Can superposition states be created by thermal excitation?

Question

I was wondering if pure superposition states can be created thermally. Are only energy eigenstates occupied when we provide heat or are superposition states also a part of the statistical mix of states of a system in thermal equilibrium, proportional to a weight, based on their energy expectation value ?

Is for example a superposition state $|\psi\rangle =\frac{1}{\sqrt2}(|n_1\rangle +|n_2\rangle ) $ with energy expectation value $ \langle E\rangle_\psi = (E_1+E_2)/2 $ found with a probability of $p_\psi = e^{\beta \langle E\rangle _\psi}/Z$ or is it not part of a thermal ensemble ?

Answer 1

The density matrix of a thermal state in the energy basis is diagonal. This means that \begin{align} \langle n_m \vert \hat{\rho}_{th} \vert n_n \rangle &= \delta_{m,n} e^{-\beta E_m}/Z. \end{align}

As a consequence \begin{align} \langle \psi \vert \hat{\rho}_{th} \vert \psi \rangle &= \frac{1}{2}\left(\langle n_1 \vert + \langle n_2 \vert \right) \hat{\rho}_{th} \left( \vert n_1 \rangle + \vert n_2 \rangle \right) \\ &= \frac{1}{2} ~Z^{-1} \left[e^{-\beta E_1} + e^{-\beta E_2}\right] \\ &\neq Z^{-1} e^{-\beta(E_1 + E_2)/2}. \end{align}

We can see that although the probability attributed to that state is not what you expected, it is not zero. It is different from your expression because the Gibbs distribution tells you only about the probabilities of energy eigenstates, as \begin{align} \hat{\rho}_{th} &= e^{-\beta \hat{H}}. \end{align}

Answer 2

Is it somehow weird to ask about the thermal states of just two components. Many of the results of statistical mechanics (including the equation of the canonical ensemble) are derived from the $N\rightarrow \infty$ limit. For example, the theory of the equivalence between ensembles doesn't apply and you don't know if this partition function would be compatible with Boltzmann's Entropy.

However, I don't see why these coherent states shouldn't take part in the summation for $Z$. Since, by definition, the summation of $Z$ go through all the states of the system.