State degeneracy

This counts microstates sharing a macrostate; when the shared quantity is energy, it gives the multiplicity of degenerate energy levels. The degeneracy $g(l)$ of a macrostate labelled by $l$ is the number of distinct microstates having that value of $l$.

When a single-particle energy is independent of its spin projection and all projections are available, the state degeneracy due to spin is $g_s=2S+1$. A spin-half gas has $g_s=2$ if both projections are available, while a fully polarized gas can have one populated spin species. A magnetic field that resolves Zeeman splitting requires counting the separate energies instead of multiplying one common occupancy by $g_s$.

A degeneracy factor counts the distinct microscopic states that share the same macroscopic quantum numbers or energy.

A microstate is a complete microscopic specification within the chosen model. For two distinguishable, noninteracting two-level systems, label the units $A,B$ and their states $0,1$, with energies $0,\Delta$ and $\Delta >0$. The four microstates arewith total energies $0,\Delta ,\Delta ,2\Delta$. Distinguishability matters: $(0,1)$ and $(1,0)$ describe different assignments to the labeled units. Noninteraction matters too: their energies add without an extra coupling term.

For a classical particle model, a microstate can instead be a point in phase space, specifying every position and momentum. In the quantum canonical ensemble, the sum uses energy eigenstates; a general quantum state may also contain quantum coherence, which an energy label alone does not specify.

A macrostate groups microstates sharing prescribed coarse data. The grouping depends on what is retained: here retain only total energy. For the two distinguishable two-level systems, the macrostates $E=0,\Delta ,2\Delta$ have state degeneracies $1,2,1$. Although two units are not macroscopic in size, they make this loss of information visible: knowing $E=\Delta$ does not tell us which unit is excited.

At inverse temperature $\beta \Delta =\ln 3$, each unit's excited-state probability is $p=1/4$. Their independence gives microstate probabilities $9/16,3/16,3/16,1/16$ in the order above. Adding within each macrostate givesThe total expected value of energy is therefore $\Delta (6/16)+2\Delta (1/16)=\Delta /2$, matching two units each with expected value of energy $\Delta /4$. Equal state degeneracies would give the wrong answer: macrostate probabilities combine counting with the Boltzmann factor. At very high positive temperature, all four microstates become equally likely, but these three macrostate probabilities approach $1/4,1/2,1/4$.