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Fermi momentum is the magnitude of the momentum at the boundary of filled one-particle states in an isotropic zero-temperature Fermi gas. For the uniform three-dimensional two-component Electron model, a short state count givesTo see the factors, impose periodic boundary conditions on a cubic box of side $L$, so $V=L^3$ and each momentum component is $p_i=2\pi \hslash j_i/L$ for an integer $j_i$. Each allowed momentum point occupies a cell of volume $(2\pi \hslash /L{)}^3$ in momentum space. Divide the sphere volume $4\pi p_F^3/3$ by this cell volume and multiply by two for the two spin states. Replacing the discrete count with a volume is the large-box approximation; surface and shell effects are neglected.^[ref]

The factor two is not automatic for every Fermi gas. If only one internal state is populated, its number density $n_{\sigma }$ instead satisfies $p_{F,\sigma }=\hslash (6{\pi }^2{n}_{\sigma }{)}^{1/3}$. For two equally populated states, $n_{\sigma }=n/2$ recovers the displayed result. A Fermi surface in a crystal can have a more complicated shape, so a single spherical Fermi momentum is a property of this isotropic model.

The Fermi sea is the zero-temperature many-fermion quantum state with every one-particle state below the Fermi energy occupied and every state above it empty. For a uniform isotropic Fermi gas, the occupied region in momentum space is a solid sphere; its boundary is the Fermi surface. These are momenta, not positions: filling the sphere does not put the particles in a ball in ordinary space. Opposite momentum contributions cancel, so the gas at rest has zero total momentum even though its total kinetic energy is positive.

The filled states are forced upward in energy by the Pauli exclusion principle. This is different from the zero-point energy of one particle confined by a boundary. At fixed number density, the Fermi energy remains positive as the box grows to the thermodynamic limit, while the lowest box confinement energy tends to zero. Low-energy excitations move particles from occupied states near the Fermi surface into nearby empty states; deep states have no nearby available destination. This explains why only a small fraction of Electrons contribute to the low-temperature electron heat capacity.^[ref]

A vector space whose coordinates are the components of a particle's momentum; a distribution in momentum space describes momenta rather than positions.

Take a normalized spin one-half state in the $S_z$ eigenbasis. Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, direct multiplication of the Pauli matrices givesSubstitution of the half-angle parametrization givesThus $\theta$ and $\varphi$ are precisely the polar and azimuthal angles of the mean spin. Two complex coefficients contain four real parameters; normalization removes one and an unobservable common quantum phase removes another. The resulting pure states form the Bloch sphere. One may choose $\alpha$ real and nonnegative away from the patch where $\alpha =0$. At the poles the azimuth is redundant. The length $|⟨\mathbf{S}⟩|=\hslash /2$ should not be confused with $\sqrt{⟨S^2⟩}=\sqrt{3}\hslash /2$.

For a spatial plane wave, $-i\hslash \nabla$ becomes $\hslash \mathbf{k}$, so the Hamiltonian operator reduces to the Weyl Hamiltonian matrixDefine $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and ${\chi }_{+}=(\cos (\theta /2),e^{i\varphi }\sin (\theta /2){)}^{\mathsf{T}}$. The matrixsatisfies $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{+}={\chi }_{+}$ by the angle-addition identities. Consequently the displayed spinor is an energy eigenstate for either $\mathbf{k}=+k\mathbf{n}$ or $\mathbf{k}=-k\mathbf{n}$, with respective energies $+\hslash vk$ and $-\hslash vk$, for $k\ge 0$. Equivalently, at each fixed nonzero wavevector the two branches areIf $v<0$, the two energy labels exchange; the ordered eigenvalues are $\pm \hslash |v||\mathbf{k}|$. Taking $v>0$ sets the convention in the sketches. An orthogonal spinor ${\chi }_{-}=(-e^{-i\varphi }\sin (\theta /2),\cos (\theta /2){)}^{\mathsf{T}}$ obeys $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{-}=-{\chi }_{-}$ and supplies the other branch at $+k\mathbf{n}$.

The Pauli matrix multiplication law independently gives $H(\mathbf{k}{)}^2=(\hslash v{)}^2|\mathbf{k}{|}^{2}I$. The dispersion relation is therefore a pair of cones touching at zero energy and momentum; a signed straight-line cut has upper and lower branches $\pm \hslash v|k_x|$. In the $k_z=0$ plane, $\theta =\pi /2$, and the upper-branch mean spin points radially outwards, the lower-branch spin radially inwards. This is spin-momentum locking, not a tangential spin texture. At the touching point the two states are degenerate and the spin direction is not selected by the Hamiltonian operator.

Now let a spatially constant perturbation have matrix $V=\left(\begin{array}{cc}u& w\\ w^{\ast }& z\end{array}\right)$, with $u,z$ real. The Hermitian two-by-two matrix in the Pauli basis decomposition follows explicitly fromwhere $d=(u+z)/2$, $c=(u-z)/2$, $a=\mathrm{Re}w$, and $b=-\mathrm{Im}w$ are real. Equivalently, $d=\mathrm{tr}V/2$ and $(a,b,c{)}_i=\mathrm{tr}(V{\sigma }_i)/2$. The four matrices are linearly independent under the trace inner product and span the four-dimensional real vector space of Hermitian two-by-two matrices.

Spatial constancy means the perturbation does not mix different wavevectors, so each momentum sector can still be diagonalized independently. Write $\mathbf{b}=(a,b,c)$. The full matrix and its exact eigenvalues areThis constant matrix perturbation of a Weyl Hamiltonian translates the cone toIn three-dimensional momentum space the energy bands still touch; the perturbation produces no full spectral gap. The new branch spin directions are parallel and antiparallel to $\hslash v\mathbf{k}+\mathbf{b}$, so their radial texture is centered on the shifted point.

A restricted cut must be distinguished from the full dispersion relation. In the fixed plane $k_z=0$, the energies are $d\pm \sqrt{(\hslash v{k}_x+a{)}^2+(\hslash v{k}_y+b{)}^2+c^2}$, with minimum branch separation $2|c|$. This plane misses the crossing if $c\ne 0$. Along any line through ${\mathbf{k}}_{\ast }$ the dispersion relation still crosses linearly. Thus a sketch of a gapped fixed-plane cut alone would not describe the gapless three-dimensional Hamiltonian operator. The cone after a general perturbation is also no longer a function of $|\mathbf{k}|$ about the old origin alone.