# Explain entropy (again)

I think I understand entropy finally. Will you verify for me?

$$S = k_B \ln( \Omega)$$

where $\Omega$ (the multiplicity) is the degeneracy of the system at some energy (E)?

So if the system is a quantum one, this equation only holds if the system is in thermodynamic equilibrium? Which is to say, if I am correct, that the probability of observing each possible eigenstate is the same for each? So is entropy ill defined for any other cases?

In any event, for bound state situations such as the particle in the well, degeneracy increases with energy. So entropy increases with energy until it reaches its maximum? So in a bound quantum system, is maximum entropy equivalent to maximum energy? Or is there some free energy equation which needs to be satisfied?

• This is just a definition, really. The logarithm makes it coincide with the definition of heat and temperature, and $k_B$ sets the scale of the Kelvin. The real meat is in $\Omega$ What is your question? – lionelbrits Nov 14 '13 at 23:51
• Is $\Omega$ the degeneracy of the system at some energy? I.E., for a particle in a well in its ground state, the entropy is precisely zero because $\Omega = 1$ – user28823 Nov 14 '13 at 23:56
• Ground states can be degenerate, e.g., in frustrated magnets. In the canonical ensemble, $\Omega$ is the degeneracy of a certain macrostate. – lionelbrits Nov 15 '13 at 0:15
• The expression that I gave at the end of my answer is general and becomes the von Neumann entropy for a quantum system. See the Wiki article en.wikipedia.org/wiki/Von_Neumann_entropy to understand how. – WetSavannaAnimal Nov 15 '13 at 5:10
• @lionelbrits I had to look up "frustrated magnets" - never heard of them - very interesting, although the name made me giggle. Do their frustrations make them bitter, twisted and dangerous beings? – WetSavannaAnimal Nov 15 '13 at 9:05

You're on the right track. Your statement can be true in special cases. More generally, the entropy is defined by the Boltzmann-headstone-equation you write down if $\Omega$ is the number of different detailed states (called the "microstates" or "compatible microcomplexions" in the thermodynamicist's jargon) that (i) the state could be in if (ii) its macroscopic properties are as observed. More succinctly, $\Omega$ is the number of "microcomplexions compatible (consistent) with a given macrostate". Moreover, all of the microcomplexions have to be equiprobable for the Boltzmann formula you quote to hold. So, if you have a system whose thermodynamic state is wholly defined by its internal energy $E$, your statement is true. More generally, however, several parameters are needed to define a macroscopic state of a system. For example, an ideal gas at thermodynamic equilibrium needs two macroscopic quantities, say pressure $P$ and volume $V$ or, alternatively, internal energy $E$ and either pressure or volume to wholly specify its macroscopic state.
In general (i.e. if equiprobability cannot be assumed), if the list of all possible microstates consistent with a system's macrostate is indexed by, say, $i$, and if, in that macrostate, the probability of the $i^{th}$ microstate is $p_i$, then the system's (Shannon) entropy is $k_B\,\sum\limits_i p_i\,\log p_i$. Try working this out if you assume all the $p_i$ are the same: you will find you will recover your formula.