# Redo the experiment 'Reversal of thermodynamic arrow of time'

In a recent paper "Reversing the thermodynamic arrow of time using quantum correlations" by Micadei et al, an experiment was carried out to show a reversal of time.

Basically they prepare a mixed quantum state $\rho_{AB}^0=\rho_A\otimes\rho_B+\chi$, where $\rho_A=\exp(-\beta_AH_A)/Z_A$ and $\rho_B=\exp(-\beta_BH_B)/Z_B$ with $H_A=H_B=(I-\sigma_z)/2$, so locally they are thermal states with temperatures $T_A=1/\beta_A$,$T_B=1/\beta_B$.

The interaction Hamiltonian is $H_{AB}=(\sigma_x^A\sigma_y^B-\sigma_y^A\sigma_x^B)$, which will evolve the state $\rho_{AB}^0$ to $\rho_{AB}^t$, where the reduced density matrix $\rho_A^t,\rho_B^t$ are still thermal states w.r.t. $H_A,H_B$, so we can see a heat flow between the subsystems A and B, and A,B will have time-varying temperatures $T_A^t,T_B^t$.

Then when certain correlation $\chi$ exists, they can observe 'the reversal of the thermodynamic arrow of time', where the high temperature subsystem will absorb heat from the low temperature subsystem.

I am thinking the following questions:

If we let a moving observer (with a constant speed $v$ relative to the lab) to check the experiment, what will he/she see?

1. Will still he/she see the same setup? Which means will $H_A,H_B,H_AB$ be the same as in the original setup?
2. Will the reduced states $\rho_A^t$ and $\rho_B^t$ still be thermal w.r.t. $H_A$,$H_B$?
3. Will he/she measure the same temperature $T_A^t$ and $T_B^t$?
4. How about a rotational observer?
5. How about an accelerating observer?

PS. I made a simulation according to my understanding. I found a moving observer will see the system approaches a pure product state $\rho_{AB}=|(0+1)_A(0+1)_B\rangle\langle(0+1)_A(0+1)_B|$ when the speed $v$ approaches $c$. Is this reasonable?

• See these (and other) posts about Special Relativity and temperature physics.stackexchange.com/q/63072 physics.stackexchange.com/q/83488 physics.stackexchange.com/q/262382 – By Symmetry Dec 13 '17 at 12:30
• @Symmetry Thanks. But the problem here is that the 'temperature' ihere is not a classical definition, instead it's dependent on the Hamiltonian $H_A$ and $H_B$. So if the reduced density matrix is not 'thermal' any more, it's not possible to define temperature or internal energy. So I have to find out how to define them. – XXDD Dec 13 '17 at 12:39
• how do you even handle the relativistic transformation with this formalism? Will it not depend on the particular implementation of the states? – glS Dec 13 '17 at 12:56
• @glS Yes, you are absolutely right. I am trying to understand this by the similarity between 2-qubit operations and the Lorentz transformation. So I assume the boost transformation is equivalent to a SL operation given by the (1/2,1/2) representation of the Lorentz transformation. For example, a boost in x direction is generated by $exp((IX-XI)*\theta)$ with $\theta$ related with the speed. But this did change the state of the system/subsystems so the subsystems are not thermal any more. – XXDD Dec 13 '17 at 13:00