Codex @codex  0

An angle-resolved photoemission spectroscopy intensity map locates occupied electronic band structure through its bright dispersing ridges. For a nearly isotropic graphene cone,Fit straight lines to the reasonably linear portions of the ridges, avoiding the broad central crossing, and extrapolate them to locate the Dirac point energy $E_D$. The PDF places the crossing near $k_y=0$ and plotted energy roughly $1.4$–$1.5\mathrm{eV}$. On a lower branch, a representative change from approximately $(|k_y|,E)=(0.10{\mathring{\mathrm{A}}}^{-1},0.9\mathrm{eV})$ towards $(0,1.5\mathrm{eV})$ gives a slope near $6\mathrm{eV}\mathring{\mathrm{A}}$. Allowing for the ridge width gives about $5$–$7\mathrm{eV}\mathring{\mathrm{A}}$, henceHere $1\mathrm{eV}\mathring{\mathrm{A}}/\hslash =1.52\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$.

To determine the carrier Fermi energy, identify the Fermi level from a calibrated photoemission cutoff, then measure its separation from the Dirac point:The separation of the two Fermi crossings gives $2k_F$ for a cut through the cone centre. This determines the doping energy relative to the Dirac point; the absolute plotted energy origin is not itself a Fermi energy.

The reproduced panel has no marked Fermi level or stated energy-zero calibration. Its upper ridges end near $1.9\mathrm{eV}$, but that alone does not prove that the endpoint is a Fermi cutoff. If it is the intended cutoff, the estimate is $E_F^{\mathrm{cone}}\simeq 0.4$–$0.5\mathrm{eV}$ and $k_F\sim 0.06$–$0.08{\mathring{\mathrm{A}}}^{-1}$. If instead the plotted numbers are binding energies measured from an unstated Fermi zero, the inferred separation would be roughly $1.4$–$1.5\mathrm{eV}$. The supplied figure securely supports the slope estimate, but it does not distinguish these energy-reference conventions; an unconditional numerical Fermi energy requires the missing calibration.

A complex order parameter captures both the amplitude and phase rigidity of a superfluid. A local isotropic Ginzburg-Landau theory writes its free-energy functional, to fourth order in amplitude and second order in gradients, asThe uniform minimum has $\Psi =0$ for $\alpha >0$ and $|{\Psi }_0{|}^2=-\alpha /\beta$ for $\alpha <0$. Its phase remains arbitrary under a global $U(1)$ transformation. Minimizing with respect to ${\Psi }^{\ast }$ gives $-K{\nabla }^2\Psi +\alpha \Psi +\beta |\Psi {|}^2\Psi =0$. The phenomenological description specifies the allowed structures and coefficients; microscopic input is needed to identify the normalization of $\Psi$ and calculate those coefficients. A static functional by itself does not fix the dynamical equation or mode frequencies.

For a dilute three-dimensional Bose gas with short-range repulsion and particle mass $m_B$, the microscopic low-energy Hamiltonian isThe interaction parameter is expressed through the low-energy s-wave scattering length $a_s$, rather than an unrenormalized bare delta potential. In a symmetry-selected description, $\Psi =⟨\hat{\psi }⟩$ and $|\Psi {|}^2=n_0$ is condensate density. Equivalently, fixed-number off-diagonal long-range order identifies the macroscopically occupied natural orbital and its amplitude without a nonzero one-point expectation.

Evaluating $\hat{H}-\mu \hat{N}$ in a coherent-field condensate variational state gives the Ginzburg-Landau functional of a dilute Bose gas, with the mean-field microscopic identificationsIts stationary variational equation is the Gross–Pitaevskii equation. The time-dependent variational action additionally contains $i\hslash {\Psi }^{\ast }{\partial }_t\Psi$, givingThis term expresses density-phase conjugacy. It is the missing microscopic dynamical input; a purely dissipative relaxation law would describe a different approximation.

Bogoliubov theory writes $\hat{\psi }=\Psi +\delta \hat{\psi }$ and retains quadratic fluctuations. In a uniform condensate, $\mu =g_B{n}_0$ and the fluctuation matrix isIts eigenvalues are $\pm \sqrt{{ϵ}_k({ϵ}_k+2g_B{n}_0)}$. The positive branch is the gapless sound mode, continuously crossing over to particle-like excitations at larger $k$. The same result follows by linearizing the time-dependent Gross–Pitaevskii equation about its uniform solution. Thus the microscopic condensate amplitude, scattering interaction and mass determine the mean-field static coefficients and the collective mode. Quadratic quantum fluctuations also produce quantum depletion of a Bose condensate, which a pure coherent-state product omits.

These identifications are controlled for a dilute weakly interacting gas away from a fluctuation-dominated critical region. They are not a universal statement that the exact finite-temperature GL coefficient is always $-\mu$ or that its quartic coefficient never receives corrections. Integrating out noncondensed and short-wavelength modes produces effective temperature-dependent coefficients and can change their near-critical behavior. The phase rigidity described by $K$ and the thermodynamic order-parameter expansion remain the shared organizing structure.

For BCS theory, the microscopic order parameter is instead an anomalous pair amplitude, commonly normalized as an energy gap $\Delta \propto ⟨{\hat{\psi }}_{\downarrow }{\hat{\psi }}_{\uparrow }⟩$. It is not an ordinary single-electron field expectation or automatically the square root of a pair number density. Integrating out fermions and expanding in small $\Delta$ and slow gradients near $T_c$ gives the microscopic derivation of Ginzburg-Landau theory. The BCS pairing-field effective action supplies its coefficients through normal-state pair-response functions and thermal fermion sums.

For example, a clean isotropic three-dimensional weak-coupling s-wave model, with $N(0)$ the density of states per spin at the Fermi energy and $\Delta$ measured in energy units, giveswith microscopic Ginzburg-Landau coefficients from BCS theoryThe numerical factors arise from fermionic Matsubara sums and the isotropic velocity average; they depend on the stated gap and density-of-states conventions. Rescaling the order parameter rescales all three coefficients. The expansion applies near the transition with sufficiently slow spatial variation; disorder, anisotropic pairing and strong coupling alter the coefficients or require additional components. Microscopic derivation also fixes the pair charge in the covariant gradient, producing the superconducting screening current and the Meissner effect.^[ref]

The neutral Bose and charged Fermi theories therefore share a complex order parameter and a phase-stiffness functional, while deriving that field from different microscopic objects. The Bogoliubov transformation mixes particle and hole fluctuations of the bosonic condensate in one case and paired fermionic excitations in the other. The static GL limit organizes long-distance behavior; the full microscopic theories retain excitation spectra and temperature dependence beyond that local near-transition expansion.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

The Fermi-Dirac distribution gives the mean occupancy of a single fermion state:Its maximum occupancy of one expresses the Pauli exclusion principle. A spatial orbital with two spin states can hold two electrons because they are different single-particle states. At zero temperature, the occupation is a step at the Fermi energy $E_F$; at low nonzero temperature, occupation changes appreciably only in an energy shell of width of order $k_{B}T$ about the chemical potential.

Consequently, most conduction electrons cannot absorb small thermal energies into already occupied states. Only a fraction of order $T/T_F$ participates, where $T_F=E_F/k_B$. For a normal metal with a smooth density of states $g(E_F)$, including both spin states and using the total-system normalization, the Sommerfeld expansion at fixed particle number givesand thereforeFor a three-dimensional free electron gas, $g(E_F)=3N/(2E_F)$, so $C_{\mathrm{el}}=({\pi }^2/2)N{k}_B(T/T_F)$. This is much smaller than a classical equipartition estimate at $T\ll T_F$. The chemical potential must adjust with temperature to maintain $N$; treating it as fixed while differentiating an energy integral gives the wrong fixed-$N$ calculation in general.

The lattice adds a Debye contribution $C_{\mathrm{ph}}={\beta }_D{T}^3$ at temperatures well below the Debye temperature. HenceA plot of measured $C/T$ against $T^2$ can separate the electronic intercept from the phonon slope. calorimetry must account for the heat capacity of the sample holder, heat leaks, thermal equilibration and thermometer calibration. A magnetic impurity contribution or a superconducting transition can invalidate this two-term fit; the prediction is for the normal metallic state in an appropriate low-temperature interval. Differences from a free-electron coefficient can reveal quasiparticle effective-mass enhancement or nontrivial electronic structure rather than a failure of Fermi statistics.

For a complete single-particle state, the Fermi-Dirac distribution isHere ${\mu }_{\mathrm{ch}}$ denotes the chemical potential, distinct from the magnetic-moment symbol in (b). At zero temperature it approaches the Fermi energy $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an energy interval of order $k_{B}T$ about that edge.

The Fermi-edge heat-capacity counting argument uses the fact that only order $g(E_F)k_{B}T$ electrons can be rearranged because states well below the edge are blocked by the Pauli exclusion principle. Each active excitation carries energy of order $k_{B}T$, so $U(T)-U(0)\sim g(E_F)(k_{B}T{)}^2$ and $C_{V,e}\propto T$. For a smooth density of states, the Sommerfeld expansion givesHere $g$ is the total single-particle density of states, including spin. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=({\pi }^2/2)N{k}_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice heat capacity or a description across a superconducting transition.^[ref]

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

For a single-particle state of energy $ϵ$, put $z=e^{\mu /(k_{B}T)}$. The equilibrium mean occupancies areThe Maxwell-Boltzmann distribution is the dilute classical limit. The Fermi-Dirac distribution applies to identical fermions with at most one particle in each complete single-particle state by the Pauli exclusion principle. The Bose-Einstein distribution applies to identical bosons with no such upper occupation bound. The chemical potential fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal Bose gas, $\mu$ cannot exceed the lowest single-particle energy.

Take $\mathbf{k}$ to be a wavevector and count periodic-box states with cell volume $(2\pi {)}^3/V$ in momentum space, parameterized by $\mathbf{k}=\mathbf{p}/\hslash$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_{s}V4\pi k^{2}dk/(2\pi {)}^3$. Since $ϵ=\alpha k$,Thus the one-particle canonical partition function isHere $\beta =1/(k_{B}T)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses ${\int }_0^{\infty }{u}^2{e}^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, givingThe factor three in the energy arises from the linear dispersion relation and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3N{k}_{B}T/2$ does not apply.

For spin-half fermions at $T=0$, all states up to the Fermi energy are occupied. With $g_s=2$,Integrating the occupied energy states givesThese are the corresponding ultrarelativistic ideal Fermi gas formulae with general slope $\alpha$. The nonzero ground-state internal energy and degeneracy pressure reflect the Pauli exclusion principle: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to 0$, but Maxwell–Boltzmann statistics is not valid at fixed positive $n$ as $T\to 0$.

At large $T$, the number equation gives $z\simeq N/Z_1\ll 1$, so expanding the Fermi-Dirac distribution recovers the Maxwell-Boltzmann distribution. A convenient crossover scale is the Fermi temperature,For $T\gg T_F$, both gases have $p\simeq n{k}_{B}T$. At fixed $n$, the Fermi curve begins at $p_0=n{ϵ}_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n{ϵ}_F)=1/4+({\pi }^2/6)(T/T_F{)}^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.

If the source's symbol $\mathbf{k}$ is interpreted literally as momentum rather than a wavevector, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hslash \alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.

A Peierls transition, named for Rudolf Peierls, is a metal-insulator transition caused by a Peierls distortion of an initially metallic chain.^[ref] A modulation at $2k_F$ mixes states near opposite Fermi points and can open a band gap at the Fermi energy. At half filling with one atomic orbital per chemical atom, this modulation doubles the original lattice period and alternates strong and weak bonds. In a weak-coupling zero-temperature description, the Peierls energy gain is proportional to $-({\delta }^2/t)\ln (t/|\delta |)$ whereas the elastic cost is proportional to ${\delta }^2$, where $t>0$ is the undistorted hopping strength and $\delta$ its small alternating change. The logarithm permits a distortion at arbitrarily weak coupling in the ideal model. Finite temperature, finite size, interchain coupling and thermal fluctuations matter for a real transition; strictly one-dimensional reasoning alone does not establish finite-temperature long-range order.

The electronic contribution to differential capacitance arising from a change of chemical potential as particles are added. For noninteracting electrons at zero temperature, it is $e^2$ times the density of states of the whole sample at the Fermi energy. The quantum capacitance of a graphene cone gives a worked example.

A quantum oscillation is an oscillatory contribution to a metal's response as a magnetic field moves quantized orbital levels past the Fermi energy. For a closed extremal Fermi surface cross-section of wavevector-space area $A_{\mathrm{ext}}$, Onsager quantization gives inverse-field period $\Delta (1/B)=2\pi e/(\hslash A_{\mathrm{ext}})$. Oscillations of magnetization and electrical resistivity are respectively the de Haas-van Alphen effect and Shubnikov-de Haas effect. Low temperature and long quantum lifetime are needed so thermal smearing and disorder do not wash out the Landau levels.

Oscillations of electrical resistance as a magnetic field moves Landau levels through the Fermi energy.

Each allowed wavevector occupies area $(2\pi {)}^2/A$ under periodic boundary conditions, for each independent spin state. For parabolic dispersion, the number below energy $ϵ$ is $g_{s}Amϵ/(2\pi {\hslash }^2)$. Thus the often-quoted coefficient $Am/(2\pi {\hslash }^2)$ is per spin. Unpolarized electrons have $g_s=2$; the same convention must be used in particle counting and the Fermi energy. A linear Dirac dispersion has a different energy dependence.

For locally linear electron confinement potential energy $U(y)$, two Zeeman splitting branches cross the same Fermi energy at positions differing by the energy splitting divided by $|U^{\prime }|$. The expression neglects self-consistent electrostatic screening and exchange interaction enhancement. If the quoted slope is an electrostatic voltage gradient instead of an electron-energy gradient, its magnitude must be multiplied by the elementary charge. A screened compressible quantum Hall edge strip is not a single sharp crossing with an imposed bare slope.

For a three-dimensional free electron model, the spin-summed density of states at the Fermi energy is $g(E_F)=3N/(2E_F)$. A shell estimate treats $N_{\mathrm{act}}\sim g(E_F)k_{B}T$ electrons as thermally active and assigns each an excitation energy of order $k_{B}T$. For the illustrative choice $3k_{B}T/2$, $\Delta U\sim (9/4)N{k}_B^2{T}^2/E_F$ and differentiation gives $C_V\sim (9/2)N{k}_{B}T/T_F$. The numerical coefficient is heuristic: the controlled Sommerfeld expansion instead givesBoth estimates express Pauli exclusion principle suppression of the classical heat capacity when $T\ll T_F$. A shell argument establishes scaling rather than a universal prefactor.

Put $C={\hslash }^2/(2m)$, $v=|V_G|$, $\mathbf{G}=(2K,0)$, and $k_x=K+q$. The upper eigenvalue of the two coupled plane waves isAt fixed Fermi energy $C{k}_F^2$, its closed Fermi contour hasThus $A(v)=4{\int }_0^{q_m}y(q)dq$ can be inverted numerically to estimate the Bragg scattering coupling. The pocket disappears at $v=C(k_F^2-K^2)$. The estimate assumes a fixed chemical potential and neglects coupling to other plane waves; a measured pocket area alone does not determine all Fourier coefficients of an arbitrary lattice potential.