Fermi Momentum and Fermi Energy

The Fermi momentum $p_{\mathrm F}=\hbar k_{\mathrm F}$ is the momentum at the boundary of the filled zero-temperature states of a uniform ideal Fermi gas. The Fermi energy is the one-particle energy at that boundary:

$$E_{\mathrm F} \equiv \epsilon(k_{\mathrm F}).$$

For a nonrelativistic particle of mass $m$,

$$E_{\mathrm F} = \frac{\hbar^2k_{\mathrm F}^2}{2m} = \frac{p_{\mathrm F}^2}{2m}.$$

These quantities are not independent thermodynamic parameters. Once the spatial dimension, number density, internal degeneracy, and dispersion are specified, state counting fixes $k_{\mathrm F}$ and hence every other Fermi scale.

The most common source of wrong factors is a silent change between:

• total density, summed over all internal components;
• density per spin or hyperfine component;
• spinless or fully polarized conventions;
• continuum momentum and crystal momentum;
• kinetic energy and total relativistic energy.

The formulas below keep those conventions explicit. The ideal Fermi gas develops the thermodynamic model; degenerate Fermi gases explains the low-temperature physics.

Helpful background. Pauli exclusion explains the filling rule; plane-wave normalization explains the reciprocal-space state density.

The Fermi boundary and energy convention

Unless stated otherwise, assume:

• a uniform continuum system in $d$ spatial dimensions;
• a large periodic box of $d$-dimensional measure $\mathcal V_d=L^d$;
• $N$ particles and total density $n=N/\mathcal V_d$;
• $g$ degenerate internal modes at every wave vector;
• an isotropic, monotonically increasing dispersion $\epsilon(k)$;
• zero temperature when defining the occupied Fermi region;
• one fermion at most in each complete one-particle mode.

Internal modes may be spin, hyperfine, or flavor states sharing the same dispersion and Fermi boundary. Equivalent valleys can also supply a multiplicity when each identical pocket is counted once using momentum relative to its own minimum; they are generally centered at different crystal momenta. Unequal populations or inequivalent dispersions require separate counts.

This page writes the Fermi energy as $E_{\mathrm F}$. Many texts and neighboring pages write $\epsilon_{\mathrm F}$ for the same quantity. The notation does not imply a different physical definition:

$$E_{\mathrm F} = \epsilon_{\mathrm F} = \epsilon(k_{\mathrm F}).$$

For the nonrelativistic formulas, the energy zero is the bottom of the band or free-particle spectrum. The physical degeneracy scale is the energy above that bottom. If an offset $\epsilon_0$ is retained in $\epsilon(k)$, use $[\epsilon(k_{\mathrm F})-\epsilon_0]/k_{\mathrm B}$ for the Fermi temperature and the corresponding energy difference for a microscopic Fermi time. At finite temperature, this scale remains defined from zero-temperature filling. It is generally not equal to the temperature-dependent chemical potential $\mu(T)$.

Formula sheet

For a quadratic dispersion,

$$\epsilon(k) = \frac{\hbar^2k^2}{2m},$$

the standard formulas are:

Dimension	Total density	Fermi wave number
1D	$n=gk_{\mathrm F}/\pi$	$k_{\mathrm F}=\pi n/g$
2D	$n=gk_{\mathrm F}^2/(4\pi)$	$k_{\mathrm F}=\sqrt{4\pi n/g}$
3D	$n=gk_{\mathrm F}^3/(6\pi^2)$	$k_{\mathrm F}=(6\pi^2n/g)^{1/3}$

The corresponding Fermi energies are:

Dimension	Fermi energy
1D	$E_{\mathrm F}=\hbar^2\pi^2n^2/(2mg^2)$
2D	$E_{\mathrm F}=2\pi\hbar^2n/(mg)$
3D	$E_{\mathrm F}=\dfrac{\hbar^2}{2m}(6\pi^2n/g)^{2/3}$

For a balanced spin-$1/2$ gas, $g=2$ and $n$ is the total density:

$$\begin{array}{c|c} d & k_{\mathrm F} \\ \hline 1 & \pi n/2 \\ 2 & \sqrt{2\pi n} \\ 3 & (3\pi^2n)^{1/3} \end{array}$$

For a spinless or fully polarized gas, set $g=1$.

The 1D Fermi boundary consists of the two points $\pm k_{\mathrm F}$; in 2D it is a circle; in 3D it is a sphere. These formulas require monotone isotropic filling from the minimum. The corresponding densities have units of inverse length, inverse area, and inverse volume, respectively. The geometric derivation below explains all three coefficients at once.

Geometric origin of the formulas

Periodic boundary conditions quantize each wave-vector component in steps of $2\pi/L$. One allowed $\mathbf k$ point per internal mode therefore occupies $d$-dimensional wave-vector volume

$$\left( \frac{2\pi}{L} \right)^d.$$

Equivalently, the number of wave-vector states per internal mode in an element $d^dk$ is

$$\frac{\mathcal V_d}{(2\pi)^d} \,d^dk.$$

Let $\Omega_d$ denote the volume of the unit ball in $d$ Euclidean dimensions:

$$\Omega_d = \frac{\pi^{d/2}} {\Gamma(d/2+1)}.$$

The ball of radius $k$ has volume $\Omega_d k^d$. Including $g$ internal modes, the continuum approximation to the integrated state count is

$$\mathcal N_d(k) = g \frac{\mathcal V_d}{(2\pi)^d} \Omega_d k^d.$$

At $T=0$, the ground state fills the $N$ lowest one-particle modes. For a monotone isotropic dispersion, this occupied region is the ball $|\mathbf k|\leq k_{\mathrm F}$, so

$$N = g \frac{\mathcal V_d}{(2\pi)^d} \Omega_d k_{\mathrm F}^d.$$

Dividing by $\mathcal V_d$ gives the master relation

$$n = g \frac{\Omega_d}{(2\pi)^d} k_{\mathrm F}^d.$$

Solving for the Fermi wave number,

$$k_{\mathrm F} = \left[ \frac{(2\pi)^d}{g\Omega_d} n \right]^{1/d}.$$

The numerical coefficients in 1D, 2D, and 3D are therefore geometric coefficients, not additional dynamical assumptions.

For an isotropic monotone dispersion, the zero-temperature occupied region is the $d$-ball $\mathcal B_d(k_{\mathrm F})$. Its geometric measure is $2k_{\mathrm F}$ in 1D, $\pi k_{\mathrm F}^2$ in 2D, and $4\pi k_{\mathrm F}^3/3$ in 3D. Multiplication by $g\mathcal V_d/(2\pi)^d$ converts that measure into a state count.

Internal degeneracy conventions

The degeneracy factor counts independent one-particle modes at the same $\mathbf k$. It is part of the state count, not a correction applied after the count.

Spin degeneracy

If spin is conserved and all $2s+1$ spin projections are degenerate and equally populated, then

$$g_s = 2s+1.$$

Examples include:

• spinless fermions or a fully polarized sample: $g=1$;
• balanced spin-$1/2$ fermions: $g=2$;
• an idealized balanced spin-$3/2$ multiplet: $g=4$.

This formula does not say that every spin multiplet is physically populated. Zeeman splitting, spin–orbit coupling, preparation constraints, interactions, or selection rules may remove the degeneracy.

Several independent degeneracies

If spin, valley, and another internal label are all independent and exactly degenerate, their multiplicities multiply:

$$g = g_s g_v g_{\mathrm{other}}.$$

One must not multiply by a label that is already counted as a separate band or included explicitly in the sum over states.

Total density and component density

Suppose $g$ components are balanced, each with density $n_a$. Then

$$n = \sum_{a=1}^{g}n_a = g n_a.$$

The per-component state count contains no additional degeneracy factor:

$$n_a = \frac{\Omega_d}{(2\pi)^d} k_{\mathrm F}^d.$$

Substituting $n_a=n/g$ recovers the total-density formula.

Imbalanced components

If components have unequal densities, there is generally no single common Fermi wave number. Each component has

$$k_{{\mathrm F},a} = \left[ \frac{(2\pi)^d}{\Omega_d} n_a \right]^{1/d}.$$

For two spin components in three dimensions,

$$k_{{\mathrm F},\uparrow} = (6\pi^2n_\uparrow)^{1/3}, \qquad k_{{\mathrm F},\downarrow} = (6\pi^2n_\downarrow)^{1/3}.$$

Writing one $g=2$ formula for an imbalanced gas hides the two distinct Fermi surfaces.

Fixed-density polarization effect

At fixed total density,

$$k_{\mathrm F} \propto g^{-1/d},$$

and, for a quadratic dispersion,

$$E_{\mathrm F} \propto g^{-2/d}.$$

Reducing the number of available internal components forces particles to occupy a larger region of momentum space. In three dimensions, fully polarizing a previously balanced spin-$1/2$ ideal gas changes the scales by

$$\frac{k_{\mathrm F}^{(g=1)}} {k_{\mathrm F}^{(g=2)}} = 2^{1/3},$$

and

$$\frac{E_{\mathrm F}^{(g=1)}} {E_{\mathrm F}^{(g=2)}} = 2^{2/3}.$$

Derived Fermi scales

Once $k_{\mathrm F}$ is known, several useful scales follow.

Fermi momentum

Wave number and momentum differ by $\hbar$:

$$p_{\mathrm F} = \hbar k_{\mathrm F}.$$

Thus $k_{\mathrm F}$ has units of inverse length, while $p_{\mathrm F}$ has units of momentum. The phrase Fermi momentum is often used informally for either quantity; a careful formula should display which one is meant.

Fermi velocity

For a general isotropic dispersion, the group velocity at the Fermi boundary is

$$v_{\mathrm F} = \frac{1}{\hbar} \left. \frac{d\epsilon}{dk} \right|_{k=k_{\mathrm F}}.$$

For a quadratic dispersion,

$$v_{\mathrm F} = \frac{\hbar k_{\mathrm F}}{m} = \sqrt{ \frac{2E_{\mathrm F}}{m} }.$$

Fermi temperature

The Fermi temperature is the energy scale expressed in kelvin:

$$T_{\mathrm F} \equiv \frac{E_{\mathrm F}}{k_{\mathrm B}}.$$

It is not the temperature of the zero-temperature gas. It defines the reduced temperature

$$\theta = \frac{T}{T_{\mathrm F}},$$

which distinguishes the degenerate regime $\theta\ll1$ from the classical regime $\theta\gg1$ under the appropriate density conditions.

Fermi wavelength and inverse Fermi scale

The de Broglie wavelength associated with $p_{\mathrm F}$ is

$$\lambda_{\mathrm F} = \frac{h}{p_{\mathrm F}} = \frac{2\pi}{k_{\mathrm F}}.$$

Many-body estimates also use the shorter length $k_{\mathrm F}^{-1}$. These differ by $2\pi$ and should not be interchanged silently.

Fermi time

A natural microscopic time is

$$t_{\mathrm F} = \frac{\hbar}{E_{\mathrm F}}.$$

For a quadratic dispersion,

$$t_{\mathrm F} = \frac{2m}{\hbar k_{\mathrm F}^2}.$$

This is a scale, not automatically a collision time or equilibration time.

Inverse formulas

If the nonrelativistic Fermi energy is known, then

$$k_{\mathrm F} = \frac{\sqrt{2mE_{\mathrm F}}}{\hbar}.$$

The corresponding density in $d$ dimensions is

$$n = g \frac{\Omega_d}{(2\pi)^d} \left( \frac{2mE_{\mathrm F}}{\hbar^2} \right)^{d/2}.$$

Explicitly,

$$\begin{aligned} n_{1\mathrm D} &= \frac{g}{\pi} \frac{\sqrt{2mE_{\mathrm F}}}{\hbar}, \\ n_{2\mathrm D} &= \frac{gmE_{\mathrm F}} {2\pi\hbar^2}, \\ n_{3\mathrm D} &= \frac{g}{6\pi^2} \left( \frac{2mE_{\mathrm F}}{\hbar^2} \right)^{3/2}. \end{aligned}$$

The densities have different dimensions: inverse length in 1D, inverse area in 2D, and inverse volume in 3D.

Density-of-states checks

The integrated number of states below energy $E$ for a quadratic dispersion is

$$\frac{\mathcal N_d(E)}{\mathcal V_d} = g \frac{\Omega_d}{(2\pi)^d} \left( \frac{2mE}{\hbar^2} \right)^{d/2}.$$

Differentiation gives the density of states per $d$-dimensional measure:

$$\rho_d(E) = \frac{1}{\mathcal V_d} \frac{d\mathcal N_d}{dE}.$$

At the Fermi energy,

$$\rho_d(E_{\mathrm F}) = \frac{d}{2} \frac{n}{E_{\mathrm F}}.$$

This identity is a fast consistency check. It uses a density of states that includes the same internal factor $g$ as the density $n$.

The energy dependence is

$$\rho_d(E) \propto E^{d/2-1}.$$

Thus the free-particle density of states diverges as $E^{-1/2}$ in 1D, is constant in 2D, and grows as $E^{1/2}$ in 3D. The general density-of-states normalization is explained in Density of States: First Encounter.

Zero-temperature consistency identities

For a quadratic dispersion, integrating all occupied energies gives

$$\frac{U_0}{N} = \frac{d}{d+2} E_{\mathrm F}.$$

Scale invariance of the nonrelativistic kinetic energy gives

$$P_0 = \frac{2}{d} \frac{U_0}{\mathcal V_d} = \frac{2}{d+2} nE_{\mathrm F}.$$

In one dimension, $P_0$ is a force; in two dimensions, it is force per length; in three dimensions, it is the usual force per area. These identities are useful checks, but the full equation-of-state derivation belongs to the Ideal Fermi Gas page.

Worked example: a three-dimensional electron density

Consider nonrelativistic electrons with total density

$$n = 8.5\times10^{28}\ \mathrm{m}^{-3}$$

and spin degeneracy $g=2$. Then

$$k_{\mathrm F} = (3\pi^2n)^{1/3} \approx 1.36\times10^{10}\ \mathrm{m}^{-1}.$$

The momentum scale is

$$p_{\mathrm F} = \hbar k_{\mathrm F} \approx 1.43\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}.$$

Using the electron mass,

$$E_{\mathrm F} = \frac{\hbar^2k_{\mathrm F}^2}{2m_e} \approx 7.0\ \mathrm{eV}.$$

Consequently,

$$v_{\mathrm F} \approx 1.6\times10^6\ \mathrm{m\,s^{-1}},$$

and

$$T_{\mathrm F} \approx 8.1\times10^4\ \mathrm K.$$

An ordinary laboratory temperature can therefore be much smaller than $T_{\mathrm F}$ even when it is hundreds of kelvin. This is why conduction electrons are often strongly degenerate. A real metal may require band-dependent effective masses and a nonspherical Fermi surface, so the numerical calculation is an ideal free-electron estimate rather than a complete material model.

Worked example: a two-dimensional electron gas

Take a two-dimensional spin-degenerate gas with sheet density

$$n_{2\mathrm D} = 1.0\times10^{15}\ \mathrm{m}^{-2},$$

$g=2$, and effective mass $m^*=0.067m_e$. The Fermi wave number is

$$k_{\mathrm F} = \sqrt{2\pi n_{2\mathrm D}} \approx 7.93\times10^7\ \mathrm{m}^{-1}.$$

The quadratic-band Fermi energy is

$$E_{\mathrm F} = \frac{\pi\hbar^2n_{2\mathrm D}}{m^*} \approx 3.6\ \mathrm{meV}.$$

Equivalently,

$$T_{\mathrm F} \approx 41\ \mathrm K.$$

The use of $m^*$ rather than the bare electron mass is part of the model specification. Valley degeneracy, spin splitting, or nonparabolicity would modify the count or the dispersion.

What survives for another dispersion

The state-counting relation for $k_{\mathrm F}$ depends on the occupied $k$-space volume, not on the detailed energy dispersion. For a continuum dispersion, or an isotropic pocket counted within its domain of validity, the same density relation holds:

$$n = g \frac{\Omega_d}{(2\pi)^d} k_{\mathrm F}^d.$$

Only the map from $k_{\mathrm F}$ to energy and velocity changes:

$$E_{\mathrm F} = \epsilon(k_{\mathrm F}),$$

and

$$v_{\mathrm F} = \frac{1}{\hbar} \epsilon'(k_{\mathrm F}).$$

For a power-law dispersion

$$\epsilon(k) = A k^s,$$

one obtains

$$E_{\mathrm F} \propto n^{s/d},$$

and

$$\rho_d(E_{\mathrm F}) = \frac{d}{s} \frac{n}{E_{\mathrm F}}.$$

The familiar factor $d/2$ is the special case $s=2$.

Relativistic convention

Relativistic state counting gives the same $k_{\mathrm F}$ for a specified density and degeneracy. What changes is the energy relation. With

$$p_{\mathrm F} = \hbar k_{\mathrm F},$$

the total one-particle energy at the Fermi boundary is

$$\mathcal E_{\mathrm F} = \sqrt{ p_{\mathrm F}^2c^2 + m^2c^4 }.$$

If Fermi energy means kinetic energy measured from the rest energy, then

$$E_{\mathrm F}^{\mathrm{kin}} = \sqrt{ p_{\mathrm F}^2c^2 + m^2c^4 } - mc^2.$$

At zero temperature, a relativistic chemical potential that includes rest energy is

$$\mu_0 = \mathcal E_{\mathrm F},$$

whereas a convention that subtracts rest energy gives

$$\mu_0^{\mathrm{kin}} = E_{\mathrm F}^{\mathrm{kin}}.$$

Both conventions are legitimate; mixing them in one calculation is not. The nonrelativistic approximation requires

$$p_{\mathrm F} \ll mc.$$

Bands, lattices, and anisotropy

The scalar $k_{\mathrm F}$ is exact for a spherical Fermi boundary. In a crystal band, the occupied region is determined by

$$E_n(\mathbf k) \leq \mu_0,$$

inside the Brillouin zone. The boundary can be warped, disconnected, open across a zone boundary, or split among bands. Then one should speak of:

• the Fermi surface $E_n(\mathbf k)=\mu_0$;
• direction-dependent Fermi wave vectors;
• extremal radii or orbit areas;
• an effective spherical $k_{\mathrm F}$ only when explicitly defined.

For an anisotropic quadratic band,

$$\epsilon(\mathbf k) = \frac{\hbar^2}{2} \left( \frac{k_x^2}{m_x} + \frac{k_y^2}{m_y} + \frac{k_z^2}{m_z} \right),$$

constant-energy surfaces are ellipsoids with semiaxes $k_{{\mathrm F},i}=\sqrt{2m_iE_{\mathrm F}}/\hbar$. Their volume gives

$$n=\frac{g}{6\pi^2}\left(\frac{2E_{\mathrm F}}{\hbar^2}\right)^{3/2} \sqrt{m_xm_ym_z}.$$

The density-of-states mass $m_{\mathrm{DOS}}=(m_xm_ym_z)^{1/3}$ reproduces this total count in the spherical formula. It does not reproduce the directional Fermi radii or velocities; those still need the individual masses.

Fermi Surface develops the general many-body concept and low-energy excitations near nonspherical boundaries. The linked band-structure discussions explain Bloch-band topology and material-specific surfaces.

Traps and inhomogeneous gases

A finite trap does not have translational invariance, so momentum states are not labeled by a uniform continuum density. Exact state counting uses trap eigenlevels.

In a slowly varying potential, a local-density approximation may define

$$n(\mathbf r) = g \frac{\Omega_d}{(2\pi)^d} k_{\mathrm F}(\mathbf r)^d,$$

with a local Fermi energy

$$E_{\mathrm F}(\mathbf r) = \frac{\hbar^2k_{\mathrm F}(\mathbf r)^2}{2m}.$$

These are local scales. They need not equal the global Fermi energy defined by the highest occupied trap level. Quantum gases in traps derives the trap conventions and local-density approximation.

Finite systems and shell filling

In a finite box, allowed momenta are discrete. The continuum formula replaces a lattice-point count by a geometric volume, so it becomes accurate when many modes are occupied and boundary corrections are small.

For a finite noninteracting system, several quantities may be called the Fermi energy:

• the highest occupied one-particle energy;
• the lowest unoccupied one-particle energy;
• the midpoint between them;
• an addition or removal chemical potential;
• the continuum estimate from the mean density.

For the gapless continuum gas, these differences vanish in the ordinary thermodynamic limit, but can be a shell spacing in a small system. A band insulator with a persistent gap does not have the same coincidence. A partially filled degenerate shell also makes the zero-temperature boundary less sharp than the continuum picture suggests.

Exercises

Translate between total and component density

A balanced two-component gas in two dimensions has total sheet density $n$. Compute $k_{\mathrm F}$ first from the total-density formula with $g=2$, then from the density $n_\sigma=n/2$ of one component. Verify agreement.

Solution

Using total density and $g=2$,

$$k_{\mathrm F} = \sqrt{ \frac{4\pi n}{2} } = \sqrt{2\pi n}.$$

For one component, no extra degeneracy factor is included:

$$n_\sigma = \frac{k_{\mathrm F}^2}{4\pi}.$$

Since $n_\sigma=n/2$,

$$k_{\mathrm F} = \sqrt{4\pi n_\sigma} = \sqrt{2\pi n}.$$

Both expressions count each spin component once and therefore agree.

Derive the density-of-states check

Starting from $\mathcal N_d(E)/\mathcal V_d\propto E^{d/2}$ for a quadratic dispersion, show that

$$\rho_d(E_{\mathrm F}) = \frac{d}{2} \frac{n}{E_{\mathrm F}}.$$

Solution

Write

$$\frac{\mathcal N_d(E)}{\mathcal V_d} = C E^{d/2},$$

where $C$ contains the mass, degeneracy, and geometric factors. Differentiation gives

$$\rho_d(E) = \frac{d}{2} C E^{d/2-1}.$$

At $E=E_{\mathrm F}$,

$$n = C E_{\mathrm F}^{d/2}.$$

Eliminating $C$ yields

$$\rho_d(E_{\mathrm F}) = \frac{d}{2} \frac{n}{E_{\mathrm F}}.$$

Obtain the mean ground-state energy

Use $\rho_d(E)\propto E^{d/2-1}$ to prove

$$\frac{U_0}{N} = \frac{d}{d+2} E_{\mathrm F}.$$

Check the results in 1D, 2D, and 3D.

Solution

The particle number and energy are proportional to

$$N \propto \int_0^{E_{\mathrm F}} E^{d/2-1}\,dE = \frac{2}{d} E_{\mathrm F}^{d/2},$$

and

$$U_0 \propto \int_0^{E_{\mathrm F}} E^{d/2}\,dE = \frac{2}{d+2} E_{\mathrm F}^{d/2+1}.$$

Taking the ratio gives

$$\frac{U_0}{N} = \frac{d}{d+2} E_{\mathrm F}.$$

Thus

$$\frac{U_0}{N} = \begin{cases} E_{\mathrm F}/3, & d=1,\\ E_{\mathrm F}/2, & d=2,\\ 3E_{\mathrm F}/5, & d=3. \end{cases}$$

Separate relativistic energy conventions

Show that the relativistic kinetic Fermi energy approaches $p_{\mathrm F}^2/(2m)$ when $p_{\mathrm F}\ll mc$, and approaches $p_{\mathrm F}c$ when $p_{\mathrm F}\gg mc$.

Solution

The kinetic Fermi energy is

$$E_{\mathrm F}^{\mathrm{kin}} = mc^2 \left[ \sqrt{ 1+ \left( \frac{p_{\mathrm F}}{mc} \right)^2 } -1 \right].$$

For $x=p_{\mathrm F}/(mc)\ll1$,

$$\sqrt{1+x^2} = 1+ \frac{x^2}{2} + O(x^4),$$

so

$$E_{\mathrm F}^{\mathrm{kin}} = \frac{p_{\mathrm F}^2}{2m} + O\!\left( \frac{p_{\mathrm F}^4}{m^3c^2} \right).$$

For $x\gg1$,

$$\sqrt{1+x^2} = x+ O(x^{-1}),$$

and therefore

$$E_{\mathrm F}^{\mathrm{kin}} = p_{\mathrm F}c - mc^2 + O\!\left( \frac{m^2c^3}{p_{\mathrm F}} \right).$$

The leading density-dependent term is $p_{\mathrm F}c$. The subtractive rest-energy constant remains because the kinetic-energy convention measures from $mc^2$.

An ellipsoidal Fermi sea

At fixed total density and degeneracy, compare the quadratic band with masses $(m,4m,9m)$ to the isotropic band with mass $m$. Find the energy ratio, the three radius ratios, and the velocity ratios.

Solution

The density-of-states mass is $m_{\mathrm{DOS}}=36^{1/3}m$, so $E_{\mathrm F}^{\mathrm{ell}}/E_{\mathrm F}^{\mathrm{sph}}=36^{-1/3}$. Since each radius is proportional to $\sqrt{m_iE_{\mathrm F}}$, the ratios are $36^{-1/6}(1,2,3)$. Their product is one, as fixed occupied volume requires. The velocities are $\hbar k_i/m_i$, so their ratios are $36^{-1/6}(1,1/2,1/3)$. One scalar mass preserves the state count but not these directional properties.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston (1976) — free-electron and band-theory Fermi scales.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003) — Fermi momentum, density of states, and many-body conventions.
• S. Giorgini, L. P. Pitaevskii, and S. Stringari, “Theory of ultracold atomic Fermi gases,” Reviews of Modern Physics 80, 1215–1274 (2008) — balanced and imbalanced atomic-gas conventions.
• G. F. Giuliani and G. Vignale, Quantum Theory of the Electron Liquid, Cambridge University Press (2005) — electron-gas conventions in two and three dimensions.
• X.-W. Guan, M. T. Batchelor, and C. Lee, “Fermi gases in one dimension: From Bethe ansatz to experiments,” Reviews of Modern Physics 85, 1633–1691 (2013) — one-dimensional Fermi scales and component conventions.
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987) — ideal Fermi-gas thermodynamics and dimensional state counting.
• M. Kardar, Statistical Physics of Particles, Cambridge University Press (2007) — phase-space counting and quantum statistics.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Elsevier (2011) — continuum state counting and ideal quantum gases.