Cultural Biases in Economic Exchange
Summary (3 min read)
Introduction
- It is suggested that the scaling of the relaxation time with the system size can serve as a signature of a nonequilibrium phase transition.
- Another interesting aspect is that in a certain thermodynamic or weak driving limit the solution can be written as a matrix product operator (MPO) with matri ces of small fixed dimension 4.
- It is shown that the model exhibits a nonequilibrium phase transition at zero dephasing, going from the NESS that is Gaussian and displays a ballistic transport, to the NESS that is nonGaussian, shows diffusive spin transport, and exhibits long-range correlations for nonzero dephasing.
II. THE MODEL
- The Lindblad dissipator is therefore a sum of two terms, Ldis = Lbath + Ldeph. (4) The dephasing part Ldeph = ∑n j=1 L deph j is a sum of Ldephj , each of which acts only on the j-th site and is described by a single Lindblad operator, Ldephj = √ γ 2 σzj .
- To see that this is indeed the case one can look for a stationary state of the bath LbathL dissipator only.
- Of course, the nonequilibrium stationary state of the whole master equation (2), which in addition includes a unitary part and a dephasing, will have a slightly different magnetization at the ends.
III. THE SOLUTION
- This simple structure is a consequence of the fact, that all other operators, that could result in jk or σ z j when operated on by L, are zero.
- The matrix of the superoperator L (B) j has only two nonzero elements.
A. Hierarchy of connected correlations
- Let us briefly argue why the NESS can be calculated term by term in the expansion over µ, Eq. (11), and why the set of equations for each term is closed.
- Similar expressions hold for the bath at the right end.
- The only superoperator that does not conserve the number of operators is that of the bath, which can create σz out of an identity.
- If the authors look at the coefficient in front of an operator that is a product of r non-identity operators, then this coefficient is a linear function of coefficients in the set S(r) and coefficients in the set S(r−1) (these come from the action of the bath Lbath).
- The largest non-connected term in µrR(r) on the other hand scales as ∼ brnr and comes from the product of r operators σzj , see also the explicitly ansatz below.
B. No magnetization offset, µ̄ = 0
- Let us first find the NESS for a bath with a zero offset of magnetization, µ̄ = 0.
- It is instructive to first find the equilibrium stationary state in the absence of driving, when µ = 0.
- Inducing no net magnetization.
- This equilibrium state can therefore be thought of as an infinite temperature state.
1. First two orders
- Higher order terms in µ are more complicated in this case (although of the same form) and the authors do not discuss inhomogeneous dephasing in the present work.
- The dependence of the maximal current on γ is trivial, the smaller the dephasing the larger is the current.
- If nγ is on the other hand smaller than ΓL,ΓR, for instance if the dephasing is zero, then in the limit of weak coupling, ΓL,ΓR → 0, κ diverges.
C. Nonzero offset, µ̄ 6= 0
- Let us now go to the case where there is an offset of magnetization in the baths.
- One should note though that integrable systems, such as their XX chain, do not thermalize for generic local Lindblad baths [26].
- Out of equilibrium, when µ 6= 0, the solution is actually very similar to the one with zero µ̄.
IV. THE NESS IS NON-GAUSSIAN
- I.e., that the Wick theorem does not hold.the authors.
- A system of spin-1/2 particles can be mapped to spinless fermions using the Jordan-Wigner transformation.
- Whether it is nevertheless equivalent to some existing solvable model is at present unknown.
A. Matrix product operator ansatz
- The authors have numerical indications though that the Schmidt rank, i.e., a number of nonzero Schmidt coefficients, for a bipartite cut after the first m spins is 4m.
- Looking at the series solution for NESS (11) the authors can see that the largest connected term in the r-th order µrR(r) (i.e., r-point connected correlation function) scales for large n as ∼ br n = 1/nr−1 and comes from the connected correlation of r σzs.
- Another interesting point is that for µn > νc the second largest Schmidt coefficient is independent of n and scales as λ2 ∼ µ 2/n0.
V. NONEQUILIBRIUM PHASE TRANSITION
- From the exact solution the authors can see that the NESS undergoes a transition from a state without long-range correlations for γ = 0, to the one with long-range correlations for γ.
- In the phase with long-range correlations two-point z − z correlations scale as ∼ µ2/n [9] and are therefore of purely nonequilibrium origin.
- The correlation function has a plateau because in the thermodynamic limit the decay of Ci,j with the distance between indices |i− j| gets increasingly slower (25).
- Of interest for nonequilibrium phase transitions is therefore the second largest eigenvalue λ2 of the Lindblad superoperator L, or in particular the gap ∆, ∆ = −λ2.
- Because the authors in general expect the relaxation time, in the case of local coupling to baths at chain ends, to grow with the system size at least as ∝ n – larger system simply needs more time to relax –, the gap is expected to decrease with n, even if they are not at the nonequilibrium phase transition point.
VI. SUMMARY
- The authors have provided exact expressions for all one-point, two-point and three-point connected correlations in the nonequilibrium stationary state of the XX model with dephasing.
- The nonequilibrium stationary state is non-Gaussian because the Wick theorem does not apply.
- In the thermodynamic and weak-driving limit the solution can be written in terms of a matrix product ansatz with matrices of fixed dimension 4.
- At zero dephasing the model exhibits a nonequilibrium phase transition from a state 10 with only nearest-neighbor correlations to a state possessing long-range correlations.
- It is conjectured that at a quantum nonequilibrium phase transition point the gap of the superoperator closes with the system size more rapidly than in the vicinity of the transition point.
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Q2. What are the future works mentioned in the paper "Nber working paper series cultural biases in economic exchange" ?
Only future research will be able to tell.
Q3. What is the effect of trust on differentiated goods?
The effect of trust is consistently stronger for differentiated goods: trade increases 36 percent versus 16 percent in response to one-standard deviation increase in trust.
Q4. How much of the variability in trust can be explained by the characteristics of a country?
Characteristics of the country expressing and receiving trust can (controlling for time variation) at most explain between 44 and 64% of the variability in trust depending on how the aggregate trust of a country’s citizens is computed.
Q5. How do you find the significance level of the instrument?
The authors find that the significance level is higher than 3% , where Stock and Yogo (2002) recommend an F test with at least a 5% significance level.
Q6. What is the role of trust in economic decisions?
In such a case, the role played by trust would be second order: except for very high level of risk aversion, trust modelled in this way is bound to have very little impact on decisions.
Q7. What is the effect of fixed effects on the trust of citizens of a country?
These fixed effects will capture any variable that is specific to the country and affects its average trust and trustworthiness, such as the level of protection that contracts receive, the enforcement granted by social punishment, the constraints that individuals in a country have in their behaviors due to binding cultural norms.
Q8. How much of the effect of cultural biases against an enemy is significant?
As one would expect if these cultural biases against an enemy fade over time, the impact of recent wars is five times that of distant wars.