{\Delta }}}{2}({\sigma }_{+}+{\sigma }_{-}){e}^{-\chi {\sigma }_{z}\mathrm{/2}}$$, where $${\sigma }_{+}$$ and $${\sigma }_{-}$$ are the standard spin half raising and lowering operators. Furthermore, $$[\chi {\sigma }_{z}\mathrm{/2,}{\sigma }_{+}]={\sigma }_{+}\chi$$, leading to $${e}^{\chi {\sigma }_{z}\mathrm{/2}}{\sigma }_{+}{e}^{-\chi {\sigma }_{z}\mathrm{/2}}={\sigma }_{+}{e}^{\chi }$$. Similarly, $${e}^{\chi {\sigma }_{z}\mathrm{/2}}{\sigma }_{-}{e}^{-\chi {\sigma }_{z}\mathrm{/2}}={\sigma }_{-}{e}^{-\chi }$$. Thus, we finally have the required Hamiltonian in the polaron frame.

For the large spin case, the calculation is very similar40. The major difference is that now the c-number term that we dropped before cannot be dropped any longer since this term is proportional to $${J}_{z}^{2}$$ (for the spin half case, this is proportional to the identity operator, so this is just a constant shift for the spin half case). Namely, we now find that

$$[\chi {J}_{z},\varepsilon {J}_{z}+\sum _{k}{\omega }_{k}{b}_{k}^{\dagger }{b}_{k}+2{J}_{z}\sum _{k}({g}_{k}^{\ast }{b}_{k}+{g}_{k}{b}_{k}^{\dagger })]=-\,2{J}_{z}\sum _{k}({g}_{k}{b}_{k}^{\dagger }+{g}_{k}^{\ast }{b}_{k})-8{J}_{z}^{2}\sum _{k}\frac{{|{g}_{k}|}^{2}}{{\omega }_{k}}\mathrm{.}$$
(25)

Also,

$$[\chi {J}_{z},-2{J}_{z}\sum _{k}({g}_{k}{b}_{k}^{\dagger }+{g}_{k}^{\ast }{b}_{k})-8{J}_{z}^{2}\sum _{k}\frac{{|{g}_{k}|}^{2}}{{\omega }_{k}}]=8{J}_{z}^{2}\sum _{k}\frac{{|{g}_{k}|}^{2}}{{\omega }_{k}}\mathrm{.}$$
(26)

The rest of the calculation is very similar to the spin half case, and leads to the Hamiltonian in the polaron frame.

### Finding the system density matrix in the polaron frame

Here we describe how to obtain the system density matrix in the polaron frame $${\rho }_{S}(\tau )$$ just before performing the measurement at time τ. We define $${U}_{{\rm{tot}}}(\tau )={e}^{-iH\tau }={U}_{0}(\tau ){U}_{I}(\tau )$$, where $${U}_{0}(\tau )$$ is the unitary time-evolution operator corresponding to H S and H B , while $${U}_{I}(\tau )$$ is the ‘left over’ part that we can find using time-dependent perturbation theory. Writing the system-environment coupling in the polaron frame as $${\sum }_{\mu }{F}_{\mu }\otimes {B}_{\mu }$$, with $${F}_{1}=\frac{{\rm{\Delta }}}{2}{\sigma }_{+}$$, $${B}_{1}=X$$, $${F}_{2}=\frac{{\rm{\Delta }}}{2}{\sigma }_{-}$$ and $${B}_{2}={X}^{\dagger }$$, $${U}_{I}(\tau )$$ can be found to be $${U}_{I}(\tau )=1+{A}_{1}+{A}_{2}$$, with $${A}_{1}=-i{\sum }_{\mu }{\int }_{0}^{\tau }{\tilde{F}}_{\mu }({t}_{1}){\tilde{B}}_{\mu }({t}_{1})d{t}_{1}$$ and $${A}_{2}=-\,{\sum }_{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{\tilde{F}}_{\mu }({t}_{1}){\tilde{F}}_{\nu }({t}_{2}){\tilde{B}}_{\mu }({t}_{1}){\tilde{B}}_{\nu }({t}_{2})$$. Correct to second order in the tunneling amplitude Δ (in particular, we assume that Δτ is small enough such that higher order terms can be ignored), we can then write

$$\begin{array}{rcl}{\rho }_{S}(\tau ) & \approx & {{\rm{Tr}}}_{B}\{{U}_{0}(\tau )[{\rho }_{{\rm{tot}}}\mathrm{(0)}+{\rho }_{{\rm{tot}}}\mathrm{(0)}{A}_{1}^{\dagger }+{\rho }_{{\rm{tot}}}\mathrm{(0)}{A}_{2}^{\dagger }\\ & & +\,{A}_{1}{\rho }_{{\rm{tot}}}\mathrm{(0)}+{A}_{2}{\rho }_{{\rm{tot}}}\mathrm{(0)}+{A}_{1}{\rho }_{{\rm{tot}}}\mathrm{(0)}{A}_{1}^{\dagger }]{U}_{0}^{\dagger }(\tau )\},\end{array}$$
(27)

where $${\rho }_{{\rm{tot}}}\mathrm{(0)}={\rho }_{S}\mathrm{(0)}\otimes {\rho }_{B}$$. Eq. (27) can now be simplified term by term. First, we find that $${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){\rho }_{{\rm{tot}}}\mathrm{(0)}{U}_{0}^{\dagger }(\tau )\}={\tilde{\rho }}_{S}(\tau )$$, where $${\tilde{\rho }}_{S}(\tau )={U}_{S}(\tau ){\rho }_{S}\mathrm{(0)}{U}_{S}^{\dagger }(\tau )$$ is the system density matrix if the tunneling amplitude is zero. Next, we find that $${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){\rho }_{{\rm{tot}}}\mathrm{(0)}{A}_{1}^{\dagger }{U}_{0}^{\dagger }(\tau )\}=i{\sum }_{\mu }{\int }_{0}^{\tau }d{t}_{1}\,{U}_{S}(\tau ){\rho }_{S}\mathrm{(0)}{\tilde{F}}_{\mu }({t}_{1}){U}_{S}^{\dagger }(\tau ){\langle {\tilde{B}}_{\mu }({t}_{1})\rangle }_{B}$$, where $${\langle {B}_{\mu }({t}_{1})\rangle }_{B}={{\rm{Tr}}}_{B}\{{U}_{B}(\tau ){\rho }_{B}{\tilde{B}}_{\mu }({t}_{1}){U}_{B}^{\dagger }(\tau )\}$$. Similarly, $${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){A}_{1}{\rho }_{{\rm{tot}}}\mathrm{(0)}{U}_{0}^{\dagger }(\tau )\}=-\,i{\sum }_{\mu }{\int }_{0}^{\tau }d{t}_{1}\,{U}_{S}(\tau )$$ $${\tilde{F}}_{\mu }({t}_{1}){\rho }_{S}\mathrm{(0)}{U}_{S}^{\dagger }(\tau ){\langle {\tilde{B}}_{\mu }({t}_{1})\rangle }_{B}$$. Carrying on,

$${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){A}_{2}{\rho }_{{\rm{tot}}}\mathrm{(0)}{U}_{0}^{\dagger }(\tau )\}=-\,\sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}\,{U}_{S}(\tau ){\tilde{F}}_{\mu }({t}_{1}){\tilde{F}}_{\nu }({t}_{2}){\rho }_{S}\mathrm{(0)}{U}_{S}^{\dagger }(\tau ){C}_{\mu \nu }({t}_{1},{t}_{2}),$$
(28)

with the environment correlation function $${C}_{\mu \nu }({t}_{1},{t}_{2})$$ defined as $${C}_{\mu \nu }({t}_{1},{t}_{2})={\langle {\tilde{B}}_{\mu }({t}_{1}){\tilde{B}}_{\nu }({t}_{2})\rangle }_{B}={{\rm{Tr}}}_{B}\{{\tilde{B}}_{\mu }({t}_{1}){\tilde{B}}_{\nu }({t}_{2}){\rho }_{B}\}$$. Similarly,

$${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){\rho }_{{\rm{tot}}}\mathrm{(0)}{A}_{2}^{\dagger }{U}_{0}^{\dagger }(\tau )\}=-\,\sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}\,{U}_{S}(\tau ){\rho }_{S}\mathrm{(0)}{\tilde{F}}_{\nu }({t}_{2}){\tilde{F}}_{\mu }({t}_{1}){U}_{S}^{\dagger }(\tau ){C}_{\nu \mu }({t}_{2},{t}_{1}\mathrm{).}$$
(29)

Finally,

$${{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){A}_{1}{\rho }_{{\rm{tot}}}{A}_{1}^{\dagger }{U}_{0}^{\dagger }(\tau )\}=\sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{\tau }d{t}_{2}{U}_{S}(\tau ){\tilde{F}}_{\mu }({t}_{1}){\rho }_{S}\mathrm{(0)}{\tilde{F}}_{\nu }({t}_{2}){U}_{S}^{\dagger }(\tau ){C}_{\nu \mu }({t}_{2},{t}_{1}\mathrm{).}$$
(30)

Using the fact that $${\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{\tau }d{t}_{2}={\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}+{\int }_{0}^{\tau }d{t}_{2}{\int }_{0}^{{t}_{2}}d{t}_{1}$$,

$$\begin{array}{rcl}{{\rm{Tr}}}_{B}\{{U}_{0}(\tau ){A}_{1}{\rho }_{{\rm{tot}}}{A}_{1}^{\dagger }{U}_{0}^{\dagger }(\tau )\} & = & \sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{U}_{S}(\tau ){\tilde{F}}_{\mu }({t}_{1}){\rho }_{S}\mathrm{(0)}\\ & & \times \,{\tilde{F}}_{\nu }({t}_{2}){U}_{S}^{\dagger }(\tau ){C}_{\nu \mu }({t}_{2},{t}_{1})+{\rm{h}}{\rm{.c}}\mathrm{.,}\end{array}$$
(31)

where h.c. denotes hermitian conjugate. Putting all the terms back together, the system density matrix can be written as Eq. (5).

### Finding the effective decay rate

We now explain how to find the effective decay rate given by Eq. (6). With the system density matrix at time τ available, we first calculate the survival probability s(τ). This can be done via $$s(\tau )=1-$$〈↓|ρ s (τ)|↓〉. Since the state |↓〉 is an eigenstate of H S , and ρ S (0) = |↑〉〈↑|, it is straightforward to see that

$$s(\tau )=1-2{\rm{Re}}\,(\sum _{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{C}_{\mu \nu }({t}_{1},{t}_{2})|\downarrow \rangle {\tilde{F}}_{\nu }({t}_{2})|\uparrow \rangle \langle \uparrow |{\tilde{F}}_{\mu }({t}_{1})|\downarrow \rangle )\mathrm{.}$$
(32)

We now note that, since $${F}_{1}=\frac{{\rm{\Delta }}}{2}{\sigma }_{+}$$, $${F}_{2}=\frac{{\rm{\Delta }}}{2}{\sigma }_{-}$$, and $${H}_{S}=\frac{\varepsilon }{2}{\sigma }_{z}$$, $${\tilde{F}}_{1}(t)=\frac{{\rm{\Delta }}}{2}{\sigma }_{+}{e}^{i\varepsilon t}$$ and $${\tilde{F}}_{2}(t)=\frac{{\rm{\Delta }}}{2}{\sigma }_{-}{e}^{-i\varepsilon t}$$. Therefore,

$$\begin{array}{rcl}s(\tau ) & = & 1-2{\rm{Re}}\,({\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{C}_{12}({t}_{1},{t}_{2})\langle \downarrow |{\tilde{F}}_{2}({t}_{2})|\downarrow \rangle \langle \downarrow |{\tilde{F}}_{1}({t}_{1})|\downarrow \rangle )\\ & = & 1-\frac{{{\rm{\Delta }}}^{2}}{2}{\rm{Re}}\,({\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{C}_{12}({t}_{1},{t}_{2}){e}^{i\varepsilon ({t}_{1}-{t}_{2})})\mathrm{.}\end{array}$$
(33)

What remains to be worked out is the environment correlation function $${C}_{12}({t}_{1},{t}_{2})={{\rm{Tr}}}_{B}[{\rho }_{B}\tilde{X}({t}_{1}){\tilde{X}}^{\dagger }({t}_{2})]$$. Using the cyclic invariance of the trace, it is clear that this correlation function is actually only a function of $${t}_{1}-{t}_{2}$$ only, since $${{\rm{Tr}}}_{B}\,[{\rho }_{B}\tilde{X}({t}_{1}){\tilde{X}}^{\dagger }({t}_{2})]={{\rm{Tr}}}_{B}\,[{\rho }_{B}{e}^{i{H}_{B}({t}_{1}-{t}_{2})}X{e}^{-i{H}_{B}({t}_{1}-{t}_{2})}{X}^{\dagger }]={C}_{12}({t}_{1}-{t}_{2})$$. Thus,

$$s(\tau )=1-\frac{{{\rm{\Delta }}}^{2}}{2}{\rm{Re}}\,({\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{{t}_{1}}d{t}_{2}{C}_{12}({t}_{1}-{t}_{2}){e}^{i\varepsilon ({t}_{1}-{t}_{2})})=1-\frac{{{\rm{\Delta }}}^{2}}{2}{\rm{Re}}\,({\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} {C}_{12}(t^{\prime} ){e}^{i\varepsilon t^{\prime} }),$$
(34)

where we have introduced $$t^{\prime} ={t}_{1}-{t}_{2}$$. The calculation of $${C}_{12}(t^{\prime} )$$ can be performed as follows. First, we use the useful fact that for $${{\rm{Tr}}}_{B}[{\rho }_{B}{e}^{Z}]={e}^{\langle {Z}^{2}\rangle \mathrm{/2}}$$ where Z is a linear function of the creation and annihilation operators. Second, to obtain a single exponential so that the previous identity can be used, we use the identity that for any two operators X and Y, $${e}^{X}{e}^{Y}={e}^{X+Y+\frac{1}{2}[X,Y]\ldots }$$. Fortunately for us, the series terminates for our case, so the higher order terms are zero. Using these two identities, we find that $${C}_{12}(t^{\prime} )={e}^{-{\Phi }_{R}(t^{\prime} )}{e}^{-i{\Phi }_{I}(t^{\prime} )}$$ where $${{\rm{\Phi }}}_{R}(t)$$ and $${{\rm{\Phi }}}_{I}(t)$$ have been defined in Eqs (7) and (8), and the spectral density of the environment has been introduced as $${\sum }_{k}{|{g}_{k}|}^{2}(\ldots )\to {\int }_{0}^{\infty }d\omega J(\omega )(\ldots )$$. This finally leads to $$s(\tau )=1-\frac{{{\rm{\Delta }}}^{2}}{2}{\int }_{0}^{\tau }dt{\int }_{0}^{t}dt^{\prime} {e}^{-{{\rm{\Phi }}}_{R}(t^{\prime} )}\,\cos [\varepsilon t^{\prime} -{{\rm{\Phi }}}_{I}(t^{\prime} )]$$. We can define an effective decay rate $${\rm{\Gamma }}(\tau )=-\frac{1}{\tau }\,\mathrm{ln}\,s(\tau )$$. For small Δ, we expect the deviation of the survival probability from one to be small. Thus, we end up with Eq. (6).

### Calculating the modified decay rate

Let us now briefly sketch how to obtain Eq. (12). We note that the system Hamiltonian $${H}_{S,L}$$ becomes in the polaron frame $${H}_{S,P}=\frac{\varepsilon }{2}{\sigma }_{z}+\frac{{\rm{\Delta }}}{2}({\sigma }_{+}X+{\sigma }_{-}{X}^{\dagger })$$. Then, to second order in $${\rm{\Delta }},{e}^{-i{H}_{S,P}\tau }\approx 1+{A}_{SP}^{\mathrm{(1)}}+{A}_{SP}^{\mathrm{(2)}}$$, with $${A}_{SP}^{\mathrm{(1)}}=-\,i{\int }_{0}^{\tau }d{t}_{1}{\sum }_{\mu }{\tilde{F}}_{\mu }({t}_{1}){B}_{\mu }$$, and $${A}_{SP}^{\mathrm{(2)}}=-\,{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{\tau }d{t}_{2}{\sum }_{\mu \nu }{\tilde{F}}_{\mu }({t}_{1}){B}_{\mu }{\tilde{F}}_{\nu }({t}_{2}){B}_{\nu }$$. Here, $${\tilde{F}}_{\mu }(t)={e}^{i\varepsilon {\sigma }_{z}t\mathrm{/2}}{F}_{\mu }{e}^{-i\varepsilon {\sigma }_{z}t\mathrm{/2}}$$. Substituting these expressions in the expression for the survival probability as well as the perturbation expansions for $${e}^{iH\tau }$$ and $${e}^{-iH\tau }$$, and keeping terms up to second order in Δ, we find that the new survival probability consists of the previous survival probability plus some additional terms. It can be be easily seen that most of these additional terms, once the trace with the projector |↓〉〈↓| is taken, give zero. The additional terms that need to be worked out are $${{\rm{Tr}}}_{B}[\langle \,\downarrow \,|{U}_{B}(\tau ){A}_{1}{\rho }_{S}\mathrm{(0)}{\rho }_{B}{U}_{B}^{\dagger }(\tau ){A}_{SP}^{\mathrm{(1)}}|\,\downarrow \,\rangle ]$$, $${{\rm{Tr}}}_{B}[\langle \,\downarrow \,|{A}_{SP}^{\mathrm{(1)}\dagger }{U}_{B}(\tau ){\rho }_{S}\mathrm{(0)}{\rho }_{B}{A}_{1}^{\dagger }{U}_{B}^{\dagger }(\tau )|\,\downarrow \,\rangle ]$$, and $${{\rm{Tr}}}_{B}[\langle \,\downarrow \,|{A}_{SP}^{\mathrm{(1)}\dagger }{U}_{B}(\tau ){\rho }_{S}\mathrm{(0)}{\rho }_{B}{A}_{1}^{\dagger }{U}_{B}^{\dagger }(\tau )|\,\downarrow \,\rangle ]$$. The first of these terms is equal to $$-{\sum }_{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{\tau }d{t}_{2}\langle \,\downarrow \,|{\tilde{F}}_{\nu }({t}_{1}){\rho }_{S}\mathrm{(0)}{\tilde{F}}_{\mu }({t}_{2})|\,\downarrow \,\rangle {C}_{\mu \nu }(\tau -{t}_{1})$$, while the second is simply the hermitian conjugate of the first. On the other hand, the last term is equal to $${\sum }_{\mu \nu }{\int }_{0}^{\tau }d{t}_{1}{\int }_{0}^{\tau }d{t}_{2}\langle \,\downarrow \,|{\tilde{F}}_{\nu }({t}_{1}){\rho }_{S}\mathrm{(0)}{\tilde{F}}_{\mu }({t}_{2})|\,\downarrow \,\rangle {C}_{\mu \nu }\mathrm{(0).}$$ Next, we use the fact that $${F}_{1}=\frac{{\rm{\Delta }}}{2}{\sigma }_{+}$$ and $${F}_{2}=\frac{{\rm{\Delta }}}{2}{\sigma }_{-}$$ to simply the inner products. Putting all the pieces together, we arrive at Eq. (12). The calculation of $${{\rm{\Gamma }}}_{n}(\tau )$$ for the large spin case is quite similar. One only needs to be careful about the fact that the system-environment Hamiltonian, in the polaron frame, contains a term, namely $$-\kappa {J}_{z}^{2}$$, that is not a part of the transformed system Hamiltonian $${H}_{S,P}$$.

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## Acknowledgements

Support from the LUMS startup grant is gratefully acknowledged. Support from HEC under Grant No 5917/Punjab/NRPU/R&D/HEC/2016 is also acknowledged.

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A.Z.C. contributed solely to this work.

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Chaudhry, A.Z. The quantum Zeno and anti-Zeno effects with strong system-environment coupling. Sci Rep 7, 1741 (2017). https://doi.org/10.1038/s41598-017-01844-8

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# What is the basis for a proof?

What is the basis for a proof? I am trying to understand what a proof is, what is the most simple example of a proof? like can you have a 'proof' for $1+1=2$?

• @Asaf: I don't think is quite a duplicate. The OP asks for a simple proof to understand the idea behind proofs in general. OP used $1 + 1 = 2$ in general. However, the question you link to is asking for a concrete proof of $1 + 1 = 2$. In other words, I don't think a "proof" of $1+1 = 2$ is what the OP seeks here. – JavaMan Sep 3 '12 at 23:24
• A proof is a fully convincing detailed argument that shows that a certain assertion is true. – André Nicolas Sep 4 '12 at 0:07
• Well, yes, but the proof that $1+1=2$ is actually quite hard to understand. Try looking for one of the many available proofs of Pythagoras' theorem - these should give you a good flavour of what a proof is. – user22805 Sep 4 '12 at 0:11
• Surely this counts as the simplest nontrivial proof: All men are mortal. Socrates is a man. Therefore, Socrates is mortal. – user856 Sep 4 '12 at 1:03
• I'm not sure what was the point of the edit, now this is no longer a question... – Asaf Karagila Oct 2 '12 at 10:25

What is a proof? Well, commonly in mathematics this either means "a rigorously convincing argument" or a "formal proof". I will talk about the latter here, and I will specifically take the axiomatic approach to what is a proof.

Well first we need to understand what is an "inference rule". An inference rule is a rule which allows us to deduce a formal assertion from previous assertions.

For example, "If $\alpha\land\beta$ [read: $\alpha$ and $\beta$] are true, then $\alpha$ is true". This means that if we assume something which has a certain form then it is permitted to derive a certain statement. For example:

You are wearing a black coat. Therefore you are wearing a coat.

We assumed that you are wearing a coat and that the coat is black, we deduced that you are wearing a coat. It seems trivial, but in fact inference rules come to formalize a very trivial reasoning, or to quote Mr. Spock "It's only logical".

Now what is a proof?

Well... before we talk about that we need a framework. We work with mathematical logic. In this context we have symbols to write our proof (e.g., $\land$ is the letter for "and"). We have inference rules, and we have axioms of logic which are a bit like inference rules, but slightly different. In less-mathy environment there are usually a lot of inference rules and less axioms; while in the course I took in my undergrad there were many axioms but only one inference rule.

In the mathematical logic we also have rules to tell us what sort of strings are valid formulas, sentences or terms in the language. Much like "Chair me dog" is not a valid English sentence. We only care about syntax, about the form, not about the content. Nonsense like "I ate the invisible chair for dinner last night." is syntactically correct, but has no meaning -- we don't care. So legal strings of characters are called formulas or sentences. Formulas and sentences describe some relations, or properties of objects. Terms are also valid strings of letters in the language, but those describe objects rather than properties.

So now can we get to the point? What is a mathematical proof??

Well, one last thing before we get to that. Axioms, we prove things from axioms. What are axioms? Axioms are simply sentences which we assume to be true. We may not even have axioms, in which case we prove something from the axioms of logic and the inference rules, and then whatever we proved is always true in this framework.

Now we can get to the point!

What is a proof? Well, we prove $\varphi$ from a collection of axioms $T$. This means that there is a finite sequence of sentences $\varphi_1,\ldots,\varphi_k$ such that:

1. $\varphi_k=\varphi$, we finish with what we wanted to prove.
2. For every $i<k$, either $\varphi_i$ is either an axiom (sentence from $T$) or it was derived from previous statements using axioms of logic and inference rules.

Now to give a proper example I will have to be specific and tell you what are the symbols, what are the inference rules, the logical axioms, the rules for creating sentences and formulas, what are the axioms and what we want to prove from them...

Well, this is a bit too much for a single post at this hour. However, you should definitely go to your nearest library and open up a book about mathematical logic, preferably an introductory book.

It raises a question, if that is what you need in order to write a proof, why is this site not full of such posts? Well, in mathematics there is a big difference between a formal proof and a rigorous proof. Mathematicians [usually] care about rigor, rather than absolute formality. They also interchange these two terms more often than they should.

However proving something rigorously just means explaining how the proof should look like, in words or symbols. The background framework is often agreed upon "silently" in the background, and it takes practice but one can usually make the needed switches without much notice.

Finally, when you say "can you have a proof that $1+1=2$?" the answer is yes, but in what context? Is it the context of natural numbers with their common axioms (the properties we assume natural numbers have), or is it in the context of the real numbers with the axioms of those?

The two systems are very different, and formal proofs - even of the same claim - in different frameworks can be very different.

However in this case not too different. We use the symbol $2$ to abbreviate the formal term $1+1$. So either proof would be relatively short.

• please post this great answer on the lead section on Wikipedia. this is soooooooooooo much clearer. – kittensatplay Sep 4 '12 at 1:10
• wikipedia is an "encyclopedia" environment, his response was more akin to a textbook or blog. It would violate the style in wikipedia. – Squirtle Sep 4 '12 at 1:36
• @high: Thanks... :-) – Asaf Karagila Sep 4 '12 at 5:57
• Several days ago, I had the idea of compiling a book with some answers on MSE - This answer should definitely be there. – Billy Rubina Sep 4 '12 at 7:12
• @Doug. Thank you for pointing that out. With regards to natural deduction, I think it is a good idea to start with axiomatic systems if you want to discuss what are proofs. I'll point that out in my answer as well. – Asaf Karagila Sep 7 '12 at 9:03

Belgi and Asaf have answered the question assuming "proof" means "formal proof". As André indicated in the comments, there is an informal notion of proof: A proof is a convincing argument that some fact is true.

Officially, all that counts is the formal proof. On the other hand, almost all proofs one encounters are informal (though, at least in theory, they can be translated into formal proofs.)

Here's an example of an informal proof. I'll begin with some definitions

Definition: By an "integer" we mean any number of the form $0, \pm 1, \pm2 , \pm 3,...$. (So, something like $\frac{2}{3}$ or $\pi$ is not an integer.)

Definition: If $n$ is an integer, we say "n is even" if there is another integer $k$ with the property that $2k = n$.

Theorem: The number $n=6$ is even.

Proof: In order to prove this, we need to find an integer $k$ with the property that $2k = 6$. Well, $k=3$ works, so $n=6$ is even. $\square$

(The symbol $\square$ is often used to denote the end of a proof.)

Ok, that seems like a waste of time. But what have we gained? No longer are we depending on someone's word that 6 is even. We now know it is true. Ok, it still seems like a waste of time.

Here's another one.

Theorem: If $n$ is even and $m$ is even, then so is $m+n$.

Proof: We need to show that we can write $m+n = 2k$ for some integer $k$. By assumption, since $m$ is even $m = 2a$ for some integer $a$. (I'm not using $k$ since I want to use that letter for another purpose later). Since $n$ is even, $n = 2b$ for some integer b.

Then, $m+n = 2a + 2b = 2(a+b)$. So, if I let $k = a+b$ (which is still an integer), then this is the $k$ I'm searching for. This shows that $m+n$ is even. $\square$

Now what have I gained? Before this proof, you could have experimented with many even numbers, adding them together and observing that you always got an even number out. (For example, $2+6 = 8$, $14+ 290 = 304$, etc.)

But even if you were to spend your whole life checking that adding even numbers together gives evens, there is no guarantee that there aren't even numbers that add up to a non-even number. Perhaps it happens very rarely that 2 evens add up to a non-even, perhaps you just needed to check one more case to find that elusive example.

The proof shows that no matter which 2 even numbers you pick, their sum is even. It guarantees that your examples aren't flukes, that they are happening for a reason and that they will happen every time.

Now, let me point out a potential pitfall of this proof. Right where I say "...$k = a+b$ (which is still an integer)..." - why it still an integer? The answer to this question is going to depend on what axioms you are working with and this is going to get back to what Asaf and Belgi were saying.

• I think you should change the word "fact" to "assertion" (or something similar). Facts are always true, but it's the assertions that you need to be convinced about... – Asaf Karagila Sep 4 '12 at 5:59
• @Asaf: I see what you're saying. Perhaps this is just the platonist in me coming out, but my feeling is that the "assertion" is already a "fact" before we've proven it, we just don't know it yet. If you feel strongly enough, however, feel free to change it :-) – Jason DeVito Sep 4 '12 at 12:35

I don't fully understand what you mean by "What is the most simple proof to understand what a proof is?" if you want to know what a proof is in the formal sense you will have to read some mathematical logic.

To keep it simple, a proof is a list of statements from a set of assumptions such that each statement has a previous statement that is comes from (or else it's an assumption) and the last statement is what you have proved (if, for example, the last line is $1+1=2$ then you proved it).

The answer to your second question "can you have a 'proof' for 1+1=2 ?" is yes. If you want to prove that you first need to ask yourself what is $1$ ? what is $2$ ? what is $1+1$ ? and then you would want to compare both sides. I answered most of these questions in details before here.

• "formal sense", I suppose? (Rather than "formal scence") – Asaf Karagila Sep 3 '12 at 23:29
• @AsafKaragila - I mean something like "in the formal meaning"...I'm guessing my English is wrong here ? – Belgi Sep 3 '12 at 23:31
• Well, "scence" is not a word, it's somewhat of a mix between "scene" and "sense" and maybe "science" :-) Either way, I believe you meant "sense" after all. – Asaf Karagila Sep 3 '12 at 23:33
• @AsafKaragila - thanks, I edited – Belgi Sep 3 '12 at 23:34
• umm.. ok it sounds like a proof is a conclusion – kittensatplay Sep 3 '12 at 23:42

You are probably asking about formal mathematical proof or maybe you are confused about how trivial mathematical facts like 1+1=2 can be proven. In that sense you should remember that a proof is meaningful only relative to a formal proof system. A string $\pi$ is a proof for a statement $\varphi$ if the pair is accepted by the proof system (think of a proof system as an algorithm/computer program that gets $\pi$ and $\varphi$ and returns YES or NO. Normally a proof system should be sound, i.e. false statements should not have proofs).

If your question is more philosophical then you should first say what do you mean by $1$, $2$, $+$, and $=$ (and there are various opinions about these among philosophers of mathematics and mathematical philosophers). How can we answer the question if we haven't agreed on what these symbols mean?

If you interpret them as symbols in the language of an arithmetic theory like PA or ZFC then your question is easy to answer. Here are two examples:

In the case of PA, $2$ is defined as $S(S(0))$ and $1$ is defined as $S(0)$ and $+$ is defined recursively as $x+0=x$ and $x+S(y)=S(x+y)$; and your question is to show that $S(S(0))=S(0)+S(0)$ which is not difficult. The symbol $S$ is the symbol for successor.

In case of ZFC, $0 = \emptyset, 1 = \{0\}$ and $2=\{0,1\}$ and etc and again the proof is not difficult.

To talk about a formal proof we should fix the meaning of the symbols and the theory, a formal proof outside any proof system is meaningless. Often people omit explicitly mentioning the formal system because it is clear from context, but it is there.

Many will interpret the question as asking about a formal proof inside a theory about symbols as above (somewhat formalist view of mathematics) but that is not the only way. There are other approaches where the linguistic/formal entities are not primary concern of mathematics but secondary means used for the purposes like communicating mathematical results to others, etc. Constructive mathematics and Platonism are two of the famous ones viewpoints. (And honestly, do you know anyone who believes in $1+1=2$ because of a formal proof in PA or ZFC?). Check for example the BHK interpretation of (constructive) proofs.

However, if you are asking more generally about what is a proof, then I think the following story is relevant:

In 1978, as an undergraduate, I attended Shimon's course Graph Algorithms. At some point, one student was annoyed at Shimon's "untraditional" way of analyzing algorithms, and asked whether Shimon's demonstrations constituted a proof and if so what is a proof. Shimon answer was immediate, short, and clear: A proof is whatever convinces me. [Stories about Shimon Even by Oded Goldreich]

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# Reciprocal of a Quadratic Function

• March 30th 2010, 07:28 AM
Hello!
My teacher likes to kick off a new unit by making us do a question on our own so I am a little confused..

I have a question that shows 3 graphs - and an equation of f(x)= 1/x^2-1

The question is "which graph shows f(x)= 1/x^2-1"

I've already found my answer - but it was lengthy; I found the domain - vertical asymptotes/end behaviours/horitzontal asymptotes/x+yintercepts/than graphed it on my own

My question for you guys is - is there a way to figure out which graph represents f(x)= 1/x^2-1 without having to do all those steps? By just looking at it..?

Thanks for your patience & time.
• March 30th 2010, 10:53 AM
$\frac{1}{x^2 - 1} = \frac{1}{(x+1)(x-1)}$
Hence asymptotes at $\pm 1$. Gets very small when $x \to \pm \infty$


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# Tag Info

20

The issue you're running into is often referred to as sample impoverishment. We can see why your approach suffers from it with a fairly simple example. Let's say you have 3 particles and their normalized weights are 0.1, 0.1, 0.8. Then multiplying by each weight by the 3 yields 0.3, 0.3, and 2.4. Then rounding yields 0, 0, 2. This means you would not pick ...

14

The Rao-Blackwellized Particle Filter (RBPF) as you say in your question performs a marginalization of the probability distribution of your state space. The particle filter uses sampling to represent the multivariate probability distribution of your state space. Using samples to represent a distribution is firstly only an approximation, and secondly not ...

10

As I guess you found out yourself, the resampling method you are proposing is slightly flawed, as it should not alter the number of particles (unless you want to). The principle is that the weight represents the relative probability with respect to the other particles. In the resampling step, you draw from the set of particles such that for each particle, ...

6

Your question addresses three very different problems, all of which are hard with complicated research-type algorithms. Localization: When you have a known map of the environment and an unknown robot position. The most common algorithm for this is Monte Carlo Localizataion. This is a particle filter exactly like what you're describing. Mapping: When the ...

6

For an example of python code that properly implements resampling, you might find this github project to be useful: https://github.com/mjl/particle_filter_demo Plus, it comes with its own visual representation of the resampling process, that should help you debug your own implementation. In this visualization, the green turtle shows the actual position, ...

5

Monte Carlo localization is just another name for a particle filter. Monte Carlo methods are a broader name for computational algorithms that rely on random sampling. A particle filter is a specific application of the general Monte Carlo method for localization, and so it is simply referred to sometimes as Monte Carlo localization. If you ask Lord Google, ...

4

Of course adding random particles will change the pose of the robot. That's the point: To add randomness to account for errors you can't compute yourself. You should add random particles every time the state of the robot may change, then adjust the weight of the particles based on which seem likely given sensor data. This is a very basic definition of ...

4

While this can be done it will be problematic. The issue comes from the fact that particle filters are a method for dealing with multi-modal probability distributions (PD) while Kalman filters assume your PD can be well represented with a Gaussian distribution. I can conceive of a few methods for doing so if you really must. The most naive would be to ...

4

Most particle filter implementations will use some kind of importance sampling, which does not require you to make an assumption on the underlying distribution. This is one of the main reasons for using a particle filter in the first place. Importance sampling does not sample from the estimated distribution, but from your set of weighted samples. This ...

4

Let $w_1 \dots w_n$ be the weights of $n$ particles, $p_i \triangleq \frac{w_i}{\sum\limits_{j=1}^{n}w_j}, \sum\limits_{j=1}^{n} p_j = 1$, then as you posted, the probability of the $i$th particle surviving in the resampling procedure for multinomial resampling is: $$P(Survival_i) = 1 - (1 - p_i)^n$$ In systematic resampling, one concatenate $p_1 \dots ... 4$p\left(x_t|u_t,x^{[m]}_{t-1}\right)$should be created from your motion model. See chapter 5 in your book. Algorithms can be found in table 5.3 for a velocity model or table 5.6 for a odometry model. Roughly speaking: If your robot has a speed of 1m/s and it moves straight (which may be your$u_t$) and your update rate is 1Hz, then you could move each ... 4 I have used POMDP like models on top of a localization algorithm (Adaptive Monte Carlo Localization, from ROS), and a person detector [1][2] to find and follow a person with a humanoid robot. These two algorithms generate the input (observation) for the POMDP model in [1] and [2]. Also in [3] they used a POMDP model with similar input. As next step we used ... 4 The prediction step generates a new set of states from the old set of states. The motion model of the system is used to make this best estimate of what we think the new state might be. The motion model basically uses the information about the previous state and the current control input to determine the new state. Some noise is also added for stochasticity. ... 3 A little background. You need to weight each particle by the liklihood of the particle being correct. The probability the particle is correct is given by the probability that it is correct given the measurements. Note that the "weight" (which is a terrible term) is simply the probability of the particle being correct. Therefore, each particle is really an ... 3 If you don't know the location of the obstacles and want to apply your filter, I see two solutions: Do a quick mapping using a simple algorithm easy to implement like the Occupancy Grid Algorithm (with this you localize the obstacles relatively to your robot) Apply your particles filter Move the robot Relocalize your robot using odometry Correct odometry ... 3 Once you have enough particles to resolve your position, the effect of adding more particles shrinks to zero. You are likely seeing the best possible results that your particle filter can achieve. It looks like the smallest adequate number of particles for your simulation is somewhere between 120 and 1200, and my guess is that if you plotted the Odom ... 3 Particle filter According to the OP a robot with at least a distance sensor is available and a map too. That's a nice starting point for developing a hypothesis tracker aka particle filter. At first a game engine is needed which simulate the map and the position of particles. The game-engine calculates the expected sensor-information from the distance-... 3 Grid based FastSlam relies on the same principle that Landmakr based FastSlam. The difference is that we are not working with each grid cell as a landmark, but the whole gridmap itself. For Grid based FastSlam, each particle updates its own grid-map using the data from the range sensor (Lidar, UltraSound, etc.) and its odometry. This is called "Mapping with ... 3 I am going to give a 5 minutes answer because I am still wondering how do I got to this page.... =P Assuming you know Particle Filter concept. 1 - What is "Clustered Particle Filter"? The main idea of "Clustered Particle Filter" is to keep hypotheses alive! Imagine that in your environment you have two regions that are very similar. Almost the same. But ... 2 It looks like you're missing the angle that the robot is currently looking, e.g. robot_angle_radians. Once you know where the robot is looking, you just use the atan2 function (see also: example usage) on each point: point_angle = Math.atan2(dy, dx); // in radians Then you want to check whether the angle is not more than 30° off of the robot's angle: ... 2 It is not completely clear what your question is, but I assume you wanted to ask: Why does the output not change with the number of particles above a certain number of particles? By allowing amcl to use more samples, you decrease the errors (or inaccuracies) that are caused by the fact that the particle filter approximates your probability distributions ... 2 Since you have a 2D sensor which you can not rotate in a controlled way, you can only expect to perform SLAM in a 2D plane. Your best bet is to use the IMU for attitude estimation (roll and pitch only, since you do not have a compass) and correct each slightly rotated 2D laser scan accordingly. If your IMU does not provide you with an attitude estimate ... 2 I agree that the motion models in Probabilistic Robotics are badly suited for omnidirectional robots. I always interpreted the models presented there as examples only that should enable you to devise a custom model for your own robot. First of all you need to model and solve the forward kinematics for this kind of omnidirectional drive. I guess you already ... 2 To be clear, this is a single-point range sensor, correct? It is possible to do Monte Carlo localization with such a sensor, but it will take a very long time to converge if it does so at all, and will be easily confused. Each additional point you can add when you're doing localization will improve performance, so if you can get a few more of these sensors, ... 2 Your function calculateRange() uses a variable robot_orientation which I cannot see from your code snippets that it is set or changed. I would expect something like: double particle_orientation = particleListOrientation.get(index); inside your function measurementProbability() and then call calculateRange() with that additional parameter. private int ... 2 Your description of sample impoverishment and the way to fix it seems about right. Resampling only when the variance gets low is doing exactly what you are asking for when you say the measurements come in asynchronously. You can also improve matters by selecting the right resampling strategy. Using e.g. stratified resampling you can make sure that your ... 2 As said by Jacob, sample impoverishment is inherit to the Sampling-Importance-Resampling family of particle filters. An alternative solution which does require some extra effort is to switch to a Markov Chain Monte Carlo (MCMC) particle filter. MCMC relies on constructing a Markov chain that has a stationary distribution which is equal to the distribution ... 2 This is a good question, in that it's actually two good questions. The answer to which of those two options to use, is that they are the same. Your idea of "confusion" is intuitively correct, but is not a function of how you structure the state, but is instead about how you associate the measurements to part of the state (which robot was observed). ... 2 The algorithm can be understood by taking an example (using variables used in Probabilistic Robotics and algorithm in table 4.4 in page 110 in the same book). Algorithm: (Couldn't get math mode to work inside code mode. Hence the picture.) Consider$M = 100M^{-1} = 0.01$Let$r = 0.005$So,$U = 0.005, 0.015,......., 0.995$as loop progresses. If ... 2 The third line comes from what it is called Markov Assumption and it is Stochastic Processes stuff. Basically, it says that a distribution is not altered by the insertion and/or remotion of variables that the distribution does not really depend on. It goes like this: Is assumed that$ z_t $simply does not depend on the previous reading history, inputs$ u_{...

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# Prove that a matrix can be reduced to RRE and CRE

Tags:
1. Jun 9, 2017

### Buffu

1. The problem statement, all variables and given/known data

Let $A$ be an $m \times n$ matrix. Show that by means of a finite number of elementary row/column operations $A$ can be reduced to both "row reduced echelon" and "column reduced echelon" matrix $R$. i.e $R_{ij} = 0$ if $i \ne j$, $R_{ii} = 1$, $1 \le i \le r$, $R_{ii} = 0$ if $i > r$. Also show that $R = PAQ$ where $P$ and $Q$ are invertible $m\times m$ and $n \times n$ matrices respectively.

2. Relevant equations

3. The attempt at a solution

Since I know I can pass $A$ to a row reduced echelon matrix in finite number of operations.
Lets say the row reduced echelon form of $A$ is $R^{\prime}$. Then $R^\prime = PA$.

Also since nothing is special about rows, therefore I can say that a matrix can be passed on to column reduced echelon in finite number of steps. Therefore I can pass $R^\prime$ to a column reduced form $R$ in finite number of steps. Lets say $R = QR^\prime$

From above I can say $A$ can be passed to a column and row reduced echelon form in finite number of steps and $R = QPA$.

Is this correct ? I think it is wrong since I used a lot of words and also I got $R = QPA$ not $R = PAQ$.

2. Jun 9, 2017

### Ray Vickson

So, you can have $R = QPA,$ and this can be written as $R = P_1 A Q_1$, where $P_1 = QP$ and $Q_1 = I$ (the $n \times n$ identity matrix).

3. Jun 9, 2017

### Buffu

Then the proof is correct ?

4. Jun 9, 2017

### Ray Vickson

What proof? All you did was make statements; you did not really "prove" anything.

5. Jun 9, 2017

### Buffu

No I did not get what you are saying. Isn't statements like " You can pass from from A to a row/column reduced form in finite steps " is proven to be true.
So I just need to combine these types of statements to form a proof.


text:
Sparse Recovery via Partial Regularization: Models, Theory and Algorithms

@article{Lu2015SparseRV,
title={Sparse Recovery via Partial Regularization: Models, Theory and Algorithms},
author={Zhaosong Lu and Xiaorui Li},
journal={ArXiv},
year={2015},
volume={abs/1511.07293}
}
In the context of sparse recovery, it is known that most of existing regularizers such as $\ell_1$ suffer from some bias incurred by some leading entries (in magnitude) of the associated vector. To neutralize this bias, we propose a class of models with partial regularizers for recovering a sparse solution of a linear system. We show that every local minimizer of these models is sufficiently sparse or the magnitude of all its nonzero entries is above a uniform constant depending only on the… CONTINUE READING
3

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HIGHLY INFLUENCED

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text:
# Bazaar for Subversion users, part 1 - the basics

Jul 04, 2009, 00:02 (1 Talkback[s])

[ Thanks to Alexander Sandler for this link. ]

"Way of working with BazaarBACK TO TOC There are to ways of working with Bazaar.

"Purely distributed
"Semi-centralized

"First means that there’s no central repository. Every user works in his own repository and repositories can be easily merged. Moreover, the entire concept of branch and repository has been put together. Meaning that there are no repositories per se anymore. Each branch is a repository and a repository is a branch.

"Semi-centralized approach means that repositories kept on remote computer and we checkout the repository – just like in Subversion.

"The major difference between the two is that in the first case we checkout a repository that sits on our computer, usually in the same directory as the repository (more on this later in the series). In the later case, we have to checkout the remote repository."

Complete Story

Related Stories:


text:
# A conditional Probability Question GCSE

The probability that John goes to swimming pool on Monday is 0.8. If he goes to the pool on Monday the probability that he goes on Tuesday is 0.3. If John does not go to the swimming pool on Monday the probability he goes on Tuesday is 0.9. Find the probability that John goes exactly one of the two days.

• You can draw a tree with these probabilities, and add the two braches which are (M, but not T) + (not M, but T) – Michael Paris Sep 29 '17 at 11:07
• Welcome to MSE. It will be more likely that you will get an answer if you show us that you made some effort. – José Carlos Santos Sep 29 '17 at 11:07

Let:

• $A$ = {John goes to the pool on Monday},
• $B$ = {John goes to the pool on Tuesday}.

In your post, we have the following pieces of information:

• $\mathbb{P}(A) = 0.8$,
• $\mathbb{P}(B \mid A) = 0.3$,
• $\mathbb{P}(B \mid \overline{A}) = 0.9$ .

What you are looking for is the probability that he goes only once (either Monday or Tuesday). That is the probability of the event $(A \cap \overline{B}) \cup (\overline{A} \cap B)$. [$A \cap \overline{B}$ : if he goes on Monday, he won't go on Tuesday ; $\overline{A} \cap B$ : if he did not go on Monday, he will go on Tuesday].

We have:

\begin{align*} \mathbb{P}\big( (A \cap \overline{B}) \cup (\overline{A} \cap B) \big) & = \mathbb{P}( A \cap \overline{B} ) + \mathbb{P}(\overline{A} \cap B) \\[2mm] & = \mathbb{P}(\overline{B} \mid A)\mathbb{P}(A) + \mathbb{P}(B \mid \overline{A})\mathbb{P}(\overline{A}) \\[2mm] & = \big( 1 - \mathbb{P}(B \mid A )\big) \mathbb{P}(A) + \mathbb{P}(B \mid \overline{A})\big( 1 - \mathbb{P}(A)\big) \\[2mm] & = (1 - 0.3) \times 0.8 + 0.9 \times 0.2 \\[2mm] & = 0.74. \end{align*}

Let $M=1$ denote the event that he goes on monday and $T=1$ denote tuesday. $M=0$ not going on monday, same for $T=0$. We know $P(M=1)=0.8 \Rightarrow P(M=0)=0.2$, $P(T=1|M=1)=0.3 \Rightarrow P(T=0|M=1)=0.7$, $P(T=1|M=0)=0.9$ and we have to find the probability $P(M+T=1)$. Now utilize Kolmogorov's definition of conditional probability to compute \begin{align*} P(M+T=1)&=P(M=1,T=0)+P(M=0,T=1)\\&=P(T=0|M=1)P(M=1)+P(T=1|M=0)P(M=0)\\&=0.7\cdot 0.8 + 0.9\cdot 0.2 = 0.74 \end{align*} which is your answer

• You found (correctly) that $P(M = 0) = 0.2$, but you used $P(M = 0) = 0.1$ when you did your calculation. – N. F. Taussig Sep 29 '17 at 11:27
• Thank you, it should be correct now. – O. T. Sep 29 '17 at 11:29