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Degeneracy pressure is the pressure of a fermionic system that remains at zero temperature because the Pauli exclusion principle forces particles to occupy states with nonzero momentum.

A fermionic creation operator is the adjoint of a fermionic annihilation operator and adds one fermion to a mode. Its square vanishes, implementing the Pauli exclusion principle.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is $(\square -m^2)\varphi =0$, or $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.

Resolve the free field into positive- and negative-frequency plane waves. With $E_{\mathbf{k}}=\sqrt{|\mathbf{k}{|}^2+m^2}$ and $k\cdot x=-E_{\mathbf{k}}t+\mathbf{k}\cdot \mathbf{x}$,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in $[\varphi ,\pi ]$ each supply half the Dirac delta function. Thus the normalization $1/\sqrt{2E_{\mathbf{k}}}$ is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with $a$ and $a^{†}$ its annihilation operator and creation operator.

Choose the Fock vacuum by $a(\mathbf{k})|0⟩=0$. Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume $V$ can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, $[H,a^{†}(\mathbf{k})]=E_{\mathbf{k}}{a}^{†}(\mathbf{k})$ and $[\mathbf{P},a^{†}(\mathbf{k})]=\mathbf{k}{a}^{†}(\mathbf{k})$. A one-particle state therefore has positive energy, momentum $\mathbf{k}$, and invariant mass $m$. A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.

A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase $e^{-i{E}_{\mathbf{k}}t}$; a narrow packet has group velocity $\mathbf{v}=\mathbf{k}/E_{\mathbf{k}}$, whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.

Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so $[\varphi (x),\varphi (y)]=0$ for spacelike $x-y$. Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.

Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are $(a^{†}{)}^n|0⟩/\sqrt{n!}$ with $n=0,1,2,\dots$: any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.

Treat the quarks as the fundamental triplet of approximate flavour symmetry. The flavour product follows by splitting the first two quarks into symmetric and antisymmetric pieces:ThereforeThe dimension check is $27=10+8+8+1$. The baryon decuplet is the completely symmetric flavour sector, the three-quark flavour singlet is completely antisymmetric, and the two copies of the baryon octet carry mixed permutation symmetry. The two octet copies are a multiplicity space for permutations of the three quark slots, not automatically two distinct ground-state baryon octets.

For the weight diagrams use isospin projection $I_3$ and flavour hypercharge $Y$. The quark weights areWeights add in a tensor product. In a three-quark composition, $Y=1-n_s$ and $I_3=(n_u-n_d)/2$. The baryon decuplet has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. Its upper-right weight is $uuu$, the ${\Delta }^{++}$, while its bottom weight is $sss$, the ${\Omega }^{-}$. In the baryon octet, the upper weights are $uud$ and $udd$, giving the proton and neutron. The origin has two independent states with content $uds$: the $I=1$ ${\Sigma }^0$ and the $I=0$ $\Lambda$. Their equal weights do not make them the same state. The three-quark flavour singlet has only $(I_3,Y)=(0,0)$ and content $uds$, with normalized flavour wavefunctionThis singlet is a different representation from the octet $\Lambda$, despite the same quark content and weight.

The Pauli exclusion principle requires the full three-quark wavefunction to change sign under exchange of any two quarks, including their spatial, spin, flavour and colour labels. A three-quark colour singlet has the antisymmetric colour factor ${ϵ}_{abc}/\sqrt{6}$. Consequently the remaining spatial-spin-flavour factor must be symmetric. The flavour representation alone is not the full exchange wavefunction.

For the lowest orbital state, the spatial wavefunction is symmetric. Completely symmetric decuplet flavour then requires the symmetric spin-$3/2$ wavefunction; this includes states such as $uuu$ with aligned spins and does not violate Pauli because their colours are antisymmetrized. Mixed octet flavour combines with mixed spin-$1/2$ wavefunctions to give a symmetric spin-flavour factor. More explicitly, the two-dimensional permutation representation of mixed symmetry tensored with itself contains the trivial representation, which selects the physical symmetric combination. These are the familiar ground-state spin assignments.

Antisymmetric singlet flavour in a symmetric orbital state would instead require a completely antisymmetric three-quark spin state. But ${\bigwedge }^3{\mathbb{C}}^2=0$, so three spin-$1/2$ quarks have no such state. There is no flavour-singlet three-quark ground-state S-wave baryon. A singlet is allowed with orbital excitation: mixed spatial and mixed spin symmetry can combine antisymmetrically, and their product with the antisymmetric flavour sector is symmetric. For example an $L=1$, spin-$1/2$ configuration can give negative-parity total spins $1/2$ or $3/2$. This Pauli constraint on three-quark flavour multiplets distinguishes a permitted representation in the flavour tensor product from its possible orbital-spin realization.

For a pentaquark, choose each of two quark pairs in $\overline{3}$. Their symmetric flavour combination lies in $\overline{6}$, and combining with the antiquark givesThus $3^{\otimes 4}\otimes \overline{3}$ contains a pentaquark antidecuplet. This identifies a flavour sector; it does not by itself prove binding or fix the spin and orbital structure needed for overall fermion antisymmetry.

The antidecuplet is the conjugate of the symmetric decuplet, so its rows are $(Y,I)=(2,0),(1,1/2),(0,1),(-1,3/2)$. A three-quark state only has $Y=1,0,-1,-2$, and at $Y=-1$ it contains two strange quarks and one light quark, allowing only $I_3=\pm 1/2$. Hence the exotic weights are preciselyPossible minimal contents are $uudd\bar{s}$, $ddss\bar{u}$, and $uuss\bar{d}$. By the Gell-Mann--Nishijima formula, their charges are respectively $+1,-2,+1$. Every other antidecuplet weight is also a weight of some three-quark composition, although its total isospin representation can differ.

Exotic weights do not imply a weak-decay lifetime. The strong interaction can conserve all the quantum numbers in baryon-plus-meson channels, for exampleThese are quark rearrangements into a three-quark baryon and a quark-antiquark meson, not decay into a single three-quark state. Accordingly they are generically short-lived strong resonances if these channels are kinematically open. Flavour representation theory alone gives no masses or widths; a state below all strong thresholds, or one with dynamically suppressed couplings, can be longer-lived. The quantum numbers provide no general protection against the displayed strong decays.

A stellar equation of state supplies pressure and internal energy as functions of density, temperature and composition, together with thermodynamic derivatives needed for stability and transport. Hydrostatic equilibrium fixes the pressure gradient, but does not determine which microscopic components provide the pressure. In ordinary dense interiors local thermodynamic equilibrium is a useful starting point. A consistent mixture iswhere ions, Electrons, radiation and interaction corrections are distinguished. The classical Electron pressure and electron degeneracy pressure are two limits of the same Electron contribution, and must not be added as if they belonged to different particles. The finite-temperature electron equation of state interpolates between them.

In a fully ionized, nondegenerate, nonrelativistic gas, $P_g=\rho k_{B}T/(\mu m_u)$ and specific thermal energy is $u_g=3k_{B}T/(2\mu m_u)$. With nuclear mass fractions $X_A$, charges $Z_A$ and mass numbers $A_A$, the mean molecular weight satisfies ${\mu }^{-1}={\sum }_A{X}_A(1+Z_A)/A_A$, and ${\mu }_e^{-1}={\sum }_A{X}_A{Z}_A/A_A$ is the mean molecular weight per electron. This regime describes much of an ordinary main sequence interior. Toward cooler layers, ionization and molecular dissociation change particle numbers and consume heat. The Saha equation relates ionization to both temperature and Electron density: there is no universal horizontal ionization boundary. These regions have larger heat capacity and can have a reduced stellar adiabatic exponent. The simple fully ionized formula is then insufficient.

Equilibrium photons give radiation pressure $P_{\gamma }=a_r{T}^4/3$ and energy per volume $a_r{T}^4$, or specific energy $u_{\gamma }=a_r{T}^4/\rho$. In the nondegenerate gas regime, equality with gas pressure gives the radiation-to-gas pressure boundarya line of slope $1/3$ on a $(\log \rho ,\log T)$ plot. Higher temperatures at fixed mass density favor photon support. A monatomic gas has stellar adiabatic exponent $5/3$, while radiation alone has $4/3$; their adiabatic exponents of a monatomic gas-radiation mixture are not obtained by assuming a fixed pressure fraction during compression. Radiation support is particularly important in massive stars.

For Electrons, Pauli exclusion principle and the Fermi-Dirac distribution determine occupation numbers. The net Electron density is $n_e=\rho /({\mu }_e{m}_u)$ and the Fermi momentum is $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. Define the kinetic electron Fermi temperatureFor $T\gg T_F$ the Electrons are nearly classical; for $T\ll T_F$ they are strongly degenerate and their pressure depends primarily on density. The intermediate region requires finite-temperature electron equation of state integrals, not a discontinuous switch of formulas.

The equation of state of a cold electron gas gives, in its two limits,These are respectively the $n=3/2$ and $n=3$ pressure-density powers, explaining the approximate white dwarf polytropic mass-radius relation sequence and the Chandrasekhar limit. The kinetic energy per volume is $3P_e/2$ in the nonrelativistic limit and $3P_e$ in the ultrarelativistic limit. Ions can still supply much of the heat capacity even when the Electron pressure supplies the mechanical support.

The Electron degeneracy crossover $T\sim T_F$ has slope $2/3$ at low mass density and $1/3$ at high mass density. The electron relativistic density threshold isa vertical marker where $p_F=m_{e}c$. It is different from the thermal electron relativistic threshold $k_{B}T\sim m_e{c}^2$, near $5.93\times 10^9\mathrm{K}$, a horizontal temperature scale. Hot dilute matter can have relativistic thermal Electrons without degeneracy; cold dense matter can have relativistic degenerate Electrons without reaching that temperature.

Electron degeneracy also does not automatically imply that Electrons dominate the total pressure. Comparing the cold Electron limit with radiation gives the radiation-to-degeneracy pressure boundaryIts logarithmic slopes are $5/12$ for nonrelativistic Electrons and $1/3$ for ultrarelativistic Electrons. One must compare the pressures separately from the degeneracy criterion; extrapolating the classical gas-radiation line into a degenerate region is incorrect.

The stellar equation-of-state regime diagram uses an illustrative fully ionized carbon composition, ${\mu }_e=2$, $\mu =12/7$. It displays the electron-degeneracy crossover, both radiation-pressure comparisons, and distinct thermal and density-driven relativity scales. The curves are limiting-model comparisons, not sharp phase boundaries or a calibrated complete equation of state. Partial ionization, molecular physics and interactions modify the low-temperature regions indicated on the plot.

At sufficiently high temperature, electron-positron thermal pair abundance can become important. The pair abundance depends on density and chemical potential as well as temperature; $k_{B}T=m_e{c}^2$ is not a universal onset line. In the dilute ultrarelativistic limit, both pair species together add energy density $7a_r{T}^4/4$ to the photons' $a_r{T}^4$, and have pressure one third of their energy density. While pairs are being created, thermal energy is spent on rest mass, which can reduce the stellar adiabatic exponent below $4/3$ and contribute to pair-instability supernova physics.

At high mass density and low temperature, Interactions governed by Coulomb's law invalidate the noninteracting-ion approximation. The ionic Coulomb coupling parameter $\Gamma =Z^2{e}^2/(4\pi {ϵ}_0{a}_i{k}_{B}T)$, where $a_i=(3/(4\pi n_i){)}^{1/3}$, grows as ${\rho }^{1/3}/T$. Corrections become significant when $\Gamma$ is of order one, and a sufficiently strongly coupled plasma can crystallize. At still greater mass density, electron capture alters ${\mu }_e$ and nuclear matter replaces the ideal electron-ion model; neutron star interiors require strong-interaction and relativistic equations of state. These further regimes lie beyond the simple pressure curves plotted here. A useful stellar EOS is thermodynamically consistent across the crossovers, rather than just the maximum of unrelated pressure laws.

The normalized Bloch states $|k⟩=N^{-1/2}{\sum }_{n=1}^N{e}^{ikna}|n⟩$ satisfy periodicity when $k=2\pi j/(Na)$. Acting with the hopping terms shifts the coefficients by one site, givingThe first Brillouin zone is a fundamental cell of reciprocal space, for example $[-\pi /a,\pi /a)$; wavevectors differing by $2\pi /a$ label the same state. It contains exactly $N$ distinct allowed wavevectors and hence $N$ orbital states. The written Hamiltonian has no spin index; if electron spin degeneracy is included there are $2N$ states.

For a narrow wave packet, group velocity and band-curvature effective mass areFor the usual $J>0$, the mass is negative in the outer half of the zone, $\pi /(2a)<|k|\le \pi /a$, taking one endpoint representative. At $|k|=\pi /(2a)$ the inverse mass vanishes and the mass diverges. For arbitrary nonzero $J$, the sign criterion is $J\cos ka<0$; $J=0$ is a flat band.

For electron charge $-e$ with $e>0$ in a constant electric field $\mathcal{E}$, the single-band semiclassical equations are $\hslash \dot{k}=-e\mathcal{E}$, $\dot{x}=v(k)$. Thus $k(t)=k_0-e\mathcal{E}t/\hslash$, and for $\mathcal{E}\ne 0$,These are Bloch oscillations of period $2\pi \hslash /(|e\mathcal{E}|a)$. The zero-field limit is $x-x_0=v(k_0)t$. This assumes a narrow packet and neglects scattering and interband transitions.

For many electrons obeying the Pauli exclusion principle, a partially filled band has available nearby states and can carry current under a displaced occupation distribution: it describes a metal within band theory. A completely filled band has cancelling velocities over the zone, so its current is zero; if separated from empty higher bands by a nonzero gap, it describes an insulator at sufficiently low temperature and weak electric field. With spin degeneracy this isolated band is filled at two electrons per site. The single-electron Hamiltonian alone does not specify the filling or a gap to a next band, so those extra physical assumptions are required for classifying a material.

For noninteracting fermions, each single-particle level has occupation $n_i\in \{0,1\}$ by the Pauli exclusion principle. A microstate has $E={\sum }_i{ϵ}_i{n}_i$, $N={\sum }_i{n}_i$, so its grand-canonical weight is the product of its independent level weights. The grand potential is $\Omega =-kT\log \mathcal{Z}$; the particle number, entropy and pressure follow from $N=-{\partial }_{\mu }\Omega$, $S=-{\partial }_T\Omega$ and $P=-{\partial }_V\Omega$. This occupation description is the basis for the product, continuum and adiabatic calculations.

Summing separately over the two allowed occupations of every level givesThusThe factorization uses the absence of interactions; each level, including any degeneracy labels, is counted separately.

Assume zero temperature, complete ionization, constant composition, noninteracting electrons, negligible ion thermal pressure, and Newtonian stellar gravity. Ions supply almost all the mass, while electron degeneracy pressure supplies support. By the Pauli exclusion principle, the two electron spin states fill a momentum sphere up to Fermi momentum $p_F$. Counting states giveswhere ${\mu }_e$ is mean molecular weight per electron. The momentum flux of this isotropic Fermi gas isWriting $x=p_F/(m_{e}c)$ and performing the integral gives the equation of state of a cold electron gasIn the nonrelativistic limit $p_F\ll m_{e}c$, $v(p)\simeq p/m_e$, soIn the ultrarelativistic limit $p_F\gg m_{e}c$, $v(p)\simeq c$, soThese are polytropic equations of state with indices $3/2$ and $3$, respectively. The relativistic softening underlies the Chandrasekhar mass limit; Coulomb corrections, thermal effects, rotation, and general relativity are excluded from this idealized derivation.

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.

For fermions the Pauli exclusion principle permits only $n_{\mathbf{p}}=0,1$. The two-term partition sum gives the Fermi-Dirac distribution