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If I have a quantum mechanical system (QMS) whose density matrix is at temperature $T$ and I put it in contact with another quantum mechanical system (QMS2) at temperature $T$ again.

However, if I try to model this dynamically I'd be tempted to use the sudden approximation. Let the probability of QMS being in energy state $| E_i \rangle$ is $e^{- \beta E_i}$ and QMS2 being in energy $| E_j \rangle$ is $e^{- \beta E_j}$. Hence, the probability of the below outcome

$$ \Big (H_{QMS} \otimes 1+ 1 \otimes H_{QMS2} \Big ) |E_i ,E_j\rangle \to \Big (H_{QMS} \otimes 1 + 1 \otimes H_{QMS2} + H_{int}\Big )|\lambda_l ,\lambda_k \rangle$$

(where $H_{int}$ is the interaction Hamiltonian and $|\lambda_l ,\lambda_k \rangle$ are arbitrary states of the combined system) is given by:

$$ e^{-\beta(E_i +E_j)} |\langle E_i | \lambda_l \rangle \langle E_j | \lambda_k \rangle|^2 $$

However, by the zeroth law of thermodynamics I know that these systems should remain in thermal equilibrium as well. I'm uncertain as how to show that is the case here?

More Anonymous
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2 Answers2

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The argument here is that we are dealing with thermodynamicly large systems and that there are no long ranged interactions between the 2 systems. In this case $H_{int} $ scales with the surface area where the systems were brought into contact, whilst $H_{QMS1} $ and $H_{QMS2} $ scale with volume, so $H_{int} $ can be neglected and your final Hamilton is the same as your initial Hamilton, in which case clearly the zeroth law holds.

If the previous assumptions do not hold and $H_{int} $ is comparable to the system Hamiltonians, then bringing the 2 systems into contact will involve doing a non-negligible amount of work on the systems, in which case it is not surprising the zeroth law fails.

By Symmetry
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If one applies a sudden perturbation, like suddenly turning on the interaction between two systems at different temperature, we are definitely dealing with a non-equilibrium situation. As thermodynamics predicts, there will be transfer of energy from one system to another, however what exactly happens depends on the order in which one takes the limits of infinite time and the thermodynamic limit.

A situation frequently considered when studying non-equilibrium heat and electric transport is taking the thermodynamic limit first, which results in a steady-state flow of heat or current between the two systems (in the case of electric transport the systems are usually at the same temperature but different chemical potentials). This is definitely a non-equilibrium situation, although each of the two systems is treated, as if it were in equilibrium.

Roger V.
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