What Happens if One Couples two Thermal Harmonic Oscillators at Different Temperatures?

Question

Consider two identical initially uncoupled harmonic oscillators with Hamiltonians

$$\hat{H} = \frac{p_1^2}{2m}++\frac{m\omega^2x_1^2}{2},$$

$$\hat{H}_2 = \frac{p_2^2}{2m}+\frac{m\omega^2x_2^2}{2}.$$

The Hamiltonian of the full system is then:

$$\hat{H} = H_1 + H_2$$

Each oscillator starts at a thermal state $$\rho_1 = e^{-\beta_1 H_1},$$ $$\rho_2 = e^{-\beta_2 H_2}.$$ The full state $\rho = \rho_1\otimes\rho_2$ is stationary with respect to the evolution generated by $H$.

After some time, we switch an interaction, such that the full Hamiltonian of the system is given by

$$H_I = H_1 +H_2 + \lambda^2 x_1 x_2.$$

The question is: What will happen to the state $\rho$? I don't think it will thermalize to a temperature $\beta$ because that would either violate energy or entropy conservation. However, there are different options for what might happen to the state, such as:

1. Each of the oscillators ends up in a thermal state after tracing the other one, but it is a highly entangled state.

2. The state of the system keeps oscillating temperatures, without tending to any final state.

Which one of these is more accurate? Or, is there another possibility I haven't thought of?

Answer 1

Energy will spontaneously flow from the hotter system to the colder system, until the entire combined system reaches equilibrium at some new temperature. The total energy of the system will be conserved. The entropy will increase.

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Entropy is constant for adiabatic processes in quantum mechanics. As discussed on wikipedia, a necessary condition for adiabaticity is that the Hamiltonian must commute with itself at different times, $[H(t), H(t')] = 0$ for all $t, t'$.

We can check that this condition is violated for your case. Let $t$ be a time when $\lambda=0$, and let $t'$ be a time when $\lambda\neq 0$. Then (setting $\hbar=1$ so $[x, p] = i$) \begin{eqnarray} [H(t), H(t')] &=& \left[H_1 + H_2, H_1 + H_2 + \lambda x_1 x_2 \right] \\ &=& \lambda \left( [H_1, x_1] x_2 + x_1 [H_2, x_2] \right) \\ &=& \frac{- i \lambda}{m} \left( p_1 x_2 + x_1 p_2 \right) \end{eqnarray} So there is no reason to expect the entropy to remain constant in this scenario, unless the way the interaction is turned on is done very carefully (https://arxiv.org/abs/0910.0709).

An important subtlety mentioned on the wikipedia page is that there is a distinction between the "energy entropy" and the "von Neumann entropy". The von Neumann entropy is conserved under unitary time evolution, but the energy entropy is not, and the latter is what is analogous to the thermodynamic entropy away from equilibrium (in equilibrium these two definitions of entropy agree). This is fleshed out in more detail in this paper: https://arxiv.org/abs/0806.2862. Section III.A describes an example that is conceptually very similar to yours, but even simpler: a gas is confined to one side of a box by a divider, which is then remove.