Prepared for submission to JHEP
Introduction
The effective potential is the standard tool for studying the vacuum structure of a quantum field theory. Since the work of Coleman and Weinberg and of Jackiw [1 , 2 ] , it has been used to analyse radiative symmetry breaking, vacuum stability and phase transitions, and its radiative corrections often decide the qualitative picture: whether a symmetry is broken, whether a vacuum is stable, and at which scale.
In de Sitter space these questions are of direct physical interest. De Sitter space is the geometry of inflation and, asymptotically, of the late-time accelerating Universe, and radiative corrections in this background govern vacuum stability, curvature-induced symmetry breaking and the dynamics of light scalar fields during cosmological evolution [3 , 4 , 5 , 6 , 7 ] . Curvature enters on the same footing as the masses, so a reliable potential should be exact in both. Perturbation theory in de Sitter space is, however, considerably harder than in flat spacetime, and beyond one loop most results rely on approximations, asymptotic expansions, numerical methods or partial resummations [8 , 9 ] . Exact effective potentials are known at one loop, and at two loops in three dimensions [10 ] .
In flat space the two-loop effective potential has long been known: Ford, Jack and Jones obtained it for the Standard Model [11 ] . At two loops the new ingredient is the sunset, or watermelon, vacuum diagram. Rather than integrating it directly, they used integration by parts to derive differential equations for it in the masses, and solved them, obtaining its ultraviolet poles and its finite part in closed form. Integration-by-parts identities [12 , 13 ] , which follow from the translation invariance of momentum space, have since become the standard route to such computations.
None of this carries over to de Sitter space. There are no global momenta, and integration-by-parts identities in their flat-space form are not available. The watermelon itself can be evaluated: it was obtained in closed form in Ref. [14 ] through the Källén–Lehmann representation of the product of two propagators [15 , 16 ] . In four dimensions, however, extracting its finite part from that representation proved exceedingly difficult, and the two-loop effective potential in four-dimensional de Sitter space remained out of reach. Curved-space structures related to integration-by-parts identities have recently been explored from several directions [17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 25 ] , but none had produced the finite parts this computation requires.
The obstruction was removed in the companion Letter [26 ] , which introduced the curved-space counterpart of integration-by-parts identities: exact finite-difference equations in the spectral labels of the lines, generated by two local, coordinate-free identities, propagator contiguity and Green’s identity, shared by de Sitter and anti-de Sitter space. Where the flat-space computation relies on differential equations in the masses, here the masses enter through the spectral labels of the propagators, which can only be shifted by integers, and the equations are finite-difference equations in those labels. For the watermelon they give an exact mass-shift recurrence. Its physical solution, selected by the appropriate boundary conditions, is a new representation in which all ultraviolet poles sit in Gamma-function factors while the remaining hypergeometric series are regular near four dimensions, so that both the poles and the finite part are explicit. This makes possible, for the first time, a computation in de Sitter space parallel to the flat-space one; the same equations hold in anti-de Sitter space, which we do not treat here.
In this paper we carry out that computation for the O ( N ) O(N) O ( N )
the physical solution of the recurrence, with a proof of the boundary condition;
the dimensional expansion of the watermelon near four dimensions, including its finite part;
the fully renormalised two-loop effective potential of the O ( N ) O(N) O ( N )
the corresponding renormalisation-group functions.
To our knowledge, this is the first exact two-loop effective potential in four-dimensional de Sitter space.
The paper is organised as follows. In § 2 we introduce the O ( N ) O(N) O ( N ) § 3 reviews the one-loop effective potential. In § 4 we derive the finite-difference equations for the watermelon and their exact solution. § 5 develops the four-dimensional expansion and renormalisation, and presents the two-loop effective potential. In § 6 we analyse the ultraviolet structure and derive the beta functions. Technical details on the finite-difference solution are collected in Appendix § 8 .
The O ( N ) O(N) O ( N )
Let M d + 1 M_{d+1} M d + 1 ( d + 1 ) (d+1) ( d + 1 ) M d + 1 ( c ) M_{d+1}^{(c)} M d + 1 ( c )
Z 1 ⋅ Z 2 = Z 1 0 Z 2 0 − Z 1 1 Z 2 1 − ⋯ − Z 1 d Z 2 d . (1) Z_1 \cdot Z_2 = Z_1^0 Z_2^0 - Z_1^1 Z_2^1 - \cdots- Z_1^d Z_2^d. \tag*{(1)} Z 1 ⋅ Z 2 = Z 1 0 Z 2 0 − Z 1 1 Z 2 1 − ⋯ − Z 1 d Z 2 d . (1) The complex de Sitter manifold is the hyperboloid
d S d ( c ) = { Z ∈ M d + 1 ( c ) : Z ⋅ Z = − R 2 } . (2) dS_d^{(c)} = \{Z \in M_{d+1}^{(c)} : Z \cdot Z = -R^2\}. \tag*{(2)} d S d ( c ) = { Z ∈ M d + 1 ( c ) : Z ⋅ Z = − R 2 } . (2) We initially set R = 1 R=1 R = 1 § 5 . The invariant complex variable is
z = Z 1 ⋅ Z 2 . (3) z = Z_1 \cdot Z_2. \tag*{(3)} z = Z 1 ⋅ Z 2 . (3) The future and past tuboids are
T ± = { Z = x + i y ∈ d S d ( c ) : y ∈ ± V + } , (4) T_{\pm} = \{Z=x+iy \in dS_d^{(c)} : y \in\pm V_+\}, \tag*{(4)} T ± = { Z = x + i y ∈ d S d ( c ) : y ∈ ± V + } , (4) where V + V_+ V + [27 , 28 ] : the two-point distributions are boundary values of functions holomorphic in T − × T + T_- \times T_+ T − × T + [14 , 27 , 28 ] , with h = d − 1 2 h=\frac{d-1}{2} h = 2 d − 1
W ν d ( Z 1 , Z 2 ) = Γ ( h + i ν ) Γ ( h − i ν ) 2 ( 2 π ) d / 2 ( z 2 − 1 ) − d − 2 4 P − 1 2 + i ν − d − 2 2 ( z ) (5) W_\nu^d(Z_1,Z_2) = \frac{\Gamma(h+i\nu)\Gamma(h-i\nu)}{2(2\pi)^{d/2}}(z^2-1)^{-\frac{d-2}{4}}P_{-\frac{1}{2}+i\nu}^{-\frac{d-2}{2}}(z) \tag*{(5)} W ν d ( Z 1 , Z 2 ) = 2 ( 2 π ) d /2 Γ ( h + i ν ) Γ ( h − i ν ) ( z 2 − 1 ) − 4 d − 2 P − 2 1 + i ν − 2 d − 2 ( z ) (5) = Γ ( h + i ν ) Γ ( h − i ν ) ( 4 π ) d / 2 Γ ( d / 2 ) 2 F 1 ( h + i ν , h − i ν ; d 2 ; 1 − z 2 ) . (6) = \frac{\Gamma(h+i\nu)\Gamma(h-i\nu)}{(4\pi)^{d/2}\Gamma(d/2)}\,{}_2F_1\left(h+i\nu,h-i\nu;\frac{d}{2};\frac{1-z}{2}\right). \tag*{(6)} = ( 4 π ) d /2 Γ ( d /2 ) Γ ( h + i ν ) Γ ( h − i ν ) 2 F 1 ( h + i ν , h − i ν ; 2 d ; 2 1 − z ) . (6) The spectral parameter is related to the squared mass by
M 2 = h 2 + ν 2 . (7) M^2 = h^2 + \nu^2. \tag*{(7)} M 2 = h 2 + ν 2 . (7) For real positive squared mass, ν \nu ν ∣ ν ∣ < h |\nu| < h ∣ ν ∣ < h [29 , 30 ] .
The Schwinger function is the restriction to the Euclidean sphere
S d = { Z ∈ d S d ( c ) : Z 0 = i X 0 , Z i = X i , X μ ∈ R } . (8) S^d = \left\{ Z \in dS_d^{(c)} : Z^0 = iX^0,\ Z^i = X^i,\ X^\mu\in\mathbb{R} \right\}. \tag*{(8)} S d = { Z ∈ d S d ( c ) : Z 0 = i X 0 , Z i = X i , X μ ∈ R } . (8) On this section, X ∈ S d ⊂ R d + 1 X \in S^d \subset\mathbb{R}^{d+1} X ∈ S d ⊂ R d + 1
Z 1 ⋅ Z 2 = − X 1 ⋅ E X 2 = − cos s , (9) Z_1 \cdot Z_2 = -X_1 \cdot_E X_2 = -\cos s, \tag*{(9)} Z 1 ⋅ Z 2 = − X 1 ⋅ E X 2 = − cos s , (9) where s s s
q = i ν , u = X 1 ⋅ E X 2 = cos s . (10) q = i\nu,\qquad u = X_1 \cdot_E X_2 = \cos s. \tag*{(10)} q = i ν , u = X 1 ⋅ E X 2 = cos s . (10) With all the above understood the propagator is
G q ( d ) ( u ) = Γ ( h + q ) Γ ( h − q ) 2 ( 2 π ) d / 2 ( 1 − u 2 ) − d − 2 4 P − 1 2 + q − d − 2 2 ( − u ) , (11) G_q^{(d)}(u) = \frac{\Gamma(h+q)\Gamma(h-q)}{2(2\pi)^{d/2}} (1-u^2)^{-\frac{d-2}{4}} P_{-\frac{1}{2}+q}^{-\frac{d-2}{2}}(-u), \tag*{(11)} G q ( d ) ( u ) = 2 ( 2 π ) d /2 Γ ( h + q ) Γ ( h − q ) ( 1 − u 2 ) − 4 d − 2 P − 2 1 + q − 2 d − 2 ( − u ) , (11) where P β α ( x ) P_\beta^\alpha(x) P β α ( x ) [31 ] or Ferrers function of the first kind.1
G q ( d ) ( u ) = Γ ( h + q ) Γ ( h − q ) ( 4 π ) d / 2 Γ ( d / 2 ) 2 F 1 ( h + q , h − q ; d 2 ; 1 + u 2 ) , M 2 = h 2 − q 2 . (12) G_q^{(d)}(u) = \frac{\Gamma(h+q)\Gamma(h-q)}{(4\pi)^{d/2}\Gamma(d/2)}\,{}_2F_1\left(h+q,h-q;\frac{d}{2};\frac{1+u}{2}\right),\qquad M^2 = h^2-q^2. \tag*{(12)} G q ( d ) ( u ) = ( 4 π ) d /2 Γ ( d /2 ) Γ ( h + q ) Γ ( h − q ) 2 F 1 ( h + q , h − q ; 2 d ; 2 1 + u ) , M 2 = h 2 − q 2 . (12) When the dimension is understood, we abbreviate G q ( d ) G_q^{(d)} G q ( d ) G q G_q G q
Consider a scalar field ϕ : S d → R N \phi:S^d \to\mathbb{R}^N ϕ : S d → R N
S [ ϕ ] = ∫ S d [ 1 2 ⟨ ∂ μ ϕ ∣ ∂ μ ϕ ⟩ + m 0 2 + ξ R 2 ⟨ ϕ ∣ ϕ ⟩ + c 0 4 ⟨ ϕ ∣ ϕ ⟩ 2 + L g ] d X , (13) S[\phi] = \int_{S^d} \left[ \frac{1}{2}\langle\partial_\mu\phi\mid\partial^\mu\phi\rangle+ \frac{m_0^2+\xi R}{2}\langle\phi\mid\phi\rangle+ \frac{c_0}{4}\langle\phi\mid\phi\rangle^2 + \mathcal{L}_g \right]\,dX, \tag*{(13)} S [ ϕ ] = ∫ S d [ 2 1 ⟨ ∂ μ ϕ ∣ ∂ μ ϕ ⟩ + 2 m 0 2 + ξ R ⟨ ϕ ∣ ϕ ⟩ + 4 c 0 ⟨ ϕ ∣ ϕ ⟩ 2 + L g ] d X , (13) where m 0 , c 0 m_0,c_0 m 0 , c 0 d X dX d X S d S^d S d
L g = ρ + κ R + ( b R + b E ( d − 2 ) ( d − 3 ) d ( d − 1 ) ) R 2 . (14) \mathcal{L}_g = \rho+\kappa R+\left(b_R+b_E\frac{(d-2)(d-3)}{d(d-1)}\right)R^2. \tag*{(14)} L g = ρ + κ R + ( b R + b E d ( d − 1 ) ( d − 2 ) ( d − 3 ) ) R 2 . (14) We expand about ϕ ˉ = φ u 0 \bar{\phi}=\varphi\mathbf{u}_0 ϕ ˉ = φ u 0 φ \varphi φ u 0 \mathbf{u}_0 u 0 { u j } j = 0 N − 1 \{\mathbf{u}_j\}_{j=0}^{N-1} { u j } j = 0 N − 1 R N \mathbb{R}^N R N u 0 \mathbf{u}_0 u 0
ϕ = ( φ + ψ 0 ) u 0 + ∑ j = 1 N − 1 ψ j u j . (15) \phi= (\varphi+\psi_0)\mathbf{u}_0+\sum_{j=1}^{N-1}\psi_j\mathbf{u}_j. \tag*{(15)} ϕ = ( φ + ψ 0 ) u 0 + j = 1 ∑ N − 1 ψ j u j . (15) Δ 2 = C ∖ { ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) } . (16) \Delta_2 = \mathbb{C}\setminus\{(-\infty,-1]\cup[1,\infty)\}. \tag*{(16)} Δ 2 = C ∖ {( − ∞ , − 1 ] ∪ [ 1 , ∞ )} . (16) The action becomes [14 ]
S [ ϕ ] = ∫ S d [ m 0 2 + ξ R 2 ϕ 2 + c 0 4 ϕ 4 + L g ] d X + ∑ a = 1 4 S a [ ψ ; ϕ ] , (17) S[\phi]=\int_{S^d}\left[\frac{m_0^2+\xi\mathcal{R}}{2}\phi^2+\frac{c_0}{4}\phi^4+\mathcal{L}_g\right]\mathrm{d}X+\sum_{a=1}^{4}S_a[\psi;\phi], \tag*{(17)} S [ ϕ ] = ∫ S d [ 2 m 0 2 + ξ R ϕ 2 + 4 c 0 ϕ 4 + L g ] d X + a = 1 ∑ 4 S a [ ψ ; ϕ ] , (17) with
S 1 = ∫ S d [ ( m 0 2 + ξ R ) ψ ϕ 0 + c 0 ϕ 3 ψ 0 ] d X , (18) S_1=\int_{S^d}\left[(m_0^2+\xi\mathcal{R})\psi\phi_0+c_0\phi^3\psi_0\right]\mathrm{d}X, \tag*{(18)} S 1 = ∫ S d [ ( m 0 2 + ξ R ) ψ ϕ 0 + c 0 ϕ 3 ψ 0 ] d X , (18) S 2 = 1 2 ∫ S d [ ∑ j = 0 N − 1 ∂ μ ψ j ∂ μ ψ j + M 0 2 ψ 0 2 + M 1 2 ∑ j = 1 N − 1 ψ j 2 ] d X , (19) S_2=\frac{1}{2}\int_{S^d}\left[\sum_{j=0}^{N-1}\partial_\mu\psi_j\partial^\mu\psi_j+M_0^2\psi_0^2+M_1^2\sum_{j=1}^{N-1}\psi_j^2\right]\mathrm{d}X, \tag*{(19)} S 2 = 2 1 ∫ S d [ j = 0 ∑ N − 1 ∂ μ ψ j ∂ μ ψ j + M 0 2 ψ 0 2 + M 1 2 j = 1 ∑ N − 1 ψ j 2 ] d X , (19) S 3 = ∫ S d c 0 ϕ ψ 0 ∑ j = 0 N − 1 ψ j 2 d X , (20) S_3=\int_{S^d}c_0\phi\psi_0\sum_{j=0}^{N-1}\psi_j^2\mathrm{d}X, \tag*{(20)} S 3 = ∫ S d c 0 ϕ ψ 0 j = 0 ∑ N − 1 ψ j 2 d X , (20) S 4 = c 0 4 ∫ S d ( ∑ j = 0 N − 1 ψ j 2 ) 2 d X , (21) S_4=\frac{c_0}{4}\int_{S^d}\left(\sum_{j=0}^{N-1}\psi_j^2\right)^2\mathrm{d}X, \tag*{(21)} S 4 = 4 c 0 ∫ S d ( j = 0 ∑ N − 1 ψ j 2 ) 2 d X , (21) where
M 0 2 = m 0 2 + ξ R + 3 c 0 ϕ 2 , M 1 2 = m 0 2 + ξ R + c 0 ϕ 2 . (22) M_0^2=m_0^2+\xi\mathcal{R}+3c_0\phi^2,\qquad M_1^2=m_0^2+\xi\mathcal{R}+c_0\phi^2. \tag*{(22)} M 0 2 = m 0 2 + ξ R + 3 c 0 ϕ 2 , M 1 2 = m 0 2 + ξ R + c 0 ϕ 2 . (22) These are the radial and transverse Hessian masses. At a general background, the transverse modes need not be massless Goldstone modes.
For a constant configuration, the Legendre transform gives the effective potential in the form
V ( ϕ ) = L g + m 0 2 + ξ R 2 ϕ 2 + c 0 4 ϕ 4 − 1 V d log ∫ [ ∏ j D ψ j ] e − S 2 − 1 V d log ∫ [ ∏ j D ψ j ] e − S 2 − S 3 − S 4 ∫ [ ∏ j D ψ j ] e − S 2 ∣ 1 P I . (23) V(\phi)=\mathcal{L}_g+\frac{m_0^2+\xi\mathcal{R}}{2}\phi^2+\frac{c_0}{4}\phi^4-\frac{1}{V_d}\log\int\left[\prod_jD\psi_j\right]e^{-S_2}
-\left.\frac{1}{V_d}\log\frac{\int[\prod_jD\psi_j]e^{-S_2-S_3-S_4}}{\int[\prod_jD\psi_j]e^{-S_2}}\right|_{\mathrm{1PI}}. \tag*{(23)} V ( ϕ ) = L g + 2 m 0 2 + ξ R ϕ 2 + 4 c 0 ϕ 4 − V d 1 log ∫ [ j ∏ D ψ j ] e − S 2 − V d 1 log ∫ [ ∏ j D ψ j ] e − S 2 ∫ [ ∏ j D ψ j ] e − S 2 − S 3 − S 4 1PI . (23) Here V d V_d V d S d S^d S d V d V_d V d V V V S 1 S_1 S 1 S 4 S_4 S 4
The one-loop bare potential
The tree-level plus one-loop potential can be written as
V 1 − l o o p ( ϕ ) = L g + m 0 2 + ξ R 2 ϕ 2 + c 0 4 ϕ 4 + V 0 ( ϕ ) + ( N − 1 ) V 1 ( ϕ ) . (24) V_{\mathrm{1-loop}}(\phi)=\mathcal{L}_g+\frac{m_0^2+\xi\mathcal{R}}{2}\phi^2+\frac{c_0}{4}\phi^4+V_0(\phi)+(N-1)V_1(\phi). \tag*{(24)} V 1 − loop ( ϕ ) = L g + 2 m 0 2 + ξ R ϕ 2 + 4 c 0 ϕ 4 + V 0 ( ϕ ) + ( N − 1 ) V 1 ( ϕ ) . (24) For a mode of mass M M M
V ( M ) = 1 2 V d log det ( − Δ + M 2 ) , ∂ V ∂ M 2 = 1 2 V d Tr ( − Δ + M 2 ) − 1 = 1 2 T d ( q ) . (25) V(M)=\frac{1}{2V_d}\log\det(-\Delta+M^2),\qquad\frac{\partial V}{\partial M^2}=\frac{1}{2V_d}\operatorname{Tr}(-\Delta+M^2)^{-1}=\frac{1}{2}T_d(q). \tag*{(25)} V ( M ) = 2 V d 1 log det ( − Δ + M 2 ) , ∂ M 2 ∂ V = 2 V d 1 Tr ( − Δ + M 2 ) − 1 = 2 1 T d ( q ) . (25) The unregularised determinant is divergent. Its mass derivative, however, converges for Re ( d ) < 2 \operatorname{Re}(d)<2 Re ( d ) < 2 [10 ] determines the mass dependence of V j V_j V j d d d u = 1 u=1 u = 1 (12) gives
T d ( q ) = Γ ( 1 − d / 2 ) ( 4 π ) d / 2 Γ ( h + q ) Γ ( h − q ) Γ ( 1 / 2 + q ) Γ ( 1 / 2 − q ) . (26) T_d(q)=\frac{\Gamma(1-d/2)}{(4\pi)^{d/2}}\frac{\Gamma(h+q)\Gamma(h-q)}{\Gamma(1/2+q)\Gamma(1/2-q)}. \tag*{(26)} T d ( q ) = ( 4 π ) d /2 Γ ( 1 − d /2 ) Γ ( 1/2 + q ) Γ ( 1/2 − q ) Γ ( h + q ) Γ ( h − q ) . (26) We write T q = T d ( q ) T_q=T_d(q) T q = T d ( q ) ν \nu ν T d ( i ν ) T_d(i\nu) T d ( i ν ) d d d r 2 − d r^{2-d} r 2 − d r = 0 r=0 r = 0 T d T_d T d Γ ( 1 − d / 2 ) \Gamma(1-d/2) Γ ( 1 − d /2 ) ν 2 + 1 / 4 = M 2 − R / 6 \nu^2+1/4=M^2-\mathcal{R}/6 ν 2 + 1/4 = M 2 − R /6 d = 4 d=4 d = 4 d M 2 = 2 ν d ν dM^2=2\nu\,d\nu d M 2 = 2 ν d ν d d d
V j ( φ ) = Γ ( 1 / 2 − h ) ( 4 π ) d / 2 ∫ ν j ( φ ) Γ ( h + i x ) Γ ( h − i x ) Γ ( 1 / 2 + i x ) Γ ( 1 / 2 − i x ) x d x , (27) V_j(\varphi)=\frac{\Gamma(1/2-h)}{(4\pi)^{d/2}}\int^{\nu_j(\varphi)}\frac{\Gamma(h+ix)\Gamma(h-ix)}{\Gamma(1/2+ix)\Gamma(1/2-ix)}x\,dx, \tag*{(27)} V j ( φ ) = ( 4 π ) d /2 Γ ( 1/2 − h ) ∫ ν j ( φ ) Γ ( 1/2 + i x ) Γ ( 1/2 − i x ) Γ ( h + i x ) Γ ( h − i x ) x d x , (27) where ν j ( φ ) = M j 2 ( φ ) − h 2 \nu_j(\varphi)=\sqrt{M_j^2(\varphi)-h^2} ν j ( φ ) = M j 2 ( φ ) − h 2 d = 4 − 2 ε d=4-2\varepsilon d = 4 − 2 ε ν \nu ν
T 4 − 2 ε ( i ν ) = − ( 1 + 4 ν 2 ) ( ε − 1 + 1 − γ E + log 4 π ) 64 π 2 + ( ν 2 + 1 / 4 ) [ ψ ( − 1 / 2 − i ν ) + ψ ( − 1 / 2 + i ν ) ] 16 π 2 + O ( ε ) . (28) T_{4-2\varepsilon}(i\nu)=-\frac{(1+4\nu^2)(\varepsilon^{-1}+1-\gamma_E+\log4\pi)}{64\pi^2}+\frac{(\nu^2+1/4)[\psi(-1/2-i\nu)+\psi(-1/2+i\nu)]}{16\pi^2}+O(\varepsilon). \tag*{(28)} T 4 − 2 ε ( i ν ) = − 64 π 2 ( 1 + 4 ν 2 ) ( ε − 1 + 1 − γ E + log 4 π ) + 16 π 2 ( ν 2 + 1/4 ) [ ψ ( − 1/2 − i ν ) + ψ ( − 1/2 + i ν )] + O ( ε ) . (28) Integrating yields
V j ( 4 ) ( φ ) = [ − ( x 4 + x 2 / 2 ) ( ε − 1 − γ E + 1 + log 4 π ) + ( x 2 + 1 / 4 ) 2 64 π 2 + ( x 2 + 1 / 4 ) 2 64 π 2 [ ψ ( − 1 2 + i x ) + ψ ( − 1 2 − i x ) ] + B ( x ) 16 π 2 ] x = M j 2 ( φ ) − 9 / 4 + O ( ε ) , (29) V_j^{(4)}(\varphi)=\left[-\frac{(x^4+x^2/2)(\varepsilon^{-1}-\gamma_E+1+\log4\pi)+(x^2+1/4)^2}{64\pi^2}+\frac{(x^2+1/4)^2}{64\pi^2}\left[\psi\left(-\frac{1}{2}+ix\right)+\psi\left(-\frac{1}{2}-ix\right)\right]+\frac{B(x)}{16\pi^2}\right]_{x=\sqrt{M_j^2(\varphi)-9/4}}+O(\varepsilon), \tag*{(29)} V j ( 4 ) ( φ ) = [ − 64 π 2 ( x 4 + x 2 /2 ) ( ε − 1 − γ E + 1 + log 4 π ) + ( x 2 + 1/4 ) 2 + 64 π 2 ( x 2 + 1/4 ) 2 [ ψ ( − 2 1 + i x ) + ψ ( − 2 1 − i x ) ] + 16 π 2 B ( x ) ] x = M j 2 ( φ ) − 9/4 + O ( ε ) , (29) where
B ( x ) = ( x 2 + 1 / 4 ) 2 4 − ∫ 1 / 2 x i 4 ( y 2 + 1 / 4 ) 2 [ ψ ′ ( − 1 2 + i y ) − ψ ′ ( − 1 2 − i y ) ] d y = − 1 2 ∫ 1 / 4 x 2 + 1 / 4 d y y 2 [ ∑ n = 0 ∞ 2 n − 1 2 ( n ( n − 1 ) + y ) 2 − 1 y ] . (30) \begin{aligned}
B(x)=\frac{(x^2+1/4)^2}{4}-\int_{1/2}^{x}\frac{i}{4}(y^2+1/4)^2\left[\psi'\left(-\frac{1}{2}+iy\right)-\psi'\left(-\frac{1}{2}-iy\right)\right]\,dy \\
&=-\frac{1}{2}\int_{1/4}^{x^2+1/4}dy\,y^2\left[\sum_{n=0}^{\infty}\frac{2n-1}{2(n(n-1)+y)^2}-\frac{1}{y}\right]. \tag*{(30)}
\end{aligned} B ( x ) = 4 ( x 2 + 1/4 ) 2 − ∫ 1/2 x 4 i ( y 2 + 1/4 ) 2 [ ψ ′ ( − 2 1 + i y ) − ψ ′ ( − 2 1 − i y ) ] d y = − 2 1 ∫ 1/4 x 2 + 1/4 d y y 2 [ n = 0 ∑ ∞ 2 ( n ( n − 1 ) + y ) 2 2 n − 1 − y 1 ] . (30) An alternative expression in terms of antiderivatives of ψ \psi ψ [32 ]
B ( x ) = − i x ( 4 x 2 + 1 ) 2 [ log Γ ( − 1 2 − i x ) − log Γ ( − 1 2 + i x ) ] + 12 [ ψ ( − 4 ) ( − 1 2 − i x ) + ψ ( − 4 ) ( − 1 2 + i x ) ] + 12 i x [ ψ ( − 3 ) ( − 1 2 − i x ) − ψ ( − 3 ) ( − 1 2 + i x ) ] − 12 x 2 + 1 2 [ ψ ( − 2 ) ( − 1 2 − i x ) + ψ ( − 2 ) ( − 1 2 + i x ) ] + ( 4 x 2 + 1 ) 2 32 [ ψ ( − 1 2 − i x ) + ψ ( − 1 2 + i x ) ] . (31) B(x)=-\frac{ix(4x^2+1)}{2}\left[\log\Gamma\left(-\frac{1}{2}-ix\right)-\log\Gamma\left(-\frac{1}{2}+ix\right)\right]
+12\left[\psi^{(-4)}\left(-\frac{1}{2}-ix\right)+\psi^{(-4)}\left(-\frac{1}{2}+ix\right)\right]
+12ix\left[\psi^{(-3)}\left(-\frac{1}{2}-ix\right)-\psi^{(-3)}\left(-\frac{1}{2}+ix\right)\right]
-\frac{12x^2+1}{2}\left[\psi^{(-2)}\left(-\frac{1}{2}-ix\right)+\psi^{(-2)}\left(-\frac{1}{2}+ix\right)\right]
+\frac{(4x^2+1)^2}{32}\left[\psi\left(-\frac{1}{2}-ix\right)+\psi\left(-\frac{1}{2}+ix\right)\right]. \tag*{(31)} B ( x ) = − 2 i x ( 4 x 2 + 1 ) [ log Γ ( − 2 1 − i x ) − log Γ ( − 2 1 + i x ) ] + 12 [ ψ ( − 4 ) ( − 2 1 − i x ) + ψ ( − 4 ) ( − 2 1 + i x ) ] + 12 i x [ ψ ( − 3 ) ( − 2 1 − i x ) − ψ ( − 3 ) ( − 2 1 + i x ) ] − 2 12 x 2 + 1 [ ψ ( − 2 ) ( − 2 1 − i x ) + ψ ( − 2 ) ( − 2 1 + i x ) ] + 32 ( 4 x 2 + 1 ) 2 [ ψ ( − 2 1 − i x ) + ψ ( − 2 1 + i x ) ] . (31) The antiderivative constants are fixed consistently with Eq. (30) .
Remark 3.1. The one-loop calculation was considered in Ref. [10 ] at fixed H H H R \mathcal{R} R H 2 = R / [ d ( d − 1 ) ] H^2=\mathcal{R}/[d(d-1)] H 2 = R / [ d ( d − 1 )] ε \varepsilon ε
For the two-loop potential we also require the watermelon integral. Although a closed-form representation was obtained in Ref. [14 ] , extracting its four-dimensional finite part was difficult. The finite-difference method of Ref. [26 ] provides the representation used in the next section.
The watermelon from finite-difference equations
On the unit Euclidean sphere, define
I 3 ( q ) = ∫ S d G q 1 ( X , Y ) G q 2 ( X , Y ) G q 3 ( X , Y ) d Y , q = ( q 1 , q 2 , q 3 ) . (32) \mathcal{I}_3(\mathbf{q})=\int_{S^d}G_{q_1}(X,Y)G_{q_2}(X,Y)G_{q_3}(X,Y)\,\mathrm{d}Y,\qquad\mathbf{q}=(q_1,q_2,q_3). \tag*{(32)} I 3 ( q ) = ∫ S d G q 1 ( X , Y ) G q 2 ( X , Y ) G q 3 ( X , Y ) d Y , q = ( q 1 , q 2 , q 3 ) . (32) where G q G_q G q (12) , with u = X ⋅ E Y u=X\cdot_EY u = X ⋅ E Y h = ( d − 1 ) / 2 h=(d-1)/2 h = ( d − 1 ) /2
a = h − 1 , μ q = h 2 − q 2 , κ q = a 2 − q 2 . (33) a=h-1,\qquad\mu_q=h^2-q^2,\qquad\kappa_q=a^2-q^2. \tag*{(33)} a = h − 1 , μ q = h 2 − q 2 , κ q = a 2 − q 2 . (33) Our first aim is the fixed-label Laurent expansion
( 16 π 2 ) I 3 ( q ) = C − 2 ( q ) ε 2 + C − 1 ( q ) ε + C 0 ( q ) + O ( ε ) , d = 4 − 2 ε . (34) (16\pi^2)\mathcal{I}_3(\mathbf{q})=\frac{C_{-2}(\mathbf{q})}{\varepsilon^2}+\frac{C_{-1}(\mathbf{q})}{\varepsilon}+C_0(\mathbf{q})+O(\varepsilon),\qquad d=4-2\varepsilon. \tag*{(34)} ( 16 π 2 ) I 3 ( q ) = ε 2 C − 2 ( q ) + ε C − 1 ( q ) + C 0 ( q ) + O ( ε ) , d = 4 − 2 ε . (34) Fixed labels do not imply fixed masses, since μ q = h 2 − q 2 \mu_q=h^2-q^2 μ q = h 2 − q 2 d d d T q = T d ( q ) T_q=T_d(q) T q = T d ( q ) (26) , obeys
T q + 1 = h + q q − a T q . (35) T_{q+1}=\frac{h+q}{q-a}T_q. \tag*{(35)} T q + 1 = q − a h + q T q . (35) We denote the canonical basis of R 3 \mathbb{R}^3 R 3 { e 1 , e 2 , e 3 } \{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\} { e 1 , e 2 , e 3 }
The recurrence and its solution
Proposition 4.1 (Mass-shift recurrence). For { i , j , k } = { 1 , 2 , 3 } \{i,j,k\}=\{1,2,3\} { i , j , k } = { 1 , 2 , 3 }
P k − I 3 ( q + e k ) − P k + I 3 ( q − e k ) = S k (36) P_k^{-}\mathcal{I}_3(q+e_k)-P_k^{+}\mathcal{I}_3(q-e_k)=S_k \tag*{(36)} P k − I 3 ( q + e k ) − P k + I 3 ( q − e k ) = S k (36) where
P k ± = ∏ ϵ 1 , ϵ 2 = ± 1 ( a ± q k + ϵ 1 q i + ϵ 2 q j ) = λ − 2 a ( a ± q k ) ∂ λ ∂ κ k (37) P_k^{\pm}=\prod_{\epsilon_1,\epsilon_2=\pm1}(a\pm q_k+\epsilon_1q_i+\epsilon_2q_j)=\lambda-2a(a\pm q_k)\frac{\partial\lambda}{\partial\kappa_k} \tag*{(37)} P k ± = ϵ 1 , ϵ 2 = ± 1 ∏ ( a ± q k + ϵ 1 q i + ϵ 2 q j ) = λ − 2 a ( a ± q k ) ∂ κ k ∂ λ (37) with λ = κ 1 2 + κ 2 2 + κ 3 2 − 2 ( κ 1 κ 2 + κ 2 κ 3 + κ 3 κ 1 ) \lambda=\kappa_1^2+\kappa_2^2+\kappa_3^2-2(\kappa_1\kappa_2+\kappa_2\kappa_3+\kappa_3\kappa_1) λ = κ 1 2 + κ 2 2 + κ 3 2 − 2 ( κ 1 κ 2 + κ 2 κ 3 + κ 3 κ 1 ) κ i = κ q i \kappa_i=\kappa_{q_i} κ i = κ q i
S k = 2 q k ( 2 h − 1 ) [ 2 T i T j − ( T i + T j ) T k + κ i − κ j κ k ( T i − T j ) T k ] . (38) S_k=2q_k(2h-1)\left[2T_iT_j-(T_i+T_j)T_k+\frac{\kappa_i-\kappa_j}{\kappa_k}(T_i-T_j)T_k\right]. \tag*{(38)} S k = 2 q k ( 2 h − 1 ) [ 2 T i T j − ( T i + T j ) T k + κ k κ i − κ j ( T i − T j ) T k ] . (38) Here T i = T q i T_i=T_{q_i} T i = T q i
Proof. It suffices to take ( i , j , k ) = ( 1 , 2 , 3 ) (i,j,k)=(1,2,3) ( i , j , k ) = ( 1 , 2 , 3 ) u u u
Step 1: reduction to one radial integral. The integrand of I 3 \mathcal{I}_3 I 3 Y Y Y u = X ⋅ E Y u=X\mathbin{\cdot_E}Y u = X ⋅ E Y
I 3 ( q ) = ω d ∫ − 1 1 w d ( u ) G q 1 G q 2 G q 3 d u , ω d = 2 π d / 2 Γ ( d / 2 ) , w d ( u ) = ( 1 − u 2 ) d / 2 − 1 (39) \mathcal{I}_3(q)=\omega_d\int_{-1}^{1}w_d(u)G_{q_1}G_{q_2}G_{q_3}\,\mathrm{d}u,\qquad\omega_d=\frac{2\pi^{d/2}}{\Gamma(d/2)},\qquad w_d(u)=(1-u^2)^{d/2-1} \tag*{(39)} I 3 ( q ) = ω d ∫ − 1 1 w d ( u ) G q 1 G q 2 G q 3 d u , ω d = Γ ( d /2 ) 2 π d /2 , w d ( u ) = ( 1 − u 2 ) d /2 − 1 (39) with ω d \omega_d ω d S d − 1 S^{d-1} S d − 1
( q − a ) G q + 1 = ( q + h ) u G q − ( 1 − u 2 ) G q ′ , (q-a)G_{q+1}=(q+h)uG_q-(1-u^2)G'_q, ( q − a ) G q + 1 = ( q + h ) u G q − ( 1 − u 2 ) G q ′ , ( q + a ) G q − 1 = ( q − h ) u G q + ( 1 − u 2 ) G q ′ (40) (q+a)G_{q-1}=(q-h)uG_q+(1-u^2)G'_q \tag*{(40)} ( q + a ) G q − 1 = ( q − h ) u G q + ( 1 − u 2 ) G q ′ (40) which follow from the hypergeometric contiguous relations and the normalisation of the coincidence behaviour, express G q 3 ± 1 G_{q_3\pm1} G q 3 ± 1 G q 3 G_{q_3} G q 3 G q 3 ′ G'_{q_3} G q 3 ′ (39) , gives
P 3 − I 3 ( q + e 3 ) − P 3 + I 3 ( q − e 3 ) = ω d ∫ − 1 1 w d ( u ) [ A u G q 1 G q 2 G q 3 + B ( 1 − u 2 ) G q 1 G q 2 G q 3 ′ ] d u (41) P_3^{-}\mathcal{I}_3(q+e_3)-P_3^{+}\mathcal{I}_3(q-e_3)=\omega_d\int_{-1}^{1}w_d(u)\left[A\,uG_{q_1}G_{q_2}G_{q_3}+B(1-u^2)G_{q_1}G_{q_2}G'_{q_3}\right]\mathrm{d}u \tag*{(41)} P 3 − I 3 ( q + e 3 ) − P 3 + I 3 ( q − e 3 ) = ω d ∫ − 1 1 w d ( u ) [ A u G q 1 G q 2 G q 3 + B ( 1 − u 2 ) G q 1 G q 2 G q 3 ′ ] d u (41) where
A = ( q 3 + h ) P 3 − q 3 − a − ( q 3 − h ) P 3 + q 3 + a , B = − P 3 − q 3 − a − P 3 + q 3 + a (42) A=\frac{(q_3+h)P_3^{-}}{q_3-a}-\frac{(q_3-h)P_3^{+}}{q_3+a},\qquad B=-\frac{P_3^{-}}{q_3-a}-\frac{P_3^{+}}{q_3+a} \tag*{(42)} A = q 3 − a ( q 3 + h ) P 3 − − q 3 + a ( q 3 − h ) P 3 + , B = − q 3 − a P 3 − − q 3 + a P 3 + (42) Step 2: the integrand is a total derivative. Consider the current
J ( u ) = w d ( u ) ( 1 − u 2 ) [ b 0 G q 1 G q 2 G q 3 + u ∑ i b i G q i ′ ∏ j ≠ i G q j + ( 1 − u 2 ) ∑ i < j c i j G q i ′ G q j ′ G q k ] (43) J(u)=w_d(u)(1-u^2)\left[b_0G_{q_1}G_{q_2}G_{q_3}+u\sum_i b_iG'_{q_i}\prod_{j\ne i}G_{q_j}+(1-u^2)\sum_{i<j}c_{ij}G'_{q_i}G'_{q_j}G_{q_k}\right] \tag*{(43)} J ( u ) = w d ( u ) ( 1 − u 2 ) b 0 G q 1 G q 2 G q 3 + u i ∑ b i G q i ′ j = i ∏ G q j + ( 1 − u 2 ) i < j ∑ c ij G q i ′ G q j ′ G q k (43) with constant coefficients b 0 b_0 b 0 b i b_i b i c i j = c j i c_{ij}=c_{ji} c ij = c j i k k k J J J
( 1 − u 2 ) G q ′ ′ − d u G q ′ − μ q G q = 0. (44) (1-u^2)G_q^{\prime\prime}-duG_q^\prime-\mu_qG_q=0. \tag*{(44)} ( 1 − u 2 ) G q ′′ − d u G q ′ − μ q G q = 0. (44) Then J ′ J^\prime J ′ w d w_d w d u G q 1 G q 2 G q 3 uG_{q_1}G_{q_2}G_{q_3} u G q 1 G q 2 G q 3 (41) if and only if
− d b 0 + ∑ i μ q i b i = A , b 0 + b i + ∑ j ≠ i μ q j c i j = B δ i 3 , -d b_0+\sum_i\mu_{q_i}b_i=A,\qquad b_0+b_i+\sum_{j\ne i}\mu_{q_j}c_{ij}=B\delta_{i3}, − d b 0 + i ∑ μ q i b i = A , b 0 + b i + j = i ∑ μ q j c ij = B δ i 3 , ( d − 2 ) c i j + b i + b j = 0 , ∑ i < j c i j = 0. (45) (d-2)c_{ij}+b_i+b_j=0,\qquad\sum_{i<j}c_{ij}=0. \tag*{(45)} ( d − 2 ) c ij + b i + b j = 0 , i < j ∑ c ij = 0. (45) The first two conditions match the terms without derivatives and with one derivative; the last two cancel the terms with two and three derivatives. The system is overdetermined. With P 3 − P_3^- P 3 − (37) , one verifies by direct substitution that it is solved by
b 0 = 2 q 3 ( κ 3 − κ 1 − κ 2 ) , b_0=2q_3(\kappa_3-\kappa_1-\kappa_2), b 0 = 2 q 3 ( κ 3 − κ 1 − κ 2 ) , b 1 = 2 q 3 ( d − 2 ) ( − 1 − κ 1 − κ 2 κ 3 ) , b 2 = 2 q 3 ( d − 2 ) ( − 1 + κ 1 − κ 2 κ 3 ) , b_1=2q_3(d-2)\left(-1-\frac{\kappa_1-\kappa_2}{\kappa_3}\right),\qquad
b_2=2q_3(d-2)\left(-1+\frac{\kappa_1-\kappa_2}{\kappa_3}\right), b 1 = 2 q 3 ( d − 2 ) ( − 1 − κ 3 κ 1 − κ 2 ) , b 2 = 2 q 3 ( d − 2 ) ( − 1 + κ 3 κ 1 − κ 2 ) , b 3 = 4 q 3 ( d − 2 ) , c i j = b k d − 2 ( { i , j , k } = { 1 , 2 , 3 } ) . (46) b_3=4q_3(d-2),\qquad c_{ij}=\frac{b_k}{d-2}\quad(\{i,j,k\}=\{1,2,3\}). \tag*{(46)} b 3 = 4 q 3 ( d − 2 ) , c ij = d − 2 b k ({ i , j , k } = { 1 , 2 , 3 }) . (46) Hence
P 3 − I 3 ( q + e 3 ) − P 3 + I 3 ( q − e 3 ) = ω d [ J ( 1 − ) − J ( − 1 + ) ] . (47) P_3^-\mathcal{I}_3(\mathbf q+\mathbf e_3)-P_3^+\mathcal{I}_3(\mathbf q-\mathbf e_3)=\omega_d\left[J(1^-)-J(-1^+)\right]. \tag*{(47)} P 3 − I 3 ( q + e 3 ) − P 3 + I 3 ( q − e 3 ) = ω d [ J ( 1 − ) − J ( − 1 + ) ] . (47) Step 3: boundary values. At u = − 1 u=-1 u = − 1 (12) , and the prefactor w d ( 1 − u 2 ) w_d(1-u^2) w d ( 1 − u 2 ) J ( − 1 + ) = 0 J(-1^+)=0 J ( − 1 + ) = 0 u = 1 u=1 u = 1 0 < Re ( d ) < 2 0<\operatorname{Re}(d)<2 0 < Re ( d ) < 2 G q ( 1 ) = T q G_q(1)=T_q G q ( 1 ) = T q
lim u → 1 − ( 1 − u 2 ) w d ( u ) G q ′ ( u ) = 1 ω d . (48) \lim_{u\to1^-}(1-u^2)w_d(u)G_q^\prime(u)=\frac{1}{\omega_d}. \tag*{(48)} u → 1 − lim ( 1 − u 2 ) w d ( u ) G q ′ ( u ) = ω d 1 . (48) In J J J u = 1 u=1 u = 1 ( 1 − u ) 1 − d / 2 (1-u)^{1-d/2} ( 1 − u ) 1 − d /2 b i T j T k / ω d b_iT_jT_k/\omega_d b i T j T k / ω d
ω d J ( 1 − ) = b 1 T 2 T 3 + b 2 T 1 T 3 + b 3 T 1 T 2 = 2 q 3 ( d − 2 ) [ 2 T 1 T 2 − ( T 1 + T 2 ) T 3 + κ 1 − κ 2 κ 3 ( T 1 − T 2 ) T 3 ] , (49) \begin{aligned}
\omega_dJ(1^-)=b_1T_2T_3+b_2T_1T_3+b_3T_1T_2 \\
&=2q_3(d-2)\left[2T_1T_2-(T_1+T_2)T_3+\frac{\kappa_1-\kappa_2}{\kappa_3}(T_1-T_2)T_3\right], \tag*{(49)}
\end{aligned} ω d J ( 1 − ) = b 1 T 2 T 3 + b 2 T 1 T 3 + b 3 T 1 T 2 = 2 q 3 ( d − 2 ) [ 2 T 1 T 2 − ( T 1 + T 2 ) T 3 + κ 3 κ 1 − κ 2 ( T 1 − T 2 ) T 3 ] , (49) which is S 3 S_3 S 3 d − 2 = 2 h − 1 d-2=2h-1 d − 2 = 2 h − 1 d d d (36) in general. □
To solve the homogeneous equation, define
H k ( q ) = ∏ ϵ Γ ( B ϵ ) Γ ( A ϵ ) , B ϵ = h + q k + ϵ 1 q i + ϵ 2 q j 2 , A ϵ = B ϵ − a . (50) \mathcal{H}_k(\mathbf q)=\prod_{\epsilon}\frac{\Gamma(B_\epsilon)}{\Gamma(A_\epsilon)},\qquad
B_\epsilon=\frac{h+q_k+\epsilon_1q_i+\epsilon_2q_j}{2},\qquad
A_\epsilon=B_\epsilon-a. \tag*{(50)} H k ( q ) = ϵ ∏ Γ ( A ϵ ) Γ ( B ϵ ) , B ϵ = 2 h + q k + ϵ 1 q i + ϵ 2 q j , A ϵ = B ϵ − a . (50) The product runs over four sign choices. Since Γ ( B + 1 / 2 ) / Γ ( B − 1 / 2 ) = B − 1 / 2 \Gamma(B + 1/2)/\Gamma(B - 1/2) = B - 1/2 Γ ( B + 1/2 ) /Γ ( B − 1/2 ) = B − 1/2
H k ( q + e k ) H k ( q − e k ) = ∏ ϵ B ϵ − 1 / 2 A ϵ − 1 / 2 = P k + P k − . (51) \frac{\mathcal{H}_k(q + e_k)}{\mathcal{H}_k(q - e_k)} = \prod_{\epsilon} \frac{B_\epsilon- 1/2}{A_\epsilon- 1/2} = \frac{P_k^+}{P_k^-}. \tag*{(51)} H k ( q − e k ) H k ( q + e k ) = ϵ ∏ A ϵ − 1/2 B ϵ − 1/2 = P k − P k + . (51) Thus the general homogeneous solution in channel k k k H k p \mathcal{H}_k p H k p p p p q k q_k q k q n = q + 2 n e k q_n = q + 2n e_k q n = q + 2 n e k q ^ n = q + ( 2 n + 1 ) e k \widehat{q}_n = q + (2n + 1)e_k q n = q + ( 2 n + 1 ) e k H k \mathcal{H}_k H k
I 3 ( q n + 1 ) H k ( q n + 1 ) − I 3 ( q n ) H k ( q n ) = S k ( q ^ n ) P k + ( q ^ n ) H k ( q n ) . (52) \frac{\mathcal{I}_3(q_{n+1})}{\mathcal{H}_k(q_{n+1})} - \frac{\mathcal{I}_3(q_n)}{\mathcal{H}_k(q_n)} = \frac{S_k(\widehat{q}_n)}{P_k^+(\widehat{q}_n)\mathcal{H}_k(q_n)}. \tag*{(52)} H k ( q n + 1 ) I 3 ( q n + 1 ) − H k ( q n ) I 3 ( q n ) = P k + ( q n ) H k ( q n ) S k ( q n ) . (52) Telescoping yields
I 3 ( q ) = H k ( q ) I 3 ( q N + 1 ) H k ( q N + 1 ) − ∑ n = 0 N H k ( q ) H k ( q + 2 n e k ) S k ( q + ( 2 n + 1 ) e k ) P k + ( q + ( 2 n + 1 ) e k ) . (53) \mathcal{I}_3(q) = \mathcal{H}_k(q)\frac{\mathcal{I}_3(q_{N+1})}{\mathcal{H}_k(q_{N+1})} - \sum_{n=0}^{N}\frac{\mathcal{H}_k(q)}{\mathcal{H}_k(q + 2n e_k)}\frac{S_k(q + (2n + 1)e_k)}{P_k^+(q + (2n + 1)e_k)}. \tag*{(53)} I 3 ( q ) = H k ( q ) H k ( q N + 1 ) I 3 ( q N + 1 ) − n = 0 ∑ N H k ( q + 2 n e k ) H k ( q ) P k + ( q + ( 2 n + 1 ) e k ) S k ( q + ( 2 n + 1 ) e k ) . (53) The infinite-lattice boundary value is essential. Fix k = 3 k = 3 k = 3 ζ = q 3 \zeta= q_3 ζ = q 3
b h = Γ ( h − 1 / 2 ) 2 Γ ( 3 / 2 − h ) Γ ( 2 − 2 h ) ( 4 π ) 2 h + 1 Γ ( h + 1 / 2 ) . (54) b_h = \frac{\Gamma(h - 1/2)^2\Gamma(3/2 - h)\Gamma(2 - 2h)}{(4\pi)^{2h+1}\Gamma(h + 1/2)}. \tag*{(54)} b h = ( 4 π ) 2 h + 1 Γ ( h + 1/2 ) Γ ( h − 1/2 ) 2 Γ ( 3/2 − h ) Γ ( 2 − 2 h ) . (54) This is the coefficient of the flat-space sunset with two massless lines and one line of mass m m m b h ( m 2 ) d − 3 b_h(m^2)^{d-3} b h ( m 2 ) d − 3
Lemma 4.2. For 2 < d < 3 2 < d < 3 2 < d < 3 generic ζ \zeta ζ
I 3 ( 1 / 2 , 1 / 2 , ζ ) = b h Γ ( h + ζ ) Γ ( h − ζ ) Γ ( 2 − h + ζ ) Γ ( 2 − h − ζ ) . (55) \mathcal{I}_3(1/2,1/2,\zeta) = b_h\frac{\Gamma(h + \zeta)\Gamma(h - \zeta)}{\Gamma(2 - h + \zeta)\Gamma(2 - h - \zeta)}. \tag*{(55)} I 3 ( 1/2 , 1/2 , ζ ) = b h Γ ( 2 − h + ζ ) Γ ( 2 − h − ζ ) Γ ( h + ζ ) Γ ( h − ζ ) . (55) Proof. For t = ( 1 − u ) / 2 t = (1 - u)/2 t = ( 1 − u ) /2
2 F 1 ( h + 1 / 2 , h − 1 / 2 ; d / 2 ; ( 1 + u ) / 2 ) = t 1 − d / 2 . (56) {}_2F_1(h + 1/2,h - 1/2;d/2;(1 + u)/2) = t^{1-d/2}. \tag*{(56)} 2 F 1 ( h + 1/2 , h − 1/2 ; d /2 ; ( 1 + u ) /2 ) = t 1 − d /2 . (56) The beta integral gives, initially for Re ( ρ ) > − 1 \operatorname{Re}(\rho) > -1 Re ( ρ ) > − 1
∫ S d t ρ G ζ ( u ) d Y = Γ ( d / 2 + ρ ) Γ ( 1 + ρ ) Γ ( d / 2 ) Γ ( h + ζ ) Γ ( h − ζ ) Γ ( ( d + 1 ) / 2 + ρ + ζ ) Γ ( ( d + 1 ) / 2 + ρ − ζ ) . (57) \int_{S^d} t^\rho G_\zeta(u)\,\mathrm{d}Y = \frac{\Gamma(d/2 + \rho)\Gamma(1 + \rho)}{\Gamma(d/2)}\frac{\Gamma(h + \zeta)\Gamma(h - \zeta)}{\Gamma((d + 1)/2 + \rho+ \zeta)\Gamma((d + 1)/2 + \rho- \zeta)}. \tag*{(57)} ∫ S d t ρ G ζ ( u ) d Y = Γ ( d /2 ) Γ ( d /2 + ρ ) Γ ( 1 + ρ ) Γ (( d + 1 ) /2 + ρ + ζ ) Γ (( d + 1 ) /2 + ρ − ζ ) Γ ( h + ζ ) Γ ( h − ζ ) . (57) Substitution of ρ = 2 − d \rho= 2 - d ρ = 2 − d □ \square □
Lemma 4.3. Let d d d 2 < d < 3 2 < d < 3 2 < d < 3 q 1 q_1 q 1 q 2 q_2 q 2 ζ \zeta ζ Im ( ζ ) ≠ 0 \operatorname{Im}(\zeta) \ne0 Im ( ζ ) = 0
I 3 ( q 1 , q 2 , ζ + 2 N ) − I 3 ( 1 / 2 , 1 / 2 , ζ + 2 N ) = O ( N d − 4 log N ) . (58) \mathcal{I}_3(q_1,q_2,\zeta+ 2N) - \mathcal{I}_3(1/2,1/2,\zeta+ 2N) = O(N^{d-4}\log N). \tag*{(58)} I 3 ( q 1 , q 2 , ζ + 2 N ) − I 3 ( 1/2 , 1/2 , ζ + 2 N ) = O ( N d − 4 log N ) . (58) Proof. With c d = Γ ( d / 2 − 1 ) / ( 4 π ) d / 2 c_d = \Gamma(d/2 - 1)/(4\pi)^{d/2} c d = Γ ( d /2 − 1 ) / ( 4 π ) d /2
G q ( 1 − 2 t ) = c d t 1 − d / 2 2 F 1 ( 1 / 2 − q , 1 / 2 + q ; 2 − d / 2 ; t ) + T q 2 F 1 ( h + q , h − q ; d / 2 ; t ) . (59) G_q(1 - 2t) = c_d t^{1-d/2}{}_2F_1(1/2 - q,1/2 + q;2 - d/2;t) + T_q{}_2F_1(h + q,h - q;d/2;t). \tag*{(59)} G q ( 1 − 2 t ) = c d t 1 − d /2 2 F 1 ( 1/2 − q , 1/2 + q ; 2 − d /2 ; t ) + T q 2 F 1 ( h + q , h − q ; d /2 ; t ) . (59) Consequently,
f ( t ) = G q 1 ( 1 − 2 t ) G q 2 ( 1 − 2 t ) − c d 2 t 2 − d = c d ( T q 1 + T q 2 ) t 1 − d / 2 + t 3 − d F 1 ( t ) + t 2 − d / 2 F 2 ( t ) + A ( t ) . (60) \begin{aligned}
f(t) &= G_{q_1}(1 - 2t)G_{q_2}(1 - 2t) - c_d^2t^{2-d}\\
&= c_d(T_{q_1} + T_{q_2})t^{1-d/2} + t^{3-d}F_1(t) + t^{2-d/2}F_2(t) + A(t). \tag*{(60)}
\end{aligned} f ( t ) = G q 1 ( 1 − 2 t ) G q 2 ( 1 − 2 t ) − c d 2 t 2 − d = c d ( T q 1 + T q 2 ) t 1 − d /2 + t 3 − d F 1 ( t ) + t 2 − d /2 F 2 ( t ) + A ( t ) . (60) where F 1 F_1 F 1 F 2 F_2 F 2 A A A f f f d > 0 d>0 d > 0 [33 , 34 ] . The normalised zonal polynomials are
P ℓ ( t ) = ℓ ! ( d / 2 ) ℓ P ℓ ( d / 2 − 1 , d / 2 − 1 ) ( 1 − 2 t ) = 2 F 1 ( − ℓ , ℓ + d − 1 ; d / 2 ; t ) , P ℓ ( 0 ) = 1. (61) P_\ell(t)=\frac{\ell!}{(d/2)_\ell}P_\ell^{(d/2-1,d/2-1)}(1-2t)={}_2F_1(-\ell,\ell+d-1;d/2;t),\qquad P_\ell(0)=1. \tag*{(61)} P ℓ ( t ) = ( d /2 ) ℓ ℓ ! P ℓ ( d /2 − 1 , d /2 − 1 ) ( 1 − 2 t ) = 2 F 1 ( − ℓ , ℓ + d − 1 ; d /2 ; t ) , P ℓ ( 0 ) = 1. (61) Their eigenvalues, multiplicities and sphere volume are
λ ℓ = ℓ ( ℓ + d − 1 ) , D ℓ = ( 2 ℓ + d − 1 ) Γ ( ℓ + d − 1 ) Γ ( d ) Γ ( ℓ + 1 ) , V d = 2 π ( d + 1 ) / 2 Γ ( ( d + 1 ) / 2 ) . (62) \lambda_\ell=\ell(\ell+d-1),\qquad D_\ell=\frac{(2\ell+d-1)\Gamma(\ell+d-1)}{\Gamma(d)\Gamma(\ell+1)},\qquad V_d=\frac{2\pi^{(d+1)/2}}{\Gamma((d+1)/2)}. \tag*{(62)} λ ℓ = ℓ ( ℓ + d − 1 ) , D ℓ = Γ ( d ) Γ ( ℓ + 1 ) ( 2 ℓ + d − 1 ) Γ ( ℓ + d − 1 ) , V d = Γ (( d + 1 ) /2 ) 2 π ( d + 1 ) /2 . (62) They satisfy
⟨ P ℓ , P m ⟩ = V d D ℓ δ ℓ m , G q ( 1 − 2 t ) = 1 V d ∑ ℓ ≥ 0 D ℓ P ℓ ( ℓ + h ) 2 − q ~ 2 . (63) \langle P_\ell,P_m\rangle=\frac{V_d}{D_\ell}\delta_{\ell m},\qquad G_q(1-2t)=\frac{1}{V_d}\sum_{\ell\geq0}\frac{D_\ell P_\ell}{(\ell+h)^2-\tilde q^2}. \tag*{(63)} ⟨ P ℓ , P m ⟩ = D ℓ V d δ ℓ m , G q ( 1 − 2 t ) = V d 1 ℓ ≥ 0 ∑ ( ℓ + h ) 2 − q ~ 2 D ℓ P ℓ . (63) For g = ∑ ℓ g ℓ P ℓ g=\sum_\ell g_\ell P_\ell g = ∑ ℓ g ℓ P ℓ g ( t ) = t ρ g(t)=t^\rho g ( t ) = t ρ [35 ] give
g ℓ = D ℓ Γ ( d / 2 + ρ ) Γ ( d ) Γ ( d / 2 ) Γ ( d + ρ ) ( − ρ ) ℓ ( d + ρ ) ℓ = O ( ℓ − 2 ρ − 1 ) . (64) g_\ell=D_\ell\frac{\Gamma(d/2+\rho)\Gamma(d)}{\Gamma(d/2)\Gamma(d+\rho)}\frac{(-\rho)_\ell}{(d+\rho)_\ell}=O(\ell^{-2\rho-1}). \tag*{(64)} g ℓ = D ℓ Γ ( d /2 ) Γ ( d + ρ ) Γ ( d /2 + ρ ) Γ ( d ) ( d + ρ ) ℓ ( − ρ ) ℓ = O ( ℓ − 2 ρ − 1 ) . (64) Choose a smooth cutoff χ \chi χ
f ( t ) = [ c d ( T q 1 + T q 2 ) t 1 − d / 2 + ∑ j = 0 L F j t ρ j ] χ ( t ) + R M ( t ) . (65) f(t)=\left[c_d(T_{q_1}+T_{q_2})t^{1-d/2}+\sum_{j=0}^{L}F_jt^{\rho_j}\right]\chi(t)+R_M(t). \tag*{(65)} f ( t ) = [ c d ( T q 1 + T q 2 ) t 1 − d /2 + j = 0 ∑ L F j t ρ j ] χ ( t ) + R M ( t ) . (65) The sum after the displayed leading term contains the additional monomials. The terms are chosen according to the Frobenius expansion in such the way that near zero R M R_M R M O ( t ρ ∗ ) O(t^{\rho_*}) O ( t ρ ∗ ) ρ ∗ > M \rho_*>M ρ ∗ > M f f f
J = − 1 t α ( 1 − t ) α d d t [ t α + 1 ( 1 − t ) α + 1 d d t ] , α = d / 2 − 1. (66) J=-\frac{1}{t^\alpha(1-t)^\alpha}\frac{\mathrm{d}}{\mathrm{d}t}\left[t^{\alpha+1}(1-t)^{\alpha+1}\frac{\mathrm{d}}{\mathrm{d}t}\right],\qquad\alpha=d/2-1. \tag*{(66)} J = − t α ( 1 − t ) α 1 d t d [ t α + 1 ( 1 − t ) α + 1 d t d ] , α = d /2 − 1. (66) Its orthogonal eigenbasis is P ℓ P_\ell P ℓ J P ℓ = λ ℓ P ℓ JP_\ell=\lambda_\ell P_\ell J P ℓ = λ ℓ P ℓ J M R M J^M R_M J M R M M M M R M = ∑ m ≥ 0 r m P m R_M=\sum_{m\geq0}r_mP_m R M = ∑ m ≥ 0 r m P m ℓ ≥ 1 \ell\geq1 ℓ ≥ 1
⟨ R M , P ℓ ⟩ = λ ℓ − M ⟨ J M R M , P ℓ ⟩ . (67) \langle R_M,P_\ell\rangle=\lambda_\ell^{-M}\langle J^M R_M,P_\ell\rangle. \tag*{(67)} ⟨ R M , P ℓ ⟩ = λ ℓ − M ⟨ J M R M , P ℓ ⟩ . (67) so the Cauchy–Schwarz inequality gives
∣ r ℓ ∣ ≤ C λ ℓ − M ∥ P ℓ ∥ − 1 = O ( ℓ − 2 M + ( d − 1 ) / 2 ) . (68) |r_\ell|\leq C\lambda_\ell^{-M}\|P_\ell\|^{-1}=O(\ell^{-2M+(d-1)/2}). \tag*{(68)} ∣ r ℓ ∣ ≤ C λ ℓ − M ∥ P ℓ ∥ − 1 = O ( ℓ − 2 M + ( d − 1 ) /2 ) . (68) Here ∥ P ℓ ∥ 2 = V d / D ℓ = O ( ℓ − ( d − 1 ) ) \|P_\ell\|^2=V_d/D_\ell=O(\ell^{-(d-1)}) ∥ P ℓ ∥ 2 = V d / D ℓ = O ( ℓ − ( d − 1 ) ) ∣ P ℓ ( t ) ∣ ≤ 1 |P_\ell(t)|\leq1 ∣ P ℓ ( t ) ∣ ≤ 1 d > 1 d>1 d > 1 χ t ρ \chi t^\rho χ t ρ t ρ t^\rho t ρ (64) controls its leading coefficients. The dominant exponent is 1 − d / 2 1-d/2 1 − d /2 f ℓ = O ( ( 1 + ℓ ) − 3 ) f_\ell=O((1+\ell)^{-3}) f ℓ = O (( 1 + ℓ ) − 3 )
The resolvent pairing is therefore absolutely convergent for d < 3 d<3 d < 3
∫ S d f ( t ) G w ( 1 − 2 t ) d Y = ∑ ℓ = 0 ∞ f ℓ ( ℓ + h ) 2 − w 2 , w = ζ + 2 N . (69) \int_{S^d} f(t)G_w(1-2t)\,\mathrm{d}Y=\sum_{\ell=0}^{\infty}\frac{f_\ell}{(\ell+h)^2-w^2},\qquad w=\zeta+2N. \tag*{(69)} ∫ S d f ( t ) G w ( 1 − 2 t ) d Y = ℓ = 0 ∑ ∞ ( ℓ + h ) 2 − w 2 f ℓ , w = ζ + 2 N . (69) Split the sum into ℓ < N \ell<N ℓ < N N ≤ ℓ ≤ 4 N N\le\ell\le4N N ≤ ℓ ≤ 4 N ℓ > 4 N \ell>4N ℓ > 4 N ( ℓ + h − w ) ( ℓ + h + w ) (\ell+h-w)(\ell+h+w) ( ℓ + h − w ) ( ℓ + h + w ) N N N S < S_{<} S <
∣ S < ∣ ≤ C N 2 ∑ ℓ < N ( 1 + ℓ ) d − 3 = O ( N d − 4 ) . (70) |S_{<}|\le\frac{C}{N^2}\sum_{\ell<N}(1+\ell)^{d-3}=O(N^{d-4}). \tag*{(70)} ∣ S < ∣ ≤ N 2 C ℓ < N ∑ ( 1 + ℓ ) d − 3 = O ( N d − 4 ) . (70) In the last region the denominator has magnitude comparable to ℓ 2 \ell^2 ℓ 2
∣ S > ∣ ≤ C ∑ ℓ > 4 N ℓ d − 5 = O ( N d − 4 ) . (71) |S_{>}|\le C\sum_{\ell>4N}\ell^{d-5}=O(N^{d-4}). \tag*{(71)} ∣ S > ∣ ≤ C ℓ > 4 N ∑ ℓ d − 5 = O ( N d − 4 ) . (71) In the middle region ∣ ℓ + h + w ∣ ∼ N |\ell+h+w|\sim N ∣ ℓ + h + w ∣ ∼ N Im ζ ≠ 0 \operatorname{Im}\zeta\ne0 Im ζ = 0
∣ ℓ + h − w ∣ ≥ c ( 1 + ∣ ℓ − 2 N ∣ ) . (72) |\ell+h-w|\ge c(1+|\ell-2N|). \tag*{(72)} ∣ ℓ + h − w ∣ ≥ c ( 1 + ∣ ℓ − 2 N ∣ ) . (72) Therefore, the middle sum satisfies
∣ S 0 ∣ ≤ C N d − 4 ∑ ∣ k ∣ ≤ 2 N 1 1 + ∣ k ∣ = O ( N d − 4 log N ) . (73) |S_0|\le CN^{d-4}\sum_{|k|\le2N}\frac{1}{1+|k|}=O(N^{d-4}\log N). \tag*{(73)} ∣ S 0 ∣ ≤ C N d − 4 ∣ k ∣ ≤ 2 N ∑ 1 + ∣ k ∣ 1 = O ( N d − 4 log N ) . (73) Combining the three estimates proves Eq. (58) .
It remains to prove the following Lemma.
Lemma 4.4. For λ = ( d − 1 ) / 2 > 0 \lambda=(d-1)/2>0 λ = ( d − 1 ) /2 > 0 ∣ P ℓ ( t ) ∣ ≤ 1 |P_\ell(t)|\le1 ∣ P ℓ ( t ) ∣ ≤ 1 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1
Proof. Set
t = 1 − cos θ 2 . t=\frac{1-\cos\theta}{2}. t = 2 1 − cos θ . The Poisson representation of the Gegenbauer polynomials gives [36 ]
P ℓ ( t ) = Γ ( d 2 ) π Γ ( h ) ∫ 0 π ( cos θ + i sin θ cos φ ) ℓ ( sin φ ) 2 λ − 1 d φ . (74) P_\ell(t)=\frac{\Gamma\left(\frac{d}{2}\right)}{\sqrt{\pi}\Gamma(h)}\int_0^\pi(\cos\theta+i\sin\theta\cos\varphi)^\ell(\sin\varphi)^{2\lambda-1}\,\mathrm{d}\varphi. \tag*{(74)} P ℓ ( t ) = π Γ ( h ) Γ ( 2 d ) ∫ 0 π ( cos θ + i sin θ cos φ ) ℓ ( sin φ ) 2 λ − 1 d φ . (74) Since ∣ cos θ + i sin θ cos φ ∣ 2 = 1 − sin 2 θ sin 2 φ ≤ 1 |\cos\theta+i\sin\theta\cos\varphi|^2=1-\sin^2\theta\sin^2\varphi\le1 ∣ cos θ + i sin θ cos φ ∣ 2 = 1 − sin 2 θ sin 2 φ ≤ 1
Gamma-function asymptotics give H 3 ( q 1 , q 2 , ζ + 2 N ) ∼ N 4 a \mathcal{H}_3(q_1,q_2,\zeta+2N)\sim N^{4a} H 3 ( q 1 , q 2 , ζ + 2 N ) ∼ N 4 a 2 < d < 3 2<d<3 2 < d < 3
I 3 ( q 1 , q 2 , ζ + 2 N ) − I 3 ( 1 / 2 , 1 / 2 , ζ + 2 N ) H 3 ( q 1 , q 2 , ζ + 2 N ) = O ( N 2 − d log N ) ⟶ 0. (75) \frac{\mathcal{I}_3(q_1,q_2,\zeta+2N)-\mathcal{I}_3(1/2,1/2,\zeta+2N)}{\mathcal{H}_3(q_1,q_2,\zeta+2N)}=O(N^{2-d}\log N)\longrightarrow0. \tag*{(75)} H 3 ( q 1 , q 2 , ζ + 2 N ) I 3 ( q 1 , q 2 , ζ + 2 N ) − I 3 ( 1/2 , 1/2 , ζ + 2 N ) = O ( N 2 − d log N ) ⟶ 0. (75) The reflection formula and asymptotics [35 ] yield
I 3 ( 1 / 2 , 1 / 2 , ζ + 2 N ) ∼ b h s + ( ζ ) ( 2 N ) 4 a , s + ( ζ ) = − sin π ( h + ζ ) sin π ( h − ζ ) . (76) \mathcal{I}_3(1/2,1/2,\zeta+2N)\sim b_hs^+(\zeta)(2N)^{4a},\qquad s^+(\zeta)=-\frac{\sin\pi(h+\zeta)}{\sin\pi(h-\zeta)}. \tag*{(76)} I 3 ( 1/2 , 1/2 , ζ + 2 N ) ∼ b h s + ( ζ ) ( 2 N ) 4 a , s + ( ζ ) = − sin π ( h − ζ ) sin π ( h + ζ ) . (76) :::
The physical normalised boundary value is therefore
lim N → ∞ I 3 ( q 1 , q 2 , ζ + 2 N ) H 3 ( q 1 , q 2 , ζ + 2 N ) = 2 4 a b h s + ( ζ ) . (77) \lim_{N\to\infty}\frac{\mathcal{I}_3(q_1,q_2,\zeta+2N)}{\mathcal{H}_3(q_1,q_2,\zeta+2N)}=2^{4a}b_hs^+(\zeta). \tag*{(77)} N → ∞ lim H 3 ( q 1 , q 2 , ζ + 2 N ) I 3 ( q 1 , q 2 , ζ + 2 N ) = 2 4 a b h s + ( ζ ) . (77) It fixes the periodic homogeneous addition along each lattice and gives
B k ( q ) = 2 4 a b h s k + H k ( q ) , s k + = − sin π ( h + q k ) sin π ( h − q k ) . (78) B_k(\mathbf{q})=2^{4a}b_hs_k^+\mathcal{H}_k(\mathbf{q}),\qquad s_k^+=-\frac{\sin\pi(h+q_k)}{\sin\pi(h-q_k)}. \tag*{(78)} B k ( q ) = 2 4 a b h s k + H k ( q ) , s k + = − sin π ( h − q k ) sin π ( h + q k ) . (78) The factor s k + s_k^+ s k + μ q k → − ∞ \mu_{q_k}\to-\infty μ q k → − ∞
Theorem 4.5. The recurrence and physical normalised boundary value determine
I 3 = B k − Σ d , Σ d = ∑ n ≥ 0 H k ( q ) H k ( q + 2 n e k ) S k ( q + ( 2 n + 1 ) e k ) P k + ( q + ( 2 n + 1 ) e k ) . (79) \mathcal{I}_3=B_k-\Sigma_d,\qquad\Sigma_d=\sum_{n\geq0}\frac{\mathcal{H}_k(\mathbf{q})}{\mathcal{H}_k(\mathbf{q}+2n\mathbf{e}_k)}\frac{\mathcal{S}_k(\mathbf{q}+(2n+1)\mathbf{e}_k)}{\mathcal{P}_k^+(\mathbf{q}+(2n+1)\mathbf{e}_k)}. \tag*{(79)} I 3 = B k − Σ d , Σ d = n ≥ 0 ∑ H k ( q + 2 n e k ) H k ( q ) P k + ( q + ( 2 n + 1 ) e k ) S k ( q + ( 2 n + 1 ) e k ) . (79) where
B k = 2 4 a b h s k + ∏ ϵ Γ ( B ϵ ) Γ ( A ϵ ) . (80) B_k=2^{4a}b_hs_k^+\prod_{\epsilon}\frac{\Gamma(B_\epsilon)}{\Gamma(A_\epsilon)}. \tag*{(80)} B k = 2 4 a b h s k + ϵ ∏ Γ ( A ϵ ) Γ ( B ϵ ) . (80) Remark 4.6. The identity is first obtained for 2 < d < 3 2<d<3 2 < d < 3 n 3 − 2 d n^{3-2d} n 3 − 2 d n 1 − d n^{1-d} n 1 − d n − 1 − d n^{-1-d} n − 1 − d
Remark 4.7. (The boundary term is not optional). The recurrence alone permits additions H k p \mathcal{H}_kp H k p p p p (77) fix this ambiguity by comparison with conformally coupled spectators. Setting B k B_k B k
For the remainder of this section fix k = 3 k=3 k = 3
Theorem 4.8. The solution can be written as
I 3 ( q 1 , q 2 , q 3 ) = B k − Σ d , (81) \mathcal{I}_3(q_1,q_2,q_3)=B_k-\Sigma_d, \tag*{(81)} I 3 ( q 1 , q 2 , q 3 ) = B k − Σ d , (81) with
B k = 2 4 a b h [ − sin π ( h + q 3 ) sin π ( h − q 3 ) ] ∏ ϵ Γ ( B ϵ ) Γ ( B ϵ − a ) , (82) B_k=2^{4a}b_h\left[-\frac{\sin\pi(h+q_3)}{\sin\pi(h-q_3)}\right]\prod_{\epsilon}\frac{\Gamma(B_\epsilon)}{\Gamma(B_\epsilon-a)}, \tag*{(82)} B k = 2 4 a b h [ − sin π ( h − q 3 ) sin π ( h + q 3 ) ] ϵ ∏ Γ ( B ϵ − a ) Γ ( B ϵ ) , (82) Σ d = ( q 3 + 1 ) ( 2 h − 1 ) 8 ∏ ϵ B ϵ [ 2 T q 1 T q 2 F 1 − ( T q 1 + T q 2 ) T q 3 + 1 F 2 + κ q 1 − κ q 2 κ q 3 + 1 ( T q 1 − T q 2 ) T q 3 + 1 F 3 ] . (83) \Sigma_d=\frac{(q_3+1)(2h-1)}{8\prod_{\epsilon}B_\epsilon}\left[2T_{q_1}T_{q_2}\mathcal{F}_1-(T_{q_1}+T_{q_2})T_{q_3+1}\mathcal{F}_2+\frac{\kappa_{q_1}-\kappa_{q_2}}{\kappa_{q_3+1}}(T_{q_1}-T_{q_2})T_{q_3+1}\mathcal{F}_3\right]. \tag*{(83)} Σ d = 8 ∏ ϵ B ϵ ( q 3 + 1 ) ( 2 h − 1 ) [ 2 T q 1 T q 2 F 1 − ( T q 1 + T q 2 ) T q 3 + 1 F 2 + κ q 3 + 1 κ q 1 − κ q 2 ( T q 1 − T q 2 ) T q 3 + 1 F 3 ] . (83) where the tadpoles are defined in Eq. (26) , and
F 1 = 6 F 5 [ B ϵ − a , ( q 3 + 3 ) / 2 , 1 B ϵ + 1 , ( q 3 + 1 ) / 2 ; 1 ] , \mathcal{F}_1={}_6F_5\left[\begin{array}{c} B_\epsilon-a,\ (q_3+3)/2,\ 1\\ B_\epsilon+1,\ (q_3+1)/2 \end{array};1\right], F 1 = 6 F 5 [ B ϵ − a , ( q 3 + 3 ) /2 , 1 B ϵ + 1 , ( q 3 + 1 ) /2 ; 1 ] , F 2 = 8 F 7 [ B ϵ − a , ( q 3 + 3 ) / 2 , ( q 3 + 2 + a ) / 2 , ( q 3 + 3 + a ) / 2 , 1 B ϵ + 1 , ( q 3 + 1 ) / 2 , ( q 3 + 1 − a ) / 2 , ( q 3 + 2 − a ) / 2 ; 1 ] , (84) \mathcal{F}_2={}_8F_7\left[\begin{array}{c} B_\epsilon-a,\ (q_3+3)/2,\ (q_3+2+a)/2,\ (q_3+3+a)/2,\ 1\\ B_\epsilon+1,\ (q_3+1)/2,\ (q_3+1-a)/2,\ (q_3+2-a)/2 \end{array};1\right], \tag*{(84)} F 2 = 8 F 7 [ B ϵ − a , ( q 3 + 3 ) /2 , ( q 3 + 2 + a ) /2 , ( q 3 + 3 + a ) /2 , 1 B ϵ + 1 , ( q 3 + 1 ) /2 , ( q 3 + 1 − a ) /2 , ( q 3 + 2 − a ) /2 ; 1 ] , (84) F 3 = 8 F 7 [ B ϵ − a , ( q 3 + 3 ) / 2 , ( q 3 + 1 + a ) / 2 , ( q 3 + 2 + a ) / 2 , 1 B ϵ + 1 , ( q 3 + 1 ) / 2 , ( q 3 + 2 − a ) / 2 , ( q 3 + 3 − a ) / 2 ; 1 ] . \mathcal{F}_3={}_8F_7\left[\begin{array}{c} B_\epsilon-a,\ (q_3+3)/2,\ (q_3+1+a)/2,\ (q_3+2+a)/2,\ 1\\ B_\epsilon+1,\ (q_3+1)/2,\ (q_3+2-a)/2,\ (q_3+3-a)/2 \end{array};1\right]. F 3 = 8 F 7 [ B ϵ − a , ( q 3 + 3 ) /2 , ( q 3 + 1 + a ) /2 , ( q 3 + 2 + a ) /2 , 1 B ϵ + 1 , ( q 3 + 1 ) /2 , ( q 3 + 2 − a ) /2 , ( q 3 + 3 − a ) /2 ; 1 ] . Each occurrence of B ϵ B_\epsilon B ϵ B ϵ = ( h + q 3 + ϵ 1 q 1 + ϵ 2 q 2 ) / 2 B_\epsilon=(h+q_3+\epsilon_1q_1+\epsilon_2q_2)/2 B ϵ = ( h + q 3 + ϵ 1 q 1 + ϵ 2 q 2 ) /2
Proof. Set q 3 , n = q 3 + 1 + 2 n q_{3,n}=q_3+1+2n q 3 , n = q 3 + 1 + 2 n T q 3 , n T_{q_{3,n}} T q 3 , n
S 3 ( q 3 , n ) = 2 q 3 , n ( d − 2 ) [ 2 T 1 T 2 − ( T 1 + T 2 ) T q 3 , n + κ 1 − κ 2 κ q 3 , n ( T 1 − T 2 ) T q 3 , n ] . (85) S_3(q_{3,n})=2q_{3,n}(d-2)\left[2T_1T_2-(T_1+T_2)T_{q_{3,n}}+\frac{\kappa_1-\kappa_2}{\kappa_{q_{3,n}}}(T_1-T_2)T_{q_{3,n}}\right]. \tag*{(85)} S 3 ( q 3 , n ) = 2 q 3 , n ( d − 2 ) [ 2 T 1 T 2 − ( T 1 + T 2 ) T q 3 , n + κ q 3 , n κ 1 − κ 2 ( T 1 − T 2 ) T q 3 , n ] . (85) Define
Θ n = H 3 ( q ) H 3 ( q + 2 n e 3 ) 2 q 3 , n ( d − 2 ) P 3 + ( q 3 , n ) . (86) \Theta_n=\frac{\mathcal{H}_3(q)}{\mathcal{H}_3(q+2n e_3)}\frac{2q_{3,n}(d-2)}{P_3^+(q_{3,n})}. \tag*{(86)} Θ n = H 3 ( q + 2 n e 3 ) H 3 ( q ) P 3 + ( q 3 , n ) 2 q 3 , n ( d − 2 ) . (86) The three terms in Σ d \Sigma_d Σ d
Σ d ( 1 ) = 2 T 1 T 2 ∑ n ≥ 0 Θ n . (87) \Sigma_d^{(1)}=2T_1T_2\sum_{n\geq0}\Theta_n. \tag*{(87)} Σ d ( 1 ) = 2 T 1 T 2 n ≥ 0 ∑ Θ n . (87) Σ d ( 2 ) = − ( T 1 + T 2 ) ∑ n ≥ 0 Θ n T q 3 , n . (88) \Sigma_d^{(2)}=-(T_1+T_2)\sum_{n\geq0}\Theta_nT_{q_{3,n}}. \tag*{(88)} Σ d ( 2 ) = − ( T 1 + T 2 ) n ≥ 0 ∑ Θ n T q 3 , n . (88) Σ d ( 3 ) = ( κ 1 − κ 2 ) ( T 1 − T 2 ) ∑ n ≥ 0 Θ n T q 3 , n κ q 3 , n . (89) \Sigma_d^{(3)}=(\kappa_1-\kappa_2)(T_1-T_2)\sum_{n\geq0}\Theta_n\frac{T_{q_{3,n}}}{\kappa_{q_{3,n}}}. \tag*{(89)} Σ d ( 3 ) = ( κ 1 − κ 2 ) ( T 1 − T 2 ) n ≥ 0 ∑ Θ n κ q 3 , n T q 3 , n . (89) The Gamma-function shift formula gives
H 3 ( q ) H 3 ( q + 2 n e 3 ) = ∏ ϵ ( A ϵ ) n ( B ϵ ) n , P 3 + ( q 3 , n ) = 16 ∏ ϵ ( n + B ϵ ) . (90) \frac{\mathcal{H}_3(q)}{\mathcal{H}_3(q+2n e_3)}=\prod_{\epsilon}\frac{(A_\epsilon)_n}{(B_\epsilon)_n},\qquad P_3^+(q_{3,n})=16\prod_{\epsilon}(n+B_\epsilon). \tag*{(90)} H 3 ( q + 2 n e 3 ) H 3 ( q ) = ϵ ∏ ( B ϵ ) n ( A ϵ ) n , P 3 + ( q 3 , n ) = 16 ϵ ∏ ( n + B ϵ ) . (90) Using ( B + 1 ) n = ( n + B ) ( B ) n / B (B+1)_n=(n+B)(B)_n/B ( B + 1 ) n = ( n + B ) ( B ) n / B e = ( q 3 + 1 ) / 2 e=(q_3+1)/2 e = ( q 3 + 1 ) /2
Θ n = 4 e ( d − 2 ) 16 ∏ ϵ B ϵ ( e + 1 ) n ( e ) n ∏ ϵ ( A ϵ ) n ( B ϵ + 1 ) n . (91) \Theta_n=\frac{4e(d-2)}{16\prod_\epsilon B_\epsilon}\frac{(e+1)_n}{(e)_n}\prod_\epsilon\frac{(A_\epsilon)_n}{(B_\epsilon+1)_n}. \tag*{(91)} Θ n = 16 ∏ ϵ B ϵ 4 e ( d − 2 ) ( e ) n ( e + 1 ) n ϵ ∏ ( B ϵ + 1 ) n ( A ϵ ) n . (91) This immediately gives the 6 F 5 {}_6F_5 6 F 5 Σ d ( 1 ) \Sigma_d^{(1)} Σ d ( 1 )
α = h + q 3 + 1 2 , β = q 3 + 1 − a 2 . (92) \alpha=\frac{h+q_3+1}{2},\qquad\beta=\frac{q_3+1-a}{2}. \tag*{(92)} α = 2 h + q 3 + 1 , β = 2 q 3 + 1 − a . (92) The tadpole recurrence implies
T q 3 , n = T q 3 + 1 ( α ) n ( α + 1 / 2 ) n ( β ) n ( β + 1 / 2 ) n . (93) T_{q_{3,n}}=T_{q_3+1}\frac{(\alpha)_n(\alpha+1/2)_n}{(\beta)_n(\beta+1/2)_n}. \tag*{(93)} T q 3 , n = T q 3 + 1 ( β ) n ( β + 1/2 ) n ( α ) n ( α + 1/2 ) n . (93) Indeed,
T q + 2 T q = h + q + 1 q + 1 − a h + q q − a , (94) \frac{T_{q+2}}{T_q}=\frac{h+q+1}{q+1-a}\frac{h+q}{q-a}, \tag*{(94)} T q T q + 2 = q + 1 − a h + q + 1 q − a h + q , (94) and multiplication along the lattice proves Eq. (93) . It gives
Σ d ( 2 ) = − ( T 1 + T 2 ) T q 3 + 1 4 e ( d − 2 ) 16 ∏ ϵ B ϵ 8 F 7 [ A ϵ , e + 1 , α , α + 1 / 2 , 1 B ϵ + 1 , e , β , β + 1 / 2 ; 1 ] . (95) \Sigma_d^{(2)}=-(T_1+T_2)T_{q_3+1}\frac{4e(d-2)}{16\prod_\epsilon B_\epsilon}\,{}_8F_7\left[\begin{matrix}A_\epsilon,e+1,\alpha,\alpha+1/2,1\\ B_\epsilon+1,e,\beta,\beta+1/2\end{matrix};1\right]. \tag*{(95)} Σ d ( 2 ) = − ( T 1 + T 2 ) T q 3 + 1 16 ∏ ϵ B ϵ 4 e ( d − 2 ) 8 F 7 [ A ϵ , e + 1 , α , α + 1/2 , 1 B ϵ + 1 , e , β , β + 1/2 ; 1 ] . (95) Finally, with γ = ( a + q 3 + 1 ) / 2 \gamma=(a+q_3+1)/2 γ = ( a + q 3 + 1 ) /2
κ q 3 , n = − 4 ( β + n ) ( γ + n ) , κ q 3 , n − 1 = − 1 4 γ β ( γ ) n ( β ) n ( γ + 1 ) n ( β + 1 ) n . (96) \kappa_{q_{3,n}}=-4(\beta+n)(\gamma+n),\qquad\kappa_{q_{3,n}}^{-1}=-\frac{1}{4\gamma\beta}\frac{(\gamma)_n(\beta)_n}{(\gamma+1)_n(\beta+1)_n}. \tag*{(96)} κ q 3 , n = − 4 ( β + n ) ( γ + n ) , κ q 3 , n − 1 = − 4 γ β 1 ( γ + 1 ) n ( β + 1 ) n ( γ ) n ( β ) n . (96) Multiplication and cancellation of common Pochhammer factors give
Σ d ( 3 ) = − ( κ 1 − κ 2 ) ( T 1 − T 2 ) T q 3 + 1 4 γ β 4 e ( d − 2 ) 16 ∏ ϵ B ϵ 8 F 7 [ A ϵ , e + 1 , α , γ , 1 B ϵ + 1 , e , β + 1 / 2 , β + 1 ; 1 ] . (97) \Sigma_d^{(3)}=-\frac{(\kappa_1-\kappa_2)(T_1-T_2)T_{q_3+1}}{4\gamma\beta}\frac{4e(d-2)}{16\prod_\epsilon B_\epsilon}\,{}_8F_7\left[\begin{matrix}A_\epsilon,e+1,\alpha,\gamma,1\\ B_\epsilon+1,e,\beta+1/2,\beta+1\end{matrix};1\right]. \tag*{(97)} Σ d ( 3 ) = − 4 γ β ( κ 1 − κ 2 ) ( T 1 − T 2 ) T q 3 + 1 16 ∏ ϵ B ϵ 4 e ( d − 2 ) 8 F 7 [ A ϵ , e + 1 , α , γ , 1 B ϵ + 1 , e , β + 1/2 , β + 1 ; 1 ] . (97) These are precisely the three functions in Eq. (84) .
The four-dimensional effective potential
To obtain the finite two-loop potential, we need the tadpole Laurent expansion through its coefficient of ε \varepsilon ε q q q
The four-dimensional expansion
The tadpole
We expand the tadpole of Eq. (26) at d = 4 − 2 ε d=4-2\varepsilon d = 4 − 2 ε q q q
log Γ ( z + ε ) = log Γ ( z ) + ψ ( z ) ε + 1 2 ψ ′ ( z ) ε 2 + O ( ε 3 ) (99) \log\Gamma(z+\varepsilon)=\log\Gamma(z)+\psi(z)\varepsilon+\frac{1}{2}\psi'(z)\varepsilon^2+O(\varepsilon^3) \tag*{(99)} log Γ ( z + ε ) = log Γ ( z ) + ψ ( z ) ε + 2 1 ψ ′ ( z ) ε 2 + O ( ε 3 ) (99) and
ψ ′ ( 3 2 + q ) + ψ ′ ( 3 2 − q ) = − 8 1 + 4 q 2 ( 1 − 4 q 2 ) 2 + π 2 cos 2 π q (100) \psi'\left(\frac{3}{2}+q\right)+\psi'\left(\frac{3}{2}-q\right)=-8\frac{1+4q^2}{(1-4q^2)^2}+\frac{\pi^2}{\cos^2\pi q} \tag*{(100)} ψ ′ ( 2 3 + q ) + ψ ′ ( 2 3 − q ) = − 8 ( 1 − 4 q 2 ) 2 1 + 4 q 2 + cos 2 π q π 2 (100) to obtain
T 4 − 2 ε ( q ) = − 1 / 4 − q 2 16 π 2 ε exp [ − F 1 ( q ) ε + 1 2 F 2 ( q ) ε 2 + O ( ε 3 ) ] , (101) T_{4-2\varepsilon}(q)=-\frac{1/4-q^2}{16\pi^2\varepsilon}\exp\left[-F_1(q)\varepsilon+\frac{1}{2}F_2(q)\varepsilon^2+O(\varepsilon^3)\right], \tag*{(101)} T 4 − 2 ε ( q ) = − 16 π 2 ε 1/4 − q 2 exp [ − F 1 ( q ) ε + 2 1 F 2 ( q ) ε 2 + O ( ε 3 ) ] , (101) where
F 1 ( q ) = ψ ( 3 2 + q ) + ψ ( 3 2 − q ) + γ E − 1 − log 4 π , (102) F_1(q)=\psi\left(\frac{3}{2}+q\right)+\psi\left(\frac{3}{2}-q\right)+\gamma_{\mathrm{E}}-1-\log4\pi, \tag*{(102)} F 1 ( q ) = ψ ( 2 3 + q ) + ψ ( 2 3 − q ) + γ E − 1 − log 4 π , (102) F 2 ( q ) = 1 + π 2 6 − 8 1 + 4 q 2 ( 1 − 4 q 2 ) 2 + π 2 cos 2 π q . (103) F_2(q)=1+\frac{\pi^2}{6}-8\frac{1+4q^2}{(1-4q^2)^2}+\frac{\pi^2}{\cos^2\pi q}. \tag*{(103)} F 2 ( q ) = 1 + 6 π 2 − 8 ( 1 − 4 q 2 ) 2 1 + 4 q 2 + cos 2 π q π 2 . (103) Here ψ ′ ( 1 ) = ∑ k ≥ 0 ( k + 1 ) − 2 = π 2 / 6 \psi'(1)=\sum_{k\geq0}(k+1)^{-2}=\pi^2/6 ψ ′ ( 1 ) = ∑ k ≥ 0 ( k + 1 ) − 2 = π 2 /6 q q q
16 π 2 T d ( q ) = ( 1 4 − q 2 ) [ − 1 ε + F 1 ( q ) − ε 2 ( F 1 ( q ) 2 + F 2 ( q ) ) + O ( ε 2 ) ] . (104) 16\pi^2T_d(q)=\left(\frac{1}{4}-q^2\right)\left[-\frac{1}{\varepsilon}+F_1(q)-\frac{\varepsilon}{2}\left(F_1(q)^2+F_2(q)\right)+O(\varepsilon^2)\right]. \tag*{(104)} 16 π 2 T d ( q ) = ( 4 1 − q 2 ) [ − ε 1 + F 1 ( q ) − 2 ε ( F 1 ( q ) 2 + F 2 ( q ) ) + O ( ε 2 ) ] . (104) At fixed M j 2 M_j^2 M j 2 R \mathcal{R} R
q j ( ε ) 2 = ( d − 1 ) 2 4 − d ( d − 1 ) M j 2 R = q j ( 0 ) 2 − ( 3 8 + 7 6 q j ( 0 ) 2 ) ε + ( 1 4 + 1 3 q j ( 0 ) 2 ) ε 2 = 9 4 − 12 M j 2 R − ( 3 − 14 M j 2 R ) ε + ( 1 − 4 M j 2 R ) ε 2 . (105) \begin{aligned}
q_j(\varepsilon)^2&=\frac{(d-1)^2}{4}-d(d-1)\frac{M_j^2}{\mathcal{R}}\\
&=q_j(0)^2-\left(\frac{3}{8}+\frac{7}{6}q_j(0)^2\right)\varepsilon+\left(\frac{1}{4}+\frac{1}{3}q_j(0)^2\right)\varepsilon^2\\
&=\frac{9}{4}-12\frac{M_j^2}{\mathcal{R}}-\left(3-14\frac{M_j^2}{\mathcal{R}}\right)\varepsilon+\left(1-4\frac{M_j^2}{\mathcal{R}}\right)\varepsilon^2. \tag*{(105)}
\end{aligned} q j ( ε ) 2 = 4 ( d − 1 ) 2 − d ( d − 1 ) R M j 2 = q j ( 0 ) 2 − ( 8 3 + 6 7 q j ( 0 ) 2 ) ε + ( 4 1 + 3 1 q j ( 0 ) 2 ) ε 2 = 4 9 − 12 R M j 2 − ( 3 − 14 R M j 2 ) ε + ( 1 − 4 R M j 2 ) ε 2 . (105) Writing q = q ( 0 ) q=q(0) q = q ( 0 ) s = q 2 s=q^2 s = q 2
16 π 2 T d ( q ( ε ) ) = − 1 / 4 − q 2 ε + ( 1 4 − q 2 ) F 1 ( q ) − 3 8 − 7 6 q 2 − [ 1 / 4 − q 2 2 ( F 1 ( q ) 2 + F 2 ( q ) ) − 1 4 − 1 3 q 2 ] ε − ( 3 8 + 7 6 q 2 ) ∂ s [ ( 1 4 − q 2 ) F 1 ( q ) ] ε + O ( ε 2 ) . (106) 16\pi^2T_d(q(\varepsilon))=-\frac{1/4-q^2}{\varepsilon}+\left(\frac{1}{4}-q^2\right)F_1(q)-\frac{3}{8}-\frac{7}{6}q^2-\left[\frac{1/4-q^2}{2}\left(F_1(q)^2+F_2(q)\right)-\frac{1}{4}-\frac{1}{3}q^2\right]\varepsilon-\left(\frac{3}{8}+\frac{7}{6}q^2\right)\partial_s\left[\left(\frac{1}{4}-q^2\right)F_1(q)\right]\varepsilon+O(\varepsilon^2). \tag*{(106)} 16 π 2 T d ( q ( ε )) = − ε 1/4 − q 2 + ( 4 1 − q 2 ) F 1 ( q ) − 8 3 − 6 7 q 2 − [ 2 1/4 − q 2 ( F 1 ( q ) 2 + F 2 ( q ) ) − 4 1 − 3 1 q 2 ] ε − ( 8 3 + 6 7 q 2 ) ∂ s [ ( 4 1 − q 2 ) F 1 ( q ) ] ε + O ( ε 2 ) . (106) The last line contains a moving-label derivative term. Such terms will cancel from the forest-subtracted expression below. To form tadpole products, write
16 π 2 T d ( q ( ε ) ) = − Y ( q ) ε + A 0 ( q ) + A 1 ( q ) ε + O ( ε 2 ) . (107) 16\pi^2T_d(q(\varepsilon))=-\frac{Y(q)}{\varepsilon}+A_0(q)+A_1(q)\varepsilon+O(\varepsilon^2). \tag*{(107)} 16 π 2 T d ( q ( ε )) = − ε Y ( q ) + A 0 ( q ) + A 1 ( q ) ε + O ( ε 2 ) . (107) Then
( 16 π 2 ) 2 T q i ( ε ) T q j ( ε ) = Y ( q i ) Y ( q j ) ε 2 − Y ( q i ) A 0 ( q j ) + A 0 ( q i ) Y ( q j ) ε + A 0 ( q i ) A 0 ( q j ) − Y ( q i ) A 1 ( q j ) − A 1 ( q i ) Y ( q j ) + O ( ε ) . (108) (16\pi^2)^2T_{q_i}(\varepsilon)T_{q_j}(\varepsilon)=\frac{Y(q_i)Y(q_j)}{\varepsilon^2}-\frac{Y(q_i)A_0(q_j)+A_0(q_i)Y(q_j)}{\varepsilon}+A_0(q_i)A_0(q_j)-Y(q_i)A_1(q_j)-A_1(q_i)Y(q_j)+O(\varepsilon). \tag*{(108)} ( 16 π 2 ) 2 T q i ( ε ) T q j ( ε ) = ε 2 Y ( q i ) Y ( q j ) − ε Y ( q i ) A 0 ( q j ) + A 0 ( q i ) Y ( q j ) + A 0 ( q i ) A 0 ( q j ) − Y ( q i ) A 1 ( q j ) − A 1 ( q i ) Y ( q j ) + O ( ε ) . (108) The watermelon
The factorisation in Remark 4.9 gives, at fixed labels,
( 16 π 2 ) 2 I 3 ( q ) = 1 ε 2 ( 1 2 ∑ j = 1 3 q j 2 − 3 8 ) + 1 ε [ ∑ j = 1 3 ( 1 4 − q j 2 ) ( F 1 ( q j ) − 1 2 ) − 1 2 ] + C ( q ) + O ( ε ) . (109) (16\pi^2)^2\mathcal{I}_3(q)=\frac{1}{\varepsilon^2}\left(\frac{1}{2}\sum_{j=1}^{3}q_j^2-\frac{3}{8}\right)+\frac{1}{\varepsilon}\left[\sum_{j=1}^{3}\left(\frac{1}{4}-q_j^2\right)\left(F_1(q_j)-\frac{1}{2}\right)-\frac{1}{2}\right]+C(q)+O(\varepsilon). \tag*{(109)} ( 16 π 2 ) 2 I 3 ( q ) = ε 2 1 ( 2 1 j = 1 ∑ 3 q j 2 − 8 3 ) + ε 1 [ j = 1 ∑ 3 ( 4 1 − q j 2 ) ( F 1 ( q j ) − 2 1 ) − 2 1 ] + C ( q ) + O ( ε ) . (109) Here C = C 0 C=C_0 C = C 0 § 5.4 . Inserting the moving labels gives
( 16 π 2 ) 2 I 3 ( q ( ε ) ) = 1 ε 2 ( 1 2 ∑ j q j 2 − 3 8 ) + 1 ε [ ∑ j ( 1 4 − q j 2 ) ( F 1 ( q j ) + 1 12 ) − 3 2 ] + C ( q ) − 3 16 − 5 12 ∑ j q j 2 − ∑ j ( 3 8 + 7 6 q j 2 ) ∂ s j [ ( 1 4 − q j 2 ) F 1 ( q j ) ] + O ( ε ) . (110) (16\pi^2)^2\mathcal{I}_3(q(\varepsilon))=\frac{1}{\varepsilon^2}\left(\frac{1}{2}\sum_jq_j^2-\frac{3}{8}\right)+\frac{1}{\varepsilon}\left[\sum_j\left(\frac{1}{4}-q_j^2\right)\left(F_1(q_j)+\frac{1}{12}\right)-\frac{3}{2}\right]+C(q)-\frac{3}{16}-\frac{5}{12}\sum_jq_j^2-\sum_j\left(\frac{3}{8}+\frac{7}{6}q_j^2\right)\partial_{s_j}\left[\left(\frac{1}{4}-q_j^2\right)F_1(q_j)\right]+O(\varepsilon). \tag*{(110)} ( 16 π 2 ) 2 I 3 ( q ( ε )) = ε 2 1 ( 2 1 j ∑ q j 2 − 8 3 ) + ε 1 [ j ∑ ( 4 1 − q j 2 ) ( F 1 ( q j ) + 12 1 ) − 2 3 ] + C ( q ) − 16 3 − 12 5 j ∑ q j 2 − j ∑ ( 8 3 + 6 7 q j 2 ) ∂ s j [ ( 4 1 − q j 2 ) F 1 ( q j ) ] + O ( ε ) . (110) where q j = q j ( 0 ) q_j=q_j(0) q j = q j ( 0 ) s j = q j 2 s_j=q_j^2 s j = q j 2
Restoring physical units
Since R = d ( d − 1 ) H 2 \mathcal{R}=d(d-1)H^2 R = d ( d − 1 ) H 2 R \mathcal{R} R d = 4 − 2 ε d=4-2\varepsilon d = 4 − 2 ε H ≡ H ε H\equiv H_\varepsilon H ≡ H ε ε \varepsilon ε H ε 2 − 2 ε H_\varepsilon^{2-2\varepsilon} H ε 2 − 2 ε H ε 2 − 4 ε H_\varepsilon^{2-4\varepsilon} H ε 2 − 4 ε μ \mu μ μ 2 ε \mu^{2\varepsilon} μ 2 ε
T q j p h y s ( ε ) = R 12 exp [ ( log 12 μ 2 R + 7 6 ) ε − 59 72 ε 2 + O ( ε 3 ) ] T q j ( ε ) . (111) T_{q_j}^{\mathrm{phys}}(\varepsilon)=\frac{\mathcal{R}}{12}\exp\left[\left(\log\frac{12\mu^2}{\mathcal{R}}+\frac{7}{6}\right)\varepsilon-\frac{59}{72}\varepsilon^2+O(\varepsilon^3)\right]T_{q_j}(\varepsilon). \tag*{(111)} T q j phys ( ε ) = 12 R exp [ ( log R 12 μ 2 + 6 7 ) ε − 72 59 ε 2 + O ( ε 3 ) ] T q j ( ε ) . (111) I 3 p h y s ( q ( ε ) ) = R 12 exp [ ( 2 log 12 μ 2 R + 7 6 ) ε − 143 72 ε 2 + O ( ε 3 ) ] I 3 ( q ( ε ) ) . (112) \mathcal{I}_3^{\mathrm{phys}}(q(\varepsilon))=\frac{\mathcal{R}}{12}\exp\left[\left(2\log\frac{12\mu^2}{\mathcal{R}}+\frac{7}{6}\right)\varepsilon-\frac{143}{72}\varepsilon^2+O(\varepsilon^3)\right]\mathcal{I}_3(q(\varepsilon)). \tag*{(112)} I 3 phys ( q ( ε )) = 12 R exp [ ( 2 log R 12 μ 2 + 6 7 ) ε − 72 143 ε 2 + O ( ε 3 ) ] I 3 ( q ( ε )) . (112) For clarity, define
Y j = 1 4 − q j 2 , σ j = 3 8 + 7 6 q j 2 , τ j = 1 4 + 1 3 q j 2 , L μ = log 12 μ 2 R , Φ j = ∂ s j [ Y j F 1 ( q j ) ] . (113) Y_j=\frac{1}{4}-q_j^2,\qquad\sigma_j=\frac{3}{8}+\frac{7}{6}q_j^2,\qquad\tau_j=\frac{1}{4}+\frac{1}{3}q_j^2,\qquad L_\mu=\log\frac{12\mu^2}{\mathcal{R}},\qquad\Phi_j=\partial_{s_j}[Y_jF_1(q_j)]. \tag*{(113)} Y j = 4 1 − q j 2 , σ j = 8 3 + 6 7 q j 2 , τ j = 4 1 + 3 1 q j 2 , L μ = log R 12 μ 2 , Φ j = ∂ s j [ Y j F 1 ( q j )] . (113) These are local abbreviations for the following expansions:
16 π 2 12 R T q j p h y s ( ε ) = − Y j ε + Y j F 1 ( q j ) − 2 3 − Y j L μ + ε [ − Y j 2 ( F 1 ( q j ) 2 + F 2 ( q j ) ) + τ j + ( Y j F 1 ( q j ) − σ j ) ( L μ + 7 6 ) − Y j ( 1 2 ( L μ + 7 6 ) 2 − 59 72 ) − σ j Φ j ] + O ( ε 2 ) . (114) 16\pi^2\frac{12}{\mathcal{R}}T^{\mathrm{phys}}_{q_j}(\varepsilon)=-\frac{Y_j}{\varepsilon}+Y_jF_1(q_j)-\frac{2}{3}-Y_jL_\mu
+\varepsilon\left[-\frac{Y_j}{2}\left(F_1(q_j)^2+F_2(q_j)\right)+\tau_j+\left(Y_jF_1(q_j)-\sigma_j\right)\left(L_\mu+\frac{7}{6}\right)
-Y_j\left(\frac{1}{2}\left(L_\mu+\frac{7}{6}\right)^2-\frac{59}{72}\right)-\sigma_j\Phi_j\right]+O(\varepsilon^2). \tag*{(114)} 16 π 2 R 12 T q j phys ( ε ) = − ε Y j + Y j F 1 ( q j ) − 3 2 − Y j L μ + ε [ − 2 Y j ( F 1 ( q j ) 2 + F 2 ( q j ) ) + τ j + ( Y j F 1 ( q j ) − σ j ) ( L μ + 6 7 ) − Y j ( 2 1 ( L μ + 6 7 ) 2 − 72 59 ) − σ j Φ j ] + O ( ε 2 ) . (114) The moving-label Laurent coefficients of the watermelon are
C ^ − 2 = 1 2 ∑ j q j 2 − 3 8 , C ^ − 1 = ∑ j Y j ( F 1 ( q j ) + 1 12 ) − 3 2 , \widehat{C}_{-2}=\frac{1}{2}\sum_j q_j^2-\frac{3}{8},\qquad\widehat{C}_{-1}=\sum_jY_j\left(F_1(q_j)+\frac{1}{12}\right)-\frac{3}{2}, C − 2 = 2 1 j ∑ q j 2 − 8 3 , C − 1 = j ∑ Y j ( F 1 ( q j ) + 12 1 ) − 2 3 , C ^ 0 = C ( q ) − 3 16 − 5 12 ∑ j q j 2 − ∑ j σ j Φ j . (115) \widehat{C}_0=C(q)-\frac{3}{16}-\frac{5}{12}\sum_jq_j^2-\sum_j\sigma_j\Phi_j. \tag*{(115)} C 0 = C ( q ) − 16 3 − 12 5 j ∑ q j 2 − j ∑ σ j Φ j . (115) The physical watermelon expansion is
( 16 π 2 ) 2 12 R T 3 p h y s ( q ( ε ) ) = C ^ − 2 ε 2 + C ^ − 1 + ( 2 L μ + 7 / 6 ) C ^ − 2 ε + C ^ 0 + ( 2 L μ + 7 6 ) C ^ − 1 + [ 1 2 ( 2 L μ + 7 6 ) 2 − 143 72 ] C ^ − 2 + O ( ε ) . (116) (16\pi^2)^2\frac{12}{\mathcal{R}}T^{\mathrm{phys}}_3(q(\varepsilon))=\frac{\widehat{C}_{-2}}{\varepsilon^2}+\frac{\widehat{C}_{-1}+(2L_\mu+7/6)\widehat{C}_{-2}}{\varepsilon}+\widehat{C}_0+\left(2L_\mu+\frac{7}{6}\right)\widehat{C}_{-1}
+\left[\frac{1}{2}\left(2L_\mu+\frac{7}{6}\right)^2-\frac{143}{72}\right]\widehat{C}_{-2}+O(\varepsilon). \tag*{(116)} ( 16 π 2 ) 2 R 12 T 3 phys ( q ( ε )) = ε 2 C − 2 + ε C − 1 + ( 2 L μ + 7/6 ) C − 2 + C 0 + ( 2 L μ + 6 7 ) C − 1 + [ 2 1 ( 2 L μ + 6 7 ) 2 − 72 143 ] C − 2 + O ( ε ) . (116) These compact forms retain all terms required for the forest subtraction.
Renormalisation
We use dimensional regularisation and minimal subtraction, organised by the forest formula. We first subtract the pole parts of divergent proper subgraphs and then subtract the remaining local poles. The renormalised tadpole is
16 π 2 12 R t ( M j ) = ( 1 4 − q j 2 ) F 1 ( q j ) − 2 3 − ( 1 4 − q j 2 ) log 12 μ 2 R . (117) 16\pi^2\frac{12}{\mathcal{R}}t(M_j)=\left(\frac{1}{4}-q_j^2\right)F_1(q_j)-\frac{2}{3}-\left(\frac{1}{4}-q_j^2\right)\log\frac{12\mu^2}{\mathcal{R}}. \tag*{(117)} 16 π 2 R 12 t ( M j ) = ( 4 1 − q j 2 ) F 1 ( q j ) − 3 2 − ( 4 1 − q j 2 ) log R 12 μ 2 . (117) Its higher-order terms remain those in Eq. (114) . The watermelon has three divergent bubble subgraphs, one for each pair of lines. Each bubble contributes a pole 1 / ( 16 π 2 ε ) 1/(16\pi^2\varepsilon) 1/ ( 16 π 2 ε ) [14 ] . The first forest step is
T 3 ∗ ( q ( ε ) ) = T 3 p h y s ( q ( ε ) ) − 1 16 π 2 ε ∑ j = 1 3 T q j p h y s ( ε ) . (118) T_3^*(q(\varepsilon))=T_3^{\mathrm{phys}}(q(\varepsilon))-\frac{1}{16\pi^2\varepsilon}\sum_{j=1}^{3}T_{q_j}^{\mathrm{phys}}(\varepsilon). \tag*{(118)} T 3 ∗ ( q ( ε )) = T 3 phys ( q ( ε )) − 16 π 2 ε 1 j = 1 ∑ 3 T q j phys ( ε ) . (118) Substitution gives
( 16 π 2 ) 2 1 R I 3 ∗ ( q ( ε ) ) = − 1 ε 2 ( 1 2 ∑ j q j 2 − 3 8 ) + 1 ε ( 1 2 ∑ j q j 2 + 1 8 ) + C ( q ) + 1 2 ∑ j ( 1 4 − q j 2 ) [ F 1 ( q j ) 2 + F 2 ( q j ) ] + L μ [ ∑ j ( 1 4 − q j 2 ) F 1 ( q j ) + ∑ j q j 2 − 7 4 ] + L μ 2 ( 1 2 ∑ j q j 2 − 3 8 ) − 11 12 + O ( ε ) . (119) (16\pi^2)^2\frac{1}{\mathcal{R}}\mathcal{I}_3^*(q(\varepsilon))=-\frac{1}{\varepsilon^2}\left(\frac{1}{2}\sum_j q_j^2-\frac{3}{8}\right)+\frac{1}{\varepsilon}\left(\frac{1}{2}\sum_j q_j^2+\frac{1}{8}\right)+C(\mathbf{q})+\frac{1}{2}\sum_j\left(\frac{1}{4}-q_j^2\right)[F_1(q_j)^2+F_2(q_j)]
+\ L_\mu\left[\sum_j\left(\frac{1}{4}-q_j^2\right)F_1(q_j)+\sum_j q_j^2-\frac{7}{4}\right]
+L_\mu^2\left(\frac{1}{2}\sum_j q_j^2-\frac{3}{8}\right)-\frac{11}{12}+O(\varepsilon). \tag*{(119)} ( 16 π 2 ) 2 R 1 I 3 ∗ ( q ( ε )) = − ε 2 1 ( 2 1 j ∑ q j 2 − 8 3 ) + ε 1 ( 2 1 j ∑ q j 2 + 8 1 ) + C ( q ) + 2 1 j ∑ ( 4 1 − q j 2 ) [ F 1 ( q j ) 2 + F 2 ( q j )] + L μ [ j ∑ ( 4 1 − q j 2 ) F 1 ( q j ) + j ∑ q j 2 − 4 7 ] + L μ 2 ( 2 1 j ∑ q j 2 − 8 3 ) − 12 11 + O ( ε ) . (119) The moving-label derivative terms cancel. The remaining poles are local and are removed by the two-loop counterterms. The fully renormalised watermelon is
W ( M 1 , M 2 , M 3 ) = R 12 ( 16 π 2 ) 2 { C ( q ) + 1 2 ∑ j ( 1 4 − q j 2 ) [ F 1 ( q j ) 2 + F 2 ( q j ) ] + L μ [ ∑ j ( 1 4 − q j 2 ) F 1 ( q j ) + ∑ j q j 2 − 7 4 ] + L μ 2 ( 1 2 ∑ j q j 2 − 3 8 ) − 11 12 } . (120) \mathcal{W}(M_1,M_2,M_3)=\frac{\mathcal{R}}{12(16\pi^2)^2}\left\{C(\mathbf{q})+\frac{1}{2}\sum_j\left(\frac{1}{4}-q_j^2\right)[F_1(q_j)^2+F_2(q_j)]
+L_\mu\left[\sum_j\left(\frac{1}{4}-q_j^2\right)F_1(q_j)+\sum_j q_j^2-\frac{7}{4}\right]
+L_\mu^2\left(\frac{1}{2}\sum_j q_j^2-\frac{3}{8}\right)-\frac{11}{12}\right\}. \tag*{(120)} W ( M 1 , M 2 , M 3 ) = 12 ( 16 π 2 ) 2 R { C ( q ) + 2 1 j ∑ ( 4 1 − q j 2 ) [ F 1 ( q j ) 2 + F 2 ( q j )] + L μ [ j ∑ ( 4 1 − q j 2 ) F 1 ( q j ) + j ∑ q j 2 − 4 7 ] + L μ 2 ( 2 1 j ∑ q j 2 − 8 3 ) − 12 11 } . (120) Equivalently,
I 3 p h y s ( q ( ε ) ) = R 12 ( 16 π 2 ) 2 [ − 1 2 ∑ j q j 2 − 3 8 ε 2 + 1 2 ∑ j q j 2 + 1 8 ε ] + W ( M 1 , M 2 , M 3 ) + 1 16 π 2 ε ∑ j T q j ( ε ) p h y s + O ( ε ) . (121) \mathcal{I}_3^{\mathrm{phys}}(\mathbf{q}(\varepsilon))=\frac{\mathcal{R}}{12(16\pi^2)^2}\left[\frac{-\frac{1}{2}\sum_j q_j^2-\frac{3}{8}}{\varepsilon^2}+\frac{\frac{1}{2}\sum_j q_j^2+\frac{1}{8}}{\varepsilon}\right]+\mathcal{W}(M_1,M_2,M_3)+\frac{1}{16\pi^2\varepsilon}\sum_j\mathcal{T}_{q_j(\varepsilon)}^{\mathrm{phys}}+O(\varepsilon). \tag*{(121)} I 3 phys ( q ( ε )) = 12 ( 16 π 2 ) 2 R [ ε 2 − 2 1 ∑ j q j 2 − 8 3 + ε 2 1 ∑ j q j 2 + 8 1 ] + W ( M 1 , M 2 , M 3 ) + 16 π 2 ε 1 j ∑ T q j ( ε ) phys + O ( ε ) . (121) The cancellation concerns the displayed moving-label derivative terms after subdivergence subtraction; it does not assert that every intermediate Laurent coefficient is independent of the continuation prescription.
The effective potential
Equation (117) shows that the fixed-mass, fixed-curvature prescription adds a constant − 2 / 3 -2/3 − 2/3 q 2 / 3 q^2/3 q 2 /3
V 1 ( M ) = R 2 144 ( 16 π 2 ) [ B ~ ( q ) + 1 4 ( 1 4 − q 2 ) ( ψ ( 3 2 + q ) + ψ ( 3 2 − q ) − 2 − log 12 Q 2 R ) + q 2 3 ] = R 48 [ ( 1 4 − q 2 ) t ( M ) + R 48 π 2 ( B ~ ( q ) − q 4 4 + 7 q 2 24 + 5 192 ) ] . (122) \begin{aligned}
V_1(M)=\frac{\mathcal{R}^2}{144(16\pi^2)}\left[\widetilde{B}(q)+\frac{1}{4}\left(\frac{1}{4}-q^2\right)\left(\psi\left(\frac{3}{2}+q\right)+\psi\left(\frac{3}{2}-q\right)-2-\log\frac{12Q^2}{\mathcal{R}}\right)+\frac{q^2}{3}\right] \\
&=\frac{\mathcal{R}}{48}\left[\left(\frac{1}{4}-q^2\right)t(M)+\frac{\mathcal{R}}{48\pi^2}\left(\widetilde{B}(q)-\frac{q^4}{4}+\frac{7q^2}{24}+\frac{5}{192}\right)\right]. \tag*{(122)}
\end{aligned} V 1 ( M ) = 144 ( 16 π 2 ) R 2 [ B ( q ) + 4 1 ( 4 1 − q 2 ) ( ψ ( 2 3 + q ) + ψ ( 2 3 − q ) − 2 − log R 12 Q 2 ) + 3 q 2 ] = 48 R [ ( 4 1 − q 2 ) t ( M ) + 48 π 2 R ( B ( q ) − 4 q 4 + 24 7 q 2 + 192 5 ) ] . (122) where
B ~ ( q ) = B ( − i q ) = ( 1 4 − q 2 ) 2 4 − ∫ 1 4 q ( 1 4 − y 2 ) 2 [ ψ ′ ( − 1 2 + y ) − ψ ′ ( − 1 2 − y ) ] d y = − 1 2 ∫ 0 1 4 − q 2 d y y 2 [ ∑ n = 0 ∞ 2 n − 1 2 ( n ( n − 1 ) + y ) 2 − 1 y ] , (123) \begin{aligned}
\widetilde{B}(q)=B(-iq)=\frac{\left(\frac{1}{4}-q^{2}\right)^{2}}{4}-\int_{\frac{1}{4}}^{q}\left(\frac{1}{4}-y^{2}\right)^{2}\left[\psi'\left(-\frac{1}{2}+y\right)-\psi'\left(-\frac{1}{2}-y\right)\right]\mathrm{d}y \\
&=-\frac{1}{2}\int_{0}^{\frac{1}{4}-q^{2}}\mathrm{d}y\,y^{2}\left[\sum_{n=0}^{\infty}\frac{2n-1}{2\left(n(n-1)+y\right)^{2}}-\frac{1}{y}\right], \tag*{(123)}
\end{aligned} B ( q ) = B ( − i q ) = 4 ( 4 1 − q 2 ) 2 − ∫ 4 1 q ( 4 1 − y 2 ) 2 [ ψ ′ ( − 2 1 + y ) − ψ ′ ( − 2 1 − y ) ] d y = − 2 1 ∫ 0 4 1 − q 2 d y y 2 [ n = 0 ∑ ∞ 2 ( n ( n − 1 ) + y ) 2 2 n − 1 − y 1 ] , (123) q 2 = 9 4 − 12 M 2 R , Q 2 = 4 π μ 2 e − γ E . (124) q^{2}=\frac{9}{4}-\frac{12M^{2}}{\mathcal{R}},\qquad Q^{2}=4\pi\mu^{2}e^{-\gamma_{\mathrm{E}}}. \tag*{(124)} q 2 = 4 9 − R 12 M 2 , Q 2 = 4 π μ 2 e − γ E . (124) The subscript r r r
M 0 2 = m r 2 + ξ r R + 3 c r φ r 2 , M 1 2 = m r 2 + ξ r R + c r φ r 2 . (125) M_{0}^{2}=m_{r}^{2}+\xi_{r}\mathcal{R}+3c_{r}\varphi_{r}^{2},\qquad M_{1}^{2}=m_{r}^{2}+\xi_{r}\mathcal{R}+c_{r}\varphi_{r}^{2}. \tag*{(125)} M 0 2 = m r 2 + ξ r R + 3 c r φ r 2 , M 1 2 = m r 2 + ξ r R + c r φ r 2 . (125) The mass integration leaves a field-independent curvature normalisation. Only local ambiguities of the form allowed by the gravitational action can be absorbed into its finite couplings for varying R \mathcal{R} R
The double-tadpole contractions have weights 3 3 3 2 ( N − 1 ) 2(N-1) 2 ( N − 1 ) N 2 − 1 N^{2}-1 N 2 − 1 ( M 0 , M 0 ) (M_{0},M_{0}) ( M 0 , M 0 ) ( M 0 , M 1 ) (M_{0},M_{1}) ( M 0 , M 1 ) ( M 1 , M 1 ) (M_{1},M_{1}) ( M 1 , M 1 ) [10 ] . Thus
V T ( φ r ) = c r 4 [ 3 t ( M 0 ) 2 + 2 ( N − 1 ) t ( M 0 ) t ( M 1 ) + ( N 2 − 1 ) t ( M 1 ) 2 ] . (126) V_{T}(\varphi_{r})=\frac{c_{r}}{4}\left[3t(M_{0})^{2}+2(N-1)t(M_{0})t(M_{1})+(N^{2}-1)t(M_{1})^{2}\right]. \tag*{(126)} V T ( φ r ) = 4 c r [ 3 t ( M 0 ) 2 + 2 ( N − 1 ) t ( M 0 ) t ( M 1 ) + ( N 2 − 1 ) t ( M 1 ) 2 ] . (126) The cubic vertices give six contractions for three radial lines and 2 ( N − 1 ) 2(N-1) 2 ( N − 1 ) 1 / 2 1/2 1/2
V W ( φ r ) = − c r 2 φ r 2 [ 3 W ( M 0 , M 0 , M 0 ) + ( N − 1 ) W ( M 1 , M 1 , M 0 ) ] . (127) V_{W}(\varphi_{r})=-c_{r}^{2}\varphi_{r}^{2}\left[3W(M_{0},M_{0},M_{0})+(N-1)W(M_{1},M_{1},M_{0})\right]. \tag*{(127)} V W ( φ r ) = − c r 2 φ r 2 [ 3 W ( M 0 , M 0 , M 0 ) + ( N − 1 ) W ( M 1 , M 1 , M 0 ) ] . (127) Restoring ℏ \hbar ℏ
V e f f ( φ r ) = L g , r + 1 2 ( m r 2 + ξ r R ) φ r 2 + c r 4 φ r 4 + ℏ [ V 1 ( M 0 ) + ( N − 1 ) V 1 ( M 1 ) ] + ℏ 2 c r 4 [ 3 t ( M 0 ) 2 + 2 ( N − 1 ) t ( M 0 ) t ( M 1 ) + ( N 2 − 1 ) t ( M 1 ) 2 ] − ℏ 2 c r 2 φ r 2 [ 3 W ( M 0 , M 0 , M 0 ) + ( N − 1 ) W ( M 1 , M 1 , M 0 ) ] + O ( ℏ 3 ) . (128) \begin{aligned}
V_{\mathrm{eff}}(\varphi_{r})={}&\mathcal{L}_{g,r}+\frac{1}{2}\left(m_{r}^{2}+\xi_{r}\mathcal{R}\right)\varphi_{r}^{2}+\frac{c_{r}}{4}\varphi_{r}^{4}+\hbar\left[V_{1}(M_{0})+(N-1)V_{1}(M_{1})\right]\\
&+\hbar^{2}\frac{c_{r}}{4}\left[3t(M_{0})^{2}+2(N-1)t(M_{0})t(M_{1})+(N^{2}-1)t(M_{1})^{2}\right]\\
&-\hbar^{2}c_{r}^{2}\varphi_{r}^{2}\left[3W(M_{0},M_{0},M_{0})+(N-1)W(M_{1},M_{1},M_{0})\right]+O(\hbar^{3}). \tag*{(128)}
\end{aligned} V eff ( φ r ) = L g , r + 2 1 ( m r 2 + ξ r R ) φ r 2 + 4 c r φ r 4 + ℏ [ V 1 ( M 0 ) + ( N − 1 ) V 1 ( M 1 ) ] + ℏ 2 4 c r [ 3 t ( M 0 ) 2 + 2 ( N − 1 ) t ( M 0 ) t ( M 1 ) + ( N 2 − 1 ) t ( M 1 ) 2 ] − ℏ 2 c r 2 φ r 2 [ 3 W ( M 0 , M 0 , M 0 ) + ( N − 1 ) W ( M 1 , M 1 , M 0 ) ] + O ( ℏ 3 ) . (128) For explicit evaluation define
F ( M ) = ψ ( 3 2 + q ( M ) ) + ψ ( 3 2 − q ( M ) ) − 1 , (129) F(M)=\psi\left(\frac{3}{2}+q(M)\right)+\psi\left(\frac{3}{2}-q(M)\right)-1, \tag*{(129)} F ( M ) = ψ ( 2 3 + q ( M ) ) + ψ ( 2 3 − q ( M ) ) − 1 , (129) G ( M ) = 1 + π 2 6 + 1 6 ( M 2 R − 1 6 ) − 1 − 1 144 ( M 2 R − 1 6 ) − 2 + π 2 cos 2 ( π q ( M ) ) , (130) G(M)=1+\frac{\pi^{2}}{6}+\frac{1}{6}\left(\frac{M^{2}}{\mathcal{R}}-\frac{1}{6}\right)^{-1}-\frac{1}{144}\left(\frac{M^{2}}{\mathcal{R}}-\frac{1}{6}\right)^{-2}+\frac{\pi^{2}}{\cos^{2}(\pi q(M))}, \tag*{(130)} G ( M ) = 1 + 6 π 2 + 6 1 ( R M 2 − 6 1 ) − 1 − 144 1 ( R M 2 − 6 1 ) − 2 + cos 2 ( π q ( M )) π 2 , (130) q ( M ) 2 = 9 4 − 12 M 2 R . (131) q(M)^{2}=\frac{9}{4}-\frac{12M^{2}}{\mathcal{R}}. \tag*{(131)} q ( M ) 2 = 4 9 − R 12 M 2 . (131) The tadpole can then be written compactly as
t ( M ) = R 16 π 2 [ ( M 2 R − 1 6 ) ( F ( M ) − log 12 Q 2 R ) − 1 18 ] . (132) t(M)=\frac{\mathcal{R}}{16\pi^{2}}\left[\left(\frac{M^{2}}{\mathcal{R}}-\frac{1}{6}\right)\left(F(M)-\log\frac{12Q^{2}}{\mathcal{R}}\right)-\frac{1}{18}\right]. \tag*{(132)} t ( M ) = 16 π 2 R [ ( R M 2 − 6 1 ) ( F ( M ) − log R 12 Q 2 ) − 18 1 ] . (132) The finite-part conversion is
C Q = C + 2 λ E ∑ j ( 1 4 − q j 2 ) F ( M j ) + ( λ E 2 − λ E ) ∑ j ( 1 4 − q j 2 ) − λ E , λ E = γ E − log 4 π . (133) C_Q=C+2\lambda_E\sum_j\left(\frac{1}{4}-q_j^2\right)F(M_j)+(\lambda_E^2-\lambda_E)\sum_j\left(\frac{1}{4}-q_j^2\right)-\lambda_E,\qquad\lambda_E=\gamma_E-\log4\pi. \tag*{(133)} C Q = C + 2 λ E j ∑ ( 4 1 − q j 2 ) F ( M j ) + ( λ E 2 − λ E ) j ∑ ( 4 1 − q j 2 ) − λ E , λ E = γ E − log 4 π . (133) For any three line masses, set
x a = M a 2 R − 1 6 , L Q = log 12 Q 2 R , D i j k = C Q ( q ( M i ) , q ( M j ) , q ( M k ) ) . (134) x_a=\frac{M_a^2}{\mathcal{R}}-\frac{1}{6},\qquad L_Q=\log\frac{12Q^2}{\mathcal{R}},\qquad D_{ijk}=C_Q(q(M_i),q(M_j),q(M_k)). \tag*{(134)} x a = R M a 2 − 6 1 , L Q = log R 12 Q 2 , D ij k = C Q ( q ( M i ) , q ( M j ) , q ( M k )) . (134) Then Eq. (120) becomes
W ( M i , M j , M k ) = R ( 16 π 2 ) 2 { D i j k − 11 / 12 12 + 1 2 ∑ a ∈ ( i , j , k ) x a [ F ( M a ) 2 + G ( M a ) ] + L Q [ ∑ a ∈ ( i , j , k ) x a F ( M a ) − ∑ a ∈ ( i , j , k ) M a 2 R + 5 12 ] − L Q 2 2 ∑ a ∈ ( i , j , k ) x a } . (135) \mathcal{W}(M_i,M_j,M_k)=\frac{\mathcal{R}}{(16\pi^2)^2}\left\{\frac{D_{ijk}-11/12}{12}+\frac{1}{2}\sum_{a\in(i,j,k)}x_a[F(M_a)^2+G(M_a)]
+L_Q\left[\sum_{a\in(i,j,k)}x_aF(M_a)-\sum_{a\in(i,j,k)}\frac{M_a^2}{\mathcal{R}}+\frac{5}{12}\right]-\frac{L_Q^2}{2}\sum_{a\in(i,j,k)}x_a\right\}. \tag*{(135)} W ( M i , M j , M k ) = ( 16 π 2 ) 2 R ⎩ ⎨ ⎧ 12 D ij k − 11/12 + 2 1 a ∈ ( i , j , k ) ∑ x a [ F ( M a ) 2 + G ( M a )] + L Q a ∈ ( i , j , k ) ∑ x a F ( M a ) − a ∈ ( i , j , k ) ∑ R M a 2 + 12 5 − 2 L Q 2 a ∈ ( i , j , k ) ∑ x a ⎭ ⎬ ⎫ . (135) The sums run over the three propagator lines, counting repeated masses with multiplicity. In particular,
W ( M 1 , M 1 , M 0 ) = R ( 16 π 2 ) 2 { D 110 − 11 / 12 12 + x 1 [ F ( M 1 ) 2 + G ( M 1 ) ] + x 0 2 [ F ( M 0 ) 2 + G ( M 0 ) ] + L Q [ 2 x 1 F ( M 1 ) + x 0 F ( M 0 ) − 2 M 1 2 + M 0 2 R + 5 12 ] − L Q 2 ( M 1 2 R + M 0 2 2 R − 1 4 ) } . (136) \mathcal{W}(M_1,M_1,M_0)=\frac{\mathcal{R}}{(16\pi^2)^2}\left\{\frac{D_{110}-11/12}{12}+x_1[F(M_1)^2+G(M_1)]+\frac{x_0}{2}[F(M_0)^2+G(M_0)]
+L_Q\left[2x_1F(M_1)+x_0F(M_0)-\frac{2M_1^2+M_0^2}{\mathcal{R}}+\frac{5}{12}\right]-L_Q^2\left(\frac{M_1^2}{\mathcal{R}}+\frac{M_0^2}{2\mathcal{R}}-\frac{1}{4}\right)\right\}. \tag*{(136)} W ( M 1 , M 1 , M 0 ) = ( 16 π 2 ) 2 R { 12 D 110 − 11/12 + x 1 [ F ( M 1 ) 2 + G ( M 1 )] + 2 x 0 [ F ( M 0 ) 2 + G ( M 0 )] + L Q [ 2 x 1 F ( M 1 ) + x 0 F ( M 0 ) − R 2 M 1 2 + M 0 2 + 12 5 ] − L Q 2 ( R M 1 2 + 2 R M 0 2 − 4 1 ) } . (136) W ( M 0 , M 0 , M 0 ) = R ( 16 π 2 ) 2 { D 000 − 11 / 12 12 + 3 x 0 2 [ F ( M 0 ) 2 + G ( M 0 ) ] + L Q [ 3 x 0 F ( M 0 ) − 3 M 0 2 R + 5 12 ] − L Q 2 ( 3 M 0 2 2 R − 1 4 ) } . (137) \mathcal{W}(M_0,M_0,M_0)=\frac{\mathcal{R}}{(16\pi^2)^2}\left\{\frac{D_{000}-11/12}{12}+\frac{3x_0}{2}[F(M_0)^2+G(M_0)]
+L_Q\left[3x_0F(M_0)-\frac{3M_0^2}{\mathcal{R}}+\frac{5}{12}\right]-L_Q^2\left(\frac{3M_0^2}{2\mathcal{R}}-\frac{1}{4}\right)\right\}. \tag*{(137)} W ( M 0 , M 0 , M 0 ) = ( 16 π 2 ) 2 R { 12 D 000 − 11/12 + 2 3 x 0 [ F ( M 0 ) 2 + G ( M 0 )] + L Q [ 3 x 0 F ( M 0 ) − R 3 M 0 2 + 12 5 ] − L Q 2 ( 2 R 3 M 0 2 − 4 1 ) } . (137) The finite-part function C ( q ) C(q) C ( q )
By definition, ( 16 π 2 ) − 2 C ( q ) (16\pi^2)^{-2}C(q) ( 16 π 2 ) − 2 C ( q ) (81) . All expansions in this subsection are at fixed labels. Each hypergeometric parameter is either constant in ε \varepsilon ε ± 1 / 2 \pm1/2 ± 1/2 F r \mathcal{F}_r F r ε = 0 \varepsilon=0 ε = 0 F r \mathcal{F}_r F r (83) , multiplied by ( 16 π 2 ) 2 (16\pi^2)^2 ( 16 π 2 ) 2
p − 2 ( r ) ε 2 + p − 1 ( r ) ε + p 0 ( r ) + O ( ε ) , (138) \frac{p_{-2}^{(r)}}{\varepsilon^2}+\frac{p_{-1}^{(r)}}{\varepsilon}+p_0^{(r)}+O(\varepsilon), \tag*{(138)} ε 2 p − 2 ( r ) + ε p − 1 ( r ) + p 0 ( r ) + O ( ε ) , (138) then
C 0 = B 0 − ∑ r = 1 3 [ p − 2 ( r ) 2 d 2 F r d ϵ 2 ∣ 0 + p − 1 ( r ) d F r d ϵ ∣ 0 + p 0 ( r ) F r ( 0 ) ] . (139) C_0=\mathcal{B}_0-\sum_{r=1}^{3}\left[\frac{p_{-2}^{(r)}}{2}\left.\frac{\mathrm{d}^2\mathcal{F}_r}{\mathrm{d}\epsilon^2}\right|_0+p_{-1}^{(r)}\left.\frac{\mathrm{d}\mathcal{F}_r}{\mathrm{d}\epsilon}\right|_0+p_0^{(r)}\mathcal{F}_r(0)\right]. \tag*{(139)} C 0 = B 0 − r = 1 ∑ 3 [ 2 p − 2 ( r ) d ϵ 2 d 2 F r 0 + p − 1 ( r ) d ϵ d F r 0 + p 0 ( r ) F r ( 0 ) ] . (139) The coefficients are built from Gamma functions and rational factors. For a compact explicit expression, define the following four pairs at d = 4 d=4 d = 4
A s t ( 4 ) = 1 / 2 + q 3 + s q 1 + t q 2 2 , B s t ( 4 ) = 3 / 2 + q 3 + s q 1 + t q 2 2 , s , t = ± 1 , (140) A_{st}^{(4)}=\frac{1/2+q_3+sq_1+tq_2}{2},\qquad B_{st}^{(4)}=\frac{3/2+q_3+sq_1+tq_2}{2},\qquad s,t=\pm1, \tag*{(140)} A s t ( 4 ) = 2 1/2 + q 3 + s q 1 + t q 2 , B s t ( 4 ) = 2 3/2 + q 3 + s q 1 + t q 2 , s , t = ± 1 , (140) and
Π = ∏ s , t = ± 1 Γ ( B s t ( 4 ) ) Γ ( A s t ( 4 ) ) , P 0 = ∏ s , t = ± 1 ( 3 2 + q 3 + s q 1 + t q 2 ) . (141) \Pi=\prod_{s,t=\pm1}\frac{\Gamma\left(B_{st}^{(4)}\right)}{\Gamma\left(A_{st}^{(4)}\right)},\qquad P_0=\prod_{s,t=\pm1}\left(\frac{3}{2}+q_3+sq_1+tq_2\right). \tag*{(141)} Π = s , t = ± 1 ∏ Γ ( A s t ( 4 ) ) Γ ( B s t ( 4 ) ) , P 0 = s , t = ± 1 ∏ ( 2 3 + q 3 + s q 1 + t q 2 ) . (141) Theorem 5.1. The finite part is
C = B 0 − 4 ( q 3 + 1 ) P 0 ∑ r = 1 3 [ 2 u r ∂ d 2 F r ( 4 ) − 2 ( w r + ℓ u r ) ∂ d F r ( 4 ) + ( r r + ℓ w r + ℓ 2 + m 2 u r ) F r ( 4 ) ] , (142) C=\mathcal{B}_0-\frac{4(q_3+1)}{P_0}\sum_{r=1}^{3}\left[2u_r\partial_d^2\mathcal{F}_r(4)-2(w_r+\ell u_r)\partial_d\mathcal{F}_r(4)+\left(r_r+\ell w_r+\frac{\ell^2+m}{2}u_r\right)\mathcal{F}_r(4)\right], \tag*{(142)} C = B 0 − P 0 4 ( q 3 + 1 ) r = 1 ∑ 3 [ 2 u r ∂ d 2 F r ( 4 ) − 2 ( w r + ℓ u r ) ∂ d F r ( 4 ) + ( r r + ℓ w r + 2 ℓ 2 + m u r ) F r ( 4 ) ] , (142) where
B 0 = 2 ( X B 2 2 + Y B ) Π , (143) \mathcal{B}_0=2\left(\frac{X_B^2}{2}+Y_B\right)\Pi, \tag*{(143)} B 0 = 2 ( 2 X B 2 + Y B ) Π , (143) X B = 2 log π + 3 − 2 γ E + 2 π tan ( π q 3 ) − 1 2 ∑ s , t [ ψ ( A s t ( 4 ) ) + ψ ( B s t ( 4 ) ) ] , (144) X_B=2\log\pi+3-2\gamma_E+2\pi\tan(\pi q_3)-\frac{1}{2}\sum_{s,t}\left[\psi\left(A_{st}^{(4)}\right)+\psi\left(B_{st}^{(4)}\right)\right], \tag*{(144)} X B = 2 log π + 3 − 2 γ E + 2 π tan ( π q 3 ) − 2 1 s , t ∑ [ ψ ( A s t ( 4 ) ) + ψ ( B s t ( 4 ) ) ] , (144) Y B = π 2 2 + 5 2 + 1 8 ∑ s , t [ ψ ′ ( B s t ( 4 ) ) − ψ ′ ( A s t ( 4 ) ) ] , (145) Y_B=\frac{\pi^2}{2}+\frac{5}{2}+\frac{1}{8}\sum_{s,t}\left[\psi'\left(B_{st}^{(4)}\right)-\psi'\left(A_{st}^{(4)}\right)\right], \tag*{(145)} Y B = 2 π 2 + 2 5 + 8 1 s , t ∑ [ ψ ′ ( B s t ( 4 ) ) − ψ ′ ( A s t ( 4 ) ) ] , (145) ℓ = − 1 + ∑ s , t 1 3 / 2 + q 3 + s q 1 + t q 2 , m = − 1 + ∑ s , t 1 ( 3 / 2 + q 3 + s q 1 + t q 2 ) 2 . (146) \ell=-1+\sum_{s,t}\frac{1}{3/2+q_3+sq_1+tq_2},\qquad m=-1+\sum_{s,t}\frac{1}{(3/2+q_3+sq_1+tq_2)^2}. \tag*{(146)} ℓ = − 1 + s , t ∑ 3/2 + q 3 + s q 1 + t q 2 1 , m = − 1 + s , t ∑ ( 3/2 + q 3 + s q 1 + t q 2 ) 2 1 . (146) Writing Y q = 1 / 4 − q 2 Y_q=1/4-q^2 Y q = 1/4 − q 2
u 1 = 2 Y q 1 Y q 2 , (147) u_1=2Y_{q_1}Y_{q_2}, \tag*{(147)} u 1 = 2 Y q 1 Y q 2 , (147) u 2 = − ( Y q 1 + Y q 2 ) Y q 3 + 1 , (148) u_2=-(Y_{q_1}+Y_{q_2})Y_{q_3+1}, \tag*{(148)} u 2 = − ( Y q 1 + Y q 2 ) Y q 3 + 1 , (148) u 3 = ( q 2 2 − q 1 2 ) 2 , (149) u_3=(q_2^2-q_1^2)^2, \tag*{(149)} u 3 = ( q 2 2 − q 1 2 ) 2 , (149) w 1 = − 2 [ Y q 1 a q 2 + Y q 2 a q 1 ] , (150) w_1=-2[Y_{q_1}a_{q_2}+Y_{q_2}a_{q_1}], \tag*{(150)} w 1 = − 2 [ Y q 1 a q 2 + Y q 2 a q 1 ] , (150) w 2 = ( Y q 1 + Y q 2 ) a q 3 + 1 + Y q 3 + 1 ( a q 1 + a q 2 ) , (151) w_2=(Y_{q_1}+Y_{q_2})a_{q_3+1}+Y_{q_3+1}(a_{q_1}+a_{q_2}), \tag*{(151)} w 2 = ( Y q 1 + Y q 2 ) a q 3 + 1 + Y q 3 + 1 ( a q 1 + a q 2 ) , (151) w 3 = ( q 2 2 − q 1 2 ) 2 χ 3 − ( q 2 2 − q 1 2 ) ( a q 1 − a q 2 ) , (152) w_3=(q_2^2-q_1^2)^2\chi_3-(q_2^2-q_1^2)(a_{q_1}-a_{q_2}), \tag*{(152)} w 3 = ( q 2 2 − q 1 2 ) 2 χ 3 − ( q 2 2 − q 1 2 ) ( a q 1 − a q 2 ) , (152) r 1 = 2 [ a q 1 a q 2 − Y q 1 b q 2 − Y q 2 b q 1 ] , (153) r_1=2[a_{q_1}a_{q_2}-Y_{q_1}b_{q_2}-Y_{q_2}b_{q_1}], \tag*{(153)} r 1 = 2 [ a q 1 a q 2 − Y q 1 b q 2 − Y q 2 b q 1 ] , (153) r 2 = ( Y q 1 + Y q 2 ) b q 3 + 1 + Y q 3 + 1 ( b q 1 + b q 2 ) − ( a q 1 + a q 2 ) a q 3 + 1 , (154) r_2=(Y_{q_1}+Y_{q_2})b_{q_3+1}+Y_{q_3+1}(b_{q_1}+b_{q_2})-(a_{q_1}+a_{q_2})a_{q_3+1}, \tag*{(154)} r 2 = ( Y q 1 + Y q 2 ) b q 3 + 1 + Y q 3 + 1 ( b q 1 + b q 2 ) − ( a q 1 + a q 2 ) a q 3 + 1 , (154) r 3 = ( q 2 2 − q 1 2 ) 2 ( χ 3 2 2 + 1 2 + π 2 12 + π 2 2 cos 2 π q 3 ) − ( q 2 2 − q 1 2 ) ( a q 1 − a q 2 ) χ 3 − ( q 2 2 − q 1 2 ) ( b q 1 − b q 2 ) , (155) r_3=(q_2^2-q_1^2)^2\left(\frac{\chi_3^2}{2}+\frac{1}{2}+\frac{\pi^2}{12}+\frac{\pi^2}{2\cos^2\pi q_3}\right)-(q_2^2-q_1^2)(a_{q_1}-a_{q_2})\chi_3-(q_2^2-q_1^2)(b_{q_1}-b_{q_2}), \tag*{(155)} r 3 = ( q 2 2 − q 1 2 ) 2 ( 2 χ 3 2 + 2 1 + 12 π 2 + 2 cos 2 π q 3 π 2 ) − ( q 2 2 − q 1 2 ) ( a q 1 − a q 2 ) χ 3 − ( q 2 2 − q 1 2 ) ( b q 1 − b q 2 ) , (155) with
χ 3 = log 4 π + 1 − γ E − ψ ( 3 2 + q 3 ) − ψ ( − 1 2 − q 3 ) (156) \chi_3=\log4\pi+1-\gamma_E-\psi\left(\frac{3}{2}+q_3\right)-\psi\left(-\frac{1}{2}-q_3\right) \tag*{(156)} χ 3 = log 4 π + 1 − γ E − ψ ( 2 3 + q 3 ) − ψ ( − 2 1 − q 3 ) (156) a q = 1 − Y q [ log 4 π + 1 − γ E − ψ ( 1 2 + q ) − ψ ( 1 2 − q ) ] (157) a_q=1-Y_q\left[\log4\pi+1-\gamma_E-\psi\left(\frac{1}{2}+q\right)-\psi\left(\frac{1}{2}-q\right)\right] \tag*{(157)} a q = 1 − Y q [ log 4 π + 1 − γ E − ψ ( 2 1 + q ) − ψ ( 2 1 − q ) ] (157) b q = log 4 π − γ E − ψ ( 1 2 + q ) − ψ ( 1 2 − q ) − Y q { 1 2 [ log 4 π + 1 − γ E − ψ ( 1 2 + q ) − ψ ( 1 2 − q ) ] 2 + 1 2 + π 2 12 + π 2 2 cos 2 π q } (158) b_q=\log4\pi-\gamma_E-\psi\left(\frac{1}{2}+q\right)-\psi\left(\frac{1}{2}-q\right)-Y_q\left\{\frac{1}{2}\left[\log4\pi+1-\gamma_E-\psi\left(\frac{1}{2}+q\right)-\psi\left(\frac{1}{2}-q\right)\right]^2+\frac{1}{2}+\frac{\pi^2}{12}+\frac{\pi^2}{2\cos^2\pi q}\right\} \tag*{(158)} b q = log 4 π − γ E − ψ ( 2 1 + q ) − ψ ( 2 1 − q ) − Y q { 2 1 [ log 4 π + 1 − γ E − ψ ( 2 1 + q ) − ψ ( 2 1 − q ) ] 2 + 2 1 + 12 π 2 + 2 cos 2 π q π 2 } (158) Proof. Expand Eq. (81) directly in ε \varepsilon ε ∂ ε = − 2 ∂ d \partial_\varepsilon=-2\partial_d ∂ ε = − 2 ∂ d u r u_r u r w r w_r w r r r r_r r r ℓ \ell ℓ m m m B 0 B_0 B 0
The dimension derivatives in this formula can be evaluated by convergent series. At d = 4 d=4 d = 4
Theorem 5.2 (Evaluation at d = 4 d=4 d = 4 Π \Pi Π P 0 P_0 P 0 (141) ,
F 1 ( 4 ) = P 0 16 ( q 3 + 1 ) Y q 1 Y q 2 [ 1 4 + q 3 2 − q 1 2 − q 2 2 − 4 Π ] (159) \mathcal{F}_1(4)=\frac{P_0}{16(q_3+1)Y_{q_1}Y_{q_2}}\left[\frac{1}{4}+q_3^2-q_1^2-q_2^2-4\Pi\right] \tag*{(159)} F 1 ( 4 ) = 16 ( q 3 + 1 ) Y q 1 Y q 2 P 0 [ 4 1 + q 3 2 − q 1 2 − q 2 2 − 4Π ] (159) F 2 ( 4 ) = P 0 16 ( q 3 + 1 ) Y q 1 Y q 2 Y q 3 + 1 [ ( q 2 2 − q 1 2 ) 2 − ( Y q 1 + Y q 2 ) Y q 3 − 4 ( Y q 1 + Y q 2 ) Π ] (160) \mathcal{F}_2(4)=\frac{P_0}{16(q_3+1)Y_{q_1}Y_{q_2}Y_{q_3+1}}\left[(q_2^2-q_1^2)^2-(Y_{q_1}+Y_{q_2})Y_{q_3}-4(Y_{q_1}+Y_{q_2})\Pi\right] \tag*{(160)} F 2 ( 4 ) = 16 ( q 3 + 1 ) Y q 1 Y q 2 Y q 3 + 1 P 0 [ ( q 2 2 − q 1 2 ) 2 − ( Y q 1 + Y q 2 ) Y q 3 − 4 ( Y q 1 + Y q 2 ) Π ] (160) F 3 ( 4 ) = F 1 ( 4 ) . (161) \mathcal{F}_3(4)=\mathcal{F}_1(4). \tag*{(161)} F 3 ( 4 ) = F 1 ( 4 ) . (161) Proof. In F 3 ( 4 ) \mathcal{F}_3(4) F 3 ( 4 ) F 1 ( 4 ) \mathcal{F}_1(4) F 1 ( 4 )
S = Y q 1 + Y q 2 , D = Y q 1 − Y q 2 , P = Y q 1 Y q 2 , (162) S=Y_{q_1}+Y_{q_2},\qquad D=Y_{q_1}-Y_{q_2},\qquad P=Y_{q_1}Y_{q_2}, \tag*{(162)} S = Y q 1 + Y q 2 , D = Y q 1 − Y q 2 , P = Y q 1 Y q 2 , (162) and, with q 3 , n = q 3 + 1 + 2 n q_{3,n}=q_3+1+2n q 3 , n = q 3 + 1 + 2 n
W n = ∏ s , t ( A s t ( 4 ) ) n ( B s t ( 4 ) ) n , K n = 4 q 3 , n W n ∏ s , t ( 1 / 2 + q 3 , n + s q 1 + t q 2 ) . (163) W_n=\prod_{s,t}\frac{\left(A_{st}^{(4)}\right)_n}{\left(B_{st}^{(4)}\right)_n},\qquad K_n=\frac{4q_{3,n}W_n}{\prod_{s,t}(1/2+q_{3,n}+sq_1+tq_2)}. \tag*{(163)} W n = s , t ∏ ( B s t ( 4 ) ) n ( A s t ( 4 ) ) n , K n = ∏ s , t ( 1/2 + q 3 , n + s q 1 + t q 2 ) 4 q 3 , n W n . (163) Two rational identities telescope the sums:
W n + 1 Y q 3 , n + 1 − S 4 P − W n Y q 3 , n − 1 − S 4 P = K n , W_{n+1}\frac{Y_{q_{3,n+1}}-S}{4P}-W_n\frac{Y_{q_{3,n-1}}-S}{4P}=K_n, W n + 1 4 P Y q 3 , n + 1 − S − W n 4 P Y q 3 , n − 1 − S = K n , W n + 1 S Y q 3 , n + 1 − D 2 4 P − W n S Y q 3 , n − 1 − D 2 4 P = K n Y q 3 , n . (164) W_{n+1}\frac{SY_{q_{3,n+1}}-D^2}{4P}-W_n\frac{SY_{q_{3,n-1}}-D^2}{4P}=K_nY_{q_{3,n}}. \tag*{(164)} W n + 1 4 P S Y q 3 , n + 1 − D 2 − W n 4 P S Y q 3 , n − 1 − D 2 = K n Y q 3 , n . (164) The upper endpoint terms tend to − Π / P -\Pi/P − Π/ P − S Π / P -S\Pi/P − S Π/ P
∑ n ≥ 0 K n = S − Y q 3 − 4 Π 4 P , ∑ n ≥ 0 K n Y q 3 , n = D 2 − S Y q 3 − 4 S Π 4 P . (165) \sum_{n\geq0}K_n=\frac{S-Y_{q_3}-4\Pi}{4P},\qquad\sum_{n\geq0}K_nY_{q_{3,n}}=\frac{D^2-SY_{q_3}-4S\Pi}{4P}. \tag*{(165)} n ≥ 0 ∑ K n = 4 P S − Y q 3 − 4Π , n ≥ 0 ∑ K n Y q 3 , n = 4 P D 2 − S Y q 3 − 4 S Π . (165) The 6 F 5 {}_6F_5 6 F 5 K n / K 0 K_n/K_0 K n / K 0 K 0 = 4 ( q 3 + 1 ) / P 0 K_0=4(q_3+1)/P_0 K 0 = 4 ( q 3 + 1 ) / P 0 8 F 7 {}_8F_7 8 F 7 d = 4 d=4 d = 4 ( K n / K 0 ) ( Y q 3 , n / Y q 3 + 1 ) (K_n/K_0)(Y_{q_{3,n}}/Y_{q_3+1}) ( K n / K 0 ) ( Y q 3 , n / Y q 3 + 1 )
An integral-free convergent series. Introduce
Y q = 1 4 − q 2 , g q = log 4 π + 1 − γ E − ψ ( 1 2 + q ) − ψ ( 1 2 − q ) , Y_q=\frac{1}{4}-q^2,\qquad g_q=\log4\pi+1-\gamma_E-\psi\left(\frac{1}{2}+q\right)-\psi\left(\frac{1}{2}-q\right), Y q = 4 1 − q 2 , g q = log 4 π + 1 − γ E − ψ ( 2 1 + q ) − ψ ( 2 1 − q ) , Ω q = 1 2 + π 2 12 + π 2 2 cos 2 π q . (166) \Omega_q=\frac{1}{2}+\frac{\pi^2}{12}+\frac{\pi^2}{2\cos^2\pi q}. \tag*{(166)} Ω q = 2 1 + 12 π 2 + 2 cos 2 π q π 2 . (166) Using Eq. (140) , define the weight and its logarithmic variations:
K n = 4 ( q 3 + 1 + 2 n ) ∏ l , s , t ( 3 / 2 + q 3 + s q 1 + t q 2 + 2 n ) ∏ s , t ( A s t ( 4 ) ) n ( B s t ( 4 ) ) n , (167) K_n=\frac{4(q_3+1+2n)}{\prod_{l,s,t}\left(3/2+q_3+sq_1+tq_2+2n\right)}\prod_{s,t}\frac{\left(A_{st}^{(4)}\right)_n}{\left(B_{st}^{(4)}\right)_n}, \tag*{(167)} K n = ∏ l , s , t ( 3/2 + q 3 + s q 1 + t q 2 + 2 n ) 4 ( q 3 + 1 + 2 n ) s , t ∏ ( B s t ( 4 ) ) n ( A s t ( 4 ) ) n , (167) L n = − 1 + 1 2 ∑ s , t [ ψ ( n + A s t ( 4 ) ) − ψ ( A s t ( 4 ) ) + ψ ( n + B s t ( 4 ) + 1 ) − ψ ( B s t ( 4 ) ) ] , (168) L_n=-1+\frac{1}{2}\sum_{s,t}\left[\psi\left(n+A_{st}^{(4)}\right)-\psi\left(A_{st}^{(4)}\right)+\psi\left(n+B_{st}^{(4)}+1\right)-\psi\left(B_{st}^{(4)}\right)\right], \tag*{(168)} L n = − 1 + 2 1 s , t ∑ [ ψ ( n + A s t ( 4 ) ) − ψ ( A s t ( 4 ) ) + ψ ( n + B s t ( 4 ) + 1 ) − ψ ( B s t ( 4 ) ) ] , (168) Λ n = − 1 + 1 4 ∑ s , t [ ψ ′ ( n + A s t ( 4 ) ) − ψ ′ ( A s t ( 4 ) ) − ψ ′ ( n + B s t ( 4 ) + 1 ) + ψ ′ ( B s t ( 4 ) ) ] . (169) \Lambda_n=-1+\frac{1}{4}\sum_{s,t}\left[\psi'\left(n+A_{st}^{(4)}\right)-\psi'\left(A_{st}^{(4)}\right)-\psi'\left(n+B_{st}^{(4)}+1\right)+\psi'\left(B_{st}^{(4)}\right)\right]. \tag*{(169)} Λ n = − 1 + 4 1 s , t ∑ [ ψ ′ ( n + A s t ( 4 ) ) − ψ ′ ( A s t ( 4 ) ) − ψ ′ ( n + B s t ( 4 ) + 1 ) + ψ ′ ( B s t ( 4 ) ) ] . (169) Theorem 5.3. With q 3 , n = q 3 + 1 + 2 n q_{3,n}=q_3+1+2n q 3 , n = q 3 + 1 + 2 n
C 0 = B 0 − ∑ n = 0 ∞ K n [ Y q 1 Y q 2 Z n ( q 1 , q 2 ) + 1 2 Y q 1 ( q 2 2 − q 1 2 − Y q 3 , n ) Z n ( q 1 , q 3 , n ) − 1 2 Y q 2 ( q 2 2 − q 1 2 + Y q 3 , n ) Z n ( q 2 , q 3 , n ) − 2 ( Y q 1 + Y q 2 ) J n ( q 1 , q 2 ) + ( Y q 2 + Y q 3 , n ) J n ( q 1 , q 3 , n ) + ( Y q 1 + Y q 3 , n ) J n ( q 2 , q 3 , n ) + Y q 1 + Y q 2 − 2 Y q 3 , n ] . (170) \begin{aligned}
C_0={}&B_0-\sum_{n=0}^{\infty}K_n\biggl[Y_{q_1}Y_{q_2}Z_n(q_1,q_2)+\frac{1}{2}Y_{q_1}\left(q_2^2-q_1^2-Y_{q_{3,n}}\right)Z_n(q_1,q_{3,n})\\
&-\frac{1}{2}Y_{q_2}\left(q_2^2-q_1^2+Y_{q_{3,n}}\right)Z_n(q_2,q_{3,n})-2(Y_{q_1}+Y_{q_2})J_n(q_1,q_2)\\
&+(Y_{q_2}+Y_{q_{3,n}})J_n(q_1,q_{3,n})+(Y_{q_1}+Y_{q_{3,n}})J_n(q_2,q_{3,n})\\
&+Y_{q_1}+Y_{q_2}-2Y_{q_{3,n}}\biggr]. \tag*{(170)}
\end{aligned} C 0 = B 0 − n = 0 ∑ ∞ K n [ Y q 1 Y q 2 Z n ( q 1 , q 2 ) + 2 1 Y q 1 ( q 2 2 − q 1 2 − Y q 3 , n ) Z n ( q 1 , q 3 , n ) − 2 1 Y q 2 ( q 2 2 − q 1 2 + Y q 3 , n ) Z n ( q 2 , q 3 , n ) − 2 ( Y q 1 + Y q 2 ) J n ( q 1 , q 2 ) + ( Y q 2 + Y q 3 , n ) J n ( q 1 , q 3 , n ) + ( Y q 1 + Y q 3 , n ) J n ( q 2 , q 3 , n ) + Y q 1 + Y q 2 − 2 Y q 3 , n ] . (170) where
J n ( q , q ′ ) = L n + g q + g q ′ , Z n ( q , q ′ ) = J n ( q , q ′ ) 2 + Λ n + 2 Ω q + 2 Ω q ′ . (171) J_n(q,q')=L_n+g_q+g_{q'},\qquad Z_n(q,q')=J_n(q,q')^2+\Lambda_n+2\Omega_q+2\Omega_{q'}. \tag*{(171)} J n ( q , q ′ ) = L n + g q + g q ′ , Z n ( q , q ′ ) = J n ( q , q ′ ) 2 + Λ n + 2 Ω q + 2 Ω q ′ . (171) Proof. For d = 4 − 2 ε d=4-2\varepsilon d = 4 − 2 ε κ q = Y q − ε + ε 2 \kappa_q=Y_q-\varepsilon+\varepsilon^2 κ q = Y q − ε + ε 2
16 π 2 T q = − κ q ε E q ( ε ) , (172) 16\pi^2T_q=-\frac{\kappa_q}{\varepsilon}E_q(\varepsilon), \tag*{(172)} 16 π 2 T q = − ε κ q E q ( ε ) , (172) E q ( ε ) = Γ ( 1 + ε ) 1 − ε ( 4 π ) ε Γ ( 1 / 2 + q − ε ) Γ ( 1 / 2 − q − ε ) Γ ( 1 / 2 + q ) Γ ( 1 / 2 − q ) = exp [ ε g q + ε 2 Ω q + O ( ε 3 ) ] , (173) E_q(\varepsilon)=\frac{\Gamma(1+\varepsilon)}{1-\varepsilon}(4\pi)^\varepsilon\frac{\Gamma(1/2+q-\varepsilon)\Gamma(1/2-q-\varepsilon)}{\Gamma(1/2+q)\Gamma(1/2-q)}=\exp\left[\varepsilon g_q+\varepsilon^2\Omega_q+O(\varepsilon^3)\right], \tag*{(173)} E q ( ε ) = 1 − ε Γ ( 1 + ε ) ( 4 π ) ε Γ ( 1/2 + q ) Γ ( 1/2 − q ) Γ ( 1/2 + q − ε ) Γ ( 1/2 − q − ε ) = exp [ ε g q + ε 2 Ω q + O ( ε 3 ) ] , (173) T q η = − Y q ε + a q + b q ε + O ( ε 2 ) , η = 1 16 π 2 . (174) \frac{T_q}{\eta}=-\frac{Y_q}{\varepsilon}+a_q+b_q\varepsilon+O(\varepsilon^2),\qquad\eta=\frac{1}{16\pi^2}. \tag*{(174)} η T q = − ε Y q + a q + b q ε + O ( ε 2 ) , η = 16 π 2 1 . (174) After multiplying by ε 2 \varepsilon^2 ε 2
2 κ q 1 κ q 2 E q 1 E q 2 + κ q 1 ( q 2 2 − q 1 2 − κ q 3 , n ) E q 1 E q 3 , n − κ q 2 ( q 2 2 − q 1 2 + κ q 3 , n ) E q 2 E q 3 , n . (175) 2\kappa_{q_1}\kappa_{q_2}E_{q_1}E_{q_2}+\kappa_{q_1}\left(q_2^2-q_1^2-\kappa_{q_{3,n}}\right)E_{q_1}E_{q_{3,n}}-\kappa_{q_2}\left(q_2^2-q_1^2+\kappa_{q_{3,n}}\right)E_{q_2}E_{q_{3,n}}. \tag*{(175)} 2 κ q 1 κ q 2 E q 1 E q 2 + κ q 1 ( q 2 2 − q 1 2 − κ q 3 , n ) E q 1 E q 3 , n − κ q 2 ( q 2 2 − q 1 2 + κ q 3 , n ) E q 2 E q 3 , n . (175) For v = q 3 , n v=q_{3,n} v = q 3 , n
Table 5.78.
where S = Y q 1 + Y q 2 S=Y_{q_1}+Y_{q_2} S = Y q 1 + Y q 2 E q E q ′ E_qE_{q'} E q E q ′
2 q 3 , n ( d − 2 ) W n ( d ) P 3 + ( q 3 , n ; d ) = K n exp [ ε L n + 1 2 ε 2 Λ n + O ( ε 3 ) ] (176) \frac{2q_{3,n}(d-2)W_n(d)}{P_3^+(q_{3,n};d)}=K_n\exp\left[\varepsilon L_n+\frac{1}{2}\varepsilon^2\Lambda_n+O(\varepsilon^3)\right] \tag*{(176)} P 3 + ( q 3 , n ; d ) 2 q 3 , n ( d − 2 ) W n ( d ) = K n exp [ ε L n + 2 1 ε 2 Λ n + O ( ε 3 ) ] (176) gives Eq. (170) .
This representation uses Gamma functions, digammas, trigammas and one absolutely convergent series. Appendix § 8 gives a form whose summand contains only linear digammas.
Remark 5.4 (Convergence and interchange). On compact sets avoiding poles of the split representation, for large n n n K n = O ( n − 5 ) K_n=O(n^{-5}) K n = O ( n − 5 ) Y q 3 , n = O ( n 2 ) Y_{q_{3,n}}=O(n^2) Y q 3 , n = O ( n 2 ) g q 3 , n = O ( log n ) g_{q_{3,n}}=O(\log n) g q 3 , n = O ( log n ) L n = O ( log n ) L_n=O(\log n) L n = O ( log n ) Λ n = O ( 1 ) \Lambda_n=O(1) Λ n = O ( 1 ) O ( n − 3 log 2 n ) O(n^{-3}\log^2 n) O ( n − 3 log 2 n ) ε \varepsilon ε ∣ Re ( ε ) ∣ ≤ δ < 1 / 2 |\operatorname{Re}(\varepsilon)|\leq\delta<1/2 ∣ Re ( ε ) ∣ ≤ δ < 1/2 n − 5 + 4 δ ( 1 + log 2 n ) n^{-5+4\delta}(1+\log^2 n) n − 5 + 4 δ ( 1 + log 2 n ) n − 3 + 2 δ ( 1 + log 2 n ) n^{-3+2\delta}(1+\log^2 n) n − 3 + 2 δ ( 1 + log 2 n ) C 0 C_0 C 0
Two equal and three equal labels
For ( q 1 , q 2 , q 3 ) = ( q 1 , q 1 , q 0 ) (q_1,q_2,q_3)=(q_1,q_1,q_0) ( q 1 , q 2 , q 3 ) = ( q 1 , q 1 , q 0 ) q 2 2 − q 1 2 q_2^2-q_1^2 q 2 2 − q 1 2 q 0 , n = q 0 + 1 + 2 n q_{0,n}=q_0+1+2n q 0 , n = q 0 + 1 + 2 n
C 0 ( q 1 , q 1 , q 0 ) = B 0 ( q 1 , q 1 , q 0 ) − ∑ n ≥ 0 K n [ Y q 1 2 { ( L n + 2 g q 1 ) 2 + Λ n + 4 Ω q 1 } − Y q 1 Y q 0 , n { ( L n + g q 1 + g q 0 , n ) 2 + Λ n + 2 Ω q 1 + 2 Ω q 0 } − 4 Y q 1 ( L n + 2 g q 1 ) + 2 ( Y q 1 + Y q 0 , n ) ( L n + g q 1 + g q 0 , n ) + 2 Y q 1 − 2 Y q 0 , n ] . (177) \begin{aligned}
C_0(q_1,q_1,q_0)={}&\mathcal{B}_0(q_1,q_1,q_0)-\sum_{n\geq0}K_n\Bigl[Y_{q_1}^2\{(L_n+2g_{q_1})^2+\Lambda_n+4\Omega_{q_1}\}\\
&-Y_{q_1}Y_{q_{0,n}}\{(L_n+g_{q_1}+g_{q_{0,n}})^2+\Lambda_n+2\Omega_{q_1}+2\Omega_{q_0}\}\\
&-4Y_{q_1}(L_n+2g_{q_1})+2(Y_{q_1}+Y_{q_{0,n}})(L_n+g_{q_1}+g_{q_{0,n}})\\
&+2Y_{q_1}-2Y_{q_{0,n}}\Bigr]. \tag*{(177)}
\end{aligned} C 0 ( q 1 , q 1 , q 0 ) = B 0 ( q 1 , q 1 , q 0 ) − n ≥ 0 ∑ K n [ Y q 1 2 {( L n + 2 g q 1 ) 2 + Λ n + 4 Ω q 1 } − Y q 1 Y q 0 , n {( L n + g q 1 + g q 0 , n ) 2 + Λ n + 2 Ω q 1 + 2 Ω q 0 } − 4 Y q 1 ( L n + 2 g q 1 ) + 2 ( Y q 1 + Y q 0 , n ) ( L n + g q 1 + g q 0 , n ) + 2 Y q 1 − 2 Y q 0 , n ] . (177) Here Ω q 0 , n = Ω q 0 \Omega_{q_{0,n}}=\Omega_{q_0} Ω q 0 , n = Ω q 0 d → 4 d\to4 d → 4
A ± = 2 − h + p ± 2 q 2 , B ± = h + p ± 2 q 2 , β = p + 2 − h 2 , γ = p + h 2 , e = p + 1 2 . (178) A_\pm=\frac{2-h+p\pm2q}{2},\qquad B_\pm=\frac{h+p\pm2q}{2},\qquad\beta=\frac{p+2-h}{2},\qquad\gamma=\frac{p+h}{2},\qquad e=\frac{p+1}{2}. \tag*{(178)} A ± = 2 2 − h + p ± 2 q , B ± = 2 h + p ± 2 q , β = 2 p + 2 − h , γ = 2 p + h , e = 2 p + 1 . (178) The two nonzero hypergeometric constants are both 6 F 5 {}_6F_5 6 F 5
F 1 ( d ; q , q , p ) = 6 F 5 [ A + , β , β , A − , e + 1 , 1 B + + 1 , γ + 1 , γ + 1 , B − + 1 , e ; 1 ] , (179) \mathcal{F}_1(d;q,q,p)={}_6F_5\left[\begin{matrix}A_+,\beta,\beta,A_-,e+1,1\\B_++1,\gamma+1,\gamma+1,B_-+1,e\end{matrix};1\right], \tag*{(179)} F 1 ( d ; q , q , p ) = 6 F 5 [ A + , β , β , A − , e + 1 , 1 B + + 1 , γ + 1 , γ + 1 , B − + 1 , e ; 1 ] , (179) F 2 ( d ; q , q , p ) = 6 F 5 [ A + , β , A − , e + 1 , γ + 1 2 , 1 B + + 1 , γ + 1 , B − + 1 , e , β + 1 2 ; 1 ] . (180) \mathcal{F}_2(d;q,q,p)={}_6F_5\left[\begin{matrix}A_+,\beta,A_-,e+1,\gamma+\frac{1}{2},1\\B_++1,\gamma+1,B_-+1,e,\beta+\frac{1}{2}\end{matrix};1\right]. \tag*{(180)} F 2 ( d ; q , q , p ) = 6 F 5 [ A + , β , A − , e + 1 , γ + 2 1 , 1 B + + 1 , γ + 1 , B − + 1 , e , β + 2 1 ; 1 ] . (180) These identities hold in d d d d d d q 1 = q 0 q_1=q_0 q 1 = q 0 (177) .
Beta functions
Finally, we derive the beta functions up to two loops.
Divergences and beta functions in four dimensions
Set v = φ r v=\varphi_r v = φ r n = N − 1 n=N-1 n = N − 1 K = 16 π 2 K=16\pi^2 K = 16 π 2 r r r
A = m 2 + ( ξ − 1 6 ) R , Δ 0 = A + 3 c v 2 , Δ 1 = A + c v 2 . (181) A=m^2+\left(\xi-\frac{1}{6}\right)\mathcal{R},\qquad\Delta_0=A+3cv^2,\qquad\Delta_1=A+cv^2. \tag*{(181)} A = m 2 + ( ξ − 6 1 ) R , Δ 0 = A + 3 c v 2 , Δ 1 = A + c v 2 . (181) The matter-dependent one-loop counterterm is
δ V 1 = Δ 0 2 + n Δ 1 2 4 K ε , Δ 0 2 + n Δ 1 2 = N A 2 + 2 ( N + 2 ) c A v 2 + ( N + 8 ) c 2 v 4 . (182) \delta V_1=\frac{\Delta_0^2+n\Delta_1^2}{4K\varepsilon},\qquad\Delta_0^2+n\Delta_1^2=NA^2+2(N+2)cAv^2+(N+8)c^2v^4. \tag*{(182)} δ V 1 = 4 K ε Δ 0 2 + n Δ 1 2 , Δ 0 2 + n Δ 1 2 = N A 2 + 2 ( N + 2 ) c A v 2 + ( N + 8 ) c 2 v 4 . (182) The geometric heat-kernel term adds − N R 2 / ( 4320 K ε ) -N\mathcal{R}^2/(4320K\varepsilon) − N R 2 / ( 4320 K ε ) S 4 S^4 S 4
δ M 0 , 1 2 = ( N + 2 ) c A + 3 ( N + 8 ) c 2 v 2 K ε , (183) \delta M_{0,1}^2=\frac{(N+2)cA+3(N+8)c^2v^2}{K\varepsilon}, \tag*{(183)} δ M 0 , 1 2 = K ε ( N + 2 ) c A + 3 ( N + 8 ) c 2 v 2 , (183) δ M 1 , 1 2 = ( N + 2 ) c A + ( N + 8 ) c 2 v 2 K ε . (184) \delta M_{1,1}^2=\frac{(N+2)cA+(N+8)c^2v^2}{K\varepsilon}. \tag*{(184)} δ M 1 , 1 2 = K ε ( N + 2 ) c A + ( N + 8 ) c 2 v 2 . (184) From Eq. (119) , the subdivergence-subtracted sunset is
R ′ I i j k = 1 K 2 [ Δ i + Δ j + Δ k 2 ε 2 + R / 24 − ( Δ i + Δ j + Δ k ) / 2 ε ] + W i j k + O ( ε ) . (185) R'I_{ijk}=\frac{1}{K^2}\left[\frac{\Delta_i+\Delta_j+\Delta_k}{2\varepsilon^2}+\frac{\mathcal{R}/24-(\Delta_i+\Delta_j+\Delta_k)/2}{\varepsilon}\right]+\mathcal{W}_{ijk}+O(\varepsilon). \tag*{(185)} R ′ I ij k = K 2 1 [ 2 ε 2 Δ i + Δ j + Δ k + ε R /24 − ( Δ i + Δ j + Δ k ) /2 ] + W ij k + O ( ε ) . (185) Combining the two vacuum topologies with 1 2 ( δ M 0 , 1 2 T 0 + n δ M 1 , 1 2 T 1 ) \frac{1}{2}(\delta M_{0,1}^2T_0+n\delta M_{1,1}^2T_1) 2 1 ( δ M 0 , 1 2 T 0 + n δ M 1 , 1 2 T 1 )
δ V 2 = 1 4 K 2 ε 2 [ N ( N + 2 ) c A 2 + 2 ( N + 2 ) ( N + 5 ) c 2 A v 2 + ( N + 8 ) 2 c 3 v 4 ] − c 2 v 2 4 K 2 ε [ 6 ( N + 2 ) A + 2 ( 5 N + 22 ) c v 2 − ( N + 2 ) R 6 ] . (186) \delta V_2=\frac{1}{4K^2\varepsilon^2}\left[N(N+2)cA^2+2(N+2)(N+5)c^2Av^2+(N+8)^2c^3v^4\right]
-\frac{c^2v^2}{4K^2\varepsilon}\left[6(N+2)A+2(5N+22)cv^2-\frac{(N+2)\mathcal{R}}{6}\right]. \tag*{(186)} δ V 2 = 4 K 2 ε 2 1 [ N ( N + 2 ) c A 2 + 2 ( N + 2 ) ( N + 5 ) c 2 A v 2 + ( N + 8 ) 2 c 3 v 4 ] − 4 K 2 ε c 2 v 2 [ 6 ( N + 2 ) A + 2 ( 5 N + 22 ) c v 2 − 6 ( N + 2 ) R ] . (186) Here δ V 1 + δ V 2 \delta V_1+\delta V_2 δ V 1 + δ V 2
δ V = δ ρ + δ κ R + δ b R 2 + 1 2 ( δ m A 2 + δ ξ A R ) v 2 + 1 4 δ c A v 4 , (187) \delta V=\delta\rho+\delta\kappa\mathcal{R}+\delta b\mathcal{R}^2+\frac{1}{2}(\delta m_A^2+\delta\xi_A\mathcal{R})v^2+\frac{1}{4}\delta c_Av^4, \tag*{(187)} δ V = δ ρ + δ κ R + δ b R 2 + 2 1 ( δ m A 2 + δ ξ A R ) v 2 + 4 1 δ c A v 4 , (187) the defining relations are
ϕ B = Z ϕ 1 / 2 ϕ , m B 2 = m 2 + δ m A 2 Z ϕ , ξ B = ξ + δ ξ A Z ϕ , c B = μ 2 ε c + δ c A Z ϕ 2 . (188) \phi_B=Z_\phi^{1/2}\phi,\qquad m_B^2=\frac{m^2+\delta m_A^2}{Z_\phi},\qquad\xi_B=\frac{\xi+\delta\xi_A}{Z_\phi},\qquad c_B=\mu^{2\varepsilon}\frac{c+\delta c_A}{Z_\phi^2}. \tag*{(188)} ϕ B = Z ϕ 1/2 ϕ , m B 2 = Z ϕ m 2 + δ m A 2 , ξ B = Z ϕ ξ + δ ξ A , c B = μ 2 ε Z ϕ 2 c + δ c A . (188) The two-loop simple residues in Eq. (186) are
δ m A , 21 2 = − 3 ( N + 2 ) c 2 m 2 K 2 , (189) \delta m_{A,21}^2=-\frac{3(N+2)c^2m^2}{K^2}, \tag*{(189)} δ m A , 21 2 = − K 2 3 ( N + 2 ) c 2 m 2 , (189) δ ξ A , 21 = ( N + 2 ) c 2 K 2 [ − 3 ( ξ − 1 6 ) + 1 12 ] , (190) \delta\xi_{A,21}=\frac{(N+2)c^2}{K^2}\left[-3\left(\xi-\frac{1}{6}\right)+\frac{1}{12}\right], \tag*{(190)} δ ξ A , 21 = K 2 ( N + 2 ) c 2 [ − 3 ( ξ − 6 1 ) + 12 1 ] , (190) δ c A , 21 = − 2 ( 5 N + 22 ) c 3 K 2 . (191) \delta c_{A,21}=-\frac{2(5N+22)c^3}{K^2}. \tag*{(191)} δ c A , 21 = − K 2 2 ( 5 N + 22 ) c 3 . (191) The index 21 denotes the coefficient of 1 / ε 1/\varepsilon 1/ ε 2 ( 3 − 1 ) = 4 2(3-1)=4 2 ( 3 − 1 ) = 4 c B c_B c B μ 2 ε \mu^{2\varepsilon} μ 2 ε
A constant-field potential fixes action combinations, not the wavefunction counterterm separately. The additional local calculation is the massless momentum-dependent sunset. Successive one-loop convolutions give
J ( p ) = ( p 2 ) 1 − 2 ε ( 4 π ) 4 − 2 ε Γ ( 1 − ε ) Γ 3 ( − 1 + 2 ε ) Γ ( 3 − 3 ε ) = − p 2 4 K 2 ε + O ( 1 ) . (192) J(p)=\frac{(p^2)^{1-2\varepsilon}}{(4\pi)^{4-2\varepsilon}}\frac{\Gamma(1-\varepsilon)\Gamma^3(-1+2\varepsilon)}{\Gamma(3-3\varepsilon)}=-\frac{p^2}{4K^2\varepsilon}+O(1). \tag*{(192)} J ( p ) = ( 4 π ) 4 − 2 ε ( p 2 ) 1 − 2 ε Γ ( 3 − 3 ε ) Γ ( 1 − ε ) Γ 3 ( − 1 + 2 ε ) = − 4 K 2 ε p 2 + O ( 1 ) . (192) For the vertex tensor 2 c ( δ a b δ c d + δ a c δ b d + δ a d δ b c ) 2c(\delta_{ab}\delta_{cd}+\delta_{ac}\delta_{bd}+\delta_{ad}\delta_{bc}) 2 c ( δ ab δ c d + δ a c δ b d + δ a d δ b c ) 2 ( N + 2 ) c 2 2(N+2)c^2 2 ( N + 2 ) c 2 1 / 6 1/6 1/6 + ( N + 2 ) c 2 p 2 / ( 2 K 2 ε ) +(N+2)c^2p^2/(2K^2\varepsilon) + ( N + 2 ) c 2 p 2 / ( 2 K 2 ε )
Z ϕ = 1 − ( N + 2 ) c 2 2 K 2 ε + O ( c 3 ) , γ ϕ = 1 2 μ d log Z ϕ d μ = ( N + 2 ) c 2 K 2 + O ( c 3 ) . (193) Z_\phi=1-\frac{(N+2)c^2}{2K^2\varepsilon}+O(c^3),\qquad\gamma_\phi=\frac{1}{2}\mu\frac{\mathrm{d}\log Z_\phi}{\mathrm{d}\mu}=\frac{(N+2)c^2}{K^2}+O(c^3). \tag*{(193)} Z ϕ = 1 − 2 K 2 ε ( N + 2 ) c 2 + O ( c 3 ) , γ ϕ = 2 1 μ d μ d log Z ϕ = K 2 ( N + 2 ) c 2 + O ( c 3 ) . (193) Reading the simple poles and including field normalisation yields
β c = 2 ( N + 8 ) c 2 K − 12 ( 3 N + 14 ) c 3 K 2 + O ( c 4 ) , (194) \beta_c=\frac{2(N+8)c^2}{K}-\frac{12(3N+14)c^3}{K^2}+O(c^4), \tag*{(194)} β c = K 2 ( N + 8 ) c 2 − K 2 12 ( 3 N + 14 ) c 3 + O ( c 4 ) , (194) β m 2 = 2 ( N + 2 ) c m 2 K − 10 ( N + 2 ) c 2 m 2 K 2 + O ( c 3 m 2 ) , (195) \beta_{m^2}=\frac{2(N+2)cm^2}{K}-\frac{10(N+2)c^2m^2}{K^2}+O(c^3m^2), \tag*{(195)} β m 2 = K 2 ( N + 2 ) c m 2 − K 2 10 ( N + 2 ) c 2 m 2 + O ( c 3 m 2 ) , (195) β ξ = 2 ( N + 2 ) c K ( ξ − 1 6 ) + ( N + 2 ) c 2 K 2 [ − 10 ( ξ − 1 6 ) + 2 3 ] + O ( c 3 ) , (196) \beta_\xi=\frac{2(N+2)c}{K}\left(\xi-\frac{1}{6}\right)+\frac{(N+2)c^2}{K^2}\left[-10\left(\xi-\frac{1}{6}\right)+\frac{2}{3}\right]+O(c^3), \tag*{(196)} β ξ = K 2 ( N + 2 ) c ( ξ − 6 1 ) + K 2 ( N + 2 ) c 2 [ − 10 ( ξ − 6 1 ) + 3 2 ] + O ( c 3 ) , (196) γ ϕ = ( N + 2 ) c 2 K 2 + O ( c 3 ) . (197) \gamma_\phi=\frac{(N+2)c^2}{K^2}+O(c^3). \tag*{(197)} γ ϕ = K 2 ( N + 2 ) c 2 + O ( c 3 ) . (197) Thus ξ = 1 / 6 \xi=1/6 ξ = 1/6 β ξ \beta_\xi β ξ γ m = β m 2 / ( 2 m 2 ) \gamma_m=\beta_{m^2}/(2m^2) γ m = β m 2 / ( 2 m 2 ) λ ( ϕ 2 ) 2 / 4 ! \lambda(\phi^2)^2/4! λ ( ϕ 2 ) 2 /4 ! λ = 6 c \lambda=6c λ = 6 c
Conclusions
We have constructed the renormalised two-loop effective potential of the O ( N ) O(N) O ( N ) [26 ] . We have proved the boundary condition selecting the physical solution by comparison with conformally coupled spectator lines. The resulting representation isolates the ultraviolet poles and gives the finite part as an absolutely convergent series.
We have tracked the regulator dependence of the spectral labels when the field-dependent masses and scalar curvature are held fixed. The associated derivative terms cancel after the forest subtraction. Combining the renormalised watermelon with the double-tadpole diagrams gives the two-loop potential, and the local counterterms determine the matter-sector beta functions once the wavefunction renormalisation is supplied by a separate momentum-dependent calculation.
The result is perturbative: exact mass-to-curvature dependence does not remove infrared limitations near propagator zero modes. The corresponding anti-de Sitter calculation and extensions to higher-loop vacuum diagrams are natural applications of the same finite-difference framework.
Lowering the weight: only linear digammas remain
The summand of Eq. (170) is quadratic in digammas. An exchange of summation removes these squares because Eq. (164) supplies both tails in closed form. In this appendix use
Y i = Y q i , a i = a q i , b i = b q i , v n = q 3 + 1 + 2 n , g n = g v n , (198) Y_i = Y_{q_i}, \qquad a_i = a_{q_i}, \qquad b_i = b_{q_i}, \qquad v_n = q_3 + 1 + 2n, \qquad g_n = g_{v_n}, \tag*{(198)} Y i = Y q i , a i = a q i , b i = b q i , v n = q 3 + 1 + 2 n , g n = g v n , (198) S = Y 1 + Y 2 , D = Y 1 − Y 2 , P = Y 1 Y 2 . (199) S = Y_1 + Y_2, \qquad D = Y_1 - Y_2, \qquad P = Y_1Y_2. \tag*{(199)} S = Y 1 + Y 2 , D = Y 1 − Y 2 , P = Y 1 Y 2 . (199) The tails are
T n = ∑ m ≥ n K m = − 4 Π − W n ( Y v n − 1 − S ) 4 P , U n = ∑ m ≥ n K m Y v m = S T n − W n . (200) T_n = \sum_{m \ge n} K_m = \frac{-4\Pi- W_n(Y_{v_n-1} - S)}{4P}, \qquad U_n = \sum_{m \ge n} K_mY_{v_m} = ST_n - W_n. \tag*{(200)} T n = m ≥ n ∑ K m = 4 P − 4Π − W n ( Y v n − 1 − S ) , U n = m ≥ n ∑ K m Y v m = S T n − W n . (200) Set
A = a 1 + a 2 , B = b 1 + b 2 , C A = − 2 ( Y 1 a 2 + Y 2 a 1 ) , E = S − D ( a 1 − a 2 ) , J n = L n + g n . (201) A = a_1 + a_2, \qquad B = b_1 + b_2, \qquad C_A = -2(Y_1a_2 + Y_2a_1), \qquad E = S - D(a_1 - a_2), \qquad J_n = L_n + g_n. \tag*{(201)} A = a 1 + a 2 , B = b 1 + b 2 , C A = − 2 ( Y 1 a 2 + Y 2 a 1 ) , E = S − D ( a 1 − a 2 ) , J n = L n + g n . (201) The appendix coefficient C A C_A C A C ( q ) C(q) C ( q )
N n = − S 2 [ J n 2 + Λ n + 2 Ω q 3 ] + A J n + B , (202) N_n = -\frac{S}{2}\left[J_n^2 + \Lambda_n + 2\Omega_{q_3}\right] + AJ_n + B, \tag*{(202)} N n = − 2 S [ J n 2 + Λ n + 2 Ω q 3 ] + A J n + B , (202) V n = D 2 2 [ J n 2 + Λ n + 2 Ω q 3 ] + E J n + C A L n + P ( L n 2 + Λ n ) + 2 a 1 a 2 − A − S ( 1 + B ) . (203) V_n = \frac{D^2}{2}\left[J_n^2 + \Lambda_n + 2\Omega_{q_3}\right] + EJ_n + C_AL_n + P(L_n^2 + \Lambda_n) + 2a_1a_2 - A - S(1 + B). \tag*{(203)} V n = 2 D 2 [ J n 2 + Λ n + 2 Ω q 3 ] + E J n + C A L n + P ( L n 2 + Λ n ) + 2 a 1 a 2 − A − S ( 1 + B ) . (203) The bracket in Eq. (170) is Y v n N n + V n Y_{v_n}N_n + V_n Y v n N n + V n
p n = 1 2 ∑ s , t [ 1 n + A s t ( 4 ) + 1 n + B s t ( 4 ) + 1 ] , p_n = \frac{1}{2}\sum_{s,t}\left[\frac{1}{n + A_{st}^{(4)}} + \frac{1}{n + B_{st}^{(4)} + 1}\right], p n = 2 1 s , t ∑ [ n + A s t ( 4 ) 1 + n + B s t ( 4 ) + 1 1 ] , r n = − 2 v n + 1 / 2 − 2 v n + 3 / 2 , t n = p n + r n . (204) r_n = -\frac{2}{v_n + 1/2} - \frac{2}{v_n + 3/2}, \qquad t_n = p_n + r_n. \tag*{(204)} r n = − v n + 1/2 2 − v n + 3/2 2 , t n = p n + r n . (204) δ Λ n = Λ n + 1 − Λ n = 1 4 ∑ s , t [ − 1 ( n + A s t ( 4 ) ) 2 + 1 ( n + B s t ( 4 ) + 1 ) 2 ] . \delta\Lambda_n = \Lambda_{n+1} - \Lambda_n = \frac{1}{4}\sum_{s,t}\left[-\frac{1}{(n + A_{st}^{(4)})^2} + \frac{1}{(n + B_{st}^{(4)} + 1)^2}\right]. δ Λ n = Λ n + 1 − Λ n = 4 1 s , t ∑ [ − ( n + A s t ( 4 ) ) 2 1 + ( n + B s t ( 4 ) + 1 ) 2 1 ] . Hence
δ N n = ( A − S J n ) t n − S 2 ( t n 2 + δ Λ n ) , (205) \delta N_n = (A - SJ_n)t_n - \frac{S}{2}(t_n^2 + \delta\Lambda_n), \tag*{(205)} δ N n = ( A − S J n ) t n − 2 S ( t n 2 + δ Λ n ) , (205) δ V n = ( D 2 J n + E ) t n + ( C A + 2 P L n ) p n + D 2 2 ( t n 2 + δ Λ n ) + P ( p n 2 + δ Λ n ) . (206) \delta V_n = (D^2J_n + E)t_n + (C_A + 2PL_n)p_n + \frac{D^2}{2}(t_n^2 + \delta\Lambda_n) + P(p_n^2 + \delta\Lambda_n). \tag*{(206)} δ V n = ( D 2 J n + E ) t n + ( C A + 2 P L n ) p n + 2 D 2 ( t n 2 + δ Λ n ) + P ( p n 2 + δ Λ n ) . (206) Theorem A.1 (Lower-weight form). The sum in Eq. (170) equals
U 0 N 0 + T 0 V 0 + ∑ n = 0 ∞ [ N n + 1 δ N n + T n + 1 δ V n ] . (207) U_0N_0 + T_0V_0 + \sum_{n=0}^{\infty}\left[N_{n+1}\delta N_n + T_{n+1}\delta V_n\right]. \tag*{(207)} U 0 N 0 + T 0 V 0 + n = 0 ∑ ∞ [ N n + 1 δ N n + T n + 1 δ V n ] . (207) Its summand is O ( n − 3 log n ) O(n^{-3}\log n) O ( n − 3 log n ) and contains only linear J n J_n J n L n L_n L n and rational functions of n n n there is no moving trigamma or squared digamma.
Proof. Use T n − T n + 1 = K n T_n-T_{n+1}=K_n T n − T n + 1 = K n U n − U n + 1 = K n Y v n U_n-U_{n+1}=K_nY_{v_n} U n − U n + 1 = K n Y v n N N N V V V U N N N + T N V N U_NN_N+T_NV_N U N N N + T N V N U N = O ( N − 2 ) U_N=O(N^{-2}) U N = O ( N − 2 ) T N = O ( N − 4 ) T_N=O(N^{-4}) T N = O ( N − 4 ) N N , V N = O ( log 2 N ) N_N,V_N=O(\log^2N) N N , V N = O ( log 2 N )
References 