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 for the hydrogen
ground state energy in the Pauli-Fierz model up to the order O(α5 log α−1 ),
where α denotes the finestructure constant, and prove rigorous bounds on
the remainder term of the order o(α5 log α−1 ). As a consequence, we prove
that the ground state energy is not a real analytic function of α, and verify
the existence of logarithmic corrections to the expansion of the ground state
energy in powers of α, as conjectured in the recent literature.
1. Introduction
For a hydrogen-like atom consisting of an electron interacting with a static nucleus of charge eZ, described by the Schrödinger Hamiltonian −∆ − αZ
|x| ,
(Zα)2
αZ
) =
.
|x|
4
corresponds to the binding energy necessary to remove the electron to spatial infinity.
The interaction of the electron with the quantized electromagnetic field is accounted for by adding to −∆ − αZ
|x| the photon field energy operator Hf , and an
operator I(α) which describes the coupling of the electron to the quantized electromagnetic field; the small parameter α is the fine structure constant. Thereby, one
obtains the Pauli-Fierz Hamiltonian described in detail in Section 2. In this case,
determining the binding energy
αZ
(1)
inf σ − ∆ + Hf + I(α) − inf σ − ∆ −
+ Hf + I(α)
|x|
is a very hard problem. A main obstacle emerges from the fact that the ground state
energy is not an isolated eigenvalue of the Hamiltonian, and can not be determined
with ordinary perturbation theory. Furthermore, the photon form factor in the
quantized electromagnetic vector potential occurring in the interaction term I(α)
contains a critical frequency space singularity that is responsible for the infamous
infrared problem in quantum electrodynamics. As a consequence, quantities such
as the ground state energy do not exist as a convergent power series in the fine
structure constant α with coefficients independent of α.
In recent years, several rigorous results addressing the computation of the binding
energy have been obtained, [14, 13, 8]. In particular, the coupling to the photon
field has been shown to increase the binding energy of the electron to the nucleus,
2
[12, 14, 8].
and that up to normal ordering, the leading term is (αZ)
4
Moreover, for a model with scalar bosons, the binding energy is determined
in [13], in the first subleading order in powers of α, up to α3 , with error term
inf σ(−∆) − inf σ(−∆ −
1
2
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
α3 log α−1 . This result has inspired the question of a possible emergence of logarithmic terms in the expansion of the binding energy; however, this question has so
far remained open.
In [2], 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 of α, with an expansion of the form
(2)
ε0 +
2N
X
εk (α)αk/2 + o(αN ) ,
k=1
(for any given N ) where the coefficients εk (α) diverge as α → 0, but are smaller
in magnitude than any positive power of α−1 . The recursive algorithms developed
in [2] are highly complex, and explicitly computing the ground state energy to any
subleading order in powers of α is an extensive task. While it is expected that the
rate of divergence of these coefficient functions is proportional to a power of log α−1 ,
this is not explicitly exhibited in the current literature; for instance, it can a priori
not be ruled out that terms involving logarithmic corrections cancel mutually.
The goal of the current paper is to develop an alternative method (as a continuation of [4]) that determines the binding energy up to several subleading orders
in powers of α, with rigorous error bounds, and proving the presence of terms
logarithmic in α.
The main result established in the present paper (for Z = 1) states that the
binding energy can be estimated as
α2
+ e(1) α3 + e(2) α4 + e(3) α5 log α−1 + o(α5 log α−1 ) ,
4
where e(i) (i = 1, 2, 3) are independent of α, e(1) > 0, and e(3) 6= 0. Their explicit
values are given in Theorem 2.1.
As a consequence, we conclude that the binding energy is not analytic in α. In
addition, our proof clarifies how the logarithmic factor in α5 log α−1 is linked to the
infrared singularity of the photon form factor in the interaction term I(α). We note
that for some models with a mild infrared behavior, [10], the ground state energy
is proven to be analytic in α.
(3)
Organization of the proof. Our proof of the expression (3) for the hydrogen
binding energy involves the following main steps:
Since the binding energy (1) involves the ground state energy of the self-energy
operator T = −∆ + Hf + I(α), it is necessary to determine the value of inf σ(T ).
This was achieved in [4] (see also Lemma A.7), using in particular the fact that
inf σ(T ) = inf σ(T (0)) [9, 6], where T (0) is the restriction of T to total momentum
P = 0 (see details in Section 2, (7)).
The derivation of (3) requires both a lower and an upper bound for the binding
energy. This is accomplished in three steps.
2
The first term in the expansion for the binding energy is the Coulomb term α4 ;
it can be recovered with an approximate ground state containing no photon, and
with an electronic part given by the ground state uα of the Schrödinger-Coulomb
α
operator −∆− |x|
. The first step, described in Section 4, thus consists of estimating
the contribution to the binding energy stemming from states orthogonal to uα with
α
respect to the quadratic form for −∆ − |x|
+ Hf + I(α).
HYDROGEN GROUND STATE ENERGY IN QED
3
The next two steps consist of estimating the binding energy up to the order α3
(Theorem 5.2, Section 5), and subsequently to the order α5 log α−1 (Theorem 6.1,
Section 6).
One of the key ingredient to control some of the error terms occurring in both
the upper and lower bounds, is an estimate on the expected photon number for the
ground state Ψ of the total hamiltonian −∆ − αZ
|x| + Hf + I(α). Such a bound is
2
derived in Section 3 and yields hΨ, Nf Ψi = O(α log α−1 ), where Nf is the photon
number operator.
In Section 2, we give a detailed introduction of the model and state the main
results. In the Appendix, we provide some technical lemmata used in the derivation
of the binding energy.
2. The model
We study a scalar electron interacting with the quantized electromagnetic field
in the Coulomb gauge, and with the electrostatic potential generated by a fixed
nucleus. The Hilbert space accounting for the Schrödinger electron is given by
Hel = L2 (R3 ). The Fock space of photon states is given by
M
F =
Fn ,
n∈N
Nn
where the n-photon space Fn = s L (R3 ) ⊗ C2 is the symmetric tensor product of n copies of one-photon Hilbert spaces L2 (R3 ) ⊗ C2 . The factor C2 accounts
for the two independent transversal polarizations of the photon. On F, we introduce creation- and annihilation operators a∗λ (k), aλ (k) satisfying the distributional
commutation relations
2
[ aλ (k) , a∗λ0 (k 0 ) ] = δλ,λ0 δ(k − k 0 ) ,
[ a]λ (k) , a]λ0 (k 0 ) ] = 0,
where a]λ denotes either aλ or a∗λ . There exists a unique unit ray Ωf ∈ F, the Fock
vacuum, which satisfies aλ (k) Ωf = 0 for all k ∈ R3 and λ ∈ {1, 2}.
The Hilbert space of states of the system consisting of both the electron and the
radiation field is given by
H := Hel ⊗ F.
We use units such that ~ = c = 1, and where the√mass of the electron equals
m = 1/2. The electron charge is then given by e = α, where the fine structure
constant α will here be considered as a small parameter.
Let x ∈ R3 be the position vector of the electron and let yi ∈ R3 be the position
vector of the i-th photon.
We consider the normal ordered Pauli-Fierz Hamiltonian on H for Hydrogen,
2
√
(4)
: i∇x ⊗ If − αA(x) : +V (x) ⊗ If + Iel ⊗ Hf .
where : · · · : denotes normal ordering, corresponding to the subtraction of a normal
ordering constant cn.o. α, with cn.o. If := [A+ (x), A− (x)] is independent of x.
The electrostatic√potential V (x) is the Coulomb potential for a static point nucleus of charge e = α (i.e., Z = 1)
α
V (x) = − .
|x|
We will describe the quantized electromagnetic field by use of 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) =
X Z
λ=1,2
R3
h
i
χΛ (|k|)
ikx
−ikx
∗
ε
(k)
e
⊗
a
(k)
+
e
⊗
a
(k)
dk
λ
λ
λ
2π|k|1/2
where by the Coulomb gauge condition, divA = 0.
The vectors ελ (k) ∈ R3 are the two orthonormal polarization vectors perpendicular to k,
k
(k2 , −k1 , 0)
and
ε2 (k) =
ε1 (k) = p 2
∧ ε1 (k).
2
|k|
k1 + k2
The function χΛ implements an ultraviolet cutoff on the wavenumbers k. We assume
χΛ to be of class C 1 , with compact support in {|k| ≤ Λ}, χΛ ≤ 1 and χΛ = 1 for
|k| ≤ Λ − 1. For convenience, we shall write
A(x) = A− (x) + A+ (x),
where
−
A (x) =
X Z
λ=1,2
χΛ (|k|)
ελ (k) eikx ⊗ aλ (k) dk
2π|k|1/2
R3
is the part of A(x) containing the annihilation operators, and A+ (x) = (A− (x))∗ .
The photon field energy operator Hf is given by
X Z
Hf =
|k|a∗λ (k)aλ (k)dk.
λ=1,2
R3
We will, with exception of our discussion in Section 3, study the unitarily equivalent Hamiltonian
2
√
(5)
H = U : i∇x ⊗ If − αA(x) : +V (x) ⊗ If + Iel ⊗ Hf U ∗ ,
where the unitary transform U is defined by
U = eiPf .x ,
and
Pf =
X Z
k a∗λ (k)aλ (k)dk
λ=1,2
is the photon momentum operator. We have
U i∇x U ∗ = i∇x + Pf
and U A(x)U ∗ = A(0) .
In addition, the Coulomb operator V , the photon field energy Hf , and the photon
momentum Pf remain unchanged under the action of U . Therefore, in this new
system of variables, the Hamiltonian reads as follows
2
√
α
,
H = : (i∇x ⊗ If − Iel ⊗ Pf ) − αA(0) : +Hf −
(6)
|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 of
notations, the operators Iel and If ,
α
H = (−∆x −
) + (Hf + Pf2 ) − 2Re (i∇x .Pf )
|x|
√
− 2 α(i∇x − Pf ).A(0) + 2αA+ (0).A− (0) + 2αRe (A− (0))2 .
HYDROGEN GROUND STATE ENERGY IN QED
5
For a free spinless electron coupled to the quantized electromagnetic field, the
self-energy operator T is given by
2
√
T = : i∇x ⊗ If − αA(x) : +Iel ⊗ Hf .
We note that this system is translationally invariant; that is, T commutes with the
operator of total momentum
Ptot = pel ⊗ If + Iel ⊗ Pf ,
where pel and Pf denote respectively the electron and the photon momentum operators.
Let HP ∼
= C ⊗ F denote the fibre Hilbert space corresponding to conserved total
momentum P .
For fixed value P of the total momentum, the restriction of T to the fibre space
HP is given by (see e.g. [6])
√
(7)
T (P ) = : (P − Pf − αA(0))2 : +Hf
where by abuse of notation, we again dropped all tensor products involving the
identity operators If and Iel . Henceforth, we will write
A± ≡ A± (0) .
Moreover, we denote
Σ0 = inf σ(T )
and
Σ = inf σ(H) = inf σ(T + V ).
It is proven in [1, 6] that Σ0 = inf σ(T (0)) is an eigenvalue of the operator T (0).
Our main result is the following theorem.
Theorem 2.1. The binding energy fulfills
1
Σ0 − Σ = α2 + e(1) α3 + e(2) α4 + e(3) α5 log α−1 + o(α5 log α−1 ),
4
(8)
where
e
e(2) =
+
−
2
Re
3
3
1X
3
2
=
π
Z
0
∞
χ2Λ (t)
dt,
1+t
3
X
h(A− )i (Hf + Pf2 )−1 A+ .A+ Ωf , (Hf + Pf2 )−1 (A+ )i Ωf i
i=1
1
k(Hf + Pf2 )− 2 2A+ .Pf (Hf + Pf2 )−1 (A+ )i − Pfi (Hf + Pf2 )−1 A+ .A+ Ωf k2
i=1
3
2X
3
(1)
kA− (Hf + Pf2 )−1 (A+ )i Ωf k2 + 4a20 kQ⊥
1 (−∆ −
i=1
Z
a0 =
1
1 1
+ )− 2 ∆u1 k2 ,
|x| 4
k12 + k22
2
χΛ (|k|) dk1 dk2 dk3 ,
2
3
2
4π |k| |k| + |k|
e(3) = −
1
1
1 1
k(−∆ −
+ ) 2 ∇u1 k2 ,
3π
|x| 4
and Q⊥
1 is the projection onto the orthogonal complement to the ground state u1 of
1
the Schrödinger operator −∆ − |x|
(for α = 1).
6
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
3. Bounds on the expected photon number
Lemma 3.1. Let
K = (i∇ −
√
αA(x))2 + Hf −
α
α
= v 2 + Hf −
,
|x|
|x|
be the Pauli-Fierz operator defined without normal ordering, where v = i∇ −
√
αA(x). Let Ψ ∈ H denote the ground state of K,
KΨ = EΨ,
normalized by
kΨk = 1 .
Let
Nf :=
X Z
a∗λ (k) aλ (k) dk
λ=1,2
denote the photon number operator. Then, there exists a constant c independent of
α, such that for any sufficiently small α > 0, the estimate
hΨ , Nf Ψi ≤ c α2 log α−1
is satisfied.
Proof. Using
[aλ (k), Hf ] = |k|,
and
[aλ (k), v] =
(k)
1
2π|k| 2
χλ (k)eik.x
and the pull-through formula,
aλ (k)EΨ
= aλ (k)KΨ
h
i
1
= (Hf + |k|)aλ (k) −
aλ (k) + [v, aλ (k)]v + v[v, aλ (k)] Ψ ,
|x|
we get
(9) aλ (k) Ψ
√
= −
√
2
αχΛ (|k|)
p
i∇ − αA(x) · λ (k)eik.x Ψ .
2π |k| K + |k| − E
From (9), we obtain
2
aλ (k)Ψ ≤
2 2
√
1
α χΛ (|k|) i∇ − αA(x) Ψ |k|
K + |k| − E
α χΛ (|k|) h
α i
≤
Ψ
,
K
Ψ
+
Ψ
,
Ψ ,
|k|3
|x|
2
√
α
since K − E ≥ 0, and (i∇ − αA(x) ≤ (K + |x|
). We have
(10)
Ψ , K Ψ ≤ c α.
Moreover, for the normal ordered hamiltonian defined in (4), we have
√
α
: K : = −∆x − 4 αRe i∇x .A− (x) + α : A(x)2 : +Hf −
|x|
√
√
α
2
≥ (1 − c α)(−∆x ) + α : A(x) : +(1 − c α)Hf −
(11)
|x|
1
α
≥ −cα2 − ∆x −
2
|x|
HYDROGEN GROUND STATE ENERGY IN QED
7
where in the last inequality we used (see e.g. [11])
√
α : A(x) :2 +(1 − c α)Hf ≥ −cα2 .
Inequalities (10), (11) and
1
α
− ∆x −
≥ −4α2
4
|x|
imply
1
− h∆x Ψ, Ψi ≤ cα2 kΨk2 .
4
Thus,
(12)
Ψ,
1
1
α
Ψi ≤ h−∆x Ψ, Ψi + α2 kΨk2 ≤ (c + )α2 kΨk2 .
|x|
4
4
Collecting (10), (12) and (9) we find
c α χΛ (|k|)
(13)
.
aλ (k)Ψ ≤
3
|k| 2
This a priori bound exhibits the L2 -critical singularity in frequency space. It does
not take into consideration the exponential localization of the ground state due to
the confining Coulomb potential, and appears in a similar form for the free electron.
To account for the latter, we use the following two results from the work of
Griesemer, Lieb, and Loss, [11]. Equation (58) in [11] provides the bound
√
c α χΛ (|k|) |x|Ψ .
aλ (k)Ψ <
1
2
|k|
Moreover, Lemma 6.2 in [11] states that
2
h
exp[β|x|]Ψ ≤ c 1 +
i
1
k Ψ k2 ,
2
Σ0 − E − β
for any
β 2 < Σ0 − E = O(α2 ) .
For the 1-electron case, Σ0 is the electron self-energy, and E is the ground state of
: K :. Choosing β = O(α),
1
41
1
3
3
(4!) 4 4
4
4
k |x|Ψ k ≤ k|x| Ψk kΨk ≤
exp[β|x|]Ψ kΨk 4
β
i 18
ch
1
≤
1+
kΨk
β
Σ0 − E − β 2
5
≤ c1 α − 4 .
Notably, this bound only depends on the binding energy of the potential.
Thus,
cα− 34 χ (|k|)
Λ
(14)
.
aλ (k)Ψ <
1
|k| 2
We see that binding to the Coulomb potential weakens the infrared singularity by a
factor |k|, but at the expense of a large constant factor α−2 . For the free electron,
this estimate does not exist.
8
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
Using (13) and (14), we find
Z
2
dk aλ (k)Ψ h Ψ , Nf Ψ i =
3
c α− 2
dk
+
|k|
|k|<δ
Z
≤
≤ cα
9
dk
δ≤|k|≤Λ
c α2
|k|3
1
δ
00 2
−1
+ c α log α .
− 32
≤ c α4
Z
δ 2 + c0 α2 log
15
for δ = α 8 . This proves the lemma.
4. Estimates on the quadratic form for states orthogonal to the
ground state of the Schrödinger operator
Throughout this paper, we will denote by Γn the projection onto the n-th photon
sector (without distinction for the n-photon sector of F and the n-photon sector of
Pn−1
H). We also define Γ≥n = 1 − j=0 Γj .
Starting with this section, we study the Hamiltonian H defined in (6), written
in relative coordinates. In particular, i∇x now stands for the operator unitarily
equivalent to the operator of total momentum, which, by abuse of notation, will be
denoted by P .
Let
1
(15)
uα (x) = √ α3/2 e−α|x|/2
8π
be the normalized ground state of the Schrödinger operator
α
.
hα := −∆x −
|x|
2
2
We will also denote by −e0 = − α4 and −e1 = − α16 the two lowest eigenvalues of
this operator.
Theorem 4.1. Assume that Φ ∈ H fulfils hΓk Φ, uα iL2 (R3 ,dx) = 0, for all k ≥ 0.
Then there exists ν > 0 and α0 > 0 such that for all 0 < α < α0
1
(16)
hHΦ, Φi ≥ (Σ0 − e0 )kΦk2 + δkΦk2 + νkHf2 Φk2 ,
where δ = (e0 − e1 )/2 =
3 2
32 α .
Remark 4.1. All photons with momenta larger than the ultraviolet cutoff do not
contribute to lower the energy. More precisely, due to the cutoff function χΛ (|k|)
in the definition of A(x), and since we have
√
α
H = (i∇x − Pf )2 − 2 αRe (i∇x − Pf ).A(0) + α : A(0)2 : +Hf −
,
|x|
it follows that for any given normalized state Φ ∈ H, there exists a normalized state
Φ≤Λ such that ∀x ∈ R3 , for all n ∈
n{1, 2, . . .), for all ((k1 , λ1 ), (k2 , λ2 ), . . . , (kn , λn )) ∈
3
(R \ {k, |k| ≤ Λ + 1}) × {1, 2} , we have
(17)
Γn Φ≤Λ (x, (k1 , λ1 ), (k2 , λ2 ), . . . , (kn , λn )) = 0
and
hΦ≤Λ , HΦ≤Λ i ≤ hΦ, HΦi
and
hΦ≤Λ , T Φ≤Λ i ≤ hΦ, T Φi.
HYDROGEN GROUND STATE ENERGY IN QED
9
A key consequence of this remark is that throughout the paper, all states will be
implicitly assumed to fulfill condition (17). This is crucial for the proof of Corollary 4.2.
To prove Theorem 4.1, we first need the following Lemma.
Lemma 4.1. There exists c0 > 0 such that for all α small enough we have
1
1
H − (P − Pf )2 − Hf ≥ −c0 α2 .
2
2
A straightforward consequence of this lemma is the following result.
Corollary 4.1. Let ΨGS be the normalized ground state of H. Then
hHf ΨGS , ΨGS i ≤ 2c0 α2 kΨGS k2
(18)
Proof. This follows from hHΨGS , ΨGS i ≤ (Σ0 − e0 )kΨGS k2 < 0.
Moreover, from Theorem 4.1 and Lemma 4.1, we obtain
Corollary 4.2. Assume that Φ ∈ H is such that hΓn Φ, uα iL2 (R3 ,dx) = 0 holds for
all n ≥ 0. Then, for ν and δ defined in Theorem 4.1, there exists ζ > 0, and α0 > 0
such that for all 0 < α < α0 we have
hHΦ, Φi ≥ (Σ0 − e0 )kΦk2 + M [Φ],
(19)
where
1
δ
ν
kΦk2 + kHf2 Φk2 + ζk(P − Pf )Φk2 + ζkΓn≤4 P Φk2 .
2
2
Proof. According to Remark 4.1, there exists e
c > 1 such that the operator inequality
Pf2 Γn≤4 ≤ e
cHf Γn≤4 holds on the set of states for which (17) is satisfied. The value
of e
c only depends on the ultraviolet cutoff Λ. Thus,
(20)
M [Φ] :=
1
kΓn≤4 P Φk2 ≤ 2k(P − Pf )Φk2 + 2kΓn≤4 Pf Φk2 ≤ 2k(P − Pf )gk2 + 2e
ckHf2 Γn≤4 Φk2 ,
which yields
1
1 n≤4
kΓ
P Φk2 − e
ckHf2 Γn≤4 Φk2 .
2
Therefore, it suffices to prove Corollary 4.2 with M [Φ] replaced by
k(P − Pf )Φk2 ≥
1
f[Φ] := δ kΦk2 + 3 νkH 2 Φk2 + 2ζk(P − Pf )Φk2 ,
M
f
2
4
and ζ small enough so that e
cζ < ν4 .
Now we consider two cases. Let c1 := max{8c0 , 8δ/α2 }.
If k(P − Pf )Φk2 ≤ c1 α2 kΦk2 , Theorem 4.1 and the above remark imply (19).
If k(P − Pf )Φk2 > c1 α2 kΦk2 , Lemma 4.1 implies that
(21)
1
1
h(P − Pf )2 Φ, Φi + hHf Φ, Φi − c0 α2 kΦk2
2
2
1
1
1
≥ h(P − Pf )2 Ψ⊥ , Ψ⊥ i + hHf Φ, Φi + k(P − Pf )Φk2
4
2
8
1
1
c1 α2
≥ h(P − Pf )2 Φ, Φi + hHf Φ, Φi +
kΦk2 ,
4
2
8
which concludes the proof since Σ0 − e0 < 0.
hHΦ, Φi ≥
10
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
Proof of Lemma 4.1. Recall the notation A± ≡ A± (0). The following holds.
1
1
H − (P − Pf )2 − Hf
2
2
√
1
α
1
2
− 2 αRe ((P − Pf ).A(0)) + 2αRe (A− )2 + 2αA+ .A− + Hf ,
= (P − Pf ) −
2
|x|
2
1
α
(P − Pf )2 −
≥ −4α2 .
4
|x|
(22)
and
(23)
√
√
√
2 α|h(P − Pf ).A(0)ψ, ψi| ≤ 2 αk(P − Pf )ψk2 + 2 αkA− ψk2 .
By the Schwarz inequality, there exists c1 independent of α such that
1
kA− ψk2 ≤ c1 kHf2 ψk2 .
(24)
Inequalities (23)-(24) imply that for small α,
(25)
√
1
1
2 α|h(P − Pf ).A(0)ψ, ψi| ≤ k(P − Pf )ψk2 + hHf ψ, ψi.
4
4
Moreover using (24) and
1
kA+ ψk2 ≤ c2 kψk2 + c3 kHf2 ψk2 ,
(26)
we arrive at
αh(A− )2 ψ, ψi = αhA− ψ, A+ ψi ≤ kA− ψk2 + −1 α2 kA+ ψk2
(27)
1
1
≤ c1 kHf2 ψk2 + −1 α2 (c2 kψk2 + c3 kHf2 ψk2 ) .
Collecting the inequalities (22), (25) and (27) with < 1/(8c1 ) and α small enough,
completes the proof.
Proof of Theorem 4.1: Let Φ := Φ1 + Φ2 := χ(|P | < p2c )Φ + χ(|P | ≥ p2c )Φ,
where P = i∇x is the total momentum operator (due to the transformation (5))
and pc = 13 is a lower bound on the norm of the total momentum for which [6,
Theorem 3.2] holds.
α
Since P commutes with the translation invariant operator H + |x|
, we have for
all ∈ (0, 1),
α
hHΦ, Φi = hHΦ1 , Φ1 i + hHΦ2 , Φ2 i − 2Re h Φ1 , Φ2 i
|x|
(28)
α
α
≥ hHΦ1 , Φ1 i + hHΦ2 , Φ2 i − h Φ1 , Φ1 i − −1 h Φ2 , Φ2 i.
|x|
|x|
• First, we have the following estimate
√
: (P − Pf − αA(0))2 : +Hf
√
= (P − Pf )2 − 2Re (P − Pf ). αA(0) + α : A(0)2 : +Hf
√
√
√
≥ (1 − α)(P − Pf )2 + (α − α) : A(0)2 : +Hf − cn.o. α
√
√
√
≥ (1 − α)(P − Pf )2 + (1 − O( α))Hf − O( α)
HYDROGEN GROUND STATE ENERGY IN QED
11
where in the last inequality we used (24) and (26). Therefore
√
1− α
α
α
h(H − −1 )Φ2 , Φ2 i ≥ h(
(P − Pf )2 − (1 + −1 ) )Φ2 , Φ2 i
|x|
2
|x|
(29)
D 1 − √α
E
√
√
2
+
(P − Pf ) + (1 − O( α))Hf − O( α) Φ2 , Φ2
2
√
(1+−1 )α
The lowest eigenvalue of the Schrödinger operator −(1 − O( α)) ∆
is
2 −
|x|
larger than −c α2 . Thus, using (29) and denoting
√
√
√
1− α
L :=
(P − Pf )2 + (1 − O( α))Hf − O( α) − c α2 ,
2
we get
(30)
hHΦ2 , Φ2 ) − −1 h
α
Φ2 , Φ2 i ≥ (LΦ2 , Φ2 i.
|x|
Now we have the following alternative: Either |Pf | < p3c , in which case we have
√
p2
hLΦ2 , Φ2 i ≥ ( 24c − O( α))kΦ2 k2 , or |Pf | ≥ p3c , in which case, using Φ2 = χ(|P | >
√
pc
pc
2
2 )Φ2 and Hf ≥ |Pf |, we have L ≥ ( 6 − O( α))kΦ2 k . In both cases, this yields
the bound
(31)
hLΦ2 , Φ2 i ≥
7
p2c
kΦ2 k2 ≥ (Σ0 − e0 + (e0 − e1 ))kΦ2 k2
48
8
since, for α small enough, the right hand side tends to zero, whereas pc is a constant
independent of α. Inequalities (30) and (31) yield
(32)
hHΦ2 , Φ2 i − −1 h
α
7
Φ2 , Φ2 i ≥ (Σ0 − e0 + (e0 − e1 ))kΦ2 k2 .
|x|
8
• For T (P ) being the self-energy operator with fixed total momentum P defined
in (7), we have from [6, Theorem3.2 (B)]
2
2
inf σ(T (P )) − P − inf σ(T (0)) ≤ c0 αP .
2
2
Therefore
(33)
hT (P )Φ1 , Φ1 i ≥ (1 − oα (1))(P 2 Φ1 , Φ1 i + Σ0 kΦ1 k2 .
If kΦ2 k2 ≥ 8kΦ1 k2 , using (33) yields
hHΦ1 , Φ1 i − h
(34)
α
Φ1 , Φ1 i
|x|
≥ (1 − oα (1))(P 2 Φ1 , Φ1 i − h(1 + )
α
Φ1 , Φ1 i + Σ0 kΦ1 k2
|x|
≥ (Σ0 − (1 + O(α) + O())e0 ) kΦ1 k2 .
Therefore, together with kΦ2 k2 ≥ 8kΦ1 k2 and (32), for α and small enough this
implies
(35)
3
hHΦ, Φi ≥ (Σ0 − e0 )kΦk2 + (e0 − e1 )kΦk2 .
4
12
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
If kΦ2 k2 < 8kΦ1 k2 , we have, from (33)
α
hHΦ1 , Φ1 i − h Φ1 , Φ1 i
|x|
≥(1 − oα (1))(P 2 Φ1 , Φ1 i − h(1 + )
(36)
≥(1+oα (1)+O())
− e0
∞
X
α
Φ1 , Φ1 i + Σ0 kΦ1 k2
|x|
k hΓk Φ1 , uα iL2 (R3 ,dx) k2 +e1 (kΦ1 k2
k=0
−
∞
X
!
k
2
k hΓ Φ1 , uα iL2 (R3 ,dx) k )
+ Σ0 kΦ1 k2 .
k=0
Now, by orthogonality of Φ and uα in the sense that for all k, hΓk Φ, uα iL2 (R3 ,dx) = 0,
we get
∞
X
(37)
k
2
khuα , Γ Φ1 ik =
∞
X
khuα , Γk Φ2 ik2 ≤ kΦ2 k2 kuα χ(|P | ≥
k=0
k=0
pc 2
)k
2
pc 2
)k →α→0 0
2
Thus, for α and small enough, (36), (37) and (32) imply also (35) in that case.
• Let e
c = max{δ, |c0 |α2 }.
If hHf Φ, Φi < 8e
ckΦk2 , (16) follows from (35) with ν = δ/(16e
c).
If hHf Φ, Φi ≥ 8e
ckΦk2 , using Lemma 4.1, we obtain
≤ 8kΦ1 k2 kuα χ(|P | ≥
1
1
hHf Φ, Φi − c0 α2 kgk2 ≥ hHf Φ, Φi + e
ckΦk2
2
4
(38)
1
≥ hHf Φ, Φi + δkΦk2 + (Σ0 − e0 )kΦk2 ,
4
since Σ0 − e0 ≤ 0, which yields (16) with ν = 41 .
This concludes the proof of (16).
hHΦ, Φi ≥
5. Estimate of the binding energy up to α3 term
Definition 5.1. Let uα be the normalized ground state of
α
.
hα := −∆ −
|x|
We define the projection Ψuα ∈ F of the normalized ground state ΨGS of H, onto
uα as follows
ΨGS = uα Ψuα + Ψ⊥ ,
where for all k ≥ 0,
huα , Γk Ψ⊥ iL2 (R3 ,dx) = 0.
(39)
Definition 5.2. Let
Φ2∗
:= −(Hf + Pf2 )−1 A+ · A+ Ωf
Φ3∗
:= −(Hf + Pf2 )−1 Pf · A+ Φ2∗
Φ1∗
:= −(Hf + Pf2 )−1 Pf · A− Φ2∗
where evidently, the state Φi∗ contains i photons.
HYDROGEN GROUND STATE ENERGY IN QED
13
Definition 5.3. On F, we define the positive bilinear form
h v , w i∗ := h v , (Hf + Pf2 ) w i,
(40)
1/2
and its associated semi-norm kvk∗ = hv, vi∗ .
We will also use the same notation for this bilinear forms on Fn , H and Hn .
Similarly, we define the bilinear form h . , . i] on H as
h u , v i] := h u , (Hf + Pf2 + hα + e0 ) v i
1/2
and its associated semi-norm kvk] = hv, vi] .
Definition 5.4. Let
1
Φu∗ α := 2α 2 ∇uα .(Hf + Pf2 )−1 A+ Ωf ,
(41)
and
(42)
1
Φu] α := 2α 2 ∇uα .(Hf + Pf2 + hα + e0 )−1 A+ Ωf .
Remark 5.1. The function Φu∗ α is not a vector in the Fock space H because of the
infrared singularity of the photon form factor. However, Hfγ Φu∗ α belongs to H, for
any γ > 0.
Theorem 5.1 (A priori estimate on the binding energy). We have
Σ ≤ Σ0 − e0 − kΦu] α k2] + O(α4 )
(43)
Proof. Using the trial function in H
3
3
Φtrial := uα (Ωf + 2α 2 Φ1∗ + αΦ2∗ + 2α 2 Φ3∗ ) + Φu] α ,
and from [4, Theorem 3.1],
Σ0 = −α2 kΦ2∗ k2∗ + α3 (2kA− Φ2∗ k2 − 4kΦ1∗ k2∗ − 4kΦ3∗ k2∗ ) + O(α4 ) ,
the result follows straightforwardly.
We decompose the function Ψuα defined in Definition 5.1 as follows.
Definition 5.5. Let
3
3
Ψuα := Γ0 Ψuα + 2η1 α 2 Φ1∗ + η2 αΦ2∗ + 2η3 α 2 Φ3∗ + ∆u∗ α ,
where Γ0 ∆u∗ α = 0, hΦi∗ , Γi ∆u∗ α i∗ = 0 (i=1,2,3), and Φ1∗ , Φ2∗ , Φ3∗ are given in Definition 5.2.
We further decompose Ψ⊥ into two parts.
Definition 5.6. Let
Ψ⊥ =: κ1 Φu] α + ∆⊥
] ,
with hΦu] α , Γ1 ∆⊥
] i] = 0.
To establish the estimate of the binding energy up to the α3 term, we will
compute hHΨGS , ΨGS i according to the decomposition of ΨGS introduced in Definitions 5.1 to 5.6. Using
√
α
H = (P 2 −
) + T (0) − 2Re P · (Pf + αA(0)),
|x|
and due to the orthogonality of uα and Ψ⊥ , we obtain
(44)
√
hHΨGS , ΨGS i = hHuα Ψuα , uα Ψuα i+hHΨ⊥ , Ψ⊥ i−4Re hP.(Pf + αA(0))uα Ψuα , Ψ⊥ i.
14
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
We will estimate separately each term in (44).
5.1. Estimate of the term hHuα Ψuα , uα Ψuα i.
Lemma 5.1.
hHuα Ψuα , uα Ψuα i ≥ − e0 kΨuα k2 − α2 kΦ2∗ k2∗ kΓ0 Ψuα k2 + α2 |η2 − Γ0 Ψuα |2 kΦ2∗ k2∗
+ α3 |η2 |2 2kA− Φ2∗ k2 − 4kΦ1∗ k2∗ − 4kΦ3∗ k2∗
+ 4α3 |η1 − η2 |2 kΦ1∗ k2∗ + |η3 − η2 |2 kΦ3∗ k2∗
1
+ cα4 log α−1 |η1 |2 + |η2 |2 + |η3 |2 + kΓ0 Ψuα k2 + k∆u∗ α k2∗ .
2
Proof. The proof is a trivial modification of the one given for the lower bound in
[4, Theorem 3.1]. The only modification is that we have a slightly weaker estimate
in Lemma 3.1 on the photon number for the ground state. This is accounted for by
4
−1
replacing the term of order α4 in [4, Theorem 3.1]
√ by a term of order α log α .
In addition, we need to use the equality hP.(Pf + αA(0))uα Ψuα , uα Ψuα i = 0, due
to the symmetry of uα .
√
5.2. Estimates for the cross term −4Re hP.(Pf + αA(0))uα Ψuα , Ψ⊥ i.
Lemma 5.2 (−4Re hP.Pf uα Ψuα , Ψ⊥ i term).
1
1
2
uα 2
2
− 4Re hP.Pf uα Ψuα , Ψ⊥ i ≥ −cα4 (|η1 |2 + |η2 |2 + |η3 |2 ) − kHf2 ∆⊥
] k − cαkHf ∆∗ k .
Proof. • For n = 1 photon,
(45)
3
hP.Pf Γ1 uα Ψuα , Ψ⊥ i = hP.Pf (η1 α 2 Φ1∗ + Γ1 ∆u∗ α )uα , κ1 Φu] α + Γ1 ∆⊥
] i.
Obviously
(46)
1
3
1
1 ⊥ 2
5
2
2
1 uα 2
2
2
|hP.Pf (η1 α 2 Φ1∗ + Γ1 ∆u∗ α )uα , Γ1 ∆⊥
] i| ≤ kHf Γ ∆] k + cα |η1 | + cα kHf Γ ∆∗ k .
1
Due to Lemma A.4 holds kHf2 (Φu] α − Φu∗ α )k2 = O(α5 ), which implies
(47)
1
3
|hP.Pf (η1 α 2 Φ1∗ + Γ1 ∆u∗ α )uα , κ1 Φu] α i| ≤ |κ1 |2 cα5 + c|η1 |2 α5 + cα2 kHf2 Γ1 ∆u∗ α k2
3
+ |hP.Pf (η1 α 2 Φ1∗ + Γ1 ∆u∗ α )uα , κ1 Φu∗ α i|.
For the last term on the right hand side of (47), due to the orthogonality of
∂uα
∂uα
α
and ∂u
∂xj , i 6= j, and the equality k ∂xi k = k ∂xj k, holds
∂uα
∂xi
(48)
|hP.Pf (η1 α
3
2
Φ1∗
+Γ
1
∆u∗ α )uα , κ1 Φu∗ α i|
3
X
3
∂uα 2
k |hη1 α 2 Φ1∗ + Γ1 ∆u∗ α , κ1 Pfi (A+ Ωf )i i|
=
k
∂x
i
i=1
3
= c|hη1 α 2 Φ1∗ + Γ1 ∆u∗ α , κ1 A+ .Pf Ωf i| = 0.
• For n = 2 photons,
|hP.Pf Γ2 uα Ψuα , Ψ⊥ i| = |hP.Pf (η2 αΦ2∗ + Γ2 ∆u∗ α )uα , Γ2 ∆⊥
] i|
1
1
2
2 uα 2
2
≤ cα4 |η2 |2 + kHf2 Γ2 ∆⊥
] k + αkHf Γ ∆∗ k .
HYDROGEN GROUND STATE ENERGY IN QED
15
• For n = 3 photons, a similar estimate yields
3
|hP.Pf Γ3 uα Ψuα , Ψ⊥ i| = |hP.Pf (η3 α 2 Φ3∗ + Γ3 ∆u∗ α )uα , Γ3 ∆⊥
] i|
1
1
2
3 uα 2
2
≤ cα4 |η3 |2 + kHf2 Γ3 ∆⊥
] k + αkHf Γ ∆∗ k .
• For n ≥ 4 photons,
1
1
2
n≥4 uα 2
2
∆∗ k .
|hP.Pf Γn≥4 uα Ψuα , Ψ⊥ i| ≤ kHf2 Γn≥4 ∆⊥
] k + αkHf Γ
√
Lemma 5.3 (−4Reh αP.A(0)uα Ψuα , Ψ⊥ i term).
1
√
2
2
⊥ 2
− 4Re h αP.A(0))uα Ψuα , Ψ⊥ i ≥ −2Re κ1 Γ0 Ψuα kΦu] α k2] − kHf2 ∆⊥
] k − α k∆] k
1
− cαkHf2 ∆u∗ α k2 − cα4 (|η1 |2 + |η2 |2 + |η3 |2 + |κ1 |2 ) + O(α5 log α−1 ).
Proof. We first estimate the term
1
Re α 2 hP.A+ uα Ψuα , Ψ⊥ i = αRe
∞
X
hP.A+ uα Γn Ψuα , Γn+1 Ψ⊥ i.
n=0
• For n = 0 photon, using the orthogonality hΦu] α , Γ1 ∆⊥
] i] = 0, yields
(49)
1
1
uα
uα
0 uα
1 ⊥
Re
(Γ
Ψ
)hΦ
,
κ
Φ
+
Γ
∆
i
i
=
−
− Re α 2 hP.A+ Γ0 uα Ψuα , κ1 Φu] α + Γ1 ∆⊥
1
]
]
]
]
]
2
1
= − Re κ1 Γ0 Ψuα hΦu] α , Φu] α i] .
2
• For n ≥ 1 photons,
X
1
1
3
uα 2
⊥ 2
2
|
Re α 2 hP.A+ Γn uα ∆u∗ α , Γn+1 ∆⊥
] i| ≤ cα k∆∗ k + kHf ∆] k
n≥2
(50)
1
2
5
2
2
2
5
−1
),
≤ kHf2 ∆⊥
] k + cα (|η1 | + |η2 | + |η3 | ) + O(α log α
where we used from Lemma 3.1 that k∆u∗ α k2 ≤ O(α2 log α−1 ) + cα3 (|η1 |2 + |η3 |2 ) +
cα2 |η2 |2 . We also have
1
3
3
|Re α 2 hP (2η1 α 2 Φ1∗ + η2 αΦ2∗ + 2η3 α 2 Φ3∗ )uα , A− ∆⊥
] i|
(51)
1
2
6
2
5
2
6
2
≤ kHf2 ∆⊥
] k + c(α |η1 | + α |η2 | +α |η3 | ).
1
We next estimate the term 2Re α 2 hP.A− uα Ψuα , Ψ⊥ i. We first get
1
1
2
⊥ 2
uα 2
2
|α 2 Re hP.A− ∆u∗ α uα , ∆⊥
] i| ≤ α k∆] k + cαkHf ∆∗ k .
(52)
Then we write
(53)
3
1
3
2
⊥ 2
4
2
2
|α 2 Re h(2η1 α 2 A− Φ1∗ + 2η3 α 2 A− Φ3∗ )∇uα , ∆⊥
] i| ≤ α k∆] k + cα (|η1 | + |η3 | ).
We also have
1
(54)
|α 2 Re hη2 αA− Φ2∗ .∇uα , κ1 Φu] α i|
1
−1
1
= |α 2 Re hη2 αHf 2 A− Φ2∗ .∇uα , Hf2 κ1 Φu] α i| ≤ cα4 (|η2 |2 + |κ1 |2 ),
16
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
−1
1
3
since Hf 2 A− Φ2∗ ∈ L2 (R3 ) and kHf2 Φu] α k = O(α 2 ). Finally we get
1
(55)
|α 2 Re hη2 αA− Φ2∗ .∇uα , Γ1 ∆⊥
] i|
−1
1
1
1
1 ⊥ 2
5
2
2
= |α 2 Re hη2 αHf 2 A− Φ2∗ .∇uα , Hf2 Γ1 ∆⊥
] i| ≤ kHf Γ ∆] k + cα |η2 | .
Collecting (49) to (55) concludes the proof.
5.3. Estimate for the term hHΨ⊥ , Ψ⊥ i.
Lemma 5.4.
1
uα 2
2
4
2
⊥
2
hHΨ⊥ , Ψ⊥ i ≥ (Σ0 − e0 )k∆⊥
] k − cα |κ1 | + M [∆] ] + |κ1 | kΦ] k] ,
2
where M [ . ] is defined in Corollary 4.2
(56)
Proof. Recall that
α
) + (Hf + Pf2 ) − 2Re (P.Pf )
H = (P 2 −
|x|
√
− 2 α(P − Pf ).A(0) + 2αA+ .A− + 2αRe (A− )2 .
Due to the orthogonality hΦu] α , ∆⊥
] i] = 0 we get
(57)
hHΨ⊥ , Ψ⊥ i
uα
2
− uα 2
= hHr, ri + |κ1 |2 kΦu] α k2] − e0 |κ1 |2 kΦu] α k2 − e0 κ1 h∆⊥
] , Φ] i + 2α|κ1 | kA Φ] k
√
uα
uα
− ⊥
− 2Re hP.Pf Φu] α , Φu] α i + 2αRe hA− .A− ∆⊥
] , κ1 Φ] i − 4 αRe hP.A ∆] , κ1 Φ] i
√
√
uα
uα
uα
⊥
+
− ⊥
− 4 αRe hP.A+ ∆⊥
] , κ1 Φ] i + 4Re αhPf .A(0)∆] , κ1 Φ] i + 4αRe hA .A ∆] , κ1 Φ] i
− 4Re hP.Pf Φu] α , ∆⊥
] i.
For the first term on the right hand side of (57), we have, from Corollary 4.2
(58)
⊥
⊥ 2
⊥
hH∆⊥
] , ∆] i ≥ (Σ0 − e0 )k∆] k + M [∆] ].
According to Lemma A.3, we obtain
(59)
uα
uα 2
2
2
⊥ 2
2 5
−1
−e0 κ1 h∆⊥
.
] , Φ] i − e0 |κ1 | kΦ] k ≥ −α k∆] k − c|κ1 | α log α
The next term, namely 2α|κ1 |2 kA− Φu] α k2 , is positive.
Due to the symmetry in x-variable,
(60)
hP.Pf Φu] α , Φu] α i = 0.
uα
The term 2αRe hA− .A− ∆⊥
] , κ1 Φ] i is estimated as
(61)
1
1
1
uα
uα 2 1
⊥ 2
⊥ 2
2 5
−1
2
2
2
.
2αRe hA− .A− ∆⊥
] , κ1 Φ] i ≥ −cα kκ1 Φ] k − νkHf ∆] k = − νkHf ∆] k −c|κ1 | α log α
4
4
Due to Lemma A.3 and [11, Lemma A4],
1
1
ν
uα
2
2 6
−1
(62)
|α 2 hA− ∆⊥
kHf2 ∆⊥
.
] , P κ1 Φ] i| ≤
] k + c|κ1 | α log α
8
The next term we have to estimate fulfils
(63)
1
uα
0 ⊥ 2
2
− uα 2
⊥ 2
2 4
|α 2 hP.A+ ∆⊥
] , κ1 Φ] i| ≤ kP Γ ∆] k +cα|κ1 | kA Φ] k ≤ k(P −Pf )∆] k −c|κ1 | α .
HYDROGEN GROUND STATE ENERGY IN QED
17
We have
(64)
√
√
√
uα
uα
uα
− 2 ⊥
+ 0 ⊥
Re αhPf .A(0)∆⊥
] , κ1 Φ] i = Re αhPf .A Γ ∆] , κ1 Φ] i + Re αhPf .A Γ ∆] , κ1 Φ] i
√
√
uα
uα
+
0 ⊥
= Re αhPf .A− Γ2 ∆⊥
] , κ1 Φ] i + Re αhA .Pf Γ ∆] , κ1 Φ] i
uα
Obviously, hA+ .Pf Γ0 ∆⊥
] , κ1 Φ] i = 0, and the first term is bounded by
1
√
uα
uα 2
2 ⊥ 2
2
2
|Re αhPf .A− Γ2 ∆⊥
] , κ1 Φ] i| ≤ kHf Γ ∆] k + cα|κ1 | kPf Φ] k
(65)
1
2
2 4
≤ kHf2 Γ2 ∆⊥
] k + c|κ1 | α .
uα
For the term αRe hA+ .A− ∆⊥
] , κ1 Φ] i we obtain
(66)
1
1
1
uα
uα 2
2
2
⊥ 2
5
2
⊥ 2
2
2
2
αRe hA+ .A− ∆⊥
] , κ1 Φ] i ≥ −cα kHf κ1 Φ] k |κ1 | −kHf ∆] k = −cα |κ1 | −kHf ∆] k .
According to (214) of Lemma A.3, the last term we have to estimate fulfils
(67)
1
1
1
uα 2
1 ⊥ 2
1 ⊥ 2
5
2
2
2
2
Re hP.Pf κ1 Φu] α , ∆⊥
] i ≤ kHf Γ ∆] k + ckP |Pf | κ1 Φ] k ≤ kHf Γ ∆] k + cα |κ1 | .
Collecting the estimates (57) to (67) yields (56).
5.4. Upper bound on the binding energy up to the order α3 .
Theorem 5.2. 1) Let Σ = inf σ(H). Then
Σ = Σ0 − e0 − kΦu∗ α k2∗ + O(α4 ),
(68)
Σ0 = −α2 kΦ2∗ k2∗ + α3 (2kA− Φ2∗ k2 − 4kΦ1∗ k2∗ − 4kΦ3∗ k2∗ ) + O(α4 ),
and Φu∗ α defined by (41).
uα
uα
GS
2) For the components ∆⊥
] , Ψ , ∆∗ of the ground state Ψ , and the coefficients η1 , η2 , η3 and κ1 defined in Definitions 5.1-5.5, holds
33
1
(69)
2
⊥ 2
4
⊥ 2
4
2
16
k∆⊥
] k = O(α ), kHf ∆] k = O(α ), k(P − Pf )∆] k = O(α ),
(70)
kΨuα k2 ≥ 1 − cα2 , kΓ0 Ψuα k2 ≥ 1 − cα2 ,
(71)
k∆u∗ α k2∗ = O(α4 ), k∆u∗ α k2 = O(α 16 ),
(72)
|η1,3 − 1|2 ≤ cα, |η2 − 1|2 ≤ cα2 , |κ1 − 1|2 ≤ cα .
33
Proof. • Step 1: We first show that (68) holds with an error estimate of the order
α4 log α−1 .
Collecting Lemmata 5.1, 5.2, 5.3 and Lemma 5.4 yields
(73)
hHΨGS , ΨGS i
≥ −e0 kΨuα k2 − α2 kΦ2∗ k2∗ |Γ0 Ψuα |2 + α3 |η2 |2 (2kA− Φ2∗ k2 − 4kΦ1∗ k2∗ − 4kΦ3∗ k2∗ )
+ α2 |η2 − Γ0 Ψuα |2 kΦ2∗ k2∗ + 4α3 (|η1 − η2 |2 kΦ1∗ k2∗ + |η3 − η2 |2 kΦ3∗ k2∗ )
1
− cα4 log α−1 (|η1 |2 + |η2 |2 + |η3 |2 + kΓ0 Ψuα k2 ) + k∆u∗ α k2∗
4
1
uα 2
2
⊥ 2
⊥
+ |κ1 | kΦ] k] + (Σ0 − e0 )k∆] k + M [∆] ] − cα4 |κ1 |2
4
− 2Re (κ1 Γ0 Ψuα )kΦu] α k2] + O(α5 log α−1 ).
18
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
We first estimate
|κ1 |2 kΦu] α k2] − 2Re (κ1 Γ0 Ψuα )kΦu] α k2] − cα4 |κ1 |2
(74)
|κ1 − Γ0 Ψuα |2 uα 2
kΦ] k] + O(α4 ).
2
Moreover, since |Γ0 Ψuα | ≤ 1, we replace in (73) −α2 kΦ2∗ k2∗ |Γ0 Ψuα |2 by −α2 kΦ2∗ k2∗
and in (74) −kΦu] α k2] |Γ0 Ψuα |2 by −kΦu] α k2] . In addition, using the inequalities
≥ −kΦu] α k2] |Γ0 Ψuα |2 +
1
|ηj − Γ0 Ψuα |2 − |η2 − Γ0 Ψuα |2
2
for j = 1, 2, 3 yields that for some c > 0,
(75) |η2 − ηj |2 ≥
and |ηj |2 ≤ 2|ηj − Γ0 Ψuα |2 + 2
(76)
hHΨGS , ΨGS i
≥ −e0 kΨuα k2 − α2 kΦ2∗ k2∗ + α3 |η2 |2 (2kA− Φ2∗ k2 − 4kΦ1∗ k2∗ − 4kΦ3∗ k2∗ ) − kΦu] α k2]
+ cα2 |η2 − Γ0 Ψuα |2 kΦ2∗ k2∗ + cα3 (|η1 − Γ0 Ψuα |2 + |η3 − Γ0 Ψuα |2 ) + cα3 |κ1 − Γ0 Ψuα |2
1
1
⊥
2
4
−1
).
+ k∆u∗ α k2∗ + (Σ0 − e0 )k∆⊥
] k + M [∆] ] + O(α log α
4
4
Comparing this expression with (43) of Theorem 5.1 gives
Σ = Σ0 − e0 − kΦu] α k2] + O(α4 log α−1 ).
(77)
Finally, by Lemma A.4, we can replace kΦu] α k] by kΦu∗ α k∗ in (77).
• Step 2: We now show that the error term does not contain any log α−1 term, by
uα
deriving improved estimates on the photon number for ∆⊥
] and ∆∗ .
From (76) we obtain
(78)
1
k∆u∗ α k2∗ = O(α4 log α−1 )
2
4
−1
and kHf2 ∆⊥
).
] k = O(α log α
This last two relations enable us to improve the estimates on the expected photon
number as follows
33
1
(79)
kN 2 ∆u∗ α k2 = O(α 16 )
(80)
2
16
kN 2 ∆⊥
] k = O(α )
1
33
To see this, we note that from Definition 5.1, we have
3
kaλ (k)Ψuα k2 ≤ kaλ (k)ΨGS k2 ≤
cα− 2 χΛ (|k|)
,
|k|
where in the last inequality, we used (14). Taking into account that
3
3
∆u∗ α = Ψuα − 2α 2 Φ1∗ − αΦ2∗ − 2α 2 Φ3∗ ,
where
kaλ (k)Φ1∗ k2 ≤
cχΛ (|k|)
cχΛ (|k|)(1 + |log |k| |)
, kaλ (k)Φ2∗ k2 ≤
|k|
|k|
kaλ (k)Φ3∗ k2 ≤
we arrive at
3
kaλ (k)∆u∗ α k2
cχΛ (|k|)
,
|k|
cα− 2 χΛ (|k|)(1 + | log |k| |)
.
≤
|k|
HYDROGEN GROUND STATE ENERGY IN QED
19
For the expected photon number of ∆u∗ α thus holds
XZ
1
uα 2
2
kN ∆∗ k =
kaλ (k)∆u∗ α k2 dk
λ
≤
3
cα− 2 (1 + |log |k| |)
dk +
|k|
XZ
|k|≤α
λ
17
15
8
1
15
Z
|k|>α
15
8
|k|−1 |k| kaλ (k)∆u∗ α k2 dk
33
≤ cα 8 + cα− 8 kHf2 ∆u∗ α k2 ≤ cα 16 .
The relation (80) can be proved similarly using
kaλ (k)Φu] α k2 ≤ c
α−1
.
|k|
We are now in position to finish the proof of Theorem 5.2. First we see that
according to (79) and (80), we have
1
kN 2 Ψuα k2 = O(α2 ),
which implies that in Lemma 5.1 we can replace the term cα4 log α−1 (|η1 |2 + |η2 |2 +
|η3 |2 + |Γ0 Ψuα |2 ) with cα4 (|η1 |2 + |η2 |2 + |η3 |2 + |Γ0 Ψuα |2 ) and consequently in (76)
and (77), the term O(α4 log α−1 ) can be replaced by O(α4 ). This proves (68). The
estimates (69)-(72) follow from (68) and (76) with O(α4 log α−1 ) replaced with
O(α4 ).
6. Estimate of the binding energy up to α5 log α−1 term
Theorem 6.1 (Upper bound up to the order α5 log α−1 for the binding energy).
For α small enough, we have
1
(81)
Σ0 − Σ ≥ α2 + e(1) α3 + e(2) α4 + e(3) α5 log α−1 + o(α5 log α−1 ),
4
(1)
(2)
where e , e and e(3) are defined in Theorem 2.1.
Definition 6.1. 1) Pick
(
hΨ⊥ ,Φ2 Γ0 Ψ⊥ i
α−1 hΦ2 Γ0 Ψ⊥∗,Φ2 Γ0 Ψ]⊥ i]
∗
∗
κ2 =
0
3
if kΓ0 Ψ⊥ k > α 2 ,
3
if kΓ0 Ψ⊥ k ≤ α 2 .
⊥
n ⊥
n ⊥
n ⊥
2) We split Ψ⊥ into Ψ⊥
1 and Ψ2 as follows: ∀n ≥ 0, Γ Ψ = Γ Ψ1 + Γ Ψ2 and
for n = 0,
0 ⊥
Γ 0 Ψ⊥
and Γ0 Ψ⊥
1 =Γ Ψ
2 = 0,
for n = 1,
uα
uα
Γ1 Ψ⊥
and hΓ1 Ψ⊥
1 = κ1 Φ]
2 , Φ] i] = 0,
for n = 2,
2 0 ⊥
Γ 2 Ψ⊥
1 = ακ2 Φ∗ Γ Ψ1 +
3
X
ακ2,i (Hf + Pf2 )−1 Wi
i=1
∂uα
,
∂xi
with Wi = Pfi Φ2∗ − 2A+ .Pf (Hf + Pf2 )−1 (A+ )i Ωf ,
2 ⊥
2 ⊥
2 ⊥
2 −1
Γ 2 Ψ⊥
Wi
2 = Γ Ψ − Γ Ψ1 , with hΓ Ψ2 , (Hf + Pf )
2 0 ⊥
and hΓ2 Ψ⊥
2 , Φ∗ Γ Ψ1 i] = 0,
∂uα
i] = 0 (i = 1, 2, 3),
∂xi
20
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
for n = 3,
2 −1 +
Γ3 Ψ⊥
A .A+ Φu] α ,
1 = ακ3 (Hf + Pf )
3 ⊥
3 ⊥
3 ⊥
2 −1 +
Γ3 Ψ⊥
A .A+ Φu] α i] = 0,
2 = Γ Ψ − Γ Ψ1 , hΓ Ψ2 , (Hf + Pf )
and for n ≥ 4,
Γn Ψ⊥
1 =0
and
n ⊥
Γn Ψ⊥
2 =Γ Ψ .
Lemma 6.1. The following a priori estimates hold
1
κ1 = 1 + O(α 2 ),
|κ2 | kΓ0 Ψ⊥
1 k = O(α),
κ2,i = O(1), (i = 1, 2, 3),
kΓ0 Ψ⊥
1 k = O(α),
2
kP Γ0 Ψ⊥
1 k = O(α ).
Proof. To derive these estimates, we use Theorem 5.2.
The first equality is a consequence of (72).
To derive the next two estimates, we first notice that (69) yields
2
kP Γ2 ∆⊥
] k
≤
2
2 ⊥ 2
2k(P − Pf )∆⊥
] k + 2kPf Γ ∆] k
≤
2
2 ⊥ 2
4
2
2k(P − Pf )∆⊥
] k + 2ckHf Γ ∆] k = O(α ),
1
therefore, using again (69), we obtain
1
2
2 ⊥ 2
2 ⊥ 2
4
k(hα + e0 ) 2 Γ2 ∆⊥
] k ≤ kP Γ ∆] k + e0 kΓ ∆] k = O(α ),
1
2
4
and thus, using from (69) that kHf2 Γ2 ∆⊥
] k = O(α ), we get
2
2 ⊥ 2
4
kΓ2 ∆⊥
] k] = kΓ Ψ k] = O(α ).
(82)
2 ⊥
We then write, using (82) and the h · , · i] -orthogonality of Γ2 Ψ⊥
1 and Γ Ψ2 ,
2
2 ⊥ 2
4
kΓ2 Ψ⊥
1 k] ≤ kΓ Ψ k] = O(α ).
(83)
2 ⊥
Since kΓ2 Ψ⊥
1 k] ≤ kΓ Ψ1 k∗ , and using (206) of Lemma A.1, we obtain
2
2 0 ⊥ 2
O(α4 ) = kΓ2 Ψ⊥
1 k∗ = kακ2 Φ∗ Γ Ψ1 k∗ + kα
X
κ2,i (Hf + Pf2 )−1 Wi
i
(84)
2
= α2 kΦ2∗ k2∗ |κ2 |2 kΓ0 Ψ⊥
1k +
2
α
k∇uα k2
3
X
∂uα 2
k
∂xi ∗
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ .
i
which gives
κ2,i = O(1)
and |κ2 | kΓ0 Ψ⊥
1 k = O(α).
The last two estimates are consequences of (69).
To derive the lower bound on the quadratic form of hH(uα Ψuα +g), uα Ψuα +Ψ⊥ i,
we will follow the same strategy as in Section 5, the only difference being that now
we have better a priori estimates on Ψuα and Ψ⊥ .
HYDROGEN GROUND STATE ENERGY IN QED
21
Proposition 6.1. We have
3
X
1
2Re hHΨ⊥ , uα Ψuα i ≥ − α4 Re
κ2,i h(Hf + Pf2 )−1 Pfi Φ2∗ , Wi i
3
i=1
uα 2
0 uα
− 4αRe h∇uα .Pf Φ2∗ , Γ2 Ψ⊥
2 i − 2Re κ1 Γ Ψ kΦ] k]
(85)
3
X
1
1
2
− α4 Re
h (Hf + Pf2 )− 2 (A− )i Φ2∗ , (Hf + Pf2 )− 2 (A+ )i Ωf i
3
i=1
5
a 2
⊥
5
−1
− α2 k(Ψ⊥
− α5 |κ3 |2 − |κ1 − 1|cα4 + O(α5 ).
1 ) k − M [Ψ2 ] − α log α
8
The proof of this Proposition is detailed in Subsection 6.2
Proposition 6.2.
1
− uα 2
hHΨ⊥ , Ψ⊥ i ≥ (Σ0 − e0 )kΨ⊥ k2 − 4αk(hα + e0 )− 2 Q⊥
α P A Φ∗ k
+ |κ1 |2 kΦu] α k2] + 2αkA− Φu∗ α k2 +
(86)
3
α4 X
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗
12 i=1
3
X
2 4
+ α Re
κ2,i hPf .A− (Hf + Pf2 )−1 Wi , (Hf + Pf2 )−1 (A+ )i Ωf i
3
i=1
1
⊥
+
uα
+ 4α 2 Re hΓ2 Ψ⊥
2 , A .Pf Φ∗ i + M1 [Ψ ],
where
1
(87)
a 2
M1 [Ψ⊥ ] =(1−c0 α)k(hα + e0 ) 2 Γ0 (Ψ⊥
1) k +
|κ3 + 1|2 2 2 2 uα 2
α kΦ∗ k∗ kΦ] k
2
3
5
−1
),
− |κ1 − 1|cα4 + M [Ψ⊥
2 ] + o(α log α
4
and Q⊥
α is the projection onto the orthogonal complement to the ground state uα of
α
the Schrödinger operator hα = −∆ − |x|
The proof of this Proposition is detailed in Subsection 6.3.
6.1. Proof of Theorem 6.1 on the upper bound up to the order α5 log α−1 .
We have
(88)
hHΨGS , ΨGS i = (Σ0 − e0 )kΨuα k2 + 2Re hHΨ⊥ , uα Ψuα i + hHΨ⊥ , Ψ⊥ i.
The estimates for the last two terms are given in Propositions 6.1 and 6.2. We will
minimize this expression with respect to the parameters κ1 , κ2,i , and κ3 .
• We first estimate the second term on the right hand side of (85) together with
the seventh term on the right hand side of (86). We have
1
2 ⊥
+
uα
2
− 4αRe h∇uα .Pf Φ2∗ , Γ2 Ψ⊥
2 i + 4α Re hΓ Ψ2 , A .Pf Φ∗ i
(89)
3
X
∂uα
i
= −4αRe hΓ2 Ψ⊥
,
(
Pfi Φ2∗ − 2A+ .Pf (Hf + Pf2 )−1 (A+ )i Ωf )
2
∂xi
i=1
= −4α
3
X
i=1
2 −1
Re hΓ2 Ψ⊥
Wi
2 , (Hf + Pf )
∂uα
i∗ .
∂xi
22
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
2 −1
α
Using the h · , · i] -orthogonality of Γ2 Ψ⊥
Wi ∂u
2 and (Hf +Pf )
∂xi , the last expression
can be estimated as
− 4α
(90)
3
X
2 −1
Re hΓ2 Ψ⊥
Wi
2 , (hα + e0 )(Hf + Pf )
i=1
∂uα
i.
∂xi
By the Schwarz inequality, this term is bounded below by
2
− α2 kΓ2 Ψ⊥
2k −c
(91)
3
X
k(hα + e0 )
i=1
∂uα 2
2
6
k = −α2 kΓ2 Ψ⊥
2 k − O(α ).
∂xi
• Next, we collect all the terms involving κ1 in (85) and (86). This yields
− 2Re κ1 Γ0 Ψuα kΦu] α k2] + |κ1 |2 kΦu] α k2] − |κ1 − 1|cα4
(92)
≥ −|Γ0 Ψuα |2 kΦu] α k2] + |κ1 − Γ0 Ψuα |2 kΦu] α k2] − |κ1 − 1|cα4
Notice that from Theorem 5.2 we have |Γ0 Ψuα |2 = 1 + O(α2 ); moreover, we have
|Γ0 Ψuα | = 1 + O(α2 ). This yields
(93)
− 2Re κ1 Γ0 Ψuα kΦu] α k2] + |κ1 |2 kΦu] α k2] − |κ1 − 1|cα4
≥ −kΦu] α k2] + |κ1 − 1|2 c0 α3 − |κ1 − 1|cα4 + O(α5 ) = −kΦu] α k2] + O(α5 ).
• We now collect and estimates the terms in (85) and (86) involving κ2,i . We
get
(94)
3
3
X
α4 X
1
κ2,i h(Hf + Pf2 )−1 Wi , Pfi Φ2∗ i +
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗
− α4 Re
3
12
i=1
i=1
3
X
2
+ α4 Re
κ2,i h(Hf + Pf2 )−1 Wi , Pf .A+ (Hf + Pf2 )−1 (A+ )i Ωf i
3
i=1
3
3
X
1
α4 X
= − α4 Re
κ2,i k(Hf + Pf2 )−1 Wi k2∗ +
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗
3
12
i=1
i=1
≥−
3
α4 X
k(Hf + Pf2 )−1 Wi k2∗
3 i=1
• Collecting in (85) and (86) the terms containing κ3 yields
(95)
|κ3 + 1|2 2 2 2 uα 2
α kΦ∗ k∗ kΦ] k − α5 |κ3 |2 ≥ c1 α5 log α−1 |κ3 + 1|2 − α5 |κ3 |2 ≥ −c2 α5 ,
2
where c1 and c2 are positive constants.
• The fifth term on the right hand side of (85) and the first term on the right
hand side of (87) are estimated, for α small enough, as
1
δ
a 2
2
⊥ a 2
2
⊥ a 2
(96) (1 − c0 α)k(hα + e0 ) 2 Γ0 (Ψ⊥
1 ) k − α k(Ψ1 ) k ≥ ( − α )k(Ψ1 ) k ≥ 0,
2
with δ =
3 2
32 α ,
a
and where we used that (Ψ⊥
1 ) is orthogonal to uα .
HYDROGEN GROUND STATE ENERGY IN QED
23
• Substituting the above estimates in (88)
(97)
hHΨGS , ΨGS i
≥ (Σ0 − e0 )kΨuα k2 + (Σ0 − e0 )kΨ⊥ k2 − kΦu] α k2] − α4
3
X
k(Hf + Pf2 )−1 Wi k2∗
i=1
3
X
1
1
2
− α4 Re
h (Hf + Pf2 )− 2 (A− )i Φ2∗ , (Hf + Pf2 )− 2 (A+ )i Ωf i
3
i=1
1
− uα 2
− uα 2
5
−1
),
− 4αk(hα + e0 )− 2 Q⊥
α P A Φ∗ k + 2αkA Φ∗ k + o(α log α
where Q⊥
α is the projection onto the orthogonal complement to the ground state
α
uα of the Schrödinger operator hα = −∆ − |x|
.
To complete the proof of Theorem 6.1 we first note that
kΨuα k2 + kΨ⊥ k2 = kΨGS k2 .
(98)
Moreover, according to Lemma A.4
(99)
−kΦu] α k2] = −kΦu∗ α k2∗ +
1
1 1
k(h1 + ) 2 ∇u1 k2 α5 log α−1 + o(α5 log α−1 ),
3π
4
and
kΦu∗ α k2∗ =
(100)
α3
2π
Z
0
∞
χΛ (t)
dt = e(1) α3 .
1+t
In addition, we have the following identities (i = 1, 2, 3)
k(Hf + Pf2 )−1 Wi k2∗ =
(101)
1
k(Hf + Pf2 )− 2 2A+ .Pf (Hf + Pf2 )−1 (A+ )i − Pfi (Hf + Pf2 )−1 A+ .A+ Ωf k2 ,
and
(102)
3
X
1
1
2
− α4 Re
h (Hf + Pf2 )− 2 A− Φ2∗ , (Hf + Pf2 )− 2 (A+ )i Ωf i
3
i=1
3
X
2
hA− (Hf + Pf2 )−1 A+ .A+ Ωf , (Hf + Pf2 )−1 (A+ )i Ωf i .
= − α4 Re
3
i=1
We also have
1
− uα 2
4 2
(103) − 4αk(hα + e0 )− 2 Q⊥
α P A Φ∗ k = −4α a0 k(−∆ −
1
1 1
2
+ )− 2 Q⊥
1 ∆u1 k ,
|x| 4
with
Z
a0 =
k12 + k22
2
χΛ (|k|) dk1 dk2 dk3 ,
4π 2 |k|3 |k|2 + |k|
and
(104)
2αkA− Φu∗ α k2 =
3
2 4X −
α
kA (Hf + Pf2 )−1 (A+ )i Ωf k2 .
3 i=1
Substituting (98)-(104) into (97) finishes the proof of Theorem 6.1.
24
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
6.2. Proof of Proposition 6.1.
Lemma 6.2. The following holds
(105)
3
X
4
− 4Re hP.Pf uα Ψuα , Ψ⊥ i ≥ − α2 Re
κ2,i k∇uα k2 h(Hf + Pf2 )−1 Pfi Φ2∗ , Wi i
3
i=1
1
⊥ 2
5
2
5
2
− 4αRe h∇uα .Pf Φ2∗ , Γ2 Ψ⊥
2 i − kHf Ψ2 k − α |κ3 | + O(α )
Proof. For n 6= 2, 3, with the estimates from the proof of Lemma 5.2 and using that
due to Theorem 5.2 we have
1
kHf2 ∆u∗ α k2 = O(α4 ), |η1 | = O(1), and |κ1 | = O(1),
1 ⊥
n≥4 ⊥
and since Γ1 ∆⊥
∆] = Γn≥4 Ψ⊥
2 , we obtain
] = Γ Ψ2 and Γ
X
1
2
5
(106)
−4Re hP.Pf uα Γn Ψuα , Γn Ψ⊥ i ≥ −kHf2 Ψ⊥
2 k + O(α ).
n6=2,3
For n = 2,
(107)
− 4Re h∇uα .(αη2 Pf Φ2∗ + Pf Γ2 ∆u∗ α ), Γ2 Ψ⊥ i ≥ −4Re h∇uα .αη2 Pf Φ2∗ , Γ2 Ψ⊥
1i
1
1
2 uα 2
2 ⊥ 2
2
2
− 4Re h∇uα .αη2 Pf Φ2∗ , Γ2 Ψ⊥
2 i − cαkHf Γ ∆∗ k − cαkHf Γ Ψ k .
Using Theorem 5.2, the last two terms on the right hand side of (107) can be
estimated by O(α5 ). For the first term on the right hand side of (107), using from
Lemma A.1 that hPfi Φ2∗ , Φ2∗ i = 0, from Theorem 5.2 that η2 = 1 + O(α), and from
Lemma 6.1 that κ2,i = O(1), holds
− 4αRe h∇uα .η2 Pf Φ2∗ , Γ2 Ψ⊥
1i
(108)
= −4Re αh∇uα .η2 Pf Φ2∗ , ακ2 Φ2∗ Γ0 Ψ⊥ i
X
∂uα
− 4Re αh∇uα .η2 Pf Φ2∗ ,
ακ2,i (Hf + Pf2 )−1 Wi
i
∂xi
i
X
4
= − Re α2 k∇uα k2
κ2,i h(Hf + Pf2 )−1 Pfi Φ2∗ , Wi i
3
i
3
X
1
= − α4 Re
κ2,i h(Hf + Pf2 )−1 Pfi Φ2∗ , Wi i.
3
i=1
∂uα
∂uα
α ∂uα
We also used that h ∂u
∂xi , ∂xj i = 0 for i 6= j and k ∂xi k = k ∂xj k for all i and j.
Finally, the second term on the right hand side of (107) gives the second term
on the right hand side of (105) plus O(α5 ), using from Theorem 5.2 that |η2 − 1|2 =
1
2
O(α2 ) and kHf2 Γ2 Ψ⊥
2 k = O(α ).
To complete the proof, we shall estimate now the term for n = 3,
3
(109)
4Re hP.Pf uα Γ3 Ψuα , Γ3 Ψ⊥ i = 4Re α 2 2η3 hP.Pf uα Φ3∗ , Γ3 Ψ⊥
1i
3
3 uα
3 ⊥
+ 4Re α 2 2η3 hP.Pf uα Φ3∗ , Γ3 Ψ⊥
2 i + 4Re hP.Pf uα Γ ∆∗ , Γ Ψ i.
1
1
The inequalities kHf2 ∆u∗ α k ≤ cα2 and kHf2 Γ3 Ψ⊥ k ≤ cα2 (see Theorem 5.2) imply
that the last term on the right hand side of (109) is O(α5 ). For the second term
HYDROGEN GROUND STATE ENERGY IN QED
25
on the right hand side of (109) holds
1
3
3 ⊥ 2
5
2
Re α 2 η3 hP.Pf uα Φ3∗ , Γ3 Ψ⊥
2 i ≥ −kHf Γ Ψ2 k + O(α ),
(110)
since from Theorem 5.2 we have η3 = O(1).
Finally to estimate the first term on the right hand side of (109), we note that
1
1
1
|k1 |− 6 |k2 |− 6 |k3 |− 6 Φ3∗ (k1 , k2 , k3 ) ∈ L2 (R9 , C6 ) ,
and from Lemma A.5,
1
1
1
k |k1 | 6 |k2 | 6 |k3 | 6 (Hf + Pf2 )−1 A+ .A+ Φu] α k2 = O(α3 ).
This implies, using again |η3 | = O(1), and the explicit expression of Γ3 Ψ⊥
1
(111)
3
5
3
5
2
5
2
2
|α 2 2η3 hP.Pf uα Φ3∗ , Γ3 Ψ⊥
1 i| ≤ cα |κ3 |α |η3 | kP uα k ≤ α |κ3 | + O(α ).
Collecting (106)-(111) concludes the proof.
Lemma 6.3. The following estimate holds
√
− 4 αRe hP.A+ uα Ψuα , Ψ⊥ i
(112)
1
5
2
5
= −2Re κ1 Γ0 Ψuα kΦu] α k2] − M [Ψ⊥
2 ] − α |κ3 | + O(α ),
4
Proof. Obviously
1
(113)
− 4Re α 2 hP.A+ uα Ψuα , Ψ⊥ i
1
3
3
= −4Re α 2 hP.A+ uα (Γ0 Ψuα + 2η1 α 2 Φ1∗ + η2 αΦ2∗ + 2η3 α 2 Φ3∗ + ∆u∗ α ), Ψ⊥ i.
Step 1 From (113), let us first estimate the term
(114)
1
1
−4Re α 2 hP.A+ uα ∆u∗ α , Ψ⊥ i = −4α 2 Re
∞
X
hP.A+ Γn uα ∆u∗ α , Γn+1 Ψ⊥ i.
n=0
For
For
O(α5 )
For
n = 0, the corresponding term vanishes since Γ0 ∆u∗ α = 0.
n > 2, we can use (50) where the term O(α5 log α−1 ) can be replaced with
33
because we know from Theorem 5.2 that k∆u∗ α k2 = O(α 16 ).
n = 1, we have
1
1
2 ⊥
3
1 uα 2
2 ⊥ 2
2
|4α 2 Re hP.A+ Γ1 uα ∆u∗ α , Γ2 Ψ⊥
1 + Γ Ψ2 i| ≤ cα kΓ ∆∗ k + kHf Γ Ψ2 k
!
3
(115)
X
1
∂u
α
1
u
−
2
0
⊥
2
−1
+ |4α 2 h∇uα Γ ∆∗ α , A
ακ2 Φ∗ Γ Ψ1 + α
κ2,i (Hf + Pf ) Wi
i|.
∂xi
i=1
−1
To estimate the last term on the right hand side we note that Hf 2 A− Φ2∗ ∈ L2 and
−1
Hf 2 A− (Hf + Pf2 )−1 Wi ∈ L2 which thus gives for this term the bound
X
1
2
6
(116)
cαkHf2 ∆u∗ α k2 + α4 |κ2 |2 kΓ0 Ψ⊥
|κ2,i |2 = O(α5 ),
1 k + α
i
using Theorem 5.2 and Lemma 6.1. The inequalities (115) and (116) imply
(117)
1
1
2 ⊥
2 ⊥ 2
5
2
|Re α 2 hP.A+ Γ1 uα ∆u∗ α , Γ2 Ψ⊥
1 + Γ Ψ2 i| ≤ kHf Γ Ψ2 k + O(α ).
26
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
1
To complete the estimate of the term 4α 2 Re hP.A+ uα ∆u∗ α , Ψ⊥ i we have to esti1
3 ⊥
mate the term for n = 2 in (114), namely 4Re α 2 hP.A+ uα Γ2 ∆u∗ α , Γ3 Ψ⊥
1 + Γ Ψ2 i.
Obviously,
1
1
⊥ 2
5
2
|Re α 2 hP.A+ uα Γ2 ∆u∗ α , Γ3 Ψ⊥
2 i| ≤ kHf Ψ2 k + O(α ).
(118)
For the term involving Γ3 Ψ⊥
1 we have
1
|Re α 2 hP uα Γ2 ∆∗uα , ακ3 A− (Hf + Pf2 )−1 A+ .A+ Φu] α i|
(119)
1
1
≤ cα3− 16 kΓ2 ∆u∗ α k2 + |κ3 |2 α2+ 16 kΦu] α k2 = |κ3 |2 α5 + O(α5 ),
using Theorem 5.2 and Lemma A.5. Collecting (114)-(119) yields
1
1
2
5
2
5
|Re α 2 hP.A+ uα ∆u∗ α , gi| ≤ kHf2 Ψ⊥
2 k + α |κ3 | + O(α ).
(120)
3
1
Step 2 We next estimate in (113) the term −4Re α 2 hP.A+ uα (Γ0 Ψuα +α 2 2η1 Φ1∗ +
3
αη2 Φ2∗ + α 2 2η3 Φ3∗ ), Ψ⊥ i. First using (50) yields
1
−4Re α 2 hP.A+ uα Γ0 Ψuα , Ψ⊥ i = −2Re (κ1 Γ0 Ψuα )kΦu] α k2] .
(121)
We also have, using Theorem 5.2
3
1
3
|α 2 hP uα (α 2 2η1 Φ1∗ + α 2 2η3 Φ3∗ ), A− Ψ⊥ i|
(122)
1
1
≤ αkHf2 Γ2 Ψ⊥ k2 + αkHf2 Γ4 Ψ⊥ k2 + O(α5 ) = O(α5 ),
and
(123)
1
3 ⊥
|α 2 hP uα αη2 Φ2∗ , A− (Γ3 Ψ⊥
1 + Γ Ψ2 )i|
1
3
1
1
1
1
2
−4
5
2
≤ kHf2 Γ3 Ψ⊥
|k2 |− 4 Φ2∗ , |k1 | 4 |k2 | 4 A− Γ3 Ψ⊥
2 k + |α hP uα η2 |k1 |
1 i| + O(α )
1
2
2 5
5
≤ kHf2 Γ3 Ψ⊥
2 k + |κ3 | α + O(α ).
1
1
1
1
Here we used |k1 |− 4 |k2 |− 4 Φ2∗ ∈ L2 and k|k1 | 4 |k2 | 4 A− (Hf + Pf2 )−1 A+ .A+ Φu] α k2 =
O(α3 ) (see Lemma A.5).
Collecting (120)-(123) yields
(124)
1
1
2
5
2
5
−4Re α 2 hP.A+ uα Ψuα , Ψ⊥ i ≥ −2Re κ1 Γ0 Ψuα kΦu] α k2] −kHf2 Ψ⊥
2 k −α |κ3 | +O(α ).
Lemma 6.4.
(125)
√
− 4 αRe hP.A− uα Ψuα , Ψ⊥ i ≥
3
X
1
1
2
− α4 Re
h (Hf + Pf2 )− 2 (A− )i Φ2∗ , (Hf + Pf2 )− 2 (A+ )i Ωf i
3
i=1
1
2
⊥ a 2
5
−1
− M [Ψ⊥
(α|κ3 |2 + 1) − |κ1 − 1|cα4 + O(α5 ) ,
2 ] − α k(Ψ1 ) k − α log α
4
a
0 ⊥
0 ⊥
0 ⊥
where (Ψ⊥
1 ) (x) := (Γ Ψ1 (x) − Γ Ψ1 (−x))/2 is the odd part of Γ Ψ1 .
HYDROGEN GROUND STATE ENERGY IN QED
27
Proof. Since from Lemma A.1 we have ∇uα .A− Φ1∗ = 0, we have
3
1
4Re α 2 hP.A− uα Ψuα , Ψ⊥ i = 4α 2 Re η2 hA− Φ2∗ .P uα , Γ1 Ψ⊥ i
(126)
1
+ 4α2 Re 2η3 hA− Φ3∗ .P uα , Γ2 Ψ⊥ i + 4α 2 hA− ∆u∗ α .P uα , Ψ⊥ i.
For the first term on the right hand side of (126) we have
3
(127)
uα
4α 2 Re η2 hA− Φ2∗ .P uα , Γ1 Ψ⊥
2 + κ1 Φ] i
3
−1
1
3
uα
− 2
2
= 4α 2 Re η2 hHf 2 A− Φ2∗ .P uα , Hf2 Γ1 Ψ⊥
2 i + 4α Re η2 hA Φ∗ .P uα , κ1 Φ] i.
1
2
The first term on the right hand side of (127) is bounded from below by −kHf2 Γ1 Ψ⊥
2k +
5
O(α ).
Applying Lemma A.4, we can replace Φu] α in the second term of the right hand
side of (127) by Φu∗ α , at the expense of O(α5 ). More precisely
3
|α 2 hη2 A− Φ2∗ .P uα , κ1 (Φu] α − Φu∗ α )i|
1
≤ cα5 |η2 |2 k(Hf + Pf2 )− 2 A− Φ2∗ k2 + |κ1 |2 kΦu] α − Φu∗ α k2∗ = O(α5 ).
Moreover
3
4α 2 Re η2 κ1 hA− Φ2∗ .P uα , Φu∗ α i
=
3
X
8 2
α k∇uα k2 Re η2 κ1
h(A− )i Φ2∗ , (Hf + Pf2 )−1 (A+ )i Ωf i
3
i=1
=
3
X
2 4
h(A− )i Φ2∗ , (Hf + Pf2 )−1 (A+ )i Ωf i
α Re η2 κ1
3
i=1
≥
3
X
2 4
α Re
h(A− )i Φ2∗ , (Hf + Pf2 )−1 (A+ )i Ωf i − |κ1 − 1|cα4 ,
3
i=1
(128)
where we used κ1 = O(1) (Lemma 6.1) and η2 = 1 + O(α) (Theorem 5.2). Note
1
that the right hand side of (128) is well defined since (Hf + Pf2 )− 2 A+ Ωf ∈ F and
1
(Hf + Pf2 )− 2 A− Φ2∗ ∈ F.
Collecting the estimates for the first and the second term in the right hand side
of (127), we arrive at
(129)
3
− 4α 2 Re η2 hA− Φ2∗ .P uα , Γ1 Ψ⊥ i
1
8
2
5
≥ − α2 k∇uα k2 Re κ1 h(A− )i Φ2∗ , (Hf + Pf2 )−1 A+ Ωf i − kHf2 Γ1 Ψ⊥
2 k + O(α ).
3
Here we used also η2 = 1 + O(α).
As the next step, we return to (126) and estimate the second term on the right
hand side as
(130)
−1
1
4α2 Re 2η3 hA− Φ3∗ .P uα , Γ2 Ψ⊥ i = 8α2 Re η3 hHf 2 A− Φ3∗ .P uα , Hf2 Γ2 Ψ⊥ i = O(α5 ),
28
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
−1
1
1
2
2
⊥ 2
4
2
where we used Hf 2 A− Φ3∗ ∈ L2 and kΓ2 Hf2 ∆⊥
] k = kΓ Hf Ψ k = O(α ) from
Theorem 5.2. For the last term on the right hand side of (126), we have
1
4α 2 Re hA− ∆u∗ α · ∇uα , Ψ⊥ i
1
1
1
1
1
1
uα
− uα
2
= 4α 2 Re hA− ∆u∗ α · ∇uα , Γ0 Ψ⊥
1 i + 4α Re κ1 hA ∆∗ · ∇uα , Φ] i
(131)
− uα
2 ⊥
2
+ 4α 2 Re hA− ∆u∗ α · ∇uα , Γ1 Ψ⊥
2 i + 4α Re hA ∆∗ · ∇uα , Γ Ψ i
+ 4α 2 Re hA− ∆u∗ α · ∇uα , Γ3 Ψ⊥ i + 4α 2 Re hA− ∆u∗ α .∇uα , Γn≥4 Ψ⊥ i.
⊥ s
⊥ a
⊥ s
⊥ a
We write the function Γ0 Ψ⊥
1 = (Ψ1 ) + (Ψ1 ) where (Ψ1 ) (respectively (Ψ1 ) )
0 ⊥
denotes the even (respectively odd) part of Γ Ψ1 . Obviously, we have
(132)
1
1
2
2
⊥ a 2
2
⊥ a 2
5
2
|α 2 Re hA− ∆u∗ α ·∇uα , Γ0 Ψ⊥
1 i| ≤ cαkHf Rk +α k(Ψ1 ) k = α k(Ψ1 ) k +O(α ).
The constant can be chosen small for large c.
For the second term on the right hand side of (131), we have
(133)
1
1
|α 2 κ1 hA− ∆u∗ α ·∇uα , Φu] α i| ≤ α5 log α−1 |κ1 |2 +cαkHf2 ∆u∗ α k2 = α5 log α−1 +O(α5 ).
3 2
For the third term on the right hand side of (131), we have, since δ = 32
α
(134)
1
1
δ 1 ⊥ 2
δ 1 ⊥ 2
5
uα 2
2
|4α 2 hA− ∆u∗ α · ∇uα , Γ1 Ψ⊥
2 i| ≤ kΓ Ψ2 k + cαkHf ∆∗ k = kΓ Ψ2 k + O(α ).
8
8
Similarly
1
δ
2
5
(135)
|4α 2 hA− ∆u∗ α · ∇uα , Γn≥4 Ψ⊥ i| ≤ kΓn≥4 Ψ⊥
2 k + O(α ).
8
To complete the estimate of the last term in (131), we have to estimate two
1
1
terms: −4Re α 2 hA− ∆u∗ α .P uα , Γ2 Ψ⊥ i and −4Re α 2 hA− ∆u∗ α .P uα , Γ3 Ψ⊥ i. For the
first one we have
1
1
2
4
2
0 ⊥ 2
|Re α 2 hA− ∆u∗ α .P uα , Γ2 Ψ⊥ i| ≤ cαkHf2 ∆u∗ α k2 + α2 kΓ2 Ψ⊥
2 k + α |κ2 | kΓ Ψ k
+ α6
3
X
2
5
|κ2,i |2 = α2 kΓ2 Ψ⊥
2 k + O(α ).
i=1
Similarly,
(136)
1
1
2
6
−1
|κ3 |2 + O(α5 ).
|Re α 2 hA− ∆u∗ α .P uα , Γ3 Ψ⊥ i| ≤ cαkHf2 ∆u∗ α k2 + α2 kΓ3 Ψ⊥
2 k + α log α
Collecting the estimates (131)-(136) yields
(137)
1
δ ⊥ 2
2
5
−1
kΨ2 k + kHf2 Ψ⊥
(1 + α|κ3 |2 ) + O(α5 ).
2 k + α log α
8
Collecting (120), (124), (129), (130) and (137) concludes the proof.
1
|4Re α 2 hA− ∆u∗ α .P uα , Ψ⊥ i| ≤
We can now prove the estimate of Re hHg, uα Ψuα i of Proposition 6.1.
Proof of the Proposition 6.1. Using the orthogonality (39) of uα and Ψ⊥ , yields
√
2Re hHΨ⊥ , Ψuα i = −4Re hP.Pf uα Ψuα , Ψ⊥ i − 4 αRe hP.A(0)uα Ψuα , Ψ⊥ i.
Together with Lemmata 6.2-6.4, this concludes the proof of Proposition 6.1.
HYDROGEN GROUND STATE ENERGY IN QED
29
6.3. Proof of Proposition 6.2. We have the estimate
Proposition 6.3. We have
hHΨ⊥ , Ψ⊥ i
(138)
⊥
⊥
⊥
− ⊥ 2
− ⊥ 2
− ⊥ 2
≥ hHΨ⊥
1 , Ψ1 i + hHΨ2 , Ψ2 i + 2α kA Ψ k − kA Ψ1 k − kA Ψ2 k
1
⊥
⊥
⊥
⊥
⊥
2
− 4Re hP.Pf Ψ⊥
2 , Ψ1 i − 4α Re hP.A(0)Ψ2 , Ψ1 i + 4Re hPf .A(0)Ψ2 , Ψ1 i
1
2
5
6
−1
− M [Ψ⊥
|κ3 |2 − c0 αk(hα + e0 ) 2 Γ0 Ψ⊥
1 k + O(α ).
2 ] − cα log α
Proof. Recall that
(139)
H = P2 −
1
α
+ T (0) − 2Re P. Pf + α 2 A(0) ,
|x|
and
1
T (0) =: (Pf + α 2 A(0))2 : +Hf .
(140)
Due to the orthogonality
n ⊥
hΓn Ψ⊥
1 , Γ Ψ2 i] = 0,
n = 0, 1, . . . ,
and (139), (140), we obtain
(141)
h(H + e0 )Ψ⊥ , Ψ⊥ i
⊥
⊥
⊥
= h(H + e0 )Ψ⊥
2 , Ψ2 i + h(H + e0 )Ψ1 , Ψ1 i +
3
X
n ⊥
2αRe hA− .A− Γn+2 Ψ⊥
2 , Γ Ψ1 i
n=0
1 ⊥
− ⊥ 2
− ⊥ 2
− ⊥ 2
+ 2αRe hA− .A− Γ3 Ψ⊥
1 , Γ Ψ2 i + 2α(kA Ψ k − kA Ψ1 k − kA Ψ2 k )
1
1
⊥
⊥
⊥
⊥
⊥
2
2
− 4Re hP.Pf Ψ⊥
2 , Ψ1 i − 4α Re hP.A(0)Ψ2 , Ψ1 i + 4α Re hPf .A(0)Ψ2 , Ψ1 i.
We have
1
(142)
3 ⊥
5 ⊥ 2
2
3 ⊥ 2
2
2αRe hA− .A− Γ5 Ψ⊥
2 , Γ Ψ1 i ≥ −kHf Γ Ψ2 k − cα kΓ Ψ1 k
1
2
7
−1
≥ −kHf2 Γ5 Ψ⊥
|κ3 |2 .
2 k − cα log α
Similarly,
2 ⊥
2αRe hA− .A− Γ4 Ψ⊥
2 , Γ Ψ1 i
1
2
4
2
0 ⊥ 2
≥ −kHf2 Γ4 Ψ⊥
2 k − cα |κ2 | kΓ Ψ1 k −
3
X
cα6 |κ2,i |2
i=1
1
2
≥ −kHf Γ
4
2
Ψ⊥
2k
5
+ O(α ).
uα
+
+
1 ⊥
3 ⊥
To estimate the term 2αRe hA− .A− Γ3 Ψ⊥
2 , Γ Ψ1 i we rewrite it as 2αRe hΓ Ψ2 , A .A κ1 Φ] i
3 ⊥
2 −1 +
+ uα
and use that hΓ Ψ2 , (Hf + Pf ) A .A Φ] i] = 0. This yields, using Lemma A.5
(143)
1 ⊥
3 ⊥
2 −1 +
αRe hA− .A− Γ3 Ψ⊥
A .A+ κ1 Φu] α i
2 , Γ Ψ1 i = −αRe hΓ Ψ2 , (hα + e0 )(Hf + Pf )
2
7
−1
≥ −α2 kΓ3 Ψ⊥
.
2 k + cα log α
30
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
3
2
Similarly, if kΓ0 Ψ⊥
1k>α ,
(144)
0 ⊥
2 ⊥
2 −1 +
2αRe hA− .A− Γ2 Ψ⊥
A .A+ Γ0 Ψ⊥
2 , Γ Ψ1 i = −2αRe hΓ Ψ2 , (hα + e0 )(Hf + Pf )
1i
1
1
2
2 −1 +
2
2
A .A+ Γ0 Ψ⊥
≥ −cαk(hα + e0 ) 2 Γ2 Ψ⊥
2 k − cαk(hα + e0 ) (Hf + Pf )
1k
1
1
2
0 ⊥ 2
2
−2 2 ⊥ 2
2
Γ Ψ2 k − cαe0 kΓ2 Ψ⊥
≥ −cαkP Γ2 Ψ⊥
2 k − c0 αk(hα + e0 ) Γ Ψ1 k
2 k + cαk|x|
1
2
2
2 ⊥ 2
0 ⊥ 2
2
≥ −cαkP Γ2 Ψ⊥
2 k − α kΓ Ψ2 k − c0 αk(hα + e0 ) Γ Ψ1 k .
3
2
If kΓ0 Ψ⊥
1 k ≤ α , we have instead
1
0 ⊥
2 ⊥ 2
2
0 ⊥ 2
2
2αRe hA− .A− Γ2 Ψ⊥
2 , Γ Ψ1 i ≥ −kHf Γ Ψ2 k − cα kΓ Ψ1 k
(145)
1
2
5
≥ −kHf2 Γ2 Ψ⊥
2 k + O(α ).
Finally, using Lemma A.6 yields
1
⊥ 2
1 ⊥
5
2
2αRe hA− .A− Γ3 Ψ⊥
1 , Γ Ψ2 i ≥ −kHf Ψ2 k + O(α ).
(146)
Collecting (141)-(146) concludes the proof of the proposition.
In the rest of this subsection, we estimate further terms in (138).
⊥
6.3.1. Estimate of crossed terms involving Ψ⊥
1 and Ψ2 .
Lemma 6.5.
2
− ⊥ 2
⊥
5
2α(kA− Ψ⊥ k2 − kA− Ψ⊥
1 k − kA Ψ2 k ) ≥ −M [Ψ2 ] + O(α ).
(147)
Proof. Obviously, the left hand side of (147) is equal to
(148)
− ⊥
4αRe hA− Ψ⊥
1 , A Ψ2 i ≥ −cα
X
1
1
n ⊥
2
kHf2 Γn Ψ⊥
1 k kHf Γ Ψ2 k
n
1
2
≥ −kHf Γ
1
2
Ψ⊥
2k
1
− cα |κ1 | kHf2 Φu] α k2 −
2
2
X
1
1
2
n ⊥ 2
2
cα 3kHf2 Γn Ψ⊥
2 k + 2kHf Γ Ψ k
n6=1
≥
1
2
2
−kHf Ψ⊥
2k
5
+ O(α ),
where in the last inequality we used (210) of Lemma A.3, and (69) of Theorem 5.2.
Lemma 6.6.
⊥
⊥
7
−1
|hP.Pf Ψ⊥
|κ3 |2 + O(α5 )
2 , Ψ1 i| ≤ M [Ψ2 ] + cα log α
Proof. We have
(149)
⊥
1 ⊥
1 ⊥
2 ⊥
2 ⊥
3 ⊥
3 ⊥
hP.Pf Ψ⊥
2 , Ψ1 i = hPf Γ Ψ2 , P Γ Ψ1 i + hPf Γ Ψ2 , P Γ Ψ1 i + hPf Γ Ψ2 , P Γ Ψ1 i.
Obviously, using Lemma A.3 and the equality κ1 = O(1) from Lemma 6.1, yields
(150)
1
1
1
uα 2
1 ⊥
⊥ 2
2
⊥ 2
5
2
2
2
hPf Γ1 Ψ⊥
2 , P Γ Ψ1 i| ≤ kHf Ψ2 k + |κ1 | kP |Pf | Φ] k ≤ kHf Ψ2 k + O(α ).
We also have, by definition of Γ2 Ψ⊥
1 and using the estimates κ2,i = O(1) from
Lemma 6.1,
1
(151)
⊥ 2
2 2
0 ⊥ 2
6
2 ⊥
2
hPf Γ2 Ψ⊥
2 , P Γ Ψ1 i| ≤ kHf Ψ2 k + c|κ2 | α kP Γ Ψ1 k + O(α ).
HYDROGEN GROUND STATE ENERGY IN QED
31
We next bound the second term on the right hand side of (151). Notice that by
2
3
definition of κ2 , this term is nonzero only if kΓ0 Ψ⊥
1 k > α , which implies, with
0 ⊥ 2
0 ⊥ 2
Lemma 6.1, that kP Γ Ψ1 k ≤ cαkΓ Ψ1 k . The inequality (151) can thus be
rewritten as
(152)
1
1
⊥ 2
2 3
0 ⊥ 2
6
⊥ 2
5
2 ⊥
2
2
|hP.Pf Γ2 Ψ⊥
2 , Γ Ψ1 i| ≤ kHf Ψ2 k +c|κ2 | α kΓ Ψ1 k +O(α ) ≤ kHf Ψ2 k +O(α ),
using in the last inequality that |κ2 | kΓ0 Ψ⊥
1 k = O(α) (see Lemma 6.1).
For the second term on the right hand side of (149), using (212) from Lemma A.3
yields
1
⊥ 2
7
−1
3
⊥
3
2
|κ3 |2 .
hΓ3 Pf Ψ⊥
2 , Γ P Ψ1 i ≤ kΓ Hf Ψ2 k + cα log α
(153)
The inequalities (149), (152) and (153) prove the lemma.
Lemma 6.7.
1
⊥
⊥
5
−4α 2 Re hP.A+ Ψ⊥
2 , Ψ1 i ≥ −M [Ψ2 ] + O(α ).
0 ⊥
Proof. Since Γn>3 Ψ⊥
1 = 0, Γ Ψ2 = 0 and
1
1
1
2
n6=1 ⊥ 2
2
⊥
4
2
kHf2 Γn6=1 Ψ⊥
Ψ k + 2kHf2 Γn6=1 Ψ⊥
1 k ≤ 2kHf Γ
2 k ≤ cM [Ψ2 ] + O(α ),
(see (69) of Theorem 5.2) we obtain
1
1
2
⊥
5
⊥
n≤2
2
n≥2
P Ψ⊥
Hf2 Ψ⊥
4α 2 Re hP.A+ Ψ⊥
1 k ≤ M [Ψ2 ] + O(α ) .
2 , Ψ1 i ≤ kΓ
2 k + cαkΓ
Lemma 6.8.
1
⊥
⊥
5
−4α 2 Re hP.A− Ψ⊥
2 , Ψ1 i ≥ −M [Ψ2 ] + O(α ).
Proof.
1
1
⊥
⊥ 2
⊥ 2
2
− 4α 2 Re hP.A− Ψ⊥
2 , Ψ1 i ≥ −kHf Ψ2 k − cαkP Ψ1 k
1
2
n=0,2,3,4
2
≥ −kHf2 Ψ⊥
P Ψ⊥ k2 + kΓn=0,2,3,4 P Ψ⊥
− cα|κ1 |2 kP Φu] α k2
2 k − cα kΓ
2k
1
2
n≤4
2
6
−1
6
−1
≥ −kHf2 Ψ⊥
P Ψ⊥
) ≥ −M [Ψ⊥
),
2 k − kΓ
2 k + O(α log α
2 ] + O(α log α
using (212) of Lemma A.3.
Lemma 6.9.
1
1
⊥
2 ⊥
+ uα
⊥
5
2
4α 2 Re hPf .A(0)Ψ⊥
2 , Ψ1 i ≥ 4α Re hΓ Ψ2 , Pf .A Φ∗ i − M [Ψ2 ] + O(α ).
Proof. We have
1
⊥
4α 2 Re hPf .A− Ψ⊥
2 , Ψ1 i
1
1
uα
2
n≥2
2
− 2 ⊥
2
≥ −kHf2 Ψ⊥
Pf Ψ ⊥
2 k − cαkΓ
1 k + 4α Re hPf .A Γ Ψ2 , κ1 Φ] i
1
1
1
uα
2
n=2,3
≥ −kHf2 Ψ⊥
Hf2 Ψ⊥ k2 + 4α 2 Re hPf .A− Γ2 Ψ⊥
2 k − cαkΓ
2 , κ1 Φ] i
1
uα
− 2 ⊥
5
2
≥ −M [Ψ⊥
2 ] + 4α Re hPf .A Γ Ψ2 , κ1 Φ] i + O(α ).
32
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
We estimate the second term on the right hand side as follows,
1
uα
4α 2 Re hPf .A− Γ2 Ψ⊥
2 , κ1 Φ] i
1
1
1
1
uα 2
uα
2 ⊥ 2
2
2
≥ 4α 2 Re hPf .A− Γ2 Ψ⊥
2 , Φ] i − kHf Γ Ψ2 k − cα|κ − 1| kPf Φ] k
uα
2 ⊥ 2
5
2
≥ 4α 2 Re hPf .A− Γ2 Ψ⊥
2 , Φ∗ i − kHf Γ Ψ2 k + O(α ),
where we used |κ1 −1|2 = O(α) from Lemma 6.1, kΦu] α k2∗ = O(α3 ) from Lemma A.3,
and kPf (Φu] α − Φu∗ α )k2 = O(α4 ) from Lemma A.4.
We also have, using Pf .A+ = A+ .Pf ,
1
1
⊥
n≤2
− n≥2 ⊥
2
4α 2 Re hPf .A+ Ψ⊥
Pf Ψ ⊥
Ψ1 i
2 , Ψ1 i = 4α Re hΓ
2 ,A Γ
1
1
2
n≥2 ⊥ 2
5
2
≥ −kHf2 Ψ⊥
Ψ1 k ≥ −M [Ψ⊥
2 k − cαkHf Γ
2 ] + O(α ),
1
⊥
where kHf2 Γn≥2 Ψ⊥
1 k has been estimated as kP Ψ1 k in the proof of Lemma 6.8.
⊥
6.3.2. Estimates of the term h(H + e0 )Ψ⊥
1 , Ψ1 i.
Due to (139) and (140), one finds
1
(154)
⊥
⊥
⊥
⊥
⊥
2
h(H + e0 )Ψ⊥
1 , Ψ1 i = hΨ1 , Ψ1 i] − 2Re hP.(Pf + α A(0))Ψ1 , Ψ1 i
1
⊥
− ⊥ 2
−
− ⊥
⊥
+ 2α 2 Re hPf .A(0)Ψ⊥
1 , Ψ1 i + 2αkA Ψ1 k + 2αRe hA .A Ψ1 , Ψ1 i
We estimate the terms in (154) below.
Lemma 6.10. We have
1
⊥
− 2Re hP.(Pf + α 2 A(0))Ψ⊥
1 , Ψ1 i
1
s
⊥
4
5
≥ −4α 2 Re hA− Φu] α , P Γ0 (Ψ⊥
1 ) i − M [Ψ2 ] − |κ1 − 1|cα + O(α ),
0
⊥ a
0 ⊥
s
0 ⊥
where Γ0 (Ψ⊥
1 ) = Γ Ψ1 − Γ (Ψ1 ) is the even part of Γ Ψ1 .
Proof. Using hΦ2∗ , Pfi Φ2∗ i = 0 (see Lemma A.1), the symmetry of uα , and hP.Pf Φu] α , Φu] α i =
0, we obtain
⊥
2 0 ⊥
|2hP.Pf Ψ⊥
1 , Ψ1 i| = |2hPf ακ2 Φ∗ Γ Ψ1 , P α
3
X
κ2,i (Hf + Pf2 )−1 Wi
i=1
(155)
2
5
≤ cα3 |κ2 |2 kΓ0 Ψ⊥
1 k + cα
3
X
|κ2,i |2 = O(α5 ),
i=1
where in the last inequality we used Lemma 6.1.
∂uα
i|
∂xi
HYDROGEN GROUND STATE ENERGY IN QED
33
We also have, using again the symmetry of uα ,
1
1
⊥
− ⊥
⊥
2
− 2α 2 Re hP.A(0)Ψ⊥
1 , Ψ1 i = −4α Re hP.A Ψ1 , Ψ1 i
1
1
uα
s
−
2 0 ⊥
2
= −4α 2 Re hA− κ1 Φu] α , P Γ0 (Ψ⊥
1 ) i − 4α Re hA κ2 αΦ∗ Γ Ψ1 , P κ1 Φ] i
1
2 0 ⊥
− 4α 2 Re hA− Γ3 Ψ⊥
1 , P ακ2 Φ∗ Γ Ψ1 i
1
1
s
3 ⊥ 2
2
0 ⊥ 2
2
2
≥ −4α 2 Re hA− κ1 Φu] α , P Γ0 (Ψ⊥
1 ) i − cαkHf Γ Ψ1 k − cα kP Γ Ψ1 k |κ2 |
(156)
1
1
1
uα
6
− 4α 2 Re h|k|− 6 A− κ2 αΦ2∗ Γ0 Ψ⊥
1 , |k| κ1 P Φ] i
1
1
1
3 ⊥ 2
3 ⊥ 2
s
2
2
≥ −4α 2 Re hA− κ1 Φu] α , P Γ0 (Ψ⊥
1 ) i − cαkHf Γ Ψ k − cαkHf Γ Ψ2 k
3
1
uα
6
− cα5 − cα 2 |κ2 | kΓ0 Ψ⊥
1 k k |k| P Φ] k
1
s
⊥
5
≥ −4α 2 Re hA− κ1 Φu] α , P Γ0 (Ψ⊥
1 ) i − M [Ψ2 ] − cα ,
where we used Theorem 5.2 and Lemma A.3.
3
2
Moreover, because kA− Φu∗ α k = O(α 2 ) and kP Γ0 Ψ⊥
1 k = O(α ), we obtain
(157)
1
1
s
− uα
0
⊥ s
4
2
−4Re α 2 Re hA− κ1 Φu] α , P Γ0 (Ψ⊥
1 ) i ≥ −4α Re hA Φ] , P Γ (Ψ1 ) i − |κ1 − 1|cα .
This estimate, together with (155) and (156), proves the lemma.
Lemma 6.11. We have
1
⊥
2α 2 Re hPf .A(0)Ψ⊥
1 , Ψ1 i
X
2
κ2,i hPf .A− (Hf + Pf2 )−1 Wi , (Hf + Pf2 )−1 (A+ )i Ωf i
≥ α4 Re
3
i
− |κ1 − 1|cα4 + O(α5 ).
Proof. The following holds
(158)
1
1
uα
⊥
− 3 ⊥
2 ⊥
− 2 ⊥
2
2α 2 Re hPf .A(0)Ψ⊥
1 , Ψ1 i = 4α Re hPf .A Γ Ψ1 , Γ Ψ1 i + 4αRe hPf .A Γ Ψ1 , κ1 Φ] i
3
= 4α 2 Re
3
X
1
1
1
1
−6
κ2,i h|k1 | 6 |k2 | 6 A− Γ3 Ψ⊥
|k2 |− 6 (Hf + Pf2 )−1 Wi
1 , Pf |k1 |
i=1
3
2
1
1
1
∂uα
i
∂xi
1
−6
|k2 |− 6 Pf Φ2∗ Γ0 Ψ⊥
+ 4α Re κ2 h|k1 | 6 |k2 | 6 A− Γ3 Ψ⊥
1 , |k1 |
1i
1
uα
+ 4α 2 Re hPf .A− Γ2 Ψ⊥
1 , κ1 Φ] i.
1
1
2
Applying the Schwarz inequality and the estimates k|k1 | 6 |k2 | 6 A− Γ3 Ψ⊥
1k =
5
2
2
O(α ) (Lemma A.5), κ2,i = O(1) (Lemma 6.1), and k∇uα k = O(α ), we see that
the first term on the right hand side of (158) is O(α5 ). Applying also the estimate
|κ2 | kΓ0 Ψ⊥
1 k = O(α) (Lemma 6.1), we obtain that the second term on the right
hand side of (158) is also O(α5 ).
1
uα
Finally, we estimate 4α 2 Re hPf .A− Γ2 Ψ⊥
1 , κ1 Φ] i. The following inequality holds,
1
1
uα
uα
− 2 ⊥
4
2
(159) 4α 2 Re hPf .A− Γ2 Ψ⊥
1 , κ1 Φ] i| ≥ 4α Re hPf .A Γ Ψ1 , Φ] i − |κ1 − 1|cα ,
whose proof is similar to the one of (157). Next we get
1
(160)
1
uα
− 2 ⊥
uα
2
|4α 2 Re hPf .A− Γ2 Ψ⊥
1 , Φ] i − 4α Re hPf .A Γ Ψ1 , Φ∗ i|
1
1
1
uα
uα
6
2
≤ α 2 kHf2 Γ2 Ψ⊥
1 k kPf (Φ] − Φ∗ )k = O(α | log α| ),
34
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
7
1
1
using kPf (Φu] α − Φu∗ α )k = O(α 2 | log α| 2 ) (see Lemma A.4) and kHf2 Γ2 Ψ⊥
1k =
2
O(α ). Moreover,
1
uα
4α 2 Re hPf .A− Γ2 Ψ⊥
1 , Φ∗ i
3
= 4α 2 Re hPf .A− (κ2 Φ2∗ Γ0 Ψ⊥
1 +
X
κ2,i (Hf + Pf2 )−1 Wi
i
(161)
3
2
= 4α Re hPf .A−
X
κ2,i (Hf + Pf2 )−1 Wi
i
∂uα
), Φu∗ α i
∂xi
∂uα uα
,Φ i
∂xi ∗
X
2
= α4 Re
κ2,i hPf .A− (Hf + Pf2 )−1 Wi , (Hf + Pf2 )−1 (A+ )i Ωf i
3
i
where we used (207) of Lemma A.1 in the third equality.
Collecting (158)-(181) concludes the proof.
Lemma 6.12. We have
⊥
−
− ⊥
⊥
0 ⊥ 2
1 ⊥ 2
hΨ⊥
1 , Ψ1 i] + 2αRe hA .A Ψ1 , Ψ1 i ≥ Σ0 kΓ Ψ1 k + kΓ Ψ1 k
3
α4 X
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ + α2 |κ3 + 1|2 kΦ2∗ k2∗ kΦu] α k2
+
12 i=1
2
0 ⊥ 2
5
−1
+ kΓ1 Ψ⊥
).
1 k] + kΓ Ψ1 k] + o(α log α
Proof. Obviously we have
⊥
hΨ⊥
1 , Ψ1 i ]
(162)
=
3
X
i ⊥
hΓi Ψ⊥
1 , Γ Ψ1 i] ,
i=0
and using Lemma A.1
2 ⊥
2
2
2 0 ⊥ 2
2
(163) hΓ2 Ψ⊥
1 , Γ Ψ1 i] = α |κ2 | kΦ∗ Γ Ψ1 k] +α
3
X
|κ2,i |2 k(Hf +Pf2 )−1 Wi
i=1
∂uα 2
k ,
∂xi ]
2
2 2
0 ⊥ 2
Moreover, from the inequality kΦ2∗ Γ0 Ψ⊥
1 k] > kΦ∗ k∗ kΓ Ψ1 k we obtain,
2
−
− 2 ⊥
0 ⊥
α2 |κ2 |2 kΦ2∗ Γ0 Ψ⊥
1 k] + 2αRe hA .A Γ Ψ1 , Γ Ψ1 i
2
−
− 2 ⊥
0 ⊥
≥ α2 |κ2 |2 kΦ2∗ k2∗ kΓ0 Ψ⊥
1 k + 2αRe hA .A Γ Ψ1 , Γ Ψ1 i
(164)
2
2
2 2
0 ⊥ 2
= α2 |κ2 |2 kΦ2∗ k2∗ kΓ0 Ψ⊥
1 k − 2α Re κ2 kΦ∗ k∗ kΓ Ψ1 k
+ 2αRe
≥
3
X
αhA− .A− κ2,i (Hf + Pf2 )−1 Wi
i=1
0 ⊥ 2
Σ0 kΓ Ψ1 k
∂uα 0 ⊥
, Γ Ψ1 i
∂xi
2
5
0 ⊥ 2
5
+ |κ2 − 1|2 cα2 kΓ0 Ψ⊥
1 k + O(α ) ≥ Σ0 kΓ Ψ1 k + O(α ).
where we used Σ0 = −α2 kΦ2∗ k2∗ + O(α3 ) and hA− .A− (Hf + Pf2 )−1 Wi , Γ0 Ψ⊥
1i =
2
−Γ0 Ψ⊥
1 hWi , Φ∗ i = 0.
HYDROGEN GROUND STATE ENERGY IN QED
35
3 ⊥
Similarly kΓ3 Ψ⊥
1 k] > kΓ Ψ1 k∗ yields
(165)
2
−
− 3 ⊥
1 ⊥
2
2
2 −1 +
kΓ3 Ψ⊥
A .A+ Φu] α k2∗
1 k] + 2αRe hA .A Γ Ψ1 , Γ Ψ1 i ≥ α |κ3 | k(Hf + Pf )
+ 2Re α2 κ3 hA− .A− (Hf + Pf2 )−1 A+ .A+ Φu] α , κ1 Φu] α i
≥ −α2 k(Hf + Pf2 )−1 A+ .A+ κ1 Φu] α k2∗ + |κ3 + 1|2 α2 k(Hf + Pf2 )−1 A+ .A+ Φu] α k2∗
uα 2
2
2
2
2 2
5
−1
≥ Σ0 kΓ1 Ψ⊥
),
1 k + α |κ3 + 1| kΦ∗ k∗ kΦ] k + o(α log α
1
where in the last inequality, we used (222) from Lemma A.5, κ1 = 1 + O(α 2 )
from Lemma 6.1, kΦu] α k2 = O(α3 log α−1 ) from Lemma A.3 and −α2 kΦ2∗ k2∗ =
Σ0 + O(α3 ).
The second term on the right hand side of (163) is estimated as
(166)
3
3
X
X
∂uα 2
∂uα 2
k] ≥ α2
|κ2,i |2 k(Hf + Pf2 )−1 Wi
k
α2
|κ2,i |2 k(Hf + Pf2 )−1 Wi
∂xi
∂xi ∗
i=1
i=1
3
3
X
α4 X
1
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ =
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ .
= α2 k∇uα k2
3
12
i=1
i=1
Collecting (162)-(166) conclude the proof.
Proposition 6.4. We have
⊥
h(H + e0 )Ψ⊥
1 , Ψ1 i
1
1
0
⊥ a 2
− uα 2
2
≥ −4αk(hα + e0 )− 2 Q⊥
α P.A Φ∗ k + k(hα + e0 ) Γ (Ψ1 ) k
1
s 2
0 ⊥ 2
1 ⊥ 2
+ c0 αk(hα + e0 ) 2 Γ0 (Ψ⊥
1 ) k + Σ0 (kΓ Ψ1 k + kΓ Ψ1 k )
(167)
+
3
1
α4 X
uα
+
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ + 4α 2 Re hΓ2 Ψ⊥
1 , A .Pf Φ] i
12 i=1
+ kκ1 Φu] α k2] + 2αkA− Φu∗ α k2 + α2 |κ3 + 1|2 kΦ2∗ k2∗ kΦu] α k2
4
5
−1
− M [Ψ⊥
),
2 ] − |κ1 − 1|cα + o(α log α
⊥
⊥ a
where Q⊥
α is the orthogonal projection onto Span(uα ) , (Ψ1 ) is the odd part of
0 ⊥
Γ Ψ1 , and c0 is the same positive constant as in Proposition 6.3.
Proof. Collecting Lemmata 6.10, 6.11, and 6.12 yields
(168)
1
uα
⊥
− uα
0
⊥ s
2 ⊥
+
2
h(H + e0 )Ψ⊥
1 , Ψ1 i ≥ −4α Re hA Φ] , P Γ (Ψ1 ) i + 4αRe hΓ Ψ1 , A .Pf Φ] i
uα 2
2
1 ⊥ 2
2
2
2 2
+ Σ0 (kΓ0 Ψ⊥
1 k + kΓ Ψ1 k ) + α |κ3 + 1| kΦ∗ k∗ kΦ] k
1
uα 2
2
1 ⊥ 2
−
4
+ k(hα + e0 ) 2 Γ0 Ψ⊥
1 k + kΓ Ψ1 k] + 2αkA κ1 Φ] k − |κ1 − 1|cα
+
3
α4 X
5
−1
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗ − M [Ψ⊥
).
2 ] + o(α log α
12 i=1
Obviously,
(169)
1
1
1
2
0
⊥ s 2
0
⊥ a 2
2
2
k(hα + e0 ) 2 Γ0 Ψ⊥
1 k = k(hα + e0 ) Γ (Ψ1 ) k + k(hα + e0 ) Γ (Ψ1 ) k .
36
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
⊥ a
⊥ s
As before, we write Γ0 Ψ⊥
1 as the sum of its odd part (Ψ1 ) and its even part (Ψ1 ) .
0 ⊥
⊥
⊥ a
Since Γ Ψ1 is orthogonal to uα by definition of Ψ , and (Ψ1 ) is orthogonal to
s
uα by symmetry of uα , we also have (Ψ⊥
1 ) orthogonal to uα . Therefore, one can
0
⊥ s
⊥ 0
⊥ s
replace Γ (Ψ1 ) by Qα Γ (Ψ1 ) in (168) and (169). Thus, as the next step, given
a constant c0 > 0, we minimize
1
1
(170)
0
⊥ s 2
⊥
− uα
⊥ 0
⊥ s
2
(1 − c0 α)k(hα + e0 ) 2 Q⊥
α Γ (Ψ1 ) k − 4α Re hQα P.A Φ] , Qα Γ (Ψ1 ) i
1
4α
− uα 2
≥−
k(hα + e0 )− 2 Q⊥
α P.A Φ] k .
1 − c0 α
Obviously,
1
4α
− uα 2
k(hα + e0 )− 2 Q⊥
α P.A Φ] k
1 − c0 α
1
4(1 + α)α
− uα 2
≥−
k(hα + e0 )− 2 Q⊥
α P.A Φ∗ k
1 − c0 α
1
4(1 + α−1 )α
uα
−
uα 2
k(hα + e0 )− 2 Q⊥
−
α P.A (Φ] − Φ∗ )k .
1 − c0 α
−
(171)
There exist γ1 and γ2 positive, independent of α, such that
−1 ⊥
Q⊥
Qα ≤ (γ1 P 2 + γ2 α2 )−1 ,
α (hα + e0 )
−1 ⊥
and thus P Q⊥
Qα P is a bounded operator. In addition, since kA− (Φu] α −
α (hα +e0 )
7
−1
uα 2
Φ∗ )k = O(α log α ) (Lemma A.3), this shows that
(172)
1
4(1 + α−1 )α
uα
6
−1
−
uα 2
).
k(hα + e0 )− 2 Q⊥
α P.A (Φ] − Φ∗ )k = O(α log α
1 − c0 α
3
In addition, using kA− Φu∗ α k ≤ kA− (Φu∗ α − Φu] α )k + kA− Φu] α k ≤ cα 2 (Lemma A.3
−1 ⊥
Qα P is bounded, yields
and Lemma A.4), and the fact that P Q⊥
α (hα + e0 )
(173)
−
1
(1 + α)α
− uα 2
k(hα + e0 )− 2 Q⊥
α P.A Φ∗ k
1 − c0 α
1
− uα 2
5
= −4k(hα + e0 )− 2 Q⊥
α P.A Φ∗ k + O(α ).
Collecting (171)-(173), one gets
(174)
−
1
4α
− uα 2
− 21 ⊥
Qα P.A− Φu∗ α k2 + O(α5 ).
k(hα + e0 )− 2 Q⊥
α P.A Φ] k ≥ −4k(hα + e0 )
1 − c0 α
Finally, using kA− Φu∗ α k2 = O(α3 ), we obtain
(175)
2α|κ1 |2 kA− Φu∗ α k2 ≥ 2αkA− Φu∗ α k2 − c|κ1 − 1|α4 .
Substituting (169), (170), (174) and (175) into (168) concludes the proof.
We can now write the proof of Proposition 6.2
HYDROGEN GROUND STATE ENERGY IN QED
37
Proof of Proposition 6.2 Collecting the results of Proposition 6.3, Lemmata 6.56.9 and Proposition 6.4 yields directly the following bound,
(176)
⊥
⊥
2 4
0 ⊥ 2
6
−1
2
hHΨ⊥ , Ψ⊥ i ≥ hHΨ⊥
|κ3 |2 − e0 kΨ⊥
2 , Ψ2 i − M [Ψ2 ] − c|κ2 | α kΓ Ψ1 k − cα log α
1k
1
1
− uα 2
0
⊥ a 2
2
− 4αk(hα + e0 )− 2 Q⊥
α P.A Φ∗ k + (1 − c0 α)k(hα + e0 ) Γ (Ψ1 ) k
2
1 ⊥ 2
+ Σ0 (kΓ0 Ψ⊥
1 k + kΓ Ψ1 k ) +
1
3
α4 X
|κ2,i |2 k(Hf + Pf2 )−1 Wi k2∗
12 i=1
1
uα
uα 2
+
uα
2 ⊥
+
2
− uα 2
2
+ 4α 2 Re hΓ2 Ψ⊥
2 , A .Pf Φ∗ i + 4α Re hΓ Ψ1 , A .Pf Φ] i + |κ1 | kΦ] k] + 2αkA Φ∗ k
+ α2 |κ3 + 1|2 kΦ2∗ k2∗ kΦu] α k2 − |κ1 − 1|cα4 + o(α5 log α−1 ).
Comparing this expression with the statement of the Proposition we see that it
suffices to show that
(177)
2 4
0 ⊥ 2
6
−1
2
⊥
⊥
− M [Ψ⊥
|κ3 |2 − e0 kΨ⊥
2 ] − c|κ2 | α kΓ Ψ1 k − cα log α
1 k + hHΨ2 , Ψ2 i
uα 2
2
1 ⊥ 2
2
2
2 2
+ Σ0 (kΓ0 Ψ⊥
1 k + kΓ Ψ1 k ) + α |κ3 + 1| kΦ∗ k∗ kΦ] k
|κ3 + 1|2 2 2 2 uα 2
α kΦ∗ k∗ kΦ] k + O(α5 ).
2
⊥
⊥ 2
⊥
Using from Corollary 4.2 that hHΨ⊥
2 , Ψ2 i ≥ (Σ0 − e0 )kΨ2 k + M [Ψ2 ], we first
estimate the following terms in (177),
2
1 ⊥ 2
2
⊥
⊥
⊥
Σ0 kΓ0 Ψ⊥
− e0 kΨ⊥
1 k + kΓ Ψ1 k
1 k + hHΨ2 , Ψ2 i − M [Ψ2 ]
≥ (Σ0 − e0 )kΨ⊥ k2 + (1 − )M [Ψ⊥
2]+
⊥ 2
⊥ 2
n≥2 ⊥ 2
≥ (1 − )M [Ψ⊥
Ψ1 k
2 ] + (Σ0 − e0 )(kΨ1 k + kΨ2 k ) − Σ0 kΓ
(178)
⊥ 2
≥ (1 − )M [Ψ⊥
2 ] + (Σ0 − e0 )kΨ k
2
⊥ 2
n≥2 ⊥ 2
− (Σ0 − e0 )(kΨ⊥ k2 − kΨ⊥
Ψ1 k .
1 k − kΨ2 k ) − Σ0 kΓ
We have obviously
(179)
2
⊥ 2
1 ⊥
1 ⊥
2 ⊥
2 ⊥
3 ⊥
3 ⊥
kΨ⊥ k2 − kΨ⊥
1 k − kΨ2 k = 2Re (hΓ Ψ1 , Γ Ψ2 i + hΓ Ψ1 , Γ Ψ2 i + hΓ Ψ1 , Γ Ψ2 i).
Since |Σ0 − e0 | ≤ cα2 , by definition of Γ3 Ψ⊥
1 and Lemma A.5, we obtain
(180)
3 ⊥
2
3 ⊥ 2
4
2
2 −1 +
|(Σ0 − e0 )2Re hΓ3 Ψ⊥
A .A+ Φu] α k2
1 , Γ Ψ2 i| ≤ α kΓ Ψ2 k + cα |κ3 | k(Hf + Pf )
2
2 7
−1
≤ α2 kΓ3 Ψ⊥
.
2 k + c|κ3 | α log α
Similarly, for the two-photon sector, we find
(181)
2 ⊥
2
2 ⊥ 2
4
2
0 ⊥ 2
|(Σ0 − e0 )hΓ2 Ψ⊥
1 , Γ Ψ2 i| ≤ α kΓ Ψ2 k + cα |κ2 | kΓ Ψ1 k + c
3
X
|κ2,i |2 α6
i=1
2
= α kΓ
2
2
Ψ⊥
2k
6
+ O(α ),
1 ⊥
where we used Lemma 6.1. For the term hΓ1 Ψ⊥
1 , Γ Ψ2 i, one gets
1
1
(182)
uα
1 ⊥
−2
2
κ1 Φu] α i|
|(Σ0 − e0 )hΓ1 Ψ⊥
2 , κ1 Φ] i| = |(Σ0 − e0 )(|k| Γ Ψ2 , |k|
1
1
1
2
4
2
− 2 uα 2
2
5
≤ kHf2 Γ1 Ψ⊥
Φ] k ≤ kHf2 Γ1 Ψ⊥
2 k + cα |κ1 | k |k|
2 k + O(α ),
38
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
1
since k |k|− 2 Φu] α k2 = O(α) (Lemma A.3) and κ1 = O(1) (Lemma 6.1).
Collecting (179), (180), (181) and (182) yields
2 7
−1
2
⊥ 2
≤ M [Ψ⊥
+ O(α5 ).
(183) |Σ0 − e0 | kΨ⊥ k2 − kΨ⊥
2 ] + c|κ3 | α log α
1 k − kΨ2 k
Therefore, together with (178), one finds
(184)
2
1 ⊥ 2
2
⊥
⊥
⊥
Σ0 kΓ0 Ψ⊥
− e0 kΨ⊥
1 k + kΓ Ψ1 k
1 k + hHΨ2 , Ψ2 i − M [Ψ2 ]
⊥ 2
2 7
−1
2
5
≥ (1 − 2)M [Ψ⊥
− Σ0 kΓn≥2 Ψ⊥
2 ] + (Σ0 − e0 )kΨ k − c|κ3 | α log α
1 k + O(α ),
2
0 ⊥
By definition of Ψ⊥
1 and using Σ0 = O(α ), |κ2 |kΓ Ψ1 k = O(α) (Lemma 6.1) and
Inequality (223) of Lemma A.5, we straightforwardly obtain
2
6
2 7
−1
Σ0 kΓn≥2 Ψ⊥
.
1 k ≤ cα + c|κ3 | α log α
Substituting this in (184) yields
(185)
2
1 ⊥ 2
2
⊥
⊥
⊥
Σ0 kΓ0 Ψ⊥
− e0 kΨ⊥
1 k + kΓ Ψ1 k
1 k + hHΨ2 , Ψ2 i − M [Ψ2 ]
⊥ 2
2 7
−1
≥ (1 − 2)M [Ψ⊥
+ O(α5 ).
2 ] + (Σ0 − e0 )kΨ k − 2c|κ3 | α log α
To conclude the proof of (177), and thus of the Proposition, we first note that
according to Lemma 6.1,
2
6
−c|κ2 |2 α4 kΓ0 Ψ⊥
1 k = O(α ).
(186)
Similarly, taking into account that kΦu] α k2 = cα3 log α−1 (see (209) in Lemma A.3),
we get for some c2 > 0,
(187)
|κ3 + 1|2 2 2 uα 2
kΦ∗ k∗ kΦ] k − cα6 log α−1 |κ3 |2 − 2cα7 log α−1 |κ3 |2 ≥ −c2 α6 log α−1 .
α2
2
Collecting (185), (186) and (187) yields the bound (177), and thus concludes the
proof of the Proposition 6.2.
6.4. Lower bound up to the order α5 log α−1 for the binding energy. To
get a lower bound for Σ0 − Σ in Theorem 2.1 which coincides with the upper bound
given in (97), it suffices to compute
e trial , Φ
e trial i
h(H − Σ0 + e0 )Φ
,
e trial k2
kΦ
with the following trial function
e trial = uα Ψ0 + Ψ
e⊥
Φ
0,
where Ψ0 is ground state of the operator T (0) with the normalization Γ0 Ψ0 = Ωf ,
α
e ⊥ is defined by
, and Ψ
uα is the normalized ground state of hα = −∆ − |x|
0
1
uα
−1 ⊥
e⊥
e⊥
2
Γ0 Ψ
Qα P.A− Φu∗ α , Γ1 Ψ
0 = 2α (hα + e0 )
0 = Φ] ,
(188)
e⊥
Γ2 Ψ
0
e⊥
Γ3 Ψ
0
=
e⊥
αΦ2∗ Γ0 Ψ
0
= −α(Hf +
+
3
X
2α(Hf + Pf2 )−1 Wi
i=1
2 −1 +
Pf ) A .A+ Φu] α .
∂uα
,
∂xi
Where Φu∗ α , Φu] α , Φ2∗ and Wi are defined as in Sections 5 and 6.
HYDROGEN GROUND STATE ENERGY IN QED
39
We compute
(189)
e trial , Φ
e trial i = hHuα Ψ0 , uα Ψ0 i + 2Re hHuα Ψ0 , Ψ
e⊥
e⊥ e⊥
hH Φ
0 i + hH Ψ0 , Ψ0 i,
and we recall
(190)
1
1
H = hα + (Hf + Pf2 ) − 2Re P.Pf − 2α 2 P.A(0) + 2α 2 Pf .A(0) + 2αA+ .A− + 2α(A− )2 .
• For the first term in (189), a straightforward computation shows
hHuα Ψ0 , uα Ψ0 i = (Σ0 − e0 )kuα k2 kΨ0 k2 .
(191)
• We estimate the second term on the right hand side of (189) by computing
each term that occurs in the decomposition (190).
e ⊥ i are
Using the symmetry of uα , the only non zero terms in 2Re hHuα Ψ0 , Ψ
0
given by
(192)
e ⊥ i = −4Re hP.Pf uα Ψ0 , Ψ
e ⊥ i−4Re hP.A+ uα Ψ0 , Ψ
e ⊥ i−4Re hP.A− uα Ψ0 , Ψ
e ⊥ i.
2Re hHuα Ψ0 , Ψ
0
0
0
0
The first term on the right hand side of (192) is estimated with similar arguments as in Lemma 5.2, and using k∆u∗ α k2 = O(α3 ) (Lemma A.7) and k∆u∗ α k2∗ =
O(α4 ) ([4, Theorem 3.2]). We obtain
3
X
p
e ⊥ i = − 2 α4
(193) −4Re hP.Pf uα Ψ0 , Ψ
hPfi Φ2∗ , (Hf + Pf2 )Wi i + O(α5 log α−1 ).
0
3 i=1
The second and third terms on the right hand side of (192) are estimated as
in Lemma 5.3, and using again k∆u∗ α k2 = O(α2 ) and k∆u∗ α k2∗ = O(α4 ). This yields
uα 2
5
−1
e⊥
−4Re hP.A+ uα Ψ0 , Ψ
),
0 i = −2kΦ] k] + o(α log α
(194)
and
e ⊥i
− 4Re hP.A− uα Ψ0 , Ψ
0
(195)
3
p
2 4X
=− α
h(A− )i Φ2∗ , (Hf + Pf2 )−1 (A+ )i Ωf i + O(α5 log α−1 ).
3 i=1
• Next, we estimate the third term on the right hand side of (189). For that
sake, we also use the decomposition (190) for H.
e ⊥ k = O(α3 ) (since kP.A− Φu∗ α k =
For the term involving hα , using k(hα +e0 )Γ0 Ψ
0
5
3
α
O(α 2 )), and k(hα + e0 ) ∂u
∂xi k = O(α ), we directly obtain
(196)
e⊥
e⊥
e⊥ e⊥
e⊥ 2
hhα Ψ
0 , Ψ0 i = h(hα + e0 )Ψ0 , Ψ0 i − e0 kΨ0 k
1
uα
uα
− uα 2
5
e⊥ 2
= 4αk(hα + e0 )− 2 Q⊥
α P.A Φ∗ k + h(hα + e0 )Φ] , Φ] i − e0 kΨ0 k + O(α ).
For the term with Hf + Pf2 , we use the estimate (222) of Lemma A.5, and
the h · , · i∗ -orthogonality (see (206) of Lemma A.1) of the two vectors αΦ2∗ Γ0 g and
P3
2 −1
2 e⊥
α
Wi ∂u
i=1 2α(Hf + Pf )
∂xi that occur in Γ Ψ0 . We therefore obtain
uα
uα
2
2
2 0 e⊥ 2
e⊥
e⊥
h(Hf + Pf2 )Ψ
0 , Ψ0 i = h(Hf + Pf )Φ] , Φ] i + α kΦ∗ Γ Ψ0 k∗
(197)
+k
3
X
i=1
2α(Hf + Pf2 )−1 Wi
∂uα 2
k + kΦ2∗ k2∗ kΦu] α k2 + o(α5 log α−1 ).
∂xi ∗
40
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
e ⊥, Ψ
e ⊥ i are zero, except the
Using the symmetry of uα , all terms in Re hP.Pf Ψ
0
0
2
2 e⊥
expression Re hP.Pf Γ g, Γ Ψ0 i, which is estimated as follows
2 e⊥
2 0 e⊥
e⊥
Re hP.Pf Γ2 Ψ
0 , Γ Ψ0 i = 2Re hP.Pf αΦ∗ Γ Ψ0 ,
3
X
2α(Hf + Pf2 )−1 Wi
i=1
∂uα
i = O(α5 ),
∂xi
where we used Lemma A.1 in the third equality to prove that only the crossed term
remains. Therefore, we obtain
5
e⊥
e⊥
−2Re hP.Pf Ψ
0 , Ψ0 i = O(α ).
(198)
1
e ⊥, Ψ
e ⊥ i is estimated as in the proof of
The terms involving −2α 2 Re hP.A(0)Ψ
0
0
Lemma 6.10. This yields
1
1
e ⊥, Ψ
e ⊥ i = −4Re α 2 hP.A+ Ψ
e ⊥, Ψ
e ⊥i
− 2α 2 hP.A(0)Ψ
0
0
0
0
p
+ 0 e⊥
1 e⊥
5
−1
= −4Re hP.A Γ Ψ0 , Γ Ψ0 i + O(α log α )
p
1
− uα 2
5
= −8αk(hα + e0 )− 2 Q⊥
log α−1 ).
α P.A Φ∗ k + O(α
(199)
1
e ⊥, Ψ
e ⊥ i, we proceed as in the proof of Lemma 6.11, and
For 2α 2 Re hPf .A(0)Ψ
0
0
obtain
1
1
− e⊥ e⊥
e⊥
e⊥
2
2α 2 hPf .A(0)Ψ
0 , Ψ0 i = 4Re α hPf .A Ψ0 , Ψ0 i
(200)
1
= 4α 2 hPf .A− α
3
X
2(Hf + Pf2 )−1 Wi
i=1
p
∂uα uα
, Φ∗ i + O(α5 log α−1 ).
∂xi
e ⊥, Ψ
e ⊥ i is esti Using the symmetry of uα and Γ0 g, the term 2αRe hA− .A− Ψ
0
0
mated as follows,
e⊥
e⊥
2αRe hA− .A− Ψ
0 , Ψ0 i
e⊥ +
= 2αRe hA− .A− (αΦ2∗ Γ0 Ψ
0
3
X
2α(Hf + Pf2 )−1 Wi
i=1
(201)
∂uα
e ⊥i
), Γ0 Ψ
0
∂xi
+ 2αRe hA− .A− (−α(Hf + Pf2 )−1 A+ .A+ Φu] α ), Φu] α i
2
2 2
2
2 −1 +
e⊥
= −2α2 kΓ0 Ψ
A .A+ Φu] α k2
0 k kΦ∗ k∗ − 2α k(Hf + Pf )
uα 2
2
2 2
2
2 2
5
−1
e⊥
= −2α2 kΓ0 Ψ
),
0 k kΦ∗ k∗ − 2α kΦ∗ k∗ kΦ] k + o(α log α
where in the last inequality we used (222) of Lemma A.5.
Finally, a straightforward computation yields
− uα 2
5
− uα 2
5
e⊥
e⊥
(202) 2αRe hA− .A+ Ψ
0 , Ψ0 i = 2αkA Φ] k + O(α ) = 2αkA Φ∗ k + O(α ),
where in the last equality, we used Lemma A.4.
• Before collecting (192)-(202), we show that gathering some terms yields simpler
expressions. Namely, we have
(203)
3
3
X
1
∂uα uα
2 X i 2
hPf Φ∗ , (Hf + Pf2 )Wi i + 4α 2 hPf .A− α
2(Hf + Pf2 )−1 Wi
,Φ i
− α4
3 i=1
∂xi ∗
i=1
+k
3
X
i=1
2α(Hf + Pf2 )−1 Wi
3
1 X
∂uα 2
k∗ = − α 4
k(Hf + Pf2 )−1 Wi k2∗ .
∂xi
3 i=1
HYDROGEN GROUND STATE ENERGY IN QED
41
We also have, using −α2 kΦ2∗ k2∗ = Σ0 + O(α3 ) (see e.g. [4])
(204)
uα 2
2
2
2 2
0 e⊥ 2
2
2 2
e⊥
(Σ0 − e0 )kΨ0 k2 − e0 kΨ
0 k − α kΦ∗ k∗ kΓ Ψ0 k − α kΦ∗ k∗ kΦ] k
2
5
e⊥
= (Σ0 − e0 )(kΨ0 k2 + kΨ
0 k ) + O(α ).
Therefore, collecting (192)-(202), and using the two equalities (203)-(204), we obtain
3
1 4X
2
⊥ 2
e⊥
e
hH(uα Ψ0 + g), uα Ψ0 + Ψ
i
=
(Σ
−
e
)(kΨ
k
+
k
Ψ
k
)
−
α
k(Hf + Pf2 )−1 Wi k2∗
0
0
0
0
0
3 i=1
3
1
2 4X
− uα 2
− α
h(A− )i Φ2∗ , (Hf + Pf2 )−1 (A+ )i Ωf i − kΦu] α k2] − 4αk(hα + e0 )− 2 Q⊥
α P.A Φ∗ k
3 i=1
+ 2αkA− Φu∗ α k2 + o(α5 log α−1 ).
With the definition e(1) , e(2) , and e(3) , of Theorem 2.1 this expression can be
rewritten as
(205)
e trial , Φ
e trial i = e(1) α3 + e(2) α4 + e(3) α5 log α−1 + o(α5 log α−1 ).
h(H − Σ0 + e0 )Φ
Using Lemma A.7 yields kΨ0 k2 = 1+O(α2 ), which implies, due to the orthogonality
e ⊥ and uα in L2 (R3 , dx),
of Ψ
0
e trial k2 = kuα k2 kΨ0 k2 + kΨ
e ⊥ k2 = 1 + O(α2 ).
kΦ
0
Therefore, together with (205), this gives
e trial , Φ
e trial i
h(H − Σ0 + e0 )Φ
= e(1) α3 + e(2) α4 + e(3) α5 log α−1 + o(α5 log α−1 ).
e trial k2
kΦ
which concludes the proof of the lower bound in Theorem 2.1.
Appendix A
Lemma A.1. We have
P.A− uα Φ1∗ = 0,
and
(206)
hΦ2∗ , ζ(Hf , Pf2 )Pfi Φ2∗ i = 0 , and hΦ2∗ , ζ(Hf , Pf2 )Wi i = 0 ,
for any function ζ for which the scalar products are defined. Similarly, we have
+
uα
hΦ2∗ Γ0 Ψ⊥
1 , A .Pf Φ∗ i = 0.
(207)
Proof. Straightforward computations using the symmetries of A− Φ1∗ and Φ2∗ .
Lemma A.2. We have
P.A− Φu∗ α =
√
αa0 ∆uα ,
where
Z
a0 =
k12 + k22
2
χΛ (|k|) dk1 dk2 dk3 .
2
3
2
4π |k| |k| + |k|
Proof. Straightforward computations using the symmetries of A− Φu∗ α .
42
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
Lemma A.3.
(208)
∀ϕ ∈ F, hP.Pf Γ1 ϕuα , Φu] α i = 0 ,
(209)
kΦu] α k2 = O(α3 log α−1 ) ,
(210)
kΦu] α k2∗ = O(α3 ) ,
(211)
k |k|− 2 Φu] α k2 = O(α) .
1
(212)
kP Φu] α k2 = O(α5 log α−1 ) ,
(213)
kP Φu] α k2∗ = O(α5 ) ,
(214)
k |k| 6 P Φu] α k2 = O(α5 ) .
1
Proof. The proof of (208) is as follows
(215)
hP.Pf Γ1 ϕuα , Φu] α i
Z X
3
3 X j
X
λ (k)χΛ (|k|) ∂uα
1
i ∂uα
=
k
ϕ(k) 2
dxdk
1
∂x
k
+
|k|
+
h
+
e
∂xj
2π|k| 2
i
α
0 j=1
i=1
λ=1,2
Z
3 Z
X
∂uα
1
∂uα X k i iλ (k)χΛ (|k|)
=
dk
dx
ϕ(k)dk = 0,
1
∂xi k 2 + |k| + hα + e0 ∂xi
2π|k| 2
i=1
λ=1,2
using that the integral over x is independent of the value of i, and since k.λ (k) = 0.
To prove (211), we note that
2
Z χΛ (|k|)
− 21 uα 2
dkdx
k |k| Φ] k ≤ cα ∇u
α
|k| (|k| + k 2 + hα + e0 )
Z
χΛ (|k|)2
≤ cα3
3 2 2 dk = O(α).
|k|2 (|k| + 16
α )
The proofs of (209) and (210) are similar, but simpler.
We next prove (212).
3 Z
X
χΛ (|k|)
∂uα 2
(216)
kP Φu] α k2 = cα
kP 1
kL2 (R3 ) dk.
2
|k| 2 (|k| + k + hα + e0 ) ∂xi
i=1
The function ∂uα /∂xi is odd. On the subspace of antisymmetric functions on
α
L2 (R3 ), one has that −(1 − γ0 )∆ − |x|
> −e0 for some γ0 > 0, which implies on
2
this subspace that hα + e0 > γ0 P , and thus,
P 2 < γ0−1 (hα + e0 ).
(217)
The relation (217) yields, for all k
∂uα 2
k
|k| (|k| + + hα + e0 ) ∂xi
1 ∂uα
χΛ (|k|)
≤ γ0−1 k 1
(hα + e0 ) 2
k2
∂xi
|k| 2 (|k| + k 2 + hα + e0 )
!2
χΛ (|k|)
−1
4
≤ γ0 cα
.
1
3 2
|k| 2 (|k| + k 2 + 16
α )
kP
(218)
χΛ (|k|)
1
2
k2
HYDROGEN GROUND STATE ENERGY IN QED
Substituting (218) into (216) and integrating over k proves (212).
The proof of (214) is similar.
43
Lemma A.4. We have
1
1 1
k(h1 + ) 2 ∇u1 k2 α5 log α−1 + o(α5 log α−1 ) ,
3π
4
− Φu] α k2∗ = O(α5 ) ,
kΦu∗ α k2∗ − kΦu] α k2] =
kΦu∗ α
kA− (Φu] α − Φu∗ α )k2 = O(α7 log α−1 ).
Proof. We have
(219)
kΦu∗ α k2∗ − kΦu] α k2]
2
χΛ (|k|)
1 1
α5 4 2 ∇u )
(h
+
=
1
1
(2π)2 3 |k| 12 (|k| + k 2 ) 12 (|k| + k 2 + α2 (h1 + 41 )) 12
4
Since (h1 + 41 )∇u1 ∈ L2 , it implies that for sufficiently large c > 0 independent of
α,
1 1
kχ(h1 > c)(h1 + ) 2 ∇u1 k2 < ,
4
and
2
1 1
χΛ (|k|)
2 ∇u χ(h
>
c)(h
+
)
1
1
1
1
|k| 2 (|k| + k 2 ) 12 (|k| + k 2 + α2 (h1 + 1 )) 12
4
4
(220)
Z ∞ 2
χΛ (t)
dt = log α−1 + O(1).
≤
t + cα2
0
1
For the contribution of χ(h1 ≤ c)(h1 + 41 ) 2 ∇u1 in (219), the following inequalities
hold,
(221)
1
1 1
1
( log α−1 + O(1) ) kχ(h1 < c)(h1 + ) 2 ∇u1 k2
π
3
4
2
χΛ (|k|)
1 1
1 1
2
χ(h
<
c)(h
+
)
∇u
=
1
1
1
1
1
1
(2π)2 3 |k| 2 (|k| + k 2 ) 2 (|k| + k 2 + (c + 14 )α2 )) 2
4
2
1 1
1 1
χΛ (|k|)
2
χ(h
<
c)(h
+
)
∇u
≤
1
1
1
1
1
1
(2π)2 3 |k| 2 (|k| + k 2 ) 2 (|k| + k 2 + (h1 + 14 )α2 )) 2
4
2
1 1
1 1
χΛ (|k|)
2 ∇u χ(h
<
c)(h
+
≤
)
1
1
1
1
1
1
2
3
2
2
2
(2π) 3 |k| 2 (|k| + k ) 2 (|k| + k + 16 α )) 2
4
≤(
1
1
1 1
log α−1 + O(1) ) kχ(h1 < c)(h1 + ) 2 ∇u1 k2 .
π
3
4
The inequalities (220) and (221) prove the first equality of the Lemma.
The proofs of the last two equalities are similar but simpler.
44
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
Lemma A.5.
(222)
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2∗ − kΦ2∗ k2∗ kΦu] α k2 = o(α3 log α−1 ),
(223)
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2 = O(α3 log α−1 )
(224)
k |k1 | 6 |k2 | 6 |k3 | 6 (Hf + Pf2 )−1 A+ .A+ Φu] α k2 = O(α3 )
(225)
k(hα + e0 )(Hf + Pf2 )−1 A+ .A+ Φu] α k2 = O(α7 log α−1 ).
1
1
1
Proof. Denoting by σn the set of all permutations of {1, 2, ..., n}, we have
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2∗
P
P
X
4α 1
λ=1,2 ελ (ki ).
ν=1,2 εν (kj )
=
√
12
3
P3
P3
(2π)
6 (i,j,n)∈σ
2
n
p=1 |kp | + (
p=1 kp )
P
2
κ=1,2 εκ (kn )χΛ (|k1 |)χΛ (|k2 |)χΛ (|k3 |)
×
∇u
α
1
1
1
|ki | 2 |kj | 2 |kn | 2 (|kn | + kn2 + (hα + e0 ))
If we pick two triples (i, j, n) and (i0 , j 0 , n0 ), such that n 6= n0 , then we get a product
which is integrable at k1 = k2 = k3 = 0, even without the term (hα + e0 ). The
contribution of such terms is cαk∇uα k2 = O(α3 ). Moreover, using the symmetry
in k1 , k2 , k3 , the twelve remaining terms give the same contribution. This yields
(226)
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2∗
P
P
P
2
8α λ ελ (k3 ).
ν εν (k2 )
κ εκ (k1 )χΛ (|k1 |)χΛ (|k2 |)χΛ (|k3 |)
=
.∇u
α
1
P3
(2π)3 P3
2
1
1
1
2
2 |k | 2 |k | 2 (|k | + k 2 + (h
|k
|
+
(
|k
|
+
e
))
k
)
i
3
2
1
1
α
0
i
1
i=1
i=1
+ O(α3 ).
On the other hand, we have
kΦ2∗ k2∗ kΦu] α k2
P
P
P
2
8α λ ελ (k3 ).
ν εν (k2 )
κ εκ (k1 )χΛ (|k1 |)χΛ (|k2 |)χΛ (|k3 |)
=
.∇u
α .
(2π)3 (|k2 | + |k3 | + (k2 + k3 )2 ) 21 |k3 | 21 |k2 | 21 |k1 | 12 (|k1 | + k 2 + (hα + e0 ))
1
Therefore, we obtain
(227)
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2∗ − kΦ2∗ k2∗ kΦu] α k2
Z (|k1 |2 + 2|k1 | |k3 | + 2|k1 | |k2 |)
8α
=−
3
(2π)
|k2 | |k3 | (|k1 | + |k2 | + |k3 | + |k1 + k2 + k3 |2 )(|k2 | + |k3 | + |k2 + k3 |2 )
X
X
×
ελ (k3 ).
εν (k2 )χΛ (|k1 |)χΛ (|k2 |)2 χΛ (|k3 |)2 |u(k1 , x)|2 dk1 dk2 dk3 dx ,
λ
ν
where
P
u(k1 , x) =
1
2
κ εκ (k1 )χΛ (|k1 |)
.∇uα
1
2
2
|k1 | (|k1 | + k1 + (hα + e0 ))
.
For fixed δ, we first compute the integral in (227) over the regions |k1 | > δ. This
yields a term cδ α3 , where cδ is independent on α.
HYDROGEN GROUND STATE ENERGY IN QED
45
Next, integrating (227) over the regions |k1 | ≤ δ yields a bound O(δ)α3 log α,
with O(δ) independent of α.
This concludes the proof of (222).
The proof of (223) is a straightforward computation showing
k(Hf + Pf2 )−1 A+ .A+ Φu] α k2
P
P
P
2
8α λ ελ (k3 ).
ν εν (k2 )
κ εκ (k1 )χΛ (|k1 |)χΛ (|k2 |)χΛ (|k3 |)
.∇u
=
α
(2π)3 P3 |k | + (P3 k )2 |k | 12 |k | 21 |k | 21 (|k | + k 2 + (h + e ))
3
α
0
2
1
1
1
i=1 i
i=1 i
= O(α3 log α−1 ) .
The proofs of (224) and (225) are similar to the above.
⊥
Lemma A.6. For Ψ⊥
1 and Ψ2 given in Definition 6.1, we have
1
1 ⊥
5
⊥ 2
2
|2αRe hA− .A− Γ3 Ψ⊥
1 , Γ Ψ2 i| ≤ cα + kHf Ψ2 k .
(228)
1 ⊥
3 ⊥
+
+ 1 ⊥
Proof. Using 2αRe hA− .A− Γ3 Ψ⊥
1 , Γ Ψ2 i = 2αRe hΓ Ψ1 , A .A Γ Ψ2 i, we obtain
−
− 3 ⊥
1 ⊥
that 2αRe hA .A Γ Ψ1 , Γ Ψ2 i can be rewritten as a linear combination of the
following two integrals I1 and I2
X κ (k 0 ).ν (k 00 )
dk dk 0 dk 00 dx
0
00
2
+ |k+k +k | κ,ν |k 0 | 21 |k 00 | 12
X (k 0 ). (k 00 )
X λ (k)
1
κ
η
×
(Γ1 Ψ⊥
1 ∇uα
1
2 )(k, x),
2
0
00 21
|k|+|k| + (hα +e0 )
2
2
|k|
κ,η |k | |k |
λ
5
Z
I1 = α 2
(229)
|k|+|k 0 |+|k 00 |
and I2 is defined as I1 , except that in the last sum, we reverse the role of k and k 00 ,
namely
dk dk 0 dk 00 dxχΛ (|k|)χΛ (|k 0 |)χΛ (|k 00 |) X κ (k 0 ).ν (k 00 )
00 12
0 21
|k|+|k 0 |+|k 00 | + |k+k 0 +k 00 |2
κ,ν |k | |k |
X (k 0 ). (k)
X λ (k)
1
κ
η
00
×
∇u
(Γ1 Ψ⊥
α
1
2 )(k , x).
0 | 12 |k| 21
|k|+|k|2 + (hα +e0 )
2
|k|
|k
κ,η
λ
5
Z
I2 = α 2
(230)
To bound I2 , we use the Schwarz inequality, |ab| ≤
1 2
2 a
+ 2 b2 . This yields
Z dk dk 0 dk 00 dxχ2 (|k|)χ2 (|k 0 |)χ2 (|k 00 |)
2
Λ
Λ
Λ
0
00
0
00
2
2
0
(|k|+|k |+|k | + |k+k +k | )
|k | |k 00 |
1
1
1
2 2− 0 2− 1 2
×
∇u
|k|
|k
|
α
|k|+|k|2 + (hα +e0 ) |k| 12
|k 00 |
Z
2
2
0
2
00 1
dk dk 0 dk 00 dx 1 ⊥ 00
2 00 χΛ (|k|)χΛ (|k |)χΛ (|k |) 2
× 2
|(Γ
Ψ
)(k
,
x)|
|k
|
2
|k 0 ||k|
|k|2− |k 0 |2−
5
|I2 | ≤ α 2
1
2
≤ cα7 + kHf2 Ψ⊥
2k ,
46
J.-M. BARBAROUX, T. CHEN, V. VOUGALTER, AND S. VUGALTER
Similarly, we bound I1 as follows,
Z
5
2
dk dk 0 dk 00 dx
2
|I1 | ≤ α
0
(|k|+|k |+|k 00 | + |k+k 0 +k 00 |2 )2 |k 0 | |k 00 |
2
1
1
1
0 2− 00 2− χ2Λ (|k|)χ2Λ (|k 0 |)χ2Λ (|k 00 |) 2
×
∇u
|k
|
|k
|
α
1
|k|+|k|2 + (hα +e0 ) |k| 2
|k|
Z dk dk 0 dk 00 dx
2
2
0
2
χΛ (|k|)χΛ (|k |)χΛ (|k 00 |) 12
1 ⊥
2
× 2
|(Γ
Ψ
)(k,
x)|
|k|
2
|k 0 ||k 00 |
|k 0 |2− |k 00 |2−
Z
Z
1
χ2Λ (|k|)
2
≤ cα5 dk 0 dk 00 |k 0 |− |k 00 |− χ2Λ (|k 0 |)χ2Λ (|k 00 |) dk 2
1 2 2 k∇uα k
|k| (|k| + 16
α )
1
1
2
5
⊥ 2
2
+ kHf2 Γ1 Ψ⊥
2 k ≤ cα + kHf Ψ2 k ,
α
where we took into account that ∂u
∂xi is orthogonal to uα , and on the subspace of
1 2
such functions we have (h − α + e0 ) ≥ 16
α .
Lemma A.7. Let Ψ0 be the ground state of T (0), with |Γ0 Ψ0 | = 1. Then we have
3
3
Ψ0 = Ωf + 2η1 α 2 Φ1∗ + η2 αΦ2∗ + 2η3 α 2 Φ3∗ + ∆0∗ ,
with h∆0∗ , Φi∗ i∗ = 0 (i = 1, 2, 3), Γ0 ∆0∗ = 0
and
k∆0∗ k2 = O(α3 ).
Proof. It follows from a similar argument as in [7, Proposition 5.1] that
kaλ (k)∆0∗ k ≤
cα
.
|k|
This yields
k∆0∗ k2 ≤
Z
kaλ (k)∆0∗ k2 dk ≤
Z
|k|≤α
2 3
0 −1
≤c α +cα
1
2
kHf ∆0∗ k2
c2 α 2
dk +
|k|2
Z
|k|>α
|k| kaλ (k)∆0∗ k2 χΛ (|k|)
dk
|k|
3
= O(α ) ,
1
where in the last equality we used the estimate k∆0∗ k2∗ = k(Hf +Pf2 ) 2 ∆0∗ k2 = O(α4 )
proved in [4, Theorem 3.2].
Acknowledgements. J.-M. B. thanks V. Bach for fruitful discussions. The authors gratefully acknowledge financial support from the following institutions: The
European Union through the IHP network Analysis and Quantum HPRN-CT-200200277 (J.-M. B., T. C., and S. V.), the French Ministry of Research through the
ACI jeunes chercheurs (J.-M. B.), and the DFG grant WE 1964/2 (S. V.). The
work of T.C. was supported by NSF grant DMS-0704031.
HYDROGEN GROUND STATE ENERGY IN QED
47
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Centre de Physique Théorique, Luminy Case 907, 13288 Marseille Cedex 9, France
and Département de Mathématiques, Université du Sud Toulon-Var, 83957 La Garde
Cedex, France.
E-mail address: [email protected]
Department of Mathematics, University of Texas at Austin, 1 University Station
C1200, Austin, TX 78712, U.S.A.
E-mail address: [email protected]
University of Toronto, Department of Mathematics, Toronto, ON, M5S 2E4, Canada
E-mail address: [email protected]
Mathematisches Institut, Ludwig-Maximilians-Universität München, Theresienstrasse
39, 80333 München and Institut für Analysis, Dynamik und Modellierung, Universität
Stuttgart.
E-mail address: [email protected]