# Can superposition states be created by thermal excitation?

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 ?

• Not that simple. start with en.wikipedia.org/wiki/Quantum_thermodynamics Commented Jun 27, 2021 at 18:31
• @annav The article didn't help me. There is a small part that talks about coherence's but I don't understand it well enough to decide whether or not it is helping in answering my question. It sounded a bit like coherence's dephase due to quantum friction, which could mean that superposition states dephase into mixtures of eigenstates but I am not confident about this, my comprehension could be wrong. Commented Jun 28, 2021 at 7:14
• If you are really interested here is researchgate.net/publication/… Commented Jun 28, 2021 at 8:32
• as the ntroductory to the wiki link says, it is understanding classical thermodynamics from the underlying quantum mechanical level, and it is not simple, it is a research subject Commented Jun 28, 2021 at 8:33
• @annav I don't need a comprehensive answer at research level. I was just wondering why superposition states are never mentioned when talking about canonical density operators and wondered what role they play and if there is a simple or not so simple reason why they aren't part of a thermal ensemble. I hoped that someone familiar with the topic could give some insights into this. Commented Jun 28, 2021 at 9:03

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}

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.

• That state was just a minimal example to clarify my question. I am not particularly interested in a two state system. Commented Jun 29, 2021 at 7:08