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A Bogoliubov vacuum is annihilated by the quasiparticle linear operators that diagonalize a stable quadratic bosonic Hamiltonian. If $a_k=u_k{b}_k-v_k{b}_{-k}^{†}$, its original-mode occupation number is $|v_k{|}^2$. It is generally a multimode squeezed vacuum rather than the original Fock vacuum. In infinitely many modes, existence in the original Fock space requires a separate convergence analysis.

For $[a,a^{†}]=1$, the occupation number operator $a^{†}a$ has eigenvalues $0,1,2,\dots$ on the bosonic oscillator Fock space.

Let $b_k=\cosh {\theta }_k{a}_k+\sinh {\theta }_k{a}_{-k}^{†}$ for real even ${\theta }_k$, and let every $b_k$ annihilate a normalized vacuum state. Inverting the Bogoliubov transformation gives $a_k=\cosh {\theta }_k{b}_k-\sinh {\theta }_k{b}_{-k}^{†}$. Only $⟨b_k{b}_q^{†}⟩={\delta }_{kq}$ survives, proving the displayed vacuum expectation values. Taking the Hermitian adjoint gives the same real negative pair amplitude for two creation operators. The normal occupation number is quadratic at small ${\theta }_k$, whereas the pair amplitude is linear. For one opposite-momentum pair the two-mode squeezed vacuum is $\mathrm{sech}\theta {\sum }_{n\ge 0}(-\tanh \theta {)}^n|n,n⟩$, fixing the sign convention directly.

A normalized oscillator coherent state obeys $a|\alpha ⟩=\alpha |\alpha ⟩$, so its mean occupation number is $N=|\alpha {|}^2$. Under the quantum harmonic oscillator Hamiltonian operator $\hslash \omega (a^{†}a+1/2)$, its evolution is $|\alpha (t)⟩=e^{-i\omega t/2}|\alpha e^{-i\omega t}⟩$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=-i\sqrt{m\hslash \omega /2}(a-a^{†})$ givesThese mean values obey the classical harmonic oscillator equations. The displacement amplitude is $\sqrt{2\hslash N/(m\omega )}$, whose classical orbit has energy $\hslash \omega N$. The full quantum mean energy has the additional zero-point energy $\hslash \omega /2$. Indeed the variances are $\hslash /(2m\omega )$ and $m\hslash \omega /2$, and their kinetic and potential contributions sum to $\hslash \omega /2$.

For distinct box modes ${\phi }_1=\sqrt{2/L}\sin (p\pi x/L)$ and ${\phi }_2=\sqrt{2/L}\sin (q\pi x/L)$, the displayed contact interaction overlaps produce self terms $3g{\sum }_j{n}_j(n_j-1)/(4L)$, a density term $2g{n}_1{n}_2/L$, and pair conversion $g(a_1^{†2}{a}_2^2+a_2^{†2}{a}_1^2)/(2L)$. Pair conversion preserves total particle number but prevents diagonalization by individual occupation numbers. The additional overlaps are $I_{1112}=-{\delta }_{q,3p}/(2L)$ and $I_{1222}=-{\delta }_{p,3q}/(2L)$, following from ${\sin }^{3}u=(3\sin u-\sin 3u)/4$. Third-harmonic mode pairs therefore also permit single-particle transfer assisted by the other particles. For the first two box modes these additional terms vanish.

Exclusion constrains complete one-particle states, not a collection of little spatial boxes. In particular, two opposite-spin electrons in the same spatial orbital haveThe spatial factor is symmetric; the spin-one-half singlet state supplies the required antisymmetry. At ${\mathbf{r}}_1={\mathbf{r}}_2$, this wavefunction need not vanish for opposite spin labels. For equal spin labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint probability densities or detection in finite regions. Equal positions do not imply equal complete quantum states.

The fermionic occupation numbers $n_{\alpha }=0,1$ refer to a chosen orthonormal basis of one-particle states. A quantum superposition need not have definite occupation numbers in that basis, and an expectation value $⟨n_{\alpha }⟩$ can be fractional. For example, one electron in $(u+v)/\sqrt{2}$ has mean occupation number $1/2$ in each of two orthogonal spin orbitals, without splitting into two particles. The canonical anticommutation relations give $(a_{\alpha }^{†}{)}^2=0$ and make $a_{\alpha }^{†}{a}_{\alpha }$ a projector, so its expectation value stays between zero and one in every normalized quantum state.

For $\{f,f^{†}{\}}_{+}=1$ and $f^2=0$, the occupation number operator $f^{†}f$ has eigenvalues zero and one. This gives the single-mode exclusion principle.

Choosing $N$ different complete one-particle states from an orthonormal basis of size $M$ gives $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ independent Slater determinants. This is the binomial coefficient because only the occupied subset matters: permutations change the sign of the Slater determinant, not the physical quantum state. Summing over particle numbers gives $2^M$ states in the finite fermionic Fock space.

For the diagram's two spatial orbitals $A,B$, take energies $0,\Delta$ with $\Delta >0$, each independent of spin. In the complete-state order $(A\uparrow ,A\downarrow ,B\uparrow ,B\downarrow )$, the six $N=2$ occupation number strings are $1100,1010,1001,0110,0101,0011$. Their total energies are $0,\Delta ,\Delta ,\Delta ,\Delta ,2\Delta$, giving energy multiplicities $1,4,1$. For $N=3$, the ground state energy is $\Delta$ with two independent ground states, according to which spin orbital in $B$ is occupied. Across $N=0,1,2,3,4$, the counts are $1,4,6,4,1$, summing to $16$. These are exact counts for this model, not measured atomic energies.

A spin-$S$ Hilbert space has dimension $2S+1$. Its Holstein–Primakoff transformation therefore uses only bosonic occupation numbers $0,1,\dots ,2S$. The square root annihilates the upper endpoint. A truncated linear spin-wave approximation formally enlarges this space; it is self-consistent only when boson depletion is small compared with $S$.

The Kompaneets equation describes frequency diffusion of a mean photon occupation number $n$ through repeated nonrelativistic Compton scattering. For thermal electrons, with $x_e=h\nu /(k_B{T}_e)$ and $dy=(k_B{T}_e/m_e{c}^2)n_e{\sigma }_{T}cdt$, its full homogeneous form is ${\partial }_{y}n=x_e^{-2}{\partial }_{x_e}[x_e^4({\partial }_{x_e}n+n+n^2)]$. The drift and stimulated terms permit a thermal stationary spectrum. In the hot-electron limit $T_e\gg T_{\gamma }$, with low photon energies, they can be neglected relative to the diffusion term; any fixed linear frequency variable $x$ then gives ${\partial }_{y}n=x^{-2}{\partial }_x(x^4{\partial }_{x}n)$. The quantity $n$ is not the frequency-integrated photon number density.

The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

The Pauli matrix ${\sigma }_x$ flips the spin with unit matrix element. A nonzero oscillator matrix element changes exactly one occupation number, through a creation operator or an annihilation operator:For a spin flip, the spin energy change divided by $\hslash$ is $-s_{\sigma }\Omega$. Thus creation and annihilation in mode $j$ have total frequency changes ${\omega }_j-s_{\sigma }\Omega$ and $-{\omega }_j-s_{\sigma }\Omega$, respectively. Distinct modes or different occupation changes lead to orthogonal final quantum states, so their probabilities add. Fermi's golden rule consequently givesThe Dirac delta function is even, accounting for the sign in the annihilation term. Since all ${\omega }_j,\Omega >0$, an initially up spin can only emit an oscillator quantum, while an initially down spin can only absorb one. Writing the oscillator density of states per unit angular frequency as $g(\omega )={\sum }_j\delta (\omega -{\omega }_j)$, the continuum limit givesThe $+1$ permits emission even from the oscillator vacuum; absorption requires a quantum already present. The notation $n_{\Omega }$ assumes a common or smoothly specified occupation at the resonant frequency. If several resonant modes have different occupations, it denotes their density of states weighted mean, so that $g(\Omega )n_{\Omega }={\sum }_j{n}_j\delta (\Omega -{\omega }_j)$ after continuum smoothing. This is the golden-rule relaxation of a spin coupled to an oscillator bath.

Use $\hslash =1$ throughout. Let $N_a=a^{†}a$, $N_b=b^{†}b$ and $N=N_a+N_b$. Operators belonging to different boson species commute. The proposed Schwinger boson representation givesConsequently the first of the spin commutation relations isThe number operator commutators $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$ and their $b$ counterparts giveThus all the spin commutation relations hold. Also $[N,S^{\pm }]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its spin, calculate the Casimir operator:The irreducible spin representation with prescribed $s$ therefore requires $N=2s$, and its normalized Fock basis correspondence isConversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer occupation numbers give precisely $m=-s,-s+1,\dots ,s$, and there are $2s+1$ such states. As a check, the spin raising operator acts aswhich is the usual spin ladder operator coefficient $\sqrt{s(s+1)-m(m+1)}$.

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

Let $V=L^3$ be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:In the Fourier series for the contact interaction, the two creation operators supply phase $e^{-i({\mathbf{k}}_1+{\mathbf{k}}_2)\cdot \mathbf{r}}$ and the two annihilation operators supply $e^{i({\mathbf{k}}_3+{\mathbf{k}}_4)\cdot \mathbf{r}}$. Fourier orthogonality with periodic boundary conditions givesThe four normalization factors together contribute $V^{-2}$, leaving the momentum-space contact-interaction HamiltonianThe Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer isBoth are already in normal ordering; no extra commutator constant is introduced.

The coherent state in the zero mode has mean occupation number $N_0$. The other occupation numbers follow from part (c), giving exactlyImpose the fixed mean particle number $⟨\hat{N}⟩=N$ by choosingHere $D$ is the quantum depletion of a Bose condensate. The approximation requires $0\le D\ll N$, so $N_0$ remains positive and macroscopically large. The specified coherent state and multimode squeezed vacuum are not exact eigenstates of the total number operator; $N$ in this variational construction is their prescribed mean particle number.

Write $P={\sum }_{\mathbf{k}\ne 0}{s}_k{c}_k$. Substituting into the retained part of the interaction energy before truncating givesThe last two terms quadratic in $D,P$ have no explicit large-$N$ factor when the Bogoliubov transformation parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the number-conserving Bogoliubov quadratic Hamiltonian. Thus, defining the physical number density correctly as $\rho =N/V=N/L^3$,Replacing $N_0$ by $N$ at the outset would give the wrong coefficient $2{\sinh }^2{\theta }_k$: expanding the leading condensate interaction energy supplies the necessary subtraction.

The printed definition $\rho =N/L$ is an error. In three dimensions the contact interaction coupling $u$ has units of energy times volume, so $uN/L$ does not even have units of energy. There is a direct counterexample with no depletion: when ${\theta }_k=0$, the specified coherent state has exactly $⟨H_{\mathrm{int}}⟩=u{N}^2/(2L^3)$. In fixed length and energy units, $L=2$, $u=1$ and $N=100$ give $625$, whereas the printed $u\rho N/2$ with $\rho =N/L$ gives $2500$. The requested formula and the minimization in part (e) hold with $\rho =N/L^3$.

Put ${ϵ}_k={\hslash }^2{k}^2/(2m)>0$ for $\mathbf{k}\ne 0$ and $g=u\rho$, using $\rho =N/L^3$. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, $u>0$, each ${ϵ}_k+g>g>0$. Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution ${\theta }_k^{\ast }$, $\cosh 2{\theta }_k^{\ast }=({ϵ}_k+g)/E_k$ and $\sinh 2{\theta }_k^{\ast }=g/E_k$. HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from $\theta =0$ isIndeed $E_k<{ϵ}_k+g$. More directly, for small positive $\theta$,The pair vacuum expectation value lowers the interaction energy at first order in $\theta$, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when ${ϵ}_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-{ϵ}_k$ in some mode, $f_k(\theta )={ϵ}_k\sinh (2\theta )/2$ tends to $-\infty$ as $\theta \to -\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large $k$, $E_k-{ϵ}_k-g=-g^2/(2{ϵ}_k)+O({ϵ}_k^{-2})$, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.

The canonical anticommutation relations imply $a^2=(a^{†}{)}^2=0$. Thus the number operator satisfiesIt is a orthogonal projection, so its eigenvalues obey $n_a^2=n_a$ and are $\boxed{n_a\in \{0,1\}}$. The same argument gives $\boxed{n_b\in \{0,1\}}$. This is the Pauli exclusion principle for each orbital. The two number operators commute, so they have simultaneous eigenstates.

Take the vacuum state to have unit norm and fix the double-occupation convention $|11⟩=a^{†}{b}^{†}|0⟩$. An orthonormal basis of the four-dimensional fermionic Fock space isEach individual number operator has two-dimensional eigenspaces: for $N_a$ these are spanned by $\{|00⟩,|01⟩\}$ and $\{|10⟩,|11⟩\}$; for $N_b$ they are spanned by $\{|00⟩,|10⟩\}$ and $\{|01⟩,|11⟩\}$. Higher occupation numbers vanish because the square of a fermionic creation operator is zero. Reversing the order of the two fermionic creation operators changes only the sign convention for $|11⟩$.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

In the Maxwell-Boltzmann velocity distribution, each Cartesian component has mean-square velocity $k_B{T}_e/m_e$. Thus $⟨v^2⟩=3k_B{T}_e/m_e$. The original PDF calls this quantity an rms velocity, but the rms speed is its square root. Also, its $n_{\gamma }$ must mean the dimensionless mean occupation number $n$, since its equilibrium value is a Planck photon distribution; it is not the frequency-integrated photon number density.

Take $x=h\nu /(k_B{T}_{\gamma })$ at a fixed local radiation temperature. Because $x$ is proportional to frequency,The hot-electron Kompaneets equation therefore becomesFor traversal length $d\ell =cdt$, the Compton y parameter isIt measures electron pressure integrated along the line of sight, or the Thomson scattering optical depth weighted by $k_B{T}_e/(m_e{c}^2)$. In this hot-electron limit the mean fractional energy gain is approximately $4y$. The derivation assumes nonrelativistic electrons, Thomson-limit photon energies and $T_e\gg T_{\gamma }$. The original PDF uses $h\nu$, as required for ordinary frequency; the converted TeX incorrectly substitutes $\hslash \nu$.