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In a phase-invariant vacuum, a correlator with $r$ insertions of $\varphi$ and $s$ of ${\varphi }^{\ast }$ gains the phase $e^{i\alpha (r-s)}$. Its invariance for all real $\alpha$ forces it to vanish unless $r=s$. Balanced local interactions and the absence of same-charge free contractions give the same rule order by order in Wick's theorem. A symmetry-breaking vacuum is an essential exception to the vacuum-invariance premise.

The four-point correlation function measures a time-ordered product of four field insertions. For a centered scalar field its disconnected terms are the three products of two-point functions, and subtracting those products leaves the connected correlation function. In a free Gaussian scalar vacuum the connected four-point function vanishes, by Wick's theorem, so only the three paired products remain. The connected four-point function of phi-fourth theory starts with one quartic vertex; amputation removes its external propagators. The full correlator, its connected part and the four-point one-particle-irreducible correlation function should therefore be distinguished.

At first order the Dyson expansion inserts $i{S}_{\mathrm{int}}=\lambda \int d^{4}x{\bar{\psi }}_{\alpha }(x)({\gamma }_5{)}_{\alpha \beta }{\psi }_{\beta }(x)\varphi (x)$. Contract the scalar field at the vertex with the external scalar. The four remaining fermion fields have two admissible pairings: both external fermions attach to the vertex, or the two vertex fermions contract with each other while the external pair contracts freely. Wick's theorem givesThe minus sign in the second term is the fermion sign of the closed contraction: $⟨T{\bar{\psi }}_{\alpha }(x){\psi }_{\beta }(x)⟩=-i{S}_F(0{)}_{\beta \alpha }$. The first term is connected; the second is a scalar-attached fermion tadpole multiplied by a disconnected free external fermion propagator. The full time-ordered expectation includes both topologies, whereas the connected generating functional retains only the first. There is no order-${\lambda }^2$ contribution in this pure Yukawa expansion because the total number of scalar fields would be odd.

For a parity-preserving free propagator, the tadpole coefficient is $\mathrm{tr}[{\gamma }_5{S}_F(0)]$, whose integrand contains only $\mathrm{tr}{\gamma }_5$ and $\mathrm{tr}({\gamma }_5{\gamma }^{\mu })$. Both vanish. Thus the second term is a legitimate formal Wick contraction but is zero with a compatible regulator; normal ordering the interaction removes it from the outset.

First fix the propagator kernels in a positive-sign Gaussian generating functional. From its scalar exponent, $i^{-2}{\delta }^2{Z}_0/\delta J\delta J$ gives the contraction $D(x-y)=-i\Delta (x-y)$. The fermionic insertion rules similarly give $F(x-y)=⟨T\psi (x)\bar{\psi }(y){⟩}_0=-iS(x-y)$, writing $S$ for the kernel labelled $S_F$ in that exponent. Thus the kernels in the question do not already include the factors of $i$ in the actual time-ordered contractions.

At order $g^2$, expand the interaction operator to its second term, or equivalently apply Wick's theorem toFor a connected contribution, the external scalar insertions attach to different interaction points. The two assignments are equal after exchanging $x,y$ and cancel $1/2$. The connected fermionic contraction isThe minus sign is the closed fermion loop sign. The alternative pairing gives two separate fermion tadpoles and belongs to disconnected external components, not the requested connected function. A free external scalar line multiplying a vacuum graph also cancels in the normalized result.

Combining $(ig{)}^2=-g^2$ with the loop minus sign givesThere are four contractions, so replacing $D=-i\Delta$ and $F=-iS$ introduces $(-i{)}^4=1$. ThereforeThis is the fermion-loop diagram in the last panel of i. The trace contracts spinor indices around the loop; it is not an additional degeneracy factor. A convention using physical contractions directly obtains the preceding formula in $D,F$ with the same coefficient. Mixing one convention's Gaussian kernels with another's insertion factors would change its apparent signs.

Normalize the generating functional by its value with all sources zero, and take the connected part of the time-ordered correlation function. Source differentiation of the interaction exponential is equivalent to the Dyson series followed by Wick's theorem. Write $D(x-y)=⟨T\varphi (x)\varphi (y){⟩}_0$ and $F_{ij}(x-y)=⟨T{\psi }_i(x){\bar{\psi }}_j(y){⟩}_0$ temporarily, so that every contraction has an unambiguous meaning.

The second-order Dyson series contributesThere are two connected external scalar field attachments:They become equal after exchanging the integration variables $x,y$, cancelling $2!$. The connected fermion contractions have the closed fermion loop sign:The minus sign can also be checked directly: the pairing of the first with the fourth field and the second with the third has positive pairing parity, but $⟨T{\bar{\psi }}_i(x){\psi }_j(y)⟩=-F_{ji}(y-x)$. Multiplying it by $(ig{)}^2=-g^2$ leaves a positive result:A contraction of the two external fields with each other times a vacuum bubble cancels against the normalization. Separate fermion tadpole integrals at $x$ and $y$ belong to a product of one-point functions, removed by taking the connected part.

To match the source's generating functional convention, distinguish its displayed kernels from contractions. The positive $iJ\Delta J/2$ exponent gives $D=-i\Delta$. With the source insertion order in equation (5), the positive $i\bar{\eta }{S}_F\eta$ exponent similarly gives $F=-i{S}_F$. The two scalar propagators and two Dirac propagators supply a total phase $(-i{)}^4=1$. Thus, writing $S=S_F$ as in the requested formula,The matrix trace closes the spinor indices; no extra factor of two or additional symmetry factor remains for a Dirac field with this Yukawa interaction.

Separate the constant embedding coordinate from its nonzero modes. Its integral gives the momentum-conservation delta function; below this is understood and absorbed into the amplitude normalization on the momentum-conserving locus. For the nonzero modes the Euclidean Polyakov action is quadratic, with worldsheet Green functionTreat the tachyon vertex operators as normal ordered, removing coincident self-contractions. The Gaussian functional integral, or the same calculation by Wick's theorem, givesAll remaining determinants are independent of the insertion points and momenta and enter the overall constant. This derives the Koba-Nielsen factor and the requested position integral.

For the Möbius transformation $f(z)=(az+b)/(cz+d)$, $ad-bc=1$, one hasThe transformed Koba-Nielsen factor contributes, at insertion $r$, the power $-{\alpha }^{\prime }{\sum }_{s\ne r}{k}_r\cdot k_s$ of $|c{z}_r+d|$. Four-momentum conservation makes this power ${\alpha }^{\prime }{k}_r^2$. The complete differential integrand therefore transforms byIt is invariant when ${\alpha }^{\prime }{k}_r^2=4$ for every external tachyon. The correlator alone is covariant; the integration measures are necessary for invariance. The group acting faithfully on the sphere is $PSL(2,\mathbb{C})$, with the two elements $\pm I$ of $SL(2,\mathbb{C})$ acting identically.

The apparent integral over all positions includes the residual conformal-group gauge volume. Divide by it and fix three distinct insertions. The Faddeev-Popov determinant, equivalently the three-point worldsheet ghost correlator and its conjugate, contributesSet $z_1=\Lambda$, $z_2=0$, $z_4=1$, then send $\Lambda$ to infinity on the sphere; the sign of the real direction is irrelevant. The factors involving $z_1$ behave as $|\Lambda {|}^{{\alpha }^{\prime }{k}_1\cdot {\sum }_{s\ne 1}{k}_s}=|\Lambda {|}^{-4}$, while $J\sim |\Lambda {|}^4$. They cancel, leaving $z_3=z$ as the one integration variable.

The remaining distance exponents areFor example $(k_2+k_3{)}^2=(k_1+k_4{)}^2=-t$ and each individual squared momentum is $4/{\alpha }^{\prime }$. AlsoConsequently the reduced amplitude is proportional toApply the supplied complex beta integral with $A=4+{\alpha }^{\prime }t/2$, $B=4+{\alpha }^{\prime }s/2$, and $C=4-A-B=4+{\alpha }^{\prime }u/2$. Its numerator arguments become $-1-{\alpha }^{\prime }t/4$, $-1-{\alpha }^{\prime }s/4$, $-1-{\alpha }^{\prime }u/4$, while its denominator arguments become $2+{\alpha }^{\prime }u/4$, $2+{\alpha }^{\prime }t/4$, $2+{\alpha }^{\prime }s/4$. Thus the four-point amplitude is the Virasoro–Shapiro amplitudewhere $C_0$ is momentum-independent. The integral converges initially when $\mathrm{Re}A<2$, $\mathrm{Re}B<2$, and $\mathrm{Re}(A+B)>2$; the displayed answer elsewhere is its analytic continuation.

The gamma function has simple poles at nonpositive integers and no zeros. At generic fixed $t$, the amplitude therefore has simple $s$-channel poles at $s=4(n-1)/{\alpha }^{\prime }$, $n=0,1,2,\dots$, with the same towers in $t$ and $u$ by crossing symmetry. They describe exchange of the closed string tachyon, massless states, and the infinitely many massive levels. The reciprocal denominator factors can give zeros at negative values $s=-4(\ell +2)/{\alpha }^{\prime }$; special kinematics can also make an exchange residue vanish.

To see the structure of a generic residue, set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$, so $a+b+c=1$. At $a=-n$, the Gamma function recurrence reduces the remaining factors to a constant times ${\prod }_{j=1}^n(b-j{)}^2$. The Virasoro–Shapiro amplitude pole residue is thus a polynomial of degree $2n$ in the other channel invariant, consistent with exchange up to spin $2n$. Crossed numerator poles must not simply be multiplied: near $a=-n+\delta$ and $b=-m+\epsilon$, $c=1+n+m-\delta -\epsilon$, and the zero of $1/\Gamma (1-c)$ cancels the putative double pole. The leading singular part is proportional toThe tree amplitude is meromorphic, with no loop threshold branch cuts; its channel singularities express factorization and string dual resonance.

Put $I_0=\sqrt{2\pi /m^2}$ and assume $m^2>0$. The normalized free Gaussian integral has covariance $\Delta =1/m^2$. Expanding the interaction exponential and using Wick's theorem gives the Feynman rules: a quartic vertex contributes $-\lambda$, each internal propagator contributes $\Delta$, and each vacuum diagram is divided by its Feynman-diagram symmetry factor. Multiply the resulting sum by $I_0$. Because we are expanding $I$, rather than $\log I$, disconnected vacuum diagrams must also be included.

At order ${\lambda }^2$ there are two quartic vertices and four internal propagators. Let $r$ be the number of edges joining the two vertices. The remaining half-edges must pair at their own vertex, so $r$ is even. The possibilities $r=0,2,4$ exhaust the vacuum diagrams:

For $r=0$, each vertex has three possible pairings, giving $N_0=9$ Wick contractions. For $r=2$, choose the two half-edges at each vertex, then pair the chosen half-edges across the vertices: $N_2={\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=72$. The remaining two half-edges at each vertex form a self-loop. For $r=4$, all half-edges pair across the vertices, giving $N_4=4!=24$. The common denominator from the expansion is $2!(4!{)}^2=1152$, so the Feynman-diagram symmetry factors areEach vacuum diagram has the same factor $I_0{\lambda }^2{\Delta }^4$ before division by its Feynman-diagram symmetry factor. Consequently the coefficient of ${\lambda }^2$ isAs an independent check, the eighth moment obtained from the Gaussian moment pairing theorem is $⟨x^8{⟩}_0=7!!/m^8=105/m^8$, and $105/[2!(4!{)}^2]=35/384$. The first terms of the zero-dimensional quartic perturbation series are thereforeStrictly, the printed phrase “Taylor expansion” means a formal expansion or a right-sided asymptotic expansion. The general coefficient has absolute successive ratio $(4v+3)(4v+1)/[24(v+1)m^4]\to \infty$, so the power series has zero radius of convergence. Nevertheless, for $\lambda \ge 0$ the remainder after any fixed truncation of $e^{-\lambda x^4/4!}$ is bounded by the next absolute term, and its Gaussian integral is finite. This proves the displayed asymptotic expansion as $\lambda \downarrow 0$. For negative real $\lambda$ the defining integral diverges.

For a centered Gaussian random field, Wick's theorem says that every odd moment vanishes and every even moment is the sum over all pairings of products of two-point functions. For six fields there are $15$ pairings. In the connected three-point function at nonzero external momenta, each of the three external fields must pair with a distinct field at the cubic interaction. There are $3!=6$ such Wick contractions. Choosing the undifferentiated field gives three possibilities; interchanging the two differentiated fields gives a further factor of two. The remaining nine pairings involve an external-external pair and an internal pair and belong to tadpole/disconnected contributions. Define the background so the one-point function vanishes, or equivalently subtract these contributions.

It is useful to keep a coefficient $\mathcal{C}$ multiplying the cubic Hamiltonian: its literal printed value is $1$, whereas the standard dimensionally normalized curvature interaction has $\mathcal{C}=M_{\mathrm{Pl}}^2$. This distinction will matter for the final amplitude. Withand $dt=ad\tau$, $\dot{\zeta }={\zeta }^{\prime }/a$, the interaction entering the time integral isIn the interaction picture, the unequal-time vacuum contraction required by the in-in formalism isFor a differentiated internal field replace $u_k^{\ast }$ by $u_k^{\ast \prime }$. The equal-time power spectrum fixes its magnitude; the displayed free De Sitter curvature mode functions and vacuum choice fix its unequal-time phase.

Performing the three momentum integrations imposes ${\mathbf{p}}_a=-{\mathbf{k}}_a$ for each assignment and leaves one overall momentum delta function. Put $U=u_{k_1}(0)u_{k_2}(0)u_{k_3}(0)$, which is real here, and defineThe upper early-time contour ${\mathcal{C}}_{+}$ runs from $-\infty (1-i\delta )$ to $0$, with $\delta >0$. The Hamiltonian's minus sign and the six connected Wick contractions then giveThis is equivalently $\mathrm{Re}[-4i\mathcal{C}{ϵ}^{2}U\sum I_i]$ with unconjugated De Sitter curvature mode functions and the conjugate lower contour ${\mathcal{C}}_{-}$. Before momentum integration, the same result consists of the three cyclic delta assignments, each with the extra factor of two for the identical differentiated legs.

The PDF's intermediate formula writes only three cyclic assignments without that factor of two. Taken literally it undercounts the connected contractions. Its unconjugated modes must also use the conjugate contour, rather than the upper contour of the original in-in expression. Both points are required for a consistent in-in bispectrum conjugation rule.

Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.

Let $D_{xy}=⟨0|T\varphi (x)\varphi (y)|0{⟩}_0$ be the free Feynman propagator, and abbreviate $v_3=-i\kappa$, $v_4=-i\lambda$. Expand the normalized Dyson series, so vacuum bubbles cancel between numerator and denominator, and retain the connected correlation function. Wick's theorem yieldsThese are respectively the free line, the quartic tadpole diagram, the two-cubic-vertex bubble diagram, and an tadpole contribution to a cubic-theory two-point function, attached by a bridge to the vertex carrying both external legs. The last graph is connected but is a one-particle-reducible Feynman diagram; removing its bridge disconnects it. It must be included when all connected graphs are requested. It disappears if an additional vanishing-one-point-function renormalization condition and its corresponding tadpole subtraction are imposed; that convention was not specified here.

The factors can be counted directly. The quartic term has $4\cdot 3=12$ attachments for the two labelled external legs, divided by $4!=24$. At second order in the cubic interaction the expansion denominator is $2!(3!{)}^2=72$. There are 36 Wick contractions for the bubble, and 36 for the attached-tadpole graph, giving $1/2$ for each. Thus each loop topology shown has symmetry factor $S=2$. Disconnected products of one-point functions and vacuum bubbles are not part of $G_c$.^[ref]

The one-loop cubic-quartic scalar vertex classification follows without truncating arbitrarily in powers of the couplings. A connected graph with $E$ external legs, $V_3$ cubic vertices, $V_4$ quartic vertices and loop number $L$ satisfiesFor $E=2,L=1$, the only possibilities are $(V_3,V_4)=(2,0)$ or $(0,1)$, exactly those just listed.

Expand the normalized Euclidean path integral with weight $e^{-S_E}$ and apply Wick's theorem. Use labelled external momenta and a fixed Fourier transform convention throughout. A line carrying momentum $q$ contributes $(q^2+m_0^2{)}^{-1}$. Each $2k$-valent interaction vertex contributes $-{\lambda }_{2k}=-{\mu }^{d_k}{g}_{2k}$: its $(2k)!$ possible contractions cancel the factorial in the Lagrangian density. A vertex also carries $(2\pi {)}^D{\delta }^{(D)}(\sum q)$ with all incident momenta incoming. Integrate each independent loop momentum with $d^{D}q/(2\pi {)}^D$ and divide by the Feynman-diagram symmetry factor for fixed external labels. There are no Minkowski factors of $i$.

For the full $N$-point Green's function, include every graph in which all vertices belong to components attached to external insertions. In particular, disconnected components carrying external insertions must be included, and the zeroth-order result is the sum of free pairings. Pure vacuum bubbles cancel against the denominator. The connected Green's function restricts to connected graphs, while a one-particle-irreducible vertex imposes the stronger restriction that removing one internal line cannot disconnect the graph. These are different objects. Odd $N$ vanishes in the symmetric vacuum because every interaction has even valence. At fixed perturbative order, multiply by the appropriate power of the coupling and sum all allowed graph topologies and external attachments.

If the overall momentum-conservation delta function is removed by convention, report the remaining function as its coefficient. The ordinary unamputated Green's function includes external scalar propagators; dividing them out defines an amputated function. The definition in c performs this amputation but does not, by itself, impose the usual one-particle-irreducibility condition.

Let $C(u,v)=2D_F(u-v)$ be the action-consistent free propagator established in a. Write $S=S_0+S_I$ and putThe perturbative expansion isDivide numerator and denominator by the same free vacuum functional $Z_0$. For $\mathcal{O}=\Phi (y_1)\cdots \Phi (y_N)$, they becomeFor $\mathcal{O}=\varphi (y_1){\varphi }^{\ast }(y_2)$, abbreviate $C_{12}=C(y_1,y_2)$ and $C_{zz}=C(z,z)$. Since only opposite-charge fields contract, Wick's theorem givesandThe first multiplicity is $2!$ internal pairings. In the connected term, there are two choices for the vertex's ${\varphi }^{\ast }$ contracted to the external $\varphi$ and two for its $\varphi$ contracted to the external ${\varphi }^{\ast }$; the remaining pair forms a tadpole diagram. ThereforeExpanding the inverse denominator cancels the free-line times vacuum bubble term, leavingThe local quartic interaction vertex has factor $-i\lambda$, since $2!2!$ cancels its denominator 4. The oriented tadpole diagram has Feynman-diagram symmetry factor 1; the isolated two-loop vacuum vertex has factor $1/2$ in its coefficient. Coincident kernels and the integrals require a common regularization; none is being assigned a finite numerical value here.

More generally, vacuum bubbles exponentiate and factor from every numerator. Here global U(1) symmetry also makes all one-point functions zero. Thus a surviving graph with the two external insertions in separate connected components would have a component with one unbalanced charge and vanish. This proves that only connected graphs contribute to this particular two-point function to all perturbative orders. Vacuum normalization alone would not remove disconnected components containing external legs in a general four-point function. If one formally chooses a different free kernel, the displayed Wick multiplicities are unchanged; replacing $C$ by $C_{\mathrm{lit}}$ gives a formal oscillator series, not the Gaussian integral of the action as printed.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

The probabilistic form of Wick's theorem is Isserlis's theorem. If $(Y_1,\dots ,Y_k)$ is a centered random vector with a multivariate Gaussian distribution and covariance matrix $C$, its odd product moments vanish, and, for even $k$,Here ${\mathcal{P}}_2(k)$ is the set of perfect matchings of the $k$ labelled positions. The formula allows repeated coordinates and singular covariance matrices. To see the pairing rule, differentiate the moment-generating function $\exp (t^{T}Ct/2)$ once in each coordinate and set $t=0$: only terms pairing all differentiated coordinates survive, each with the displayed product of covariances.

Take $Y_a=M_{i_a{i}_{a+1}}$, with $i_{k+1}=i_1$. These variables are linear combinations of the original independent normal random variables, hence are jointly Gaussian. They have zero means, soSubstituting the entry covariance from part (a)(i) identifies each parenthesis with $\mathbb{E}[M_{i_a{i}_{a+1}}{M}_{i_b{i}_{b+1}}]$, giving the requested identity.

There is a state-assumption issue. For a number-conserving Gaussian quantum state with no anomalous contractions, put ${\psi }_i=\psi ({\mathbf{r}}_i)$, $n_i=⟨{\psi }_i^{†}{\psi }_i⟩$ and the one-body density matrix $g_{ij}=⟨{\psi }_i^{†}{\psi }_j⟩$. Wick's theorem pairs each creation operator with an annihilation operator. The direct pairing gives $n_1{n}_2$; the exchanged pairing gives $ϵg_{12}{g}_{21}$, with $ϵ=+1$ for bosons and $-1$ for fermions. Hencenormal ordering removes the equal-position delta-function self-counting term. Fermionic occupation-number product states are Slater determinants and obey this Wick theorem. Independent thermal bosonic modes also obey it, with $N_{\alpha }$ understood as mean occupations.

A bosonic Fock product state does not generally obey the asserted identity. The bosonic number-state corrections to Gaussian density correlations follow from $⟨a_{\alpha }^{†2}{a}_{\alpha }^2⟩=N_{\alpha }(N_{\alpha }-1)$ and giveFor one occupied mode this is $N(N-1)|{\phi }_1{|}^2|{\phi }_2{|}^2$, whereas the uncorrected Gaussian expression gives $2N^2|{\phi }_1{|}^2|{\phi }_2{|}^2$. This counterexample disproves the asserted identity for arbitrary bosonic number-product states. The formulas below give both the Gaussian result, with its additional state assumption, and the exact bosonic number-state alternative. The ensemble distinction is also explicit in a later primary treatment of ideal-gas correlations.^[ref]

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.

Set the external source to zero in solution and apply Wick's theorem. At a single quartic vertex, the four position insertions have three pairings; at a single sextic vertex, six insertions have $5!!=15$ pairings. ThusThermal time-translation invariance makes this $1-\lambda \beta g_0^2/8-\mu \beta g_0^3/48+\cdots$, with $g_0$ from solution. If desired the absolute corrections follow by multiplying by the free partition function $Z_0=[2\sinh (\beta \Omega /2){]}^{-1}$.