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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.

An integral over momentum space. For example, state counting in $d$ spatial dimensions uses a measure proportional to $d^{d}p/(2\pi \hslash {)}^d$, whereas a relativistic Feynman integral commonly uses $d^{d}p/(2\pi {)}^d$ in natural units. The convention and any regulator must be specified. Independent loop momenta supply separate integration variables in a loop Feynman diagram; ultraviolet divergences and infrared divergences can require different regularizations.

For a free intrinsic-even real scalar field, momentum reflection $Rf(\mathbf{p})=f(-\mathbf{p})$ is a self-adjoint involution on the one-particle space. Its odd orthogonal projector is $(1-R)/2$. The corresponding second quantization is $N_{\mathrm{odd}}=\frac{1}{2}\int d^{3}p/(2\pi {)}^3{a}_{\mathbf{p}}^{†}(a_{\mathbf{p}}-a_{-\mathbf{p}})$, with integration over all momentum space. Conjugation by $e^{i\pi N_{\mathrm{odd}}}$ fixes even modes and reverses odd modes, hence sends $a_{\mathbf{p}}$ to $a_{-\mathbf{p}}$ and implements the parity operator. Its square is one because $N_{\mathrm{odd}}$ has integer eigenvalues. The equivalent positive expression is $N_{\mathrm{odd}}=\frac{1}{4}\int d^{3}p/(2\pi {)}^3(a_{\mathbf{p}}^{†}-a_{-\mathbf{p}}^{†})(a_{\mathbf{p}}-a_{-\mathbf{p}})$.

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.

For four-momentum $k$ flowing along the fermion flow, the momentum space Feynman rules areThe scalar propagator and Dirac propagator invert their respective quadratic kernels. The interaction vertex follows by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$, fixing its sign. Species is conserved along each open fermion line; there is no vertex changing species 1 into species 2.

An interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum k)$ with inward four-momenta. Impose momentum conservation and integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$. Include the closed fermion loop sign $-1$ for every fermion loop, the relevant fermionic signs when rearranging external states, and the Feynman-diagram symmetry factor. External incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antifermions carry ${\bar{v}}_s(p)$ and $v_s(p)$. An external real scalar field leg has factor 1 in an amputated scattering amplitude. The Feynman i-epsilon prescription specifies the poles.

Take parity symmetry in quantum field theory to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the mode expansion of a free field into the defining parity operator relation and change the integration variable $\mathbf{p}\mapsto -\mathbf{p}$. Since $E_{-\mathbf{p}}=E_{\mathbf{p}}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even conventionHere parity symmetry in quantum field theory reverses spatial momentum and preserves time. The creation operator formulas follow by taking Hermitian adjoints. The given field relation has fixed the intrinsic phase to $+1$.

For one real scalar field, let $R$ act on one-particle wave functions by $(Rf)(\mathbf{p})=f(-\mathbf{p})$. It is a Hermitian operator with $R^2=1$, so $P_{-}=(1-R)/2$ is the orthogonal projector onto odd wave functions. Its second quantization is the odd-mode number operatorThe full momentum space is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf{p},-\mathbf{p}\}$. A suitable parity operator isFor verification, $[N_{\mathrm{odd}},a_{\mathbf{p}}]=-(a_{\mathbf{p}}-a_{-\mathbf{p}})/2$. Thus the even combination $a_{\mathbf{p}}+a_{-\mathbf{p}}$ is unchanged, and the odd combination $a_{\mathbf{p}}-a_{-\mathbf{p}}$ changes sign under conjugation by $e^{i\pi N_{\mathrm{odd}}}$, exactly reproducing $a_{\mathbf{p}}\mapsto a_{-\mathbf{p}}$. The same holds for creation operators. The odd-mode number operator has integer eigenvalues, so ${\mathcal{P}}^2=1$, and $\mathcal{P}|0⟩=|0⟩$. Multiplying by an overall constant phase would not change its conjugation action. This is parity as odd-mode number parity.

The momentum space Feynman rules use the scalar propagator and two Dirac propagatorsThe cubic scalar field interaction vertex is $-i\lambda$. Each of the two flavour-changing Yukawa coupling interaction vertices, $\varphi \bar{\Psi }\chi$ and $\varphi \bar{\chi }\Psi$, has factor $-ig{\delta }_{ab}$ in Dirac spinor indices. A fermion flow arrow continues through each interaction vertex, but its species changes. There is no $\varphi \bar{\Psi }\Psi$ or $\varphi \bar{\chi }\chi$ interaction vertex.

Every interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ with inward four-momenta. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, divide by the Feynman-diagram symmetry factor, and include a minus sign and a gamma-matrix trace for each closed fermion loop. The $i0$ is the Feynman i-epsilon prescription.

For an amputated scattering amplitude, incoming fermions supply $u^s(p)$ and outgoing fermions supply ${\bar{u}}^s(p)$; incoming antifermions supply ${\bar{v}}^s(p)$ and outgoing antifermions supply $v^s(p)$. An external real scalar field particle supplies $1$. Use the standard relativistic normalization of a one-particle state; the external factors $1/\sqrt{2E}$ belong to that normalization and the Lorentz-invariant phase-space measure, rather than an additional factor in the invariant scattering amplitude.

To include all momentum-dependent contributions to the two-point function, put one external line at each interaction vertex. Let $r$ be the number of lines connecting the vertices, and $t_A,t_B$ their numbers of tadpole diagrams. For vertex valences $v_A,v_B\in \{4,6\}$,The integer $r$ is odd, and $r=1$ gives a one-particle-reducible Feynman diagram and is excluded. Thus the complete list is

The coefficient of the mixed graph includes the reflected external attachment as well as its reciprocal Feynman-diagram symmetry factor. If both external lines attach to the same interaction vertex, no external momentum enters the connecting internal lines, and that Feynman diagram cannot contribute to wave-function renormalization.

For incoming external momentum $p$, assign the connecting lines momenta $k_1,\dots ,k_{r-1},p-{\sum }_j{k}_j$. DefineThe momentum space Feynman rules give, in the order listed,Each $C$ is a reciprocal Feynman-diagram symmetry factor, including any equivalent external attachments. There are respectively $2,3,4,4$ independent loop momenta.

If “contribute to wavefunction renormalization” means requiring a divergent kinetic counterterm after subtraction of all subdivergences, only the last topology qualifies. The first three still have nontrivial external momentum dependence and give finite contributions to a normalization condition; their raw divergent factors arise from mass counterterms and interaction vertex counterterms. Keeping all four here makes this distinction explicit. The ultraviolet divergence analysis in d establishes it directly.

In momentum space, the Feynman rules for the real scalar field areEach interaction vertex supplies $(2\pi {)}^6{\delta }^{(6)}(\sum p)$, with all its momenta counted inward. Impose momentum conservation, integrate each independent loop momentum with $d^{6}p/(2\pi {)}^6$, and divide by the Feynman-diagram symmetry factor. For a connected amputated scattering amplitude, remove the external scalar propagators and retain a single overall momentum conservation factor. These signs follow by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$; the factorials in the Lagrangian density cancel the corresponding attachments in the Wick theorem. The Feynman i-epsilon prescription specifies the scalar propagator poles.

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.

The one-loop one-particle irreducible correction is a tadpole diagram: two external lines meet one quartic interaction and the other two legs are joined. Interchanging the two ends of its loop gives the symmetry factor $2$.

Define the self-energy by an insertion $-i\Sigma (p^2)$ and a full scalar propagator $i/[p^2-m^2-\Sigma (p^2)+iϵ]$ before counterterms. The supplied cutoff regularization integral givesThusIt has no external momentum dependence. The correction to the unamputated two-point correlation function in momentum space isThe positive sign of ${\Sigma }_1$ raises the unrenormalized squared pole mass for $g_4>0$.

Uniform probability per volume does not give a uniform radial distribution. A shell of radius $r$ and thickness $dr$ occupies volume $4\pi r^{2}dr$, so the radial probability density function isIt integrates to one. The second radial moment isThe radial uncertainty scale in a uniform sphere supplies the position-uncertainty scale as instructed by the toy model. It is not literally the standard deviation of the scalar radius: $⟨r⟩=3R/4$ and $\sqrt{⟨r^2⟩-⟨r{⟩}^2}=R\sqrt{3/80}$. Nor is a uniform spatial probability density sufficient to determine a unique momentum space wavefunction. This explicit distinction keeps the order-of-magnitude assumption separate from a fully specified quantum bound-state calculation. Replacing the radial density by a uniform density in $r$ would give a different second moment because it omits the spherical volume element.

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.