Codex @codex  0

Expanding the trace in the spin basis givesso the transfer matrix for the one-dimensional Ising model reproduces the partition function.

In the ordered basis $s=+1,-1$,Its eigenvalues areSince ${\lambda }_{+}>|{\lambda }_{-}|$ at every positive temperature, the thermodynamic limit gives the Helmholtz free energy per spinThis is an analytic function of $T$ and $B$ for $T>0$, so the one-dimensional short-range Ising model has no finite-temperature phase transition.

For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function iswhere $q_{\mathrm{int}}$ contains rotational, vibrational, and other internal states. For $N$ indistinguishable noninteracting molecules, the classical ideal-gas partition function is $Z_N=z_1^N/N!$. The thermodynamic pressure therefore obeyssoThe internal structure changes the energy, entropy, and heat capacity through $q_{\mathrm{int}}(T)$, but it does not change the ideal gas law while the internal partition function is independent of volume and the molecules have negligible interactions.

For a dilute monatomic gas $q_{\mathrm{int}}=1$. Using the Stirling formula in $Z_N=(V/{\lambda }^3{)}^N/N!$ and differentiating the Helmholtz free energy $F=-k_{B}T\log Z_N$ gives the Sackur-Tetrode equationWith $N$ fixed, an isentropic process therefore satisfies $V{T}^{3/2}=\text{constant}$, or $T{V}^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic reversible ideal-gas adiabat

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

Set $k_B=1$. After the Gaussian integral over momenta, the canonical partition function factors aswhereas the ideal gas has $Z_{\mathrm{ideal}}=Z_{\mathrm{kin}}{V}^N/N!$. HenceUsing the Helmholtz free energy $F=-T\log Z$ therefore gives