Ground States for translationally invariant Pauli-Fierz Models at zero Momentum

We consider the translationally invariant Pauli-Fierz model describing a charged particle interacting with the electromagnetic field. We show under natural assumptions that the fiber Hamiltonian at zero momentum has a ground state.


Introduction
Non-relativistic qed has been a successful theory describing low energy aspects of quantum mechanical matter interacting with the quantized radiation field.Within this model many physical phenomena have been mathematically understood.In the present paper, we shall discuss an aspect which falls within the scattering problem of an electron interacting with the quantized radiation field.The main result which we prove is that the the total system composed of electron and photon field at zero momentum is free of infrared divergences.
In this situation the infrared singularity is critical.As soon as the momentum is nonzero a ground stated ceases to exist, [14].
Starting with the work of Bloch and Nordsieck [4] the so called infrared catastrophe in scattering theory has been intensively investigated.Although the physical reasons for infrared divergences were well understood at that time and did not lead to any physical problems, the formal treatment was not satisfactory.In [20] Fadeev and Kulish introduced a new space of asymptotic states and gave a theoretical discussion of this phenomenon in the framework of relativistic quantum field theory.Fröhlich studied such asymptotic states in the so called Nelson Model [9,10], which is mathematically well defined.Extending these results an iterative algorithm for constructing asymptotic states in Nelson's model was developed in [23].This construction was extended to the model of non-relativistic qed in [8].Specifically, the case of zero momentum in nonrelativistic qed has previously been investigated in [2,6].
In the present paper we consider the Hamiltonian H of non-relativistic qed describing an electron, with or without spin, coupled to the quantized radiation field.The Hamiltonian commutes with the operator of total momentum.We are interested in the operator, H(0), obtained by restricting the Hamiltonian to the subspace of total momentum zero.
Based on a natural energy inequality, we prove that for all values of the coupling constant, H(0) has a ground state.The energy inequality has been shown to hold in the spinless case for all values of the coupling constant [12], and in the case of spin the energy inequality follows in a related situation from the main theorem in [6].The existence of a ground state can for example be used to obtain expansions on the binding energy of the hydrogen atom [3].To the best of our knowledge the result, which we prove, has so far only been shown in the spinless case for small values of the coupling constant [2], see also [8] for related results.In contrast to the proofs given there, our proof is non-perturbative and independent of the magnitude of any ultraviolet cutoff parameter.The proof, which we give, uses a compactness argument and is not constructive.Nevertheless, once the existence of the ground state is established one can use other methods to obtain asymptotic expansions of the ground state as well as its energy [1,5], in this context see also [13].
The idea of the proof given in the present paper follows closely the ideas introduced in [11], which were applied in a similar case in [21].However, in the situation which we encounter the infrared singularity is more severe and subtler estimates are necessary.We believe that the estimates which we use in the present paper might establish alternative proofs of existence of ground states in other critical cases, as for example [16].
In the next section we introduce the model and state the main result.The proofs are presented in Section 3.

Model and Statement of Results
Let h be a complex Hilbert space.We introduce the symmetric Fock space where h (0) = C and where h (n) = P n ( n k=1 h) for n ≥ 1, with P n denoting the orthogonal projection onto the subspace of totally symmetric tensors.Thus we can identify ψ ∈ F (h) with the sequence (ψ (n) ) n∈N 0 with ψ (n) ∈ h (n) .The vacuum is the vector Ω := (1, 0, 0, . ..) ∈ F (h).We define for f ∈ h the creation operator a * (f ) acting on vectors ψ ∈ F by with domain D(a * (f )) := {ψ ∈ F : a * (f )ψ ∈ F }.This yields a densely defined closed operator.For f ∈ h we define the annihilation a(f ) as the adjoint of a * (f ), i.e., It follows from the definition that a(f ) is anti-linear, and a * (f ) is linear in f .Creation and annihilation operators are well known to satisfy the so called canonical commutation relations •] stands for the commutator, and f, g h denotes the inner product of h.For a self-adjoint operator A in h we define the operator dΓ(A) as follows.In h (n) we define in the sense of [26,VIII.10]and A (0) := 0. By definition ψ ∈ F (h) is in the domain of is a vector in F (h), in which case dΓ(A)ψ is defined by (1).The operator dΓ(A) is self-adjoint, see for example [26,VIII.10].
Henceforth, we shall consider specifically and write F for F (h).The Hilbert space h describes a so called transversally polarized photon.By physical interpretation the variable (λ, k) ∈ Z 2 ×R 3 consists of the wave vector k and the polarization label λ.Because of (2), the elements ψ ∈ F can be identified with , where the subscript "sym" stands for the subspace of functions wich are totally symmetric in their n arguments.Henceforth, we shall make use of this identification without mention.
The Fock space inherits a scalar product from h, explicitly We shall make use of the physics notation of the creation and annihilation operators.
One defines for (λ, k) Now (3) defines a well defined operator a λ (k) on where S stands for the Schwartz space.In the sense of quadratic forms on D S × D S its adjoint a * λ (k) is well defined.Furthermore, in the sense of quadratic forms one has for all f ∈ h the identities For details we refer the reader to [24, X.7].The field energy operator denoted by H f is given by where ω : The operator of momentum P f is defined as a three dimensional vector of operators, where the j-th component is defined by where The Hilbert space describing the system composed of a charged particle with spin s ∈ {0, 1 2 } and the quantized field is The Hamiltonian is where the ε λ (k) ∈ R 3 are vectors, depending measurably on k = k/|k|, such that (k/|k|, ε 1 (k), ε 2 (k)) forms an orthonormal basis.For the proof we shall make an explicit choice in (24), below.We shall adopt the standard convention that for v = (v 1 , v 2 , v 3 ) we write v 2 := 3 j=1 v j v j .By x we denote the position of the electron and its canonically conjugate momentum by p = −i∇ x .If s = 1/2, let S = (σ 1 , σ 2 , σ 3 ) denote the vector of Pauli-matrices.If s = 0, let S = 0.The number e ∈ R is called the coupling constant.
The so called form factor ρ : R 3 → C is a measurable function for which we shall assume the following hypothesis for the main theorem.
Hypothesis A. For some 0 < Λ < ∞ we have where We note that Hypothesis A is usually assumed in Pauli-Fierz type models.The Hamiltonian is translation invariant and commutes with the generator of translations, i.e., the operator of total momentum Note W P tot W * = p so that in the new representation p is the total momentum.One easily computes where set A := A(0) and B := B(0).Let F be the Fourier transform in the electron variable x, i.e., on L 2 (R 3 ), Then the composition U = F W is a unitary operator yielding the so called fiber decomposition of the Hamiltonian, where is an operator in the so called reduced Hilbert space The operator H(ξ) is self-adjoint on D(P 2 f ) ∩ D(H f ), see Theorem 3 in the next section.To prove that H(0) has a ground state, we use a compactness argument similar to [11].
The idea is to first introduce a positive photon mass in the field energy.For m ≥ 0 we define where and study the operator We set The proof is based on the following energy inequality.Specifically we can show our result for any e ∈ R for which there exists an m 0 > 0 such that for all m ∈ (0, m 0 ) Inequality ( 9) has been investigated in the literature.In spinless case, s = 0, Inequality (9) has in fact been shown for all values m ≥ 0 and e ∈ R using functional integration, [12,28,17,21].In case of spin s = 1/2 Inequality (9) has to the best of our knowledge not yet been shown by means of functional integration.For s = 1/2 an Inequality of the type (9) follows in a related situation for small |e| from the main theorem stated in [6], which in turn is based on perturbative arguments.We now state the main result of this paper.
Theorem 1. Suppose Hypothesis A holds, and let e ∈ R. If there exists an m 0 > 0 such that for all m ∈ (0, m 0 ) the energy inequality (9) holds, then the operator H(0) has a ground state.
By the results in the literature mentioned in the previous paragraph, we obtain immediately the following theorem as corollary.
Theorem 2. Suppose Hypothesis A holds.In the spinless case, i.e., s = 0, the operator H(0) has a ground state for all values of the coupling constant e.
We note that for small values of the coupling constant the statement in Therorem 2 has been shown previously in [2], see also [8] for related work.In the present paper we extend that result to all values of the coupling constant and provide a rather short proof.
We want to point out, that the question whether Inequality (9) holds for all values of the coupling constant in the case of spin s = 1/2 seems to be an open question.We would like to mention work [18] in that direction.

Proof of Results
We first state the following technical result about the domain of self-adjointness.
Versions of this theorem have been shown in [17], [18] and [21].Since we could not find the precise version, which we need, in the literature, we shall provide a short proof of Theorem 3 in Appendix A. The proof follows closely a proof given in [15].Moreover, we shallv use the following result to prove the main theorem.Versions of this theorem have been shown in the literature [10,9,28,11,21].We could not find in the literature the precise version, which we need, so we provide a proof of Theorem 4 in Appendix B.
Let us now outline the strategy of the proof of Theorem 1, which follows closely ideas given in [11].For m > 0 it follows from Theorem 4 that E m (0) is an eigenvalue of H m (0).
Henceforth let ψ m denote a normalized eigenvector of H m (0) with eigenvalue E m (0).We will show in Proposition 7, below, that (ψ m ) m>0 is a minimizing family for H(0) as m tends to zero, i.e., Finally we shall use a compactness argument, based on two infrared bounds, stated in Lemmas 11 and 12, to show that there exists a strongly convergent subsequence (ψ m j ) j∈N which converges to a nonzero vector, say ψ 0 .Using lower semicontinuity of nonnegative quadratic forms [27] (or alternatively the spectral theorem and Fatou's Lemma), it will then follow from (10) that i.e., that ψ 0 is a ground state of H(0).
Remark 1.We note that in contrast to [11,21] the infrared bounds which we obtain have stronger infrared singularities.Therefore it is harder to prove compactness.This difficulty will be addressed in Lemma 13 below.
Proposition 5. We have E m (0) ≥ E(0) and follows that E m (0) is monotonically decreasing as m tends to zero and E(0 This implies the existence of the limit and To show the opposite inequality we argue as follows.From Theorem 3 it follows that any core for P 2 f + H f is a core for H(0).Thus for any ǫ > 0, there exists a normalized vector On the other hand, since H m (0) ≤ H(0) + mN, it follows that for any m ≥ 0, Since ǫ is arbitrary, (11) follows from ( 12) and (13).
Let us collect a basic inequality in the following lemma.
Lemma 6.Let e ∈ R and m ≥ 0, and suppose (9) holds.Then for all ξ ∈ R 3 we have For later use we state the following Proposition.
Proof.Using that H m (0) ≥ H(0) we find from Proposition 5 that

Infrared Bounds
Throghout this section we assume that ρ To simplify the notation we write Lemma 8. Let e ∈ R und m > 0, and suppose (9) holds.Then for i = 1, 2, 3, the vector Proof.For the proof we use analytic perturbation theory.For details we refer the reader to [19,25].
is an analytic family of type (A) in each component.By Theorem 4 we know that E m (0) is an eigenvalue isolated from the essential spectrum.Let P m (0) be the orthogonal projection onto the finite dimensional eigenspace of E m (0).By first order perturbation theory and the energy inequality (9) we conclude that P m (0)vP m (0) = 0.By second order perturbation theory and an application of the energy inequality (9) we conclude that for i = 1, 2, 3, where the second inequality is understood as an operator inequality on RanP m (0).The second inequality in fact holds, since E m (ξ) is defined as an infimum.This shows the claim.
For notational convenience we set which by Lemma 6 is well defined and satisfies The formula of the next Lemma is known as a so called Pull-through relation.
Proof.The proof is similar to [7, Lemma 6.1], see also [14,Lemma 7].By Theorem 3, we know that ψ m ∈ D(H f ).Hence using the standard expression of the free field energy in terms of annihilation operators which implies a λ (k)ψ m ∈ F for a.e.k.We write By the canonical commutation relations of creation and annihlation operators we find which holds for a.e.k as an identity of measurable functions.Applying this to ψ m and using that H m ψ m = e m ψ m we find for a.e.k This implies that a λ (k)ψ m is in the domain of H m (−k) for a.e.k.Indeed, the map is bounded, since in view of ( 18) we can write , in view of Theorem 3. Hence the lemma follows by applying R m (k) to (18).
Lemma 10.Suppose there exists an m 0 > 0 such that (9) holds for all m ∈ (0, m 0 ).Then there exists a constant C such that for all e ∈ R, m ∈ (0, m 0 ), k ∈ R 3 , and j = 1, 2, 3, By Lemma 8 the second factor on the right hand side can be estimated using It remains to estimate the first factor in (19).First we use the trivial identity Estimating the last term using Now multiplying this inequality on both sides with the self-adjoint operator R m (k) we obtain Using this, we estimate Using (15) and that e m is bounded for 0 ≤ m ≤ m 0 , we see that (22) inserted in (19) implies the bound stated in (a).
(b) Using that v 2 ≤ h m we see from (20) that This implies hence (b) follows.
Estimating the expression in Lemma 9 using Lemma 10 (a) we obtain the next lemma.
Lemma 11.Suppose Hypothesis A holds.Suppose there exists an m 0 > 0 such that (9) holds for all m ∈ (0, m 0 ).Then there exists a finite constant C such that for all e ∈ R, m ∈ (0, m 0 ) we have We still need an estimate involving derivatives.To this end, we shall henceforth make an explicit choice of the polarization vectors.After a possible unitary transformation on Fock space we can always achieve that the polarization vectors are given by Lemma 12. Suppose Hypothesis A holds.Suppose there exists an m 0 > 0 such that (9) holds for all m ∈ (0, m 0 ).Then there exists a finite constant C such that for all e ∈ R, m ∈ (0, m 0 ), and a.e.k with |k| < Λ Proof.We want to calculate the derivative of the expression in Equation ( 16).Calculating the derivative with respect to the operator norm topology, we find by means of the resolvent identity, that Using this we can calculate the derivative for 0 We now use that by Lemma 10, there exists a constant C such that ω Using this together with (15) the first and second term are estimated from above by a finite constant times |k| −2 .To estimate the third term we note that by the choice (24), we have for λ = 1, 2, Lemma 13 (y-Bound).Suppose Hypothesis A holds.Suppose there exists an m 0 > 0 such that (9) holds for all m ∈ (0, m 0 ), and let e ∈ R. Then there exists a constant C, and a δ > 0 such that for all m ∈ (0, m 0 ) and all n ∈ N, where ( ψ m ) (n) denotes the Fourier transform of the n-photon component of ψ m .
Proof.We drop the subscript m.Thus by ψ (n) we denote the Fourier transform of ψ (n) in all its n-components.We define the functions .
Step 1: There exists a δ > 0 and a constant C such that for all a ∈ R 3 , The claim follows easily for |a| ≥ 1 2 Λ, since ψ is a normalized state in Fock space and |1 − e −iay | ≤ 2. Now lets consider the case |a| < 1 2 Λ.By the Fourier transform, we have the identity To estimate the second integral we use Lemma 11 and observe that the integrand vanishes Next we estimate the first integral and assume |k| < Λ − |a|.Using Lemma 12, we find where π 3 denotes the projection in R 3 along the 3-axis and const.denotes a finite constant independent of n.Let π a denote the projection in R 3 along the vector a and let π 3,a denote the projection in the (1, 2)-plane along π 3 a (with convention that π 3,a = π 3 , if π 3 a = 0).
Step 2: Step 1 implies the statement of the Lemma.
From Step 1 we know that there exists a finite constant C such that After interchanging the order of integration and a change of integration variables b = |y|a, we find where c is nonzero and does not depend on y.

Existence of the Ground State
Proof of Theorem 1. Fix a positive m 0 such that for all m ∈ (0, m 0 ) the energy inequality (9) holds.
Step 1: All ψ m , with m 0 ≥ m > 0, lie in a compact subspace of the reduce Hilbert space H.
Let T be the self-adjoint operator associated to the nonnegative and closed quadratic form q in H defined by for all φ ∈ D(q), the natural form domain of q.We choose δ > 0 such that Lemma 13 holds.By this and Lemma 11 and Proposition 7, there exists a finite C such that for all m with 0 < m < m 0 , The set K is a compact subset of H, provided T has compact resolvent [25,Theorem XIII.64].Hence it remains to show that T has compact resolvent.The operator T preserves the n-photon sectors.Let T n denote the restriction of T to the n-photon sector.
Therefore µ l (T n ) → ∞ as l tends to infinity, where µ l denotes the l-th eigenvalue obtained by the min-max principle.Moreover since µ l (T n ) ≥ n for all l, n, it follows that µ l (T ) → ∞ as l → ∞.Hence T has a compact resolvent.
Here we use the argument outlined at the beginning of this section.By Step 1, we know that all ψ m , with m 0 ≥ m > 0, lie in a compact subspace of H.It follows that there exists a subsequence (ψ m i ) i∈N , with m i → 0 as i → ∞, which converges strongly to a normalized vector ψ 0 .By lower semicontinuity of non-negative quadratic forms we see from Proposition 7 that for all κ ∈ C. Let Z 0 denote the connected component of Z ∩ Z containing 0. Since T (0) = T is self-adjoint, and hence 0 ∈ Z 0 , it follows from a Theorem of Wüst, [30, We apply the above Proposition to T (e) = T + eT (1) + e 2 T (2) , where First we note the following.Since A is divergence free we have A • P f = P f • A, and so , with natural domain.Since T and T 0 are non-negative multiplication operators, it is easy to see that the domains of T 0 and T agree, and that the operators are mutually bounded.Thus, as T (e) = H m (ξ), Theorem 3 will follow as a consequence of Proposition 14 and the following two lemmas.
Proof.Standard estimates, see for example [24], show that A 2 is H f,m -bounded and that the components of A and B are H 1/2 f,m -bounded.It follows that for some C we have f +H f,m )ψ 2 and collecting estimates shows the claim.
Proof.Let e ∈ R and m ≥ 0 be fixed.For notational compactness we set Q := P f − ξ and F := eA.
Step 1: There exist constants c 1 , c 2 , c 3 such that In the following we denote by C a constant which may change from line to line.We have The second term in (32) is estimated as follows Further, using a commutator Collecting the above estimates yields Step 1.
Step 2: There exists a constant C such that Calculating a double commutator, we see that for some b.
Step 2 follows by adding Inequality (33) to the left hand side and completing the square.
First observe that ( Inserting on the right hand side of (34) first the inequality of Step 1 and then the inequality of Step 2, we find for some constant C ( By standard estimates S • B is infinitesimally bounded with respect to H f,m .It follows by ( 35) that it is also infinitesimally bounded with respect to

B Ground State for positive Photon Mass
In this section we provide a proof of Theorem 4. It follows closely the proofs given in [11,21].Theorem 4 will follow directly from Propositions 17 and 24, below.For ξ ∈ R 3 and m ≥ 0 we define )dk) and m > 0. Then for all ξ ∈ R 3 we have To prove the proposition we need the following notation.Let h 1 and h 2 be two Hilbert spaces.If A : h 1 → h 2 is a partial isometry, we define Γ(A) to be the linear operator 1 , n ≥ 1, and which equals to the identity on h Proof.Part (b) follows directly from Lemma 20 (b).To show (36), we insert in (b) the identity from Lemma 20 (c) and find Now the inequality follows from standard estimates.To show (37) we again use (b), To estimate the second term on the right hand side we first use the canonical commutation relations and then Lemma 20 (c) and find To estimate the first term on the right hand side we find similarly Taking the difference (b) follows.Now (a) follows from (b) using standard estimates.For example using Lemmas 18 and 19 we find Right hand side of (b) = U l a * (([j 1 , h]e l , [j 2 , h]e l )a(j 1 e l , j 2 e l )Γ(j) = UdΓ(A)Γ(j), where we defined the following operator on h ⊕ h Now the bound follows since the operator preserves the n-particle sector and satisfies the following estimate Recall that H = C 2s+1 ⊗ F .We consider the map from H to H ⊗ F .By abuse of notation we shall henceforth denote this map again by J.
We introduce the operator where o(L 0 ) does not depend on ϕ.
where P Ω,2 denotes the orthogonal projection in H ⊗ F onto H ⊗ Ω.
Proof.(a) Defining the operators Now using that J is an isometry, it follows from Lemma 22 (a) (choosing for example Using again that J is an isometry it follows from Lemma 21 (a) that for ϕ ∈ D(N 1/2 ) we have, recalling the notation ( 17), Analogously, we find from Lemmas 22 and 21 for ϕ ∈ D(N) where ν : R 3 → R 3 with ν(k) = k.Now we will use that for m > 0 and that which follows by dominated convergence in Fourier space.Furthermore, we note where the first line is easy to see and the second line can be seen as follows.Let χ 1,L = χ 1 (•/L).For normalized ϕ 1 , ϕ 2 we find using the usual notation and convention for the Fourier transform To treat the term involving j 2 we apply a similar estimate to the function 1−χ 2 .Inserting the above estimates Part (a) now follows.
To show (b) we use the canonical isomorphism where the summand with n = 0 is by convention H.With respect to this fiber decomposition the Hamiltonian H m (ξ) fibrates and we have for n ∈ N 0 , This yields the expectation To calculate the summands on the right hand side we again use (38) and find for n ≥ 1 where we employed the following operator inequality.Using ω m (p) + ω m (q) ≥ ω m (p + q) we find where we used that Jϕ, (1I ⊗ P Ω )Jϕ = Γ(j 1 )ϕ 2 .Let λ ∈ σ ess (H m (ξ)).Then there exists a normalized sequence ψ n , n ∈ N, converging weakly to zero such that lim n→∞ (H m (ξ) − λ)ψ n = 0. Thus, Hm(ξ) .

1 .
The following two lemmas are straightforward to verify.Proof.The Lemma follows directly from the definition of J and the properties of Lemmas 18 and 19.Lemma 21.Let f ∈ h.