QUANTITATIVE ESTIMATES ON THE HYDROGEN GROUND STATE ENERGY IN NON-RELATIVISTIC QED

QUANTITATIVE ESTIMATES ON THE HYDROGEN GROUND STATE ENERGY IN NON-RELATIVISTIC QED J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Abstract. In this paper, we determine the exact expression or the hydrogen ground state energy in the Pauli-Fierz model up to the order O(α 5 log α 1 ), where α denotes the inestructure constant, and prove rigorous bounds on the remainder term o the order o(α 5 log α 1 ). As a consequence, we prove that the ground state energy is not a real analytic unction o α, and veriy the existence o logarithmic corrections to the expansion o the ground state energy in powers o α, as conjectured in the recent literature. 1. Introduction For a hydrogen-like atom consisting o an electron interacting with a static nucleus o charge ez, described by the Schrödinger Hamiltonian αz x, in σ( ) in σ( αz x ) = (Zα) 4 corresponds to the binding energy necessary to remove the electron to spatial ininity. The interaction o the electron with the quantized electromagnetic ield is accounted or by adding to αz x the photon ield energy operator H, and an operator I(α) which describes the coupling o the electron to the quantized electromagnetic ield; the small parameter α is the ine structure constant. Thereby, one obtains the Pauli-Fierz Hamiltonian described in detail in Section. In this case, determining the binding energy (1) in σ ( + H + I(α) ) in σ ( αz x + H + I(α) ) is a very hard problem. A main obstacle emerges rom the act that the ground state energy is not an isolated eigenvalue o the Hamiltonian, and can not be determined with ordinary perturbation theory. Furthermore, the photon orm actor in the quantized electromagnetic vector potential occurring in the interaction term I(α) contains a critical requency space singularity that is responsible or the inamous inrared problem in quantum electrodynamics. As a consequence, quantities such as the ground state energy do not exist as a convergent power series in the ine structure constant α with coeicients independent o α. In recent years, several rigorous results addressing the computation o the binding energy have been obtained, [14, 13, 8]. In particular, the coupling to the photon ield has been shown to increase the binding energy o the electron to the nucleus, and that up to normal ordering, the leading term is (αz) 4 [1, 14, 8]. Moreover, or a model with scalar bosons, the binding energy is determined in [13], in the irst subleading order in powers o α, up to α 3, with error term 1.

J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER α 3 log α 1. This result has inspired the question o a possible emergence o logarithmic terms in the expansion o the binding energy; however, this question has so ar remained open. In [], a sophisticated rigorous renormalization group analysis is developed in order to determine the ground state energy (and the renormalized electron mass) up to any arbitrary precision in powers o α, with an expansion o the orm () ε 0 + N k=1 ε k (α)α k/ + o(α N ), (or any given N) where the coeicients ε k (α) diverge as α 0, but are smaller in magnitude than any positive power o α 1. The recursive algorithms developed in [] are highly complex, and explicitly computing the ground state energy to any subleading order in powers o α is an extensive task. While it is expected that the rate o divergence o these coeicient unctions is proportional to a power o log α 1, this is not explicitly exhibited in the current literature; or instance, it can a priori not be ruled out that terms involving logarithmic corrections cancel mutually. The goal o the current paper is to develop an alternative method (as a continuation o [4]) that determines the binding energy up to several subleading orders in powers o α, with rigorous error bounds, and proving the presence o terms logarithmic in α. The main result established in the present paper (or Z = 1) states that the binding energy can be estimated as α (3) 4 + e(1) α 3 + e () α 4 + e (3) α 5 log α 1 + o(α 5 log α 1 ), where e (i) (i = 1,, 3) are independent o α, e (1) > 0, and e (3) 0. Their explicit values are given in Theorem.1. As a consequence, we conclude that the binding energy is not analytic in α. In addition, our proo clariies how the logarithmic actor in α 5 log α 1 is linked to the inrared singularity o the photon orm actor in the interaction term I(α). We note that or some models with a mild inrared behavior, [10], the ground state energy is proven to be analytic in α. Organization o the proo. Our proo o the expression (3) or the hydrogen binding energy involves the ollowing main steps: Since the binding energy (1) involves the ground state energy o the sel-energy operator T = + H + I(α), it is necessary to determine the value o in σ(t ). This was achieved in [4] (see also Lemma A.7), using in particular the act that in σ(t ) = in σ(t (0)) [9, 6], where T (0) is the restriction o T to total momentum P = 0 (see details in Section, (7)). The derivation o (3) requires both a lower and an upper bound or the binding energy. This is accomplished in three steps. The irst term in the expansion or the binding energy is the Coulomb term α 4 ; it can be recovered with an approximate ground state containing no photon, and with an electronic part given by the ground state u α o the Schrödinger-Coulomb operator α x. The irst step, described in Section 4, thus consists o estimating the contribution to the binding energy stemming rom states orthogonal to u α with respect to the quadratic orm or α x + H + I(α).

HYDROGEN GROUND STATE ENERGY IN QED 3 The next two steps consist o estimating the binding energy up to the order α 3 (Theorem 5., Section 5), and subsequently to the order α 5 log α 1 (Theorem 6.1, Section 6). One o the key ingredient to control some o the error terms occurring in both the upper and lower bounds, is an estimate on the expected photon number or the ground state Ψ o the total hamiltonian αz x + H + I(α). Such a bound is derived in Section 3 and yields Ψ, N Ψ = O(α log α 1 ), where N is the photon number operator. In Section, we give a detailed introduction o the model and state the main results. In the Appendix, we provide some technical lemmata used in the derivation o the binding energy.. The model We study a scalar electron interacting with the quantized electromagnetic ield in the Coulomb gauge, and with the electrostatic potential generated by a ixed nucleus. The Hilbert space accounting or the Schrödinger electron is given by H el = L (R 3 ). The Fock space o photon states is given by F = n N F n, where the n-photon space F n = n ( s L (R 3 ) C ) is the symmetric tensor product o n copies o one-photon Hilbert spaces L (R 3 ) C. The actor C accounts or the two independent transversal polarizations o the photon. On F, we introduce creation- and annihilation operators a λ (k), a λ(k) satisying the distributional commutation relations [ a λ (k), a λ (k ) ] = δ λ,λ δ(k k ), [ a λ (k), a λ (k ) ] = 0, where a λ denotes either a λ or a λ. There exists a unique unit ray Ω F, the Fock vacuum, which satisies a λ (k) Ω = 0 or all k R 3 and λ {1, }. The Hilbert space o states o the system consisting o both the electron and the radiation ield is given by H := H el F. We use units such that = c = 1, and where the mass o the electron equals m = 1/. The electron charge is then given by e = α, where the ine structure constant α will here be considered as a small parameter. Let x R 3 be the position vector o the electron and let y i R 3 be the position vector o the i-th photon. We consider the normal ordered Pauli-Fierz Hamiltonian on H or Hydrogen, (4) : ( i x I αa(x) ) : +V (x) I + I el H. where : : denotes normal ordering, corresponding to the subtraction o a normal ordering constant c n.o. α, with c n.o. I := [A + (x), A (x)] is independent o x. The electrostatic potential V (x) is the Coulomb potential or a static point nucleus o charge e = α (i.e., Z = 1) V (x) = α x. We will describe the quantized electromagnetic ield by use o the Coulomb gauge condition.

4 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER The operator that couples an electron to the quantized vector potential is given by A(x) = χ Λ ( k ) [ ] π k ε λ(k) e ikx a 1/ λ (k) + e ikx a λ(k) dk λ=1, R 3 where by the Coulomb gauge condition, diva = 0. The vectors ε λ (k) R 3 are the two orthonormal polarization vectors perpendicular to k, ε 1 (k) = (k, k 1, 0) k 1 + k and ε (k) = k k ε 1(k). The unction χ Λ implements an ultraviolet cuto on the wavenumbers k. We assume χ Λ to be o class C 1, with compact support in { k Λ}, χ Λ 1 and χ Λ = 1 or k Λ 1. For convenience, we shall write where A (x) = A(x) = A (x) + A + (x), λ=1, R 3 χ Λ ( k ) π k 1/ ε λ(k) e ikx a λ (k) dk is the part o A(x) containing the annihilation operators, and A + (x) = (A (x)). The photon ield energy operator H is given by H = k a λ(k)a λ (k)dk. R 3 λ=1, We will, with exception o our discussion in Section 3, study the unitarily equivalent Hamiltonian ( (5) H = U : ( i x I αa(x) ) : +V (x) I + I el H )U, where the unitary transorm U is deined by and P = λ=1, U = e ip.x, is the photon momentum operator. We have k a λ(k)a λ (k)dk Ui x U = i x + P and UA(x)U = A(0). In addition, the Coulomb operator V, the photon ield energy H, and the photon momentum P remain unchanged under the action o U. Thereore, in this new system o variables, the Hamiltonian reads as ollows H = : ( (i x I I el P ) αa(0) ) (6) : +H α x, where :... : denotes again the normal ordering. Notice that the operator H can be rewritten, taking into account the normal ordering and omitting, by abuse o notations, the operators I el and I, H = ( x α x ) + (H + P ) Re (i x.p ) α(i x P ).A(0) + αa + (0).A (0) + αre (A (0)).

HYDROGEN GROUND STATE ENERGY IN QED 5 For a ree spinless electron coupled to the quantized electromagnetic ield, the sel-energy operator T is given by T = : ( i x I αa(x) ) : +Iel H. We note that this system is translationally invariant; that is, T commutes with the operator o total momentum P tot = p el I + I el P, where p el and P denote respectively the electron and the photon momentum operators. Let H P = C F denote the ibre Hilbert space corresponding to conserved total momentum P. For ixed value P o the total momentum, the restriction o T to the ibre space H P is given by (see e.g. [6]) (7) T (P ) = : (P P αa(0)) : +H where by abuse o notation, we again dropped all tensor products involving the identity operators I and I el. Henceorth, we will write Moreover, we denote A ± A ± (0). Σ 0 = in σ(t ) and Σ = in σ(h) = in σ(t + V ). It is proven in [1, 6] that Σ 0 = in σ(t (0)) is an eigenvalue o the operator T (0). Our main result is the ollowing theorem. Theorem.1. The binding energy ulills (8) Σ 0 Σ = 1 4 α + e (1) α 3 + e () α 4 + e (3) α 5 log α 1 + o(α 5 log α 1 ), where e () = 3 Re + 1 3 3 e (1) = π 0 χ Λ (t) 1 + t dt, (A ) i (H + P ) 1 A +.A + Ω, (H + P ) 1 (A + ) i Ω ( (H + P ) 1 A +.P (H + P ) 1 (A + ) i P i (H + P ) 1 A +.A +) Ω A (H + P ) 1 (A + ) i Ω + 4a 0 Q 1 ( 1 x + 1 4 ) 1 u1, a 0 = k 1 + k 4π k 3 k + k χ Λ( k ) dk 1 dk dk 3, e (3) = 1 3π ( 1 x + 1 4 ) 1 u1, and Q 1 is the projection onto the orthogonal complement to the ground state u 1 o the Schrödinger operator 1 x (or α = 1).

6 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Lemma 3.1. Let 3. Bounds on the expected photon number K = (i αa(x)) + H α x = v + H α x, be the Pauli-Fierz operator deined without normal ordering, where v = i αa(x). Let Ψ H denote the ground state o K, normalized by Let N := K Ψ = E Ψ, Ψ = 1. λ=1, a λ(k) a λ (k) dk denote the photon number operator. Then, there exists a constant c independent o α, such that or any suiciently small α > 0, the estimate is satisied. Proo. Using Ψ, N Ψ c α log α 1 [a λ (k), H ] = k, and [a λ (k), v] = ɛ(k) χ π k 1 λ (k)e ik.x and the pull-through ormula, a λ (k)eψ = a λ (k)kψ [ = (H + k )a λ (k) 1 ] x a λ(k) + [v, a λ (k)]v + v[v, a λ (k)] Ψ, we get αχλ ( k ) (9) a λ (k) Ψ = π k From (9), we obtain a λ (k)ψ α χ Λ( k ) k α χ Λ( k ) k 3 K + k E (( i ) αa(x) ɛ λ (k)e ik.x) Ψ. ( i ) αa(x) 1 K + k E [ Ψ α ], K Ψ + Ψ, x Ψ, since K E 0, and (i αa(x) ) (K + α x ). We have (10) Ψ, K Ψ c α. Moreover, or the normal ordered hamiltonian deined in (4), we have : K : = x 4 αre ( i x.a (x) ) + α : A(x) : +H α x (1 c α)( x ) + α : A(x) : +(1 c (11) α)h α x cα 1 x α x Ψ

HYDROGEN GROUND STATE ENERGY IN QED 7 where in the last inequality we used (see e.g. [11]) Inequalities (10), (11) and imply Thus, (1) α : A(x) : +(1 c α)h cα. 1 4 x α x 4α 1 4 xψ, Ψ cα Ψ. Ψ, α x Ψ xψ, Ψ + 1 4 α Ψ (c + 1 4 )α Ψ. Collecting (10), (1) and (9) we ind (13) a λ (k)ψ c α χ Λ( k ). k 3 This a priori bound exhibits the L -critical singularity in requency space. It does not take into consideration the exponential localization o the ground state due to the conining Coulomb potential, and appears in a similar orm or the ree electron. To account or the latter, we use the ollowing two results rom the work o Griesemer, Lieb, and Loss, [11]. Equation (58) in [11] provides the bound a λ (k)ψ < c α χ Λ ( k ) x Ψ. k 1 Moreover, Lemma 6. in [11] states that exp[β x ]Ψ c or any [ 1 + β < Σ 0 E = O(α ). 1 ] Σ 0 E β Ψ, For the 1-electron case, Σ 0 is the electron sel-energy, and E is the ground state o : K :. Choosing β = O(α), x Ψ x 4 Ψ 1 3 (4!) 1 4 4 Ψ 4 β c β [ 1 + c 1 α 5 4. 1 ] 1 8 Σ 0 E β Ψ exp[β x ]Ψ 1 4 Ψ 3 4 Notably, this bound only depends on the binding energy o the potential. Thus, (14) a λ (k)ψ < cα 3 4 χ Λ ( k ). k 1 We see that binding to the Coulomb potential weakens the inrared singularity by a actor k, but at the expense o a large constant actor α. For the ree electron, this estimate does not exist.

8 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Using (13) and (14), we ind Ψ, N Ψ = dk a λ (k)ψ k <δ dk c α 3 k + c α 3 δ + c α log 1 δ δ k Λ dk c α k 3 c α 9 4 + c α log α 1. or δ = α 15 8. This proves the lemma. 4. Estimates on the quadratic orm or states orthogonal to the ground state o the Schrödinger operator Throughout this paper, we will denote by Γ n the projection onto the n-th photon sector (without distinction or the n-photon sector o F and the n-photon sector o H). We also deine Γ n = 1 n 1 j=0 Γj. Starting with this section, we study the Hamiltonian H deined in (6), written in relative coordinates. In particular, i x now stands or the operator unitarily equivalent to the operator o total momentum, which, by abuse o notation, will be denoted by P. Let (15) u α (x) = 1 α 3/ e α x / 8π be the normalized ground state o the Schrödinger operator h α := x α x. We will also denote by e 0 = α 4 and e 1 = α 16 the two lowest eigenvalues o this operator. Theorem 4.1. Assume that Φ H ulils Γ k Φ, u α L (R 3,dx) = 0, or all k 0. Then there exists ν > 0 and α 0 > 0 such that or all 0 < α < α 0 (16) HΦ, Φ (Σ 0 e 0 ) Φ + δ Φ + ν H 1 Φ, where δ = (e 0 e 1 )/ = 3 3 α. Remark 4.1. All photons with momenta larger than the ultraviolet cuto do not contribute to lower the energy. More precisely, due to the cuto unction χ Λ ( k ) in the deinition o A(x), and since we have H = (i x P ) αre (i x P ).A(0) + α : A(0) : +H α x, it ollows that or any given normalized state Φ H, there exists a normalized state Φ Λ such that x R 3, or all n {1,,...), or all ((k 1, λ 1 ), (k, λ ),..., (k n, λ n )) ( (R 3 \ {k, k Λ + 1}) {1, } ) n, we have (17) Γ n Φ Λ (x, (k 1, λ 1 ), (k, λ ),..., (k n, λ n )) = 0 and Φ Λ, HΦ Λ Φ, HΦ and Φ Λ, T Φ Λ Φ, T Φ.

HYDROGEN GROUND STATE ENERGY IN QED 9 A key consequence o this remark is that throughout the paper, all states will be implicitly assumed to ulill condition (17). This is crucial or the proo o Corollary 4.. To prove Theorem 4.1, we irst need the ollowing Lemma. Lemma 4.1. There exists c 0 > 0 such that or all α small enough we have H 1 (P P ) 1 H c 0 α. A straightorward consequence o this lemma is the ollowing result. Corollary 4.1. Let Ψ GS be the normalized ground state o H. Then (18) H Ψ GS, Ψ GS c 0 α Ψ GS Proo. This ollows rom HΨ GS, Ψ GS (Σ 0 e 0 ) Ψ GS < 0. Moreover, rom Theorem 4.1 and Lemma 4.1, we obtain Corollary 4.. Assume that Φ H is such that Γ n Φ, u α L (R 3,dx) = 0 holds or all n 0. Then, or ν and δ deined in Theorem 4.1, there exists ζ > 0, and α 0 > 0 such that or all 0 < α < α 0 we have (19) HΦ, Φ (Σ 0 e 0 ) Φ + M[Φ], where (0) M[Φ] := δ Φ + ν H 1 Φ + ζ (P P )Φ + ζ Γ n 4 P Φ. Proo. According to Remark 4.1, there exists c > 1 such that the operator inequality P Γn 4 ch Γ n 4 holds on the set o states or which (17) is satisied. The value o c only depends on the ultraviolet cuto Λ. Thus, Γ n 4 P Φ (P P )Φ + Γ n 4 P Φ (P P )g + c H 1 Γ n 4 Φ, which yields (P P )Φ 1 Γn 4 P Φ c H 1 Γ n 4 Φ. Thereore, it suices to prove Corollary 4. with M[Φ] replaced by δ (1) M[Φ] := Φ + 3 4 ν H 1 Φ + ζ (P P )Φ, and ζ small enough so that cζ < ν 4. Now we consider two cases. Let c 1 := max{8c 0, 8δ/α }. I (P P )Φ c 1 α Φ, Theorem 4.1 and the above remark imply (19). I (P P )Φ > c 1 α Φ, Lemma 4.1 implies that HΦ, Φ 1 (P P ) Φ, Φ + 1 H Φ, Φ c 0 α Φ 1 4 (P P ) Ψ, Ψ + 1 H Φ, Φ + 1 8 (P P )Φ 1 4 (P P ) Φ, Φ + 1 H Φ, Φ + c 1α which concludes the proo since Σ 0 e 0 < 0. 8 Φ,

10 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Proo o Lemma 4.1. Recall the notation A ± A ± (0). The ollowing holds. H 1 (P P ) 1 H = 1 (P P ) α x αre ((P P ).A(0)) + αre (A ) + αa +.A + 1 H, () and (3) 1 4 (P P ) α x 4α. α (P P ).A(0)ψ, ψ α (P P )ψ + α A ψ. By the Schwarz inequality, there exists c 1 independent o α such that (4) A ψ c 1 H 1 ψ. Inequalities (3)-(4) imply that or small α, (5) α (P P ).A(0)ψ, ψ 1 4 (P P )ψ + 1 4 H ψ, ψ. Moreover using (4) and (6) A + ψ c ψ + c 3 H 1 ψ, we arrive at (7) α (A ) ψ, ψ = α A ψ, A + ψ ɛ A ψ + ɛ 1 α A + ψ ɛc 1 H 1 ψ + ɛ 1 α (c ψ + c 3 H 1 ψ ). Collecting the inequalities (), (5) and (7) with ɛ < 1/(8c 1 ) and α small enough, completes the proo. Proo o Theorem 4.1: Let Φ := Φ 1 + Φ := χ( P < pc pc )Φ + χ( P )Φ, where P = i x is the total momentum operator (due to the transormation (5)) and p c = 1 3 is a lower bound on the norm o the total momentum or which [6, Theorem 3.] holds. Since P commutes with the translation invariant operator H + α x, we have or all ɛ (0, 1), (8) HΦ, Φ = HΦ 1, Φ 1 + HΦ, Φ Re α x Φ 1, Φ HΦ 1, Φ 1 + HΦ, Φ ɛ α x Φ 1, Φ 1 ɛ 1 α x Φ, Φ. First, we have the ollowing estimate : (P P αa(0)) : +H = (P P ) Re (P P ). αa(0) + α : A(0) : +H (1 α)(p P ) + (α α) : A(0) : +H c n.o. α (1 α)(p P ) + (1 O( α))h O( α)

HYDROGEN GROUND STATE ENERGY IN QED 11 where in the last inequality we used (4) and (6). Thereore (9) (H ɛ 1 α x )Φ, Φ ( 1 α (P P ) (1 + ɛ 1 ) α x )Φ, Φ ( 1 α + (P P ) + (1 O( α))h O( ) α) Φ, Φ The lowest eigenvalue o the Schrödinger operator (1 O( α)) (1+ɛ 1 )α x larger than c ɛ α. Thus, using (9) and denoting we get L := 1 α (P P ) + (1 O( α))h O( α) c ɛ α, (30) HΦ, Φ ) ɛ 1 α x Φ, Φ (LΦ, Φ. Now we have the ollowing alternative: Either P < pc 3, in which case we have LΦ, Φ ( p c 4 O( α)) Φ, or P pc 3, in which case, using Φ = χ( P > p c )Φ and H P, we have L ( pc 6 O( α)) Φ. In both cases, this yields the bound (31) LΦ, Φ p c 48 Φ (Σ 0 e 0 + 7 8 (e 0 e 1 )) Φ since, or α small enough, the right hand side tends to zero, whereas p c is a constant independent o α. Inequalities (30) and (31) yield (3) HΦ, Φ ɛ 1 α x Φ, Φ (Σ 0 e 0 + 7 8 (e 0 e 1 )) Φ. For T (P ) being the sel-energy operator with ixed total momentum P deined in (7), we have rom [6, Theorem3. (B)] P in σ(t (P )) in σ(t (0)) c 0αP. Thereore (33) T (P )Φ 1, Φ 1 (1 o α (1))(P Φ 1, Φ 1 + Σ 0 Φ 1. I Φ 8 Φ 1, using (33) yields (34) HΦ 1, Φ 1 ɛ α x Φ 1, Φ 1 (1 o α (1))(P Φ 1, Φ 1 (1 + ɛ) α x Φ 1, Φ 1 + Σ 0 Φ 1 (Σ 0 (1 + O(α) + O(ɛ))e 0 ) Φ 1. Thereore, together with Φ 8 Φ 1 and (3), or α and ɛ small enough this implies (35) HΦ, Φ (Σ 0 e 0 ) Φ + 3 4 (e 0 e 1 ) Φ. is

1 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER (36) I Φ < 8 Φ 1, we have, rom (33) HΦ 1, Φ 1 ɛ α x Φ 1, Φ 1 (1 o α (1))(P Φ 1, Φ 1 (1 + ɛ) α x Φ 1, Φ 1 + Σ 0 Φ 1 ( (1+o α (1)+O(ɛ)) e 0 Γ k Φ 1, u α L (R 3,dx) +e 1 ( Φ 1 k=0 ) Γ k Φ 1, u α L (R 3,dx) ) + Σ 0 Φ 1. k=0 Now, by orthogonality o Φ and u α in the sense that or all k, Γ k Φ, u α L (R 3,dx) = 0, we get u α, Γ k Φ 1 = u α, Γ k Φ Φ u α χ( P p c ) (37) k=0 k=0 8 Φ 1 u α χ( P p c ) α 0 0 Thus, or α and ɛ small enough, (36), (37) and (3) imply also (35) in that case. Let c = max{δ, c 0 α }. I H Φ, Φ < 8 c Φ, (16) ollows rom (35) with ν = δ/(16 c). I H Φ, Φ 8 c Φ, using Lemma 4.1, we obtain (38) HΦ, Φ 1 H Φ, Φ c 0 α g 1 4 H Φ, Φ + c Φ 1 4 H Φ, Φ + δ Φ + (Σ 0 e 0 ) Φ, since Σ 0 e 0 0, which yields (16) with ν = 1 4. This concludes the proo o (16). 5. Estimate o the binding energy up to α 3 term Deinition 5.1. Let u α be the normalized ground state o h α := α x. We deine the projection Ψ uα F o the normalized ground state Ψ GS o H, onto u α as ollows Ψ GS = u α Ψ uα + Ψ, where or all k 0, (39) u α, Γ k Ψ L (R 3,dx) = 0. Deinition 5.. Let Φ := (H + P ) 1 A + A + Ω Φ 3 := (H + P ) 1 P A + Φ Φ 1 := (H + P ) 1 P A Φ where evidently, the state Φ i contains i photons.

HYDROGEN GROUND STATE ENERGY IN QED 13 Deinition 5.3. On F, we deine the positive bilinear orm (40) v, w := v, (H + P ) w, and its associated semi-norm v = v, v 1/. We will also use the same notation or this bilinear orms on F n, H and H n. Similarly, we deine the bilinear orm.,. on H as u, v := u, (H + P + h α + e 0 ) v and its associated semi-norm v = v, v 1/. Deinition 5.4. Let (41) Φ uα := α 1 uα.(h + P ) 1 A + Ω, and (4) Φ uα := α 1 uα.(h + P + h α + e 0 ) 1 A + Ω. Remark 5.1. The unction Φ uα is not a vector in the Fock space H because o the inrared singularity o the photon orm actor. However, H γ Φuα belongs to H, or any γ > 0. Theorem 5.1 (A priori estimate on the binding energy). We have (43) Σ Σ 0 e 0 Φ uα + O(α 4 ) Proo. Using the trial unction in H and rom [4, Theorem 3.1], Φ trial := u α (Ω + α 3 Φ 1 + αφ + α 3 Φ 3 ) + Φ uα, Σ 0 = α Φ + α 3 ( A Φ 4 Φ 1 4 Φ 3 ) + O(α 4 ), the result ollows straightorwardly. We decompose the unction Ψ uα deined in Deinition 5.1 as ollows. Deinition 5.5. Let Ψ uα := Γ 0 Ψ uα + η 1 α 3 Φ 1 + η αφ + η 3 α 3 Φ 3 + uα, = 0, Φ i, Γ i uα = 0 (,,3), and Φ 1, Φ, Φ 3 are given in Dei- where Γ 0 uα nition 5.. We urther decompose Ψ into two parts. Deinition 5.6. Let with Φ uα, Γ 1 = 0. Ψ =: κ 1 Φ uα +, To establish the estimate o the binding energy up to the α 3 term, we will compute HΨ GS, Ψ GS according to the decomposition o Ψ GS introduced in Deinitions 5.1 to 5.6. Using H = (P α x ) + T (0) Re P (P + αa(0)), and due to the orthogonality o u α and Ψ, we obtain (44) HΨ GS, Ψ GS = Hu α Ψ uα, u α Ψ uα + HΨ, Ψ 4Re P.(P + αa(0))u α Ψ uα, Ψ.

14 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER We will estimate separately each term in (44). 5.1. Estimate o the term Hu α Ψ uα, u α Ψ uα. Lemma 5.1. Hu α Ψ uα, u α Ψ uα e 0 Ψ uα α Φ Γ 0 Ψ uα + α η Γ 0 Ψ uα Φ + α 3 η ( A Φ 4 Φ 1 4 Φ 3 ) + 4α 3 ( η 1 η Φ 1 + η 3 η Φ 3 ) + cα 4 log α 1 ( η 1 + η + η 3 + Γ 0 Ψ uα ) + 1 uα. Proo. The proo is a trivial modiication o the one given or the lower bound in [4, Theorem 3.1]. The only modiication is that we have a slightly weaker estimate in Lemma 3.1 on the photon number or the ground state. This is accounted or by replacing the term o order α 4 in [4, Theorem 3.1] by a term o order α 4 log α 1. In addition, we need to use the equality P.(P + αa(0))u α Ψ uα, u α Ψ uα = 0, due to the symmetry o u α. 5.. Estimates or the cross term 4Re P.(P + αa(0))u α Ψ uα, Ψ. Lemma 5. ( 4Re P.P u α Ψ uα, Ψ term). 4Re P.P u α Ψ uα, Ψ cα 4 ( η 1 + η + η 3 ) ɛ H 1 cα H 1 uα. Proo. For n = 1 photon, (45) Obviously (46) P.P Γ 1 u α Ψ uα, Ψ = P.P (η 1 α 3 Φ 1 + Γ 1 uα )u α, κ 1 Φ uα + Γ 1. P.P (η 1 α 3 Φ 1 + Γ 1 uα )u α, Γ 1 ɛ H 1 Γ 1 + cα 5 η 1 + cα H 1 Γ 1 uα. Due to Lemma A.4 holds H 1 (Φ uα (47) Φ uα ) = O(α 5 ), which implies P.P (η 1 α 3 Φ 1 + Γ 1 uα )u α, κ 1 Φ uα κ 1 cα 5 + c η 1 α 5 + cα H 1 Γ 1 uα + P.P (η 1 α 3 Φ 1 + Γ 1 uα )u α, κ 1 Φ uα. For the last term on the right hand side o (47), due to the orthogonality o uα and uα x j, i j, and the equality uα = uα x j, holds (48) P.P (η 1 α 3 Φ 1 + Γ 1 uα )u α, κ 1 Φ uα = = c η 1 α 3 Φ 1 + Γ 1 uα, κ 1 A +.P Ω = 0. For n = photons, u α η 1 α 3 Φ 1 + Γ 1 uα, κ 1 P i (A + Ω ) i P.P Γ u α Ψ uα, Ψ = P.P (η αφ + Γ uα )u α, Γ cα 4 η + ɛ H 1 Γ + α H 1 Γ uα.

HYDROGEN GROUND STATE ENERGY IN QED 15 For n = 3 photons, a similar estimate yields For n 4 photons, P.P Γ 3 u α Ψ uα, Ψ = P.P (η 3 α 3 Φ 3 + Γ 3 uα )u α, Γ 3 cα 4 η 3 + ɛ H 1 Γ 3 + α H 1 Γ 3 uα. P.P Γ n 4 u α Ψ uα, Ψ ɛ H 1 Γ n 4 + α H 1 Γ n 4 uα. Lemma 5.3 ( 4Re αp.a(0)u α Ψ uα, Ψ term). 4Re αp.a(0))u α Ψ uα, Ψ Re κ 1 Γ 0 Ψ uα Φ uα ɛ H 1 ɛα cα H 1 uα cα 4 ( η 1 + η + η 3 + κ 1 ) + O(α 5 log α 1 ). Proo. We irst estimate the term Re α 1 P.A + u α Ψ uα, Ψ = αre P.A + u α Γ n Ψ uα, Γ n+1 Ψ. For n = 0 photon, using the orthogonality Φ uα, Γ 1 = 0, yields (49) n=0 Re α 1 P.A + Γ 0 u α Ψ uα, κ 1 Φ uα + Γ 1 = 1 Re ( = 1 ) (κ Re 1 Γ 0 Ψ uα Φ uα, Φ uα For n 1 photons, (50). (Γ 0 Ψ uα ) Φ uα Re α 1 P.A + Γ n u α uα, Γ n+1 cα 3 uα + ɛ H 1 n ɛ H 1 + cα 5 ( η 1 + η + η 3 ) + O(α 5 log α 1 ),, κ 1 Φ uα + Γ 1 ) where we used rom Lemma 3.1 that uα O(α log α 1 ) + cα 3 ( η 1 + η 3 ) + cα η. We also have (51) (5) Re α 1 P (η1 α 3 Φ 1 + η αφ + η 3 α 3 Φ 3 )u α, A ɛ H 1 + c(α 6 η 1 + α 5 η +α 6 η 3 ). We next estimate the term Re α 1 P.A u α Ψ uα, Ψ. We irst get Then we write (53) α 1 Re P.A uα u α, ɛα + cα H 1 uα. α 1 Re (η1 α 3 A Φ 1 + η 3 α 3 A Φ 3 ) u α, ɛα + cα 4 ( η 1 + η 3 ). We also have (54) α 1 Re η αa Φ. u α, κ 1 Φ uα = α 1 Re η αh 1 A Φ. u α, H 1 κ 1 Φ uα cα 4 ( η + κ 1 ),

16 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER since H 1 A Φ L (R 3 ) and H 1 (55) Φ uα α 1 Re η αa Φ. u α, Γ 1 = O(α 3 ). Finally we get = α 1 Re η αh 1 A Φ. u α, H 1 Γ 1 ɛ H 1 Γ 1 + cα 5 η. Collecting (49) to (55) concludes the proo. 5.3. Estimate or the term HΨ, Ψ. Lemma 5.4. (56) HΨ, Ψ (Σ 0 e 0 ) cα 4 κ 1 + 1 M[ ] + κ 1 Φ uα, where M[. ] is deined in Corollary 4. Proo. Recall that H = (P α x ) + (H + P ) Re (P.P ) α(p P ).A(0) + αa +.A + αre (A ). Due to the orthogonality Φ uα, = 0 we get (57) HΨ, Ψ = Hr, r + κ 1 Φ uα Re P.P Φ uα, Φ uα 4 αre P.A +, κ 1 Φ uα 4Re P.P Φ uα,. e 0 κ 1 Φ uα e 0 κ 1, Φ uα + α κ 1 A Φ uα + αre A.A, κ 1 Φ uα 4 αre P.A, κ 1 Φ uα + 4Re α P.A(0), κ 1 Φ uα + 4αRe A +.A, κ 1 Φ uα For the irst term on the right hand side o (57), we have, rom Corollary 4. (58) H, (Σ 0 e 0 ) + M[ ]. According to Lemma A.3, we obtain (59) e 0 κ 1, Φ uα e 0 κ 1 Φ uα ɛα c κ 1 α 5 log α 1. The next term, namely α κ 1 A Φ uα, is positive. Due to the symmetry in x-variable, (60) P.P Φ uα, Φ uα = 0. The term αre A.A, κ 1Φ uα is estimated as (61) αre A.A, κ 1 Φ uα cα κ 1 Φ uα 1 4 ν H 1 = 1 4 ν H 1 c κ 1 α 5 log α 1. Due to Lemma A.3 and [11, Lemma A4], (6) α 1 A, P κ 1 Φ uα ν 8 H 1 + c κ 1 α 6 log α 1. The next term we have to estimate ulils (63) α 1 P.A +, κ 1 Φ uα ɛ P Γ 0 +cα κ 1 A Φ uα ɛ (P P ) c κ 1 α 4.

HYDROGEN GROUND STATE ENERGY IN QED 17 We have (64) Re α P.A(0), κ 1 Φ uα = Re α P.A Γ, κ 1 Φ uα + Re α P.A + Γ 0, κ 1 Φ uα = Re α P.A Γ, κ 1 Φ uα + Re α A +.P Γ 0, κ 1 Φ uα Obviously, A +.P Γ 0, κ 1Φ uα = 0, and the irst term is bounded by (65) Re α P.A Γ, κ 1 Φ uα ɛ H 1 Γ + cα κ 1 P Φ uα ɛ H 1 Γ + c κ 1 α 4. For the term αre A +.A, κ 1Φ uα we obtain (66) αre A +.A, κ 1 Φ uα cα H 1 κ 1 Φ uα κ 1 ɛ H 1 = cα 5 κ 1 ɛ H 1. According to (14) o Lemma A.3, the last term we have to estimate ulils (67) Re P.P κ 1 Φ uα, ɛ H 1 Γ 1 + c P P 1 κ1 Φ uα ɛ H 1 Γ 1 + cα 5 κ 1. Collecting the estimates (57) to (67) yields (56). 5.4. Upper bound on the binding energy up to the order α 3. Theorem 5.. 1) Let Σ = in σ(h). Then (68) Σ = Σ 0 e 0 Φ uα + O(α 4 ), Σ 0 = α Φ + α 3 ( A Φ 4 Φ 1 4 Φ 3 ) + O(α 4 ), and Φ uα deined by (41). ) For the components, Ψuα, uα o the ground state Ψ GS, and the coeicients η 1, η, η 3 and κ 1 deined in Deinitions 5.1-5.5, holds (69) (70) = O(α 33 1 16 ), H = O(α 4 ), (P P ) = O(α 4 ), Ψ uα 1 cα, Γ 0 Ψ uα 1 cα, = O(α 4 ), uα = O(α 33 (71) uα 16 ), (7) η 1,3 1 cα, η 1 cα, κ 1 1 cα. Proo. Step 1: We irst show that (68) holds with an error estimate o the order α 4 log α 1. Collecting Lemmata 5.1, 5., 5.3 and Lemma 5.4 yields (73) HΨ GS, Ψ GS e 0 Ψ uα α Φ Γ 0 Ψ uα + α 3 η ( A Φ 4 Φ 1 4 Φ 3 ) + α η Γ 0 Ψ uα Φ + 4α 3 ( η 1 η Φ 1 + η 3 η Φ 3 ) cα 4 log α 1 ( η 1 + η + η 3 + Γ 0 Ψ uα ) + 1 4 uα + κ 1 Φ uα + (Σ 0 e 0 ) + 1 4 M[ ] cα 4 κ 1 Re (κ 1 Γ 0 Ψ uα ) Φ uα + O(α 5 log α 1 ).

18 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER We irst estimate (74) κ 1 Φ uα Re (κ 1 Γ 0 Ψ uα ) Φ uα cα 4 κ 1 Φ uα Γ 0 Ψ uα + κ 1 Γ 0 Ψ uα Φ uα + O(α 4 ). Moreover, since Γ 0 Ψ uα 1, we replace in (73) α Φ Γ 0 Ψ uα by α Φ and in (74) Φ uα Γ0 Ψ uα by Φ uα. In addition, using the inequalities (75) η η j 1 η j Γ 0 Ψ uα η Γ 0 Ψ uα and η j η j Γ 0 Ψ uα + or j = 1,, 3 yields that or some c > 0, (76) HΨ GS, Ψ GS e 0 Ψ uα α Φ + α 3 η ( A Φ 4 Φ 1 4 Φ 3 ) Φ uα + cα η Γ 0 Ψ uα Φ + cα 3 ( η 1 Γ 0 Ψ uα + η 3 Γ 0 Ψ uα ) + cα 3 κ 1 Γ 0 Ψ uα + 1 4 uα + (Σ 0 e 0 ) + 1 4 M[ ] + O(α 4 log α 1 ). Comparing this expression with (43) o Theorem 5.1 gives (77) Σ = Σ 0 e 0 Φ uα + O(α 4 log α 1 ). Finally, by Lemma A.4, we can replace Φ uα by Φ uα in (77). Step : We now show that the error term does not contain any log α 1 term, by deriving improved estimates on the photon number or and uα. From (76) we obtain (78) uα = O(α 4 log α 1 ) and H 1 = O(α 4 log α 1 ). This last two relations enable us to improve the estimates on the expected photon number as ollows (79) N 1 u α = O(α 33 (80) N 1 = O(α 33 16 ) 16 ) To see this, we note that rom Deinition 5.1, we have a λ (k)ψ uα a λ (k)ψ GS cα 3 χ Λ ( k ), k where in the last inequality, we used (14). Taking into account that where we arrive at a λ (k)φ 1 cχ Λ( k ) k uα = Ψ uα α 3 Φ 1 αφ α 3 Φ 3,, a λ (k)φ cχ Λ( k )(1 + log k ) k a λ (k)φ 3 cχ Λ( k ) k a λ (k) uα cα 3 χ Λ ( k )(1 + log k ). k,

HYDROGEN GROUND STATE ENERGY IN QED 19 For the expected photon number o uα thus holds N 1 u α = a λ (k) uα dk λ λ k α 15 8 cα 17 8 + cα 15 cα 3 (1 + log k ) dk + k k >α 15 8 1 8 H uα 16. cα 33 The relation (80) can be proved similarly using a λ (k)φ uα c α 1 k. k 1 k a λ (k) uα dk We are now in position to inish the proo o Theorem 5.. First we see that according to (79) and (80), we have N 1 Ψ u α = O(α ), which implies that in Lemma 5.1 we can replace the term cα 4 log α 1 ( η 1 + η + η 3 + Γ 0 Ψ uα ) with cα 4 ( η 1 + η + η 3 + Γ 0 Ψ uα ) and consequently in (76) and (77), the term O(α 4 log α 1 ) can be replaced by O(α 4 ). This proves (68). The estimates (69)-(7) ollow rom (68) and (76) with O(α 4 log α 1 ) replaced with O(α 4 ). 6. Estimate o the binding energy up to α 5 log α 1 term Theorem 6.1 (Upper bound up to the order α 5 log α 1 or the binding energy). For α small enough, we have (81) Σ 0 Σ 1 4 α + e (1) α 3 + e () α 4 + e (3) α 5 log α 1 + o(α 5 log α 1 ), where e (1), e () and e (3) are deined in Theorem.1. Deinition 6.1. 1) Pick { κ = α 1 Ψ,Φ Γ0 Ψ Φ Γ0 Ψ,Φ Γ0 Ψ i Γ 0 Ψ > α 3, 0 i Γ 0 Ψ α 3. ) We split Ψ into Ψ 1 and Ψ as ollows: n 0, Γ n Ψ = Γ n Ψ 1 + Γ n Ψ and or n = 0, Γ 0 Ψ 1 = Γ 0 Ψ and Γ 0 Ψ = 0, or n = 1, or n =, Γ 1 Ψ 1 Γ Ψ 1 = ακ Φ Γ 0 Ψ 1 + Γ Ψ = κ 1 Φ uα and Γ 1 Ψ, Φ uα = 0, ακ,i (H + P ) 1 W i u α, with W i = P i Φ A +.P (H + P ) 1 (A + ) i Ω, = Γ Ψ Γ Ψ 1, with Γ Ψ, (H + P ) 1 W i u α = 0 (i = 1,, 3), and Γ Ψ, Φ Γ 0 Ψ 1 = 0,

0 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER or n = 3, Γ 3 Ψ 1 Γ 3 Ψ and or n 4, = ακ 3 (H + P ) 1 A +.A + Φ uα, = Γ 3 Ψ Γ 3 Ψ 1, Γ 3 Ψ, (H + P ) 1 A +.A + Φ uα = 0, Γ n Ψ 1 = 0 and Γ n Ψ = Γ n Ψ. Lemma 6.1. The ollowing a priori estimates hold κ 1 = 1 + O(α 1 ), κ Γ 0 Ψ 1 = O(α), κ,i = O(1), (i = 1,, 3), Γ 0 Ψ 1 = O(α), P Γ 0 Ψ 1 = O(α ). Proo. To derive these estimates, we use Theorem 5.. The irst equality is a consequence o (7). To derive the next two estimates, we irst notice that (69) yields P Γ (P P ) + P Γ thereore, using again (69), we obtain (P P ) + c H 1 Γ = O(α 4 ), (h α + e 0 ) 1 Γ P Γ + e 0 Γ = O(α 4 ), and thus, using rom (69) that H 1 Γ = O(α 4 ), we get (8) Γ = Γ Ψ = O(α 4 ). We then write, using (8) and the, -orthogonality o Γ Ψ 1 and Γ Ψ, (83) Γ Ψ 1 Γ Ψ = O(α 4 ). Since Γ Ψ 1 Γ Ψ 1, and using (06) o Lemma A.1, we obtain (84) O(α 4 ) = Γ Ψ 1 = ακ Φ Γ 0 Ψ 1 + α i = α Φ κ Γ 0 Ψ 1 + α 3 u α i κ,i (H + P ) 1 W i u α κ,i (H + P ) 1 W i. which gives κ,i = O(1) and κ Γ 0 Ψ 1 = O(α). The last two estimates are consequences o (69). To derive the lower bound on the quadratic orm o H(u α Ψ uα +g), u α Ψ uα +Ψ, we will ollow the same strategy as in Section 5, the only dierence being that now we have better a priori estimates on Ψ uα and Ψ.

HYDROGEN GROUND STATE ENERGY IN QED 1 Proposition 6.1. We have (85) Re HΨ, u α Ψ uα 1 3 α4 Re κ,i (H + P ) 1 P i Φ, W i 4αRe u α.p Φ, Γ Ψ Re κ 1 Γ 0 Ψ uα Φ uα 3 α4 Re (H + P ) 1 (A ) i Φ, (H + P ) 1 (A + ) i Ω ɛα (Ψ 1 ) a 5 8 M[Ψ ] ɛα 5 log α 1 ɛα 5 κ 3 κ 1 1 cα 4 + O(α 5 ). The proo o this Proposition is detailed in Subsection 6. Proposition 6.. (86) where (87) HΨ, Ψ (Σ 0 e 0 ) Ψ 4α (h α + e 0 ) 1 Q α P A Φ uα + κ 1 Φ uα + 3 α4 Re + α A Φ uα + α4 1 κ,i (H + P ) 1 W i κ,i P.A (H + P ) 1 W i, (H + P ) 1 (A + ) i Ω + 4α 1 Re Γ Ψ, A +.P Φ uα + M 1 [Ψ ], M 1 [Ψ ] =(1 c 0 α) (h α + e 0 ) 1 Γ 0 (Ψ 1 ) a + κ 3 + 1 κ 1 1 cα 4 + 3 4 M[Ψ ] + o(α 5 log α 1 ), α Φ Φ uα and Q α is the projection onto the orthogonal complement to the ground state u α o the Schrödinger operator h α = α x The proo o this Proposition is detailed in Subsection 6.3. 6.1. Proo o Theorem 6.1 on the upper bound up to the order α 5 log α 1. We have (88) HΨ GS, Ψ GS = (Σ 0 e 0 ) Ψ uα + Re HΨ, u α Ψ uα + HΨ, Ψ. The estimates or the last two terms are given in Propositions 6.1 and 6.. We will minimize this expression with respect to the parameters κ 1, κ,i, and κ 3. We irst estimate the second term on the right hand side o (85) together with the seventh term on the right hand side o (86). We have (89) 4αRe u α.p Φ, Γ Ψ + 4α 1 Re Γ Ψ, A +.P Φ uα = 4αRe Γ Ψ, ( = 4α P i Φ A +.P (H + P ) 1 (A + ) i Ω ) u α Re Γ Ψ, (H + P ) 1 W i u α.

J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Using the, -orthogonality o Γ Ψ and (H +P ) 1 W i u α, the last expression can be estimated as (90) 4α Re Γ Ψ, (h α + e 0 )(H + P ) 1 W i u α. By the Schwarz inequality, this term is bounded below by (91) (9) ɛα Γ Ψ c (h α + e 0 ) u α = ɛα Γ Ψ O(α 6 ). Next, we collect all the terms involving κ 1 in (85) and (86). This yields Re κ 1 Γ 0 Ψ uα Φ uα + κ 1 Φ uα κ 1 1 cα 4 Γ 0 Ψ uα Φ uα + κ 1 Γ 0 Ψ uα Φ uα κ 1 1 cα 4 Notice that rom Theorem 5. we have Γ 0 Ψ uα = 1 + O(α ); moreover, we have Γ 0 Ψ uα = 1 + O(α ). This yields (93) Re κ 1 Γ 0 Ψ uα Φ uα + κ 1 Φ uα κ 1 1 cα 4 Φ uα + κ 1 1 c α 3 κ 1 1 cα 4 + O(α 5 ) = Φ uα + O(α 5 ). We now collect and estimates the terms in (85) and (86) involving κ,i. We get (94) 1 3 α4 Re + 3 α4 Re = 1 3 α4 Re α4 3 κ,i (H + P ) 1 W i, P i Φ + α4 1 κ,i (H + P ) 1 W i κ,i (H + P ) 1 W i, P.A + (H + P ) 1 (A + ) i Ω κ,i (H + P ) 1 W i + α4 1 (H + P ) 1 W i Collecting in (85) and (86) the terms containing κ 3 yields κ,i (H + P ) 1 W i (95) κ 3 + 1 α Φ Φ uα ɛα 5 κ 3 c 1 α 5 log α 1 κ 3 + 1 ɛα 5 κ 3 c α 5, where c 1 and c are positive constants. The ith term on the right hand side o (85) and the irst term on the right hand side o (87) are estimated, or α small enough, as (96) (1 c 0 α) (h α + e 0 ) 1 Γ 0 (Ψ 1 ) a ɛα (Ψ 1 ) a ( δ ɛα ) (Ψ 1 ) a 0, with δ = 3 3 α, and where we used that (Ψ 1 ) a is orthogonal to u α.

HYDROGEN GROUND STATE ENERGY IN QED 3 Substituting the above estimates in (88) (97) HΨ GS, Ψ GS (Σ 0 e 0 ) Ψ uα + (Σ 0 e 0 ) Ψ Φ uα α 4 3 α4 Re (H + P ) 1 W i (H + P ) 1 (A ) i Φ, (H + P ) 1 (A + ) i Ω 4α (h α + e 0 ) 1 Q α P A Φ uα + α A Φ uα + o(α 5 log α 1 ), where Q α is the projection onto the orthogonal complement to the ground state u α o the Schrödinger operator h α = α x. To complete the proo o Theorem 6.1 we irst note that (98) Ψ uα + Ψ = Ψ GS. Moreover, according to Lemma A.4 (99) and Φ uα = Φ uα + 1 3π (h 1 + 1 4 ) 1 u1 α 5 log α 1 + o(α 5 log α 1 ), (100) Φ uα = α3 π 0 χ Λ (t) 1 + t dt = e(1) α 3. In addition, we have the ollowing identities (i = 1,, 3) (H + P ) 1 W i = (101) ( (H + P ) 1 A +.P (H + P ) 1 (A + ) i P i (H + P ) 1 A +.A +) Ω, and (10) We also have (103) with and 3 α4 Re = 3 α4 Re (H + P ) 1 A Φ, (H + P ) 1 (A + ) i Ω A (H + P ) 1 A +.A + Ω, (H + P ) 1 (A + ) i Ω. 4α (h α + e 0 ) 1 Q α P A Φ uα = 4α 4 a 0 ( 1 x + 1 4 ) 1 Q 1 u 1, a 0 = (104) α A Φ uα = 3 α4 k 1 + k 4π k 3 k + k χ Λ( k ) dk 1 dk dk 3, A (H + P ) 1 (A + ) i Ω. Substituting (98)-(104) into (97) inishes the proo o Theorem 6.1.

4 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER 6.. Proo o Proposition 6.1. Lemma 6.. The ollowing holds (105) 4Re P.P u α Ψ uα, Ψ 4 3 α Re κ,i u α (H + P ) 1 P i Φ, W i 4αRe u α.p Φ, Γ Ψ ɛ H 1 Ψ ɛα 5 κ 3 + O(α 5 ) Proo. For n, 3, with the estimates rom the proo o Lemma 5. and using that due to Theorem 5. we have H 1 uα = O(α 4 ), η 1 = O(1), and κ 1 = O(1), and since Γ 1 = Γ1 Ψ and Γ n 4 = Γn 4 Ψ, we obtain (106) 4Re P.P u α Γ n Ψ uα, Γ n Ψ ɛ H 1 Ψ + O(α 5 ). For n =, n,3 (107) 4Re u α.(αη P Φ + P Γ uα ), Γ Ψ 4Re u α.αη P Φ, Γ Ψ 1 4Re u α.αη P Φ, Γ Ψ cα H 1 Γ uα cα H 1 Γ Ψ. Using Theorem 5., the last two terms on the right hand side o (107) can be estimated by O(α 5 ). For the irst term on the right hand side o (107), using rom Lemma A.1 that P i Φ, Φ = 0, rom Theorem 5. that η = 1 + O(α), and rom Lemma 6.1 that κ,i = O(1), holds 4αRe u α.η P Φ, Γ Ψ 1 = 4Re α u α.η P Φ, ακ Φ Γ 0 Ψ (108) 4Re α u α.η P Φ, i = 4 3 Re α u α i ακ,i (H + P ) 1 W i u α κ,i (H + P ) 1 P i Φ, W i = 1 3 α4 Re κ,i (H + P ) 1 P i Φ, W i. We also used that uα, uα x j = 0 or i j and uα = uα x j or all i and j. Finally, the second term on the right hand side o (107) gives the second term on the right hand side o (105) plus O(α 5 ), using rom Theorem 5. that η 1 = O(α ) and H 1 Γ Ψ = O(α ). To complete the proo, we shall estimate now the term or n = 3, (109) 4Re P.P u α Γ 3 Ψ uα, Γ 3 Ψ = 4Re α 3 η3 P.P u α Φ 3, Γ 3 Ψ 1 + 4Re α 3 η3 P.P u α Φ 3, Γ 3 Ψ + 4Re P.P u α Γ 3 uα, Γ 3 Ψ. The inequalities H 1 uα cα and H 1 Γ 3 Ψ cα (see Theorem 5.) imply that the last term on the right hand side o (109) is O(α 5 ). For the second term

HYDROGEN GROUND STATE ENERGY IN QED 5 on the right hand side o (109) holds (110) Re α 3 η3 P.P u α Φ 3, Γ 3 Ψ ɛ H 1 Γ 3 Ψ + O(α 5 ), since rom Theorem 5. we have η 3 = O(1). Finally to estimate the irst term on the right hand side o (109), we note that and rom Lemma A.5, k 1 1 6 k 1 6 k3 1 6 Φ 3 (k 1, k, k 3 ) L (R 9, C 6 ), k 1 1 6 k 1 6 k3 1 6 (H + P ) 1 A +.A + Φ uα = O(α 3 ). This implies, using again η 3 = O(1), and the explicit expression o Γ 3 Ψ 1 (111) α 3 η3 P.P u α Φ 3, Γ 3 Ψ 1 cα 5 κ3 α 3 η3 P u α ɛα 5 κ 3 + O(α 5 ). Collecting (106)-(111) concludes the proo. Lemma 6.3. The ollowing estimate holds (11) 4 αre P.A + u α Ψ uα, Ψ = Re κ 1 Γ 0 Ψ uα Φ uα 1 4 M[Ψ ] α 5 κ 3 + O(α 5 ), Proo. Obviously (113) 4Re α 1 P.A + u α Ψ uα, Ψ = 4Re α 1 P.A + u α (Γ 0 Ψ uα + η 1 α 3 Φ 1 + η αφ + η 3 α 3 Φ 3 + uα ), Ψ. Step 1 From (113), let us irst estimate the term (114) 4Re α 1 P.A + u α uα, Ψ = 4α 1 Re P.A + Γ n u α uα, Γ n+1 Ψ. For n = 0, the corresponding term vanishes since Γ 0 uα = 0. For n >, we can use (50) where the term O(α 5 log α 1 ) can be replaced with O(α 5 ) because we know rom Theorem 5. that uα = O(α 33 16 ). For n = 1, we have n=0 (115) 4α 1 Re P.A + Γ 1 u α uα + 4α 1 uα Γ 1 uα, A (ακ Φ Γ 0 Ψ 1 + α, Γ Ψ 1 + Γ Ψ cα 3 Γ 1 uα + ɛ H 1 Γ Ψ ) κ,i (H + P ) 1 W i u α. To estimate the last term on the right hand side we note that H 1 A Φ L and H 1 A (H + P ) 1 W i L which thus gives or this term the bound (116) cα H 1 uα + ɛα 4 κ Γ 0 Ψ 1 + ɛα 6 i κ,i = O(α 5 ), using Theorem 5. and Lemma 6.1. The inequalities (115) and (116) imply (117) Re α 1 P.A + Γ 1 u α uα, Γ Ψ 1 + Γ Ψ ɛ H 1 Γ Ψ + O(α 5 ).

6 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER To complete the estimate o the term 4α 1 Re P.A + u α uα, Ψ we have to estimate the term or n = in (114), namely 4Re α 1 P.A + u α Γ uα, Γ 3 Ψ 1 + Γ 3 Ψ. Obviously, (118) Re α 1 P.A + u α Γ uα, Γ 3 Ψ ɛ H 1 Ψ + O(α 5 ). For the term involving Γ 3 Ψ 1 (119) we have Re α 1 P uα Γ uα, ακ 3 A (H + P ) 1 A +.A + Φ uα cα 3 1 16 Γ uα + ɛ κ 3 α + 1 16 Φ u α = κ 3 α 5 + O(α 5 ), using Theorem 5. and Lemma A.5. Collecting (114)-(119) yields (10) Re α 1 P.A + u α uα, g ɛ H 1 Ψ + α 5 κ 3 + O(α 5 ). Step We next estimate in (113) the term 4Re α 1 P.A + u α (Γ 0 Ψ uα +α 3 η 1 Φ 1 + αη Φ + α 3 η 3 Φ 3 ), Ψ. First using (50) yields (11) 4Re α 1 P.A + u α Γ 0 Ψ uα, Ψ = Re (κ 1 Γ 0 Ψ uα ) Φ uα. We also have, using Theorem 5. (1) and (13) α 1 P uα (α 3 η1 Φ 1 + α 3 η3 Φ 3 ), A Ψ α H 1 Γ Ψ + α H 1 Γ 4 Ψ + O(α 5 ) = O(α 5 ), α 1 P uα αη Φ, A (Γ 3 Ψ 1 + Γ 3 Ψ ) ɛ H 1 Γ 3 Ψ + α 3 P uα η k 1 1 4 k 1 4 Φ, k 1 1 4 k 1 4 A Γ 3 Ψ 1 + O(α 5 ) ɛ H 1 Γ 3 Ψ + ɛ κ 3 α 5 + O(α 5 ). Here we used k 1 1 4 k 1 4 Φ L and k 1 1 4 k 1 4 A (H + P ) 1 A +.A + Φ uα = O(α 3 ) (see Lemma A.5). Collecting (10)-(13) yields (14) 4Re α 1 P.A + u α Ψ uα, Ψ Re κ 1 Γ 0 Ψ uα Φ uα ɛ H 1 Ψ ɛα 5 κ 3 +O(α 5 ). Lemma 6.4. (15) 4 αre P.A u α Ψ uα, Ψ 3 α4 Re (H + P ) 1 (A ) i Φ, (H + P ) 1 (A + ) i Ω 1 4 M[Ψ ] ɛα (Ψ 1 ) a ɛα 5 log α 1 (α κ 3 + 1) κ 1 1 cα 4 + O(α 5 ), where (Ψ 1 ) a (x) := (Γ 0 Ψ 1 (x) Γ 0 Ψ 1 ( x))/ is the odd part o Γ 0 Ψ 1.

HYDROGEN GROUND STATE ENERGY IN QED 7 Proo. Since rom Lemma A.1 we have u α.a Φ 1 = 0, we have (16) 4Re α 1 P.A u α Ψ uα, Ψ = 4α 3 Re η A Φ.P u α, Γ 1 Ψ + 4α Re η 3 A Φ 3.P u α, Γ Ψ + 4α 1 A uα.p u α, Ψ. For the irst term on the right hand side o (16) we have (17) 4α 3 Re η A Φ.P u α, Γ 1 Ψ + κ 1 Φ uα = 4α 3 Re η H 1 A Φ.P u α, H 1 Γ 1 Ψ + 4α 3 Re η A Φ.P u α, κ 1 Φ uα. The irst term on the right hand side o (17) is bounded rom below by ɛ H 1 Γ 1 Ψ + O(α 5 ). Applying Lemma A.4, we can replace Φ uα side o (17) by Φ uα Moreover in the second term o the right hand, at the expense o O(α 5 ). More precisely α 3 η A Φ.P u α, κ 1 (Φ uα Φ uα ) cα 5 η (H + P ) 1 A Φ + κ 1 Φ uα Φ uα = O(α 5 ). (18) 4α 3 Re η κ 1 A Φ.P u α, Φ uα = 8 3 α u α Re η κ 1 (A ) i Φ, (H + P ) 1 (A + ) i Ω = 3 α4 Re η κ 1 (A ) i Φ, (H + P ) 1 (A + ) i Ω 3 α4 Re (A ) i Φ, (H + P ) 1 (A + ) i Ω κ 1 1 cα 4, where we used κ 1 = O(1) (Lemma 6.1) and η = 1 + O(α) (Theorem 5.). Note that the right hand side o (18) is well deined since (H + P ) 1 A + Ω F and (H + P ) 1 A Φ F. Collecting the estimates or the irst and the second term in the right hand side o (17), we arrive at (19) 4α 3 Re η A Φ.P u α, Γ 1 Ψ 8 3 α u α Re κ 1 (A ) i Φ, (H + P ) 1 A + Ω ɛ H 1 Γ 1 Ψ + O(α 5 ). Here we used also η = 1 + O(α). As the next step, we return to (16) and estimate the second term on the right hand side as (130) 4α Re η 3 A Φ 3.P u α, Γ Ψ = 8α Re η 3 H 1 A Φ 3.P u α, H 1 Γ Ψ = O(α 5 ),

8 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER where we used H 1 A Φ 3 L and Γ H 1 = Γ H 1 Ψ = O(α 4 ) rom Theorem 5.. For the last term on the right hand side o (16), we have (131) 4α 1 Re A uα u α, Ψ = 4α 1 Re A uα + 4α 1 Re A uα + 4α 1 Re A uα u α, Γ 0 Ψ 1 + 4α 1 Re κ1 A uα u α, Φ uα u α, Γ 1 Ψ + 4α 1 Re A uα u α, Γ Ψ u α, Γ 3 Ψ + 4α 1 Re A uα. u α, Γ n 4 Ψ. We write the unction Γ 0 Ψ 1 = (Ψ 1 ) s + (Ψ 1 ) a where (Ψ 1 ) s (respectively (Ψ 1 ) a ) denotes the even (respectively odd) part o Γ 0 Ψ 1. Obviously, we have (13) α 1 Re A uα u α, Γ 0 Ψ 1 cα H 1 R +ɛα (Ψ 1 ) a = ɛα (Ψ 1 ) a +O(α 5 ). The constant ɛ can be chosen small or large c. For the second term on the right hand side o (131), we have (133) α 1 κ1 A uα u α, Φ uα ɛα 5 log α 1 κ 1 +cα H 1 uα = ɛα 5 log α 1 +O(α 5 ). For the third term on the right hand side o (131), we have, since δ = 3 3 α (134) 4α 1 A uα u α, Γ 1 Ψ δ 8 Γ1 Ψ + cα H 1 uα = δ 8 Γ1 Ψ + O(α 5 ). Similarly (135) 4α 1 A uα u α, Γ n 4 Ψ δ 8 Γn 4 Ψ + O(α 5 ). To complete the estimate o the last term in (131), we have to estimate two terms: 4Re α 1 A uα.p u α, Γ Ψ and 4Re α 1 A uα.p u α, Γ 3 Ψ. For the irst one we have Re α 1 A uα.p u α, Γ Ψ cα H 1 uα + ɛα Γ Ψ + ɛα 4 κ Γ 0 Ψ + ɛα 6 κ,i = ɛα Γ Ψ + O(α 5 ). Similarly, (136) Re α 1 A uα.p u α, Γ 3 Ψ cα H 1 uα + ɛα Γ 3 Ψ + ɛα 6 log α 1 κ 3 + O(α 5 ). Collecting the estimates (131)-(136) yields (137) 4Re α 1 A uα.p u α, Ψ δ 8 Ψ + ɛ H 1 Ψ + ɛα 5 log α 1 (1 + α κ 3 ) + O(α 5 ). Collecting (10), (14), (19), (130) and (137) concludes the proo. We can now prove the estimate o Re Hg, u α Ψ uα o Proposition 6.1. Proo o the Proposition 6.1. Using the orthogonality (39) o u α and Ψ, yields Re HΨ, Ψ uα = 4Re P.P u α Ψ uα, Ψ 4 αre P.A(0)u α Ψ uα, Ψ. Together with Lemmata 6.-6.4, this concludes the proo o Proposition 6.1.

HYDROGEN GROUND STATE ENERGY IN QED 9 6.3. Proo o Proposition 6.. We have the estimate Proposition 6.3. We have (138) HΨ, Ψ HΨ 1, Ψ 1 + HΨ, Ψ + α ( A Ψ A Ψ 1 A Ψ ) 4Re P.P Ψ, Ψ 1 4α 1 Re P.A(0)Ψ, Ψ 1 + 4Re P.A(0)Ψ, Ψ 1 ɛm[ψ ] cα 6 log α 1 κ 3 c 0 α (h α + e 0 ) 1 Γ 0 Ψ 1 + O(α 5 ). Proo. Recall that (139) H = P α x + T (0) Re P. ( P + α 1 A(0) ), and (140) T (0) =: (P + α 1 A(0)) : +H. Due to the orthogonality and (139), (140), we obtain (141) (H + e 0 )Ψ, Ψ Γ n Ψ 1, Γ n Ψ = 0, n = 0, 1,..., = (H + e 0 )Ψ, Ψ + (H + e 0 )Ψ 1, Ψ 1 + αre A.A Γ n+ Ψ, Γ n Ψ 1 + αre A.A Γ 3 Ψ 1, Γ 1 Ψ + α( A Ψ A Ψ 1 A Ψ ) 4Re P.P Ψ, Ψ 1 4α 1 Re P.A(0)Ψ, Ψ 1 + 4α 1 Re P.A(0)Ψ, Ψ 1. We have (14) Similarly, n=0 αre A.A Γ 5 Ψ, Γ 3 Ψ 1 ɛ H 1 Γ 5 Ψ cα Γ 3 Ψ 1 ɛ H 1 Γ 5 Ψ cα 7 log α 1 κ 3. αre A.A Γ 4 Ψ, Γ Ψ 1 ɛ H 1 Γ 4 Ψ cα 4 κ Γ 0 Ψ 1 ɛ H 1 Γ 4 Ψ + O(α 5 ). cα 6 κ,i To estimate the term αre A.A Γ 3 Ψ, Γ 1 Ψ 1 we rewrite it as αre Γ 3 Ψ, A +.A + κ 1 Φ uα and use that Γ 3 Ψ, (H + P ) 1 A +.A + Φ uα = 0. This yields, using Lemma A.5 (143) αre A.A Γ 3 Ψ, Γ 1 Ψ 1 = αre Γ 3 Ψ, (h α + e 0 )(H + P ) 1 A +.A + κ 1 Φ uα ɛα Γ 3 Ψ + cα 7 log α 1.

30 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER Similarly, i Γ 0 Ψ 1 > α 3, (144) αre A.A Γ Ψ, Γ 0 Ψ 1 = αre Γ Ψ, (h α + e 0 )(H + P ) 1 A +.A + Γ 0 Ψ 1 cα (h α + e 0 ) 1 Γ Ψ cα (h α + e 0 ) 1 (H + P ) 1 A +.A + Γ 0 Ψ 1 cα P Γ Ψ + cα x 1 Γ Ψ cαe 0 Γ Ψ c 0 α (h α + e 0 ) 1 Γ 0 Ψ 1 cα P Γ Ψ ɛα Γ Ψ c 0 α (h α + e 0 ) 1 Γ 0 Ψ 1. I Γ 0 Ψ 1 α 3, we have instead (145) αre A.A Γ Ψ, Γ 0 Ψ 1 ɛ H 1 Γ Ψ cα Γ 0 Ψ 1 Finally, using Lemma A.6 yields ɛ H 1 Γ Ψ + O(α 5 ). (146) αre A.A Γ 3 Ψ 1, Γ 1 Ψ ɛ H 1 Ψ + O(α 5 ). Collecting (141)-(146) concludes the proo o the proposition. In the rest o this subsection, we estimate urther terms in (138). 6.3.1. Estimate o crossed terms involving Ψ 1 and Ψ. Lemma 6.5. (147) α( A Ψ A Ψ 1 A Ψ ) ɛm[ψ ] + O(α 5 ). Proo. Obviously, the let hand side o (147) is equal to (148) 4αRe A Ψ 1, A Ψ cα H 1 Γ n Ψ 1 H 1 Γ n Ψ n ɛ H 1 Γ 1 Ψ cα κ 1 H 1 Φ uα ( cα 3 H 1 Γ n Ψ + H 1 Γ n Ψ ) n 1 ɛ H 1 Ψ + O(α 5 ), where in the last inequality we used (10) o Lemma A.3, and (69) o Theorem 5.. Lemma 6.6. P.P Ψ, Ψ 1 ɛm[ψ ] + cα 7 log α 1 κ 3 + O(α 5 ) Proo. We have (149) P.P Ψ, Ψ 1 = P Γ 1 Ψ, P Γ 1 Ψ 1 + P Γ Ψ, P Γ Ψ 1 + P Γ 3 Ψ, P Γ 3 Ψ 1. Obviously, using Lemma A.3 and the equality κ 1 = O(1) rom Lemma 6.1, yields (150) P Γ 1 Ψ, P Γ 1 Ψ 1 ɛ H 1 Ψ + κ 1 P P 1 Φ u α ɛ H 1 Ψ + O(α 5 ). We also have, by deinition o Γ Ψ 1 Lemma 6.1, and using the estimates κ,i = O(1) rom (151) P Γ Ψ, P Γ Ψ 1 ɛ H 1 Ψ + c κ α P Γ 0 Ψ 1 + O(α 6 ).

HYDROGEN GROUND STATE ENERGY IN QED 31 We next bound the second term on the right hand side o (151). Notice that by deinition o κ, this term is nonzero only i Γ 0 Ψ 1 > α 3, which implies, with Lemma 6.1, that P Γ 0 Ψ 1 cα Γ 0 Ψ 1. The inequality (151) can thus be rewritten as (15) P.P Γ Ψ, Γ Ψ 1 ɛ H 1 Ψ +c κ α 3 Γ 0 Ψ 1 +O(α 6 ) ɛ H 1 Ψ +O(α 5 ), using in the last inequality that κ Γ 0 Ψ 1 = O(α) (see Lemma 6.1). For the second term on the right hand side o (149), using (1) rom Lemma A.3 yields (153) Γ 3 P Ψ, Γ 3 P Ψ 1 ɛ Γ 3 H 1 Ψ + cα 7 log α 1 κ 3. The inequalities (149), (15) and (153) prove the lemma. Lemma 6.7. 4α 1 Re P.A + Ψ, Ψ 1 ɛm[ψ ] + O(α 5 ). Proo. Since Γ n>3 Ψ 1 = 0, Γ 0 Ψ = 0 and H 1 Γ n 1 Ψ 1 H 1 Γ n 1 Ψ + H 1 Γ n 1 Ψ cm[ψ ] + O(α 4 ), (see (69) o Theorem 5.) we obtain 4α 1 Re P.A + Ψ, Ψ 1 ɛ Γ n P Ψ + cα Γ n H 1 Ψ 1 ɛm[ψ ] + O(α 5 ). Lemma 6.8. Proo. 4α 1 Re P.A Ψ, Ψ 1 ɛm[ψ ] + O(α 5 ). 4α 1 Re P.A Ψ, Ψ 1 ɛ H 1 Ψ cα P Ψ 1 ɛ H 1 Ψ cα ( Γ n=0,,3,4 P Ψ + Γ n=0,,3,4 P Ψ ) cα κ 1 P Φ uα ɛ H 1 Ψ ɛ Γ n 4 P Ψ + O(α 6 log α 1 ) ɛm[ψ ] + O(α 6 log α 1 ), using (1) o Lemma A.3. Lemma 6.9. 4α 1 Re P.A(0)Ψ, Ψ 1 4α 1 Re Γ Ψ, P.A + Φ uα ɛm[ψ ] + O(α 5 ). Proo. We have 4α 1 Re P.A Ψ, Ψ 1 ɛ H 1 Ψ cα Γ n P Ψ 1 + 4α 1 Re P.A Γ Ψ, κ 1 Φ uα ɛ H 1 Ψ cα Γ n=,3 H 1 Ψ + 4α 1 Re P.A Γ Ψ, κ 1 Φ uα ɛm[ψ ] + 4α 1 Re P.A Γ Ψ, κ 1 Φ uα + O(α 5 ).

3 J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER We estimate the second term on the right hand side as ollows, 4α 1 Re P.A Γ Ψ, κ 1 Φ uα 4α 1 Re P.A Γ Ψ, Φ uα ɛ H 1 Γ Ψ cα κ 1 P Φ uα 4α 1 Re P.A Γ Ψ, Φ uα ɛ H 1 Γ Ψ + O(α 5 ), where we used κ 1 1 = O(α) rom Lemma 6.1, Φ uα = O(α 3 ) rom Lemma A.3, and P (Φ uα Φ uα ) = O(α 4 ) rom Lemma A.4. We also have, using P.A + = A +.P, 4α 1 Re P.A + Ψ, Ψ 1 = 4α 1 Re Γ n P Ψ, A Γ n Ψ 1 ɛ H 1 Ψ cα H 1 Γ n Ψ 1 ɛm[ψ ] + O(α 5 ), where H 1 Γ n Ψ 1 has been estimated as P Ψ 1 in the proo o Lemma 6.8. 6.3.. Estimates o the term (H + e 0 )Ψ 1, Ψ 1. Due to (139) and (140), one inds (154) (H + e 0 )Ψ 1, Ψ 1 = Ψ 1, Ψ 1 Re P.(P + α 1 A(0))Ψ 1, Ψ 1 + α 1 Re P.A(0)Ψ 1, Ψ 1 + α A Ψ 1 + αre A.A Ψ 1, Ψ 1 We estimate the terms in (154) below. Lemma 6.10. We have Re P.(P + α 1 A(0))Ψ 1, Ψ 1 4α 1 Re A Φ uα, P Γ 0 (Ψ 1 ) s ɛm[ψ ] κ 1 1 cα 4 + O(α 5 ), where Γ 0 (Ψ 1 ) s = Γ 0 Ψ 1 Γ 0 (Ψ 1 ) a is the even part o Γ 0 Ψ 1. Proo. Using Φ, P iφ = 0 (see Lemma A.1), the symmetry o u α, and P.P Φ uα, Φ uα = 0, we obtain (155) P.P Ψ 1, Ψ 1 = P ακ Φ Γ 0 Ψ 1, P α cα 3 κ Γ 0 Ψ 1 + cα 5 κ,i = O(α 5 ), κ,i (H + P ) 1 W i u α where in the last inequality we used Lemma 6.1.

HYDROGEN GROUND STATE ENERGY IN QED 33 (156) We also have, using again the symmetry o u α, α 1 Re P.A(0)Ψ 1, Ψ 1 = 4α 1 Re P.A Ψ 1, Ψ 1 = 4α 1 Re A κ 1 Φ uα, P Γ 0 (Ψ 1 ) s 4α 1 Re A κ αφ Γ 0 Ψ 1, P κ 1 Φ uα 4α 1 Re A Γ 3 Ψ 1, P ακ Φ Γ 0 Ψ 1 4α 1 Re A κ 1 Φ uα, P Γ 0 (Ψ 1 ) s cα H 1 Γ 3 Ψ 1 cα P Γ 0 Ψ 1 κ 4α 1 Re k 1 6 A κ αφ Γ 0 Ψ 1, k 1 6 κ1 P Φ uα 4α 1 Re A κ 1 Φ uα, P Γ 0 (Ψ 1 ) s cα H 1 Γ 3 Ψ cα H 1 Γ 3 Ψ cα 5 cα 3 κ Γ 0 Ψ 1 k 1 6 P Φ u α 4α 1 Re A κ 1 Φ uα, P Γ 0 (Ψ 1 ) s ɛm[ψ ] cα 5, where we used Theorem 5. and Lemma A.3. Moreover, because A Φ uα = O(α 3 ) and P Γ 0 Ψ 1 = O(α ), we obtain (157) 4Re α 1 Re A κ 1 Φ uα, P Γ 0 (Ψ 1 ) s 4α 1 Re A Φ uα, P Γ 0 (Ψ 1 ) s κ 1 1 cα 4. This estimate, together with (155) and (156), proves the lemma. Lemma 6.11. We have α 1 Re P.A(0)Ψ 1, Ψ 1 3 α4 Re i κ,i P.A (H + P ) 1 W i, (H + P ) 1 (A + ) i Ω κ 1 1 cα 4 + O(α 5 ). Proo. The ollowing holds (158) α 1 Re P.A(0)Ψ 1, Ψ 1 = 4α 1 Re P.A Γ 3 Ψ 1, Γ Ψ 1 + 4αRe P.A Γ Ψ 1, κ 1 Φ uα = 4α 3 Re κ,i k 1 1 6 k 1 6 A Γ 3 Ψ 1, P k 1 1 6 k 1 6 (H + P ) 1 W i u α + 4α 3 Re κ k 1 1 6 k 1 6 A Γ 3 Ψ 1, k 1 1 6 k 1 6 P Φ Γ 0 Ψ 1 + 4α 1 Re P.A Γ Ψ 1, κ 1 Φ uα. Applying the Schwarz inequality and the estimates k 1 1 6 k 1 6 A Γ 3 Ψ 1 = O(α 5 ) (Lemma A.5), κ,i = O(1) (Lemma 6.1), and u α = O(α ), we see that the irst term on the right hand side o (158) is O(α 5 ). Applying also the estimate κ Γ 0 Ψ 1 = O(α) (Lemma 6.1), we obtain that the second term on the right hand side o (158) is also O(α 5 ). Finally, we estimate 4α 1 Re P.A Γ Ψ 1, κ 1 Φ uα. The ollowing inequality holds, (159) 4α 1 Re P.A Γ Ψ 1, κ 1 Φ uα 4α 1 Re P.A Γ Ψ 1, Φ uα κ 1 1 cα 4, whose proo is similar to the one o (157). Next we get (160) 4α 1 Re P.A Γ Ψ 1, Φ uα 4α 1 Re P.A Γ Ψ 1, Φ uα α 1 1 H Γ Ψ 1 P (Φ uα Φ uα ) = O(α 6 log α 1 ),