# Adibatic process in QM and thermodynamics?

I have come over the phrase 'Adiabatic process' in two different contexts, that of QM and Thermodynamics .

QM

A adiabatic process is one is slow compared to: $$t=\frac{\hbar}{E_n-E_m}$$ and in which the probability that the system is in a given eigenstate remains the same.

Thermo

An adiabatic process is one in which there is no heat transfer to or from the surroundings.

I was wondering how these two definitions are linked?

## 1 Answer

They are linked via the Gibbs-Shannon entropy given by: $$S_G=-\sum_i P_i \ln(P_i)$$ Where here $p_i$ is the probability that the system will be in the $i$th eigenstate. The relationship between the Gibbs-Shannon entropy and your standard thermodynamic entropy is: $$S=k_BS_G$$ In the quantum mechanical 'adiabatic process' we are told that the probabilities $P_i$ do not change, this means that $S_G$ does not change and hence $S$ also does not change (the thermodynamic definition of an adiabatic process).

So an adiabatic process in the quantum mechanic sense of the word is in essence equivalent to that of the thermodynamic meaning.

• Not quite. A unitary evolution in QM or (condensed matter) QFT, to which the procedure of "adiabatically turning on an interaction" is usually applied, is always an "adiabatic process" in the sense of your definition, since the probabilities of energy eigenstates, in general of evolved states, and the entropy itself are always conserved for any state/density matrix $\rho$: $\langle E_i|\rho(t)|E_i\rangle = \langle E_i|\rho(0)|E_i\rangle$, $\langle \Psi(t)|\rho(t)|\Psi(t)\rangle = \langle \Psi(0)|\rho(0)|\Psi(0)\rangle$, $S(\rho(t)) = S(\rho(0)$ for $\rho(t) = U(t) \rho(0) U^\dagger(t)$. – udrv Mar 16 '16 at 15:33
• Your observation would be applicable perhaps to open quantum systems, so you need to specify the context to which the condition $t = \hbar/(E_n - E_m)$ applies. – udrv Mar 16 '16 at 15:34