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Non-existence of ground states in the translation invariant Nelson model

In this paper we consider the massless translation invariant Nelson model with ultraviolet cutoff. It is proven that the fiber operators have no ground state if there is no infrared cutoff.


Introduction
In this paper we study the translation invariant massless Nelson model. The model can (after a unitary transformation) be written as a direct integral of fiber operators {H(ξ)} ξ∈R 3 . The spectral properties of these operators were first investigated by by J. Frölich in his Phd-thesis, which was published in the two papers [5] and [6]. Frölich showed, that if the field is massive or there is an infrared cut-off then H(ξ) has a ground state for ξ in an open ball around 0. He also proved, that if the field is massless, no infrared conditions are imposed and a ground state exists for sufficiently many of the H(ξ), then one can reach some physically unacceptable conclusions. The aim of this paper is to prove that H(ξ) does not have a ground state if the field is massless and no infrared conditions are assumed. We shall briefly review central results about existence of ground states in the massless Nelson model.
In the paper [10], it is proven that ground states exists in a non-equivalent Fock representation. A consequence of this result is that the usual "taking the massgap to 0" strategy for proving existence of ground states does not work. This strongly indicates that there should be no ground state.
A proof of absence of ground states in a similar model was given by I. Herbst and D. Hasler in the paper [8]. They consider the fiber operators of the massless and translation invariant Pauli-Fierz model {H(ξ)} ξ∈R 3 . They prove that H(ξ 0 ) has no ground state if ξ → inf(σ(H(ξ))) is differentiable at ξ 0 and has a nonzero derivative. One may easily work out the same problem for the Nelson model and obtain the same conclusions. However proving the existence of a non-zero derivative is an extremely hard problem and such a result has only been achieved for weak coupling and small ξ (see [1]). Furthermore, ξ = 0 is a global minimum for ξ → inf(σ(H(ξ))) and therefore the derivative is 0. However, H(0) has no ground states shall prove below.
In fact we shall prove that H(ξ) has no ground state for any non-zero coupling strength and ξ ∈ R 3 . Our proof is based on strategy used by I. Herbst and D.
Hasler, but we remove the assumption regarding the existence of a non-zero derivative. Instead we use rotation invariance of the map ξ → inf(σ(H(ξ))), non degeneracy of ground states and the HVZ-theorem.

Notation and preliminaries
We start by fixing the measure theoretic notation. Let (M, F , µ) be a σ-finite measure space and X be a separable Hilbert space. We will write L p (M, F , µ, X) for the Hilbert space valued L p -space. If X = C it will be omitted from the notation. In case M is a topological space we will write B(M) for the Borel σ-algebra.
For g ∈ H one defines the annihilation operator a(g) and creation operator a † (g) on symmetric tensors in F (H) using a(g)Ω = 0, a † (g)Ω = g and where f i means that this element is omitted. One can show that these operators extends to closed operators on F (H) and that (a(g)) * = a † (g). Furthermore, we have the canonical commutation relations which states and [a(f ), a † (g)] = f, g .
(2) If f ∈ D(ω) then ǫ(f ) ∈ D(dΓ (ω)) and dΓ (ω)ǫ(f ) = a † (ωf )ǫ(f ). In particular we find ǫ(g), dΓ (ω)ǫ(f ) = g, ωf e g,f . Let A ∈ B(R ν ). In this paper we shall mainly encounter spaces of the form which is obviously a dense subspace inside H A . We will also need the contraction is projection to the i'th coordinate and let g (n) : (R ν ) n → R ν be given by g (0) (0) = 0 and g (n) (k) = k 1 +· · ·+k n for n ≥ 1. Then for K : R ν → R we have where K A (ξ − g (n) ) is to be interpreted as the corresponding multiplication operator on H ⊗sn A . In case A = R ν we will omit A from the notation. We shall also encounter vectors of operators. Let B 1 , . . . , B n be operators on a Hilbert space H and define B = (B 1 , . . . , B n ) from ∩ n i=1 D(B i ) into H ν by Bψ = (B 1 ψ, . . . , B n ψ). Note H ν = ν k=1 H and is also a Hilbert space. For any k ∈ R ν we define In particular we find for ψ ∈ D(B)

The operator -basic properties and the main result
Fix K, ω : R ν → [0, ∞) measurable and let v ∈ H. Define for A ∈ B(R ν ) and ξ ∈ R ν the Hamiltonian where v A = P A (v). We have ) and essentially selfadjoint on any core for H 0 (ξ, A). Also, H µ (ξ, A) ≥ −µ 2 ω −1/2 v independent of A and ξ.
) + ω n . and define the In particular, H 0 (ξ, A) is selfadjoint. For ψ ∈ D(H 0 (ξ, A)) we have dΓ (ω)ψ ≤ H 0 (ξ, A)ψ and so we find via Lemma 2.1 and the Kato Rellich theorem that is selfadjoint on D(H 0 (ξ, A)) and any core for H 0 (ξ, A) is a core for H µ . Using Hypothesis 1: We assume (1) K ∈ C 2 (R ν , R) is non negative and there is C K > 0 such that ∇K 2 ≤ C K (1 + K) and D 2 K ≤ C K where D 2 K is the Hessian of K. Under these hypothesis we define maps We have the following lemma Lemma 3.2. Assume Hypothesis 1. The following holds where E ξ,A (a) ≤ C K a 2 . In particular, D(H µ (ξ, A)) is independent of ξ.
To prove (3) we note that equation (3.2) is easily obatined from statements (1) and (2). Using for any ψ ∈ D(H µ (ξ, A)) and equation (3.2) we immediately obtain continuity for ξ → H µ (ξ, A)ψ. To prove the statement regarding norm resolvent convergence we calculate using equation (3.2) which goes to 0 for a tending to 0. Continuity of ξ → inf(σ(H µ (ξ, A))) now follows from continuity of the spectral calculus and the existence of a ξ-independent lower bound by Lemma 3.1.
It only remains to prove statement (4). By Lemma 3.1 it is enough to check that J (D) and L(D) span a core for H 0 (ξ, A). Let f 1 , . . . , f n ∈ CS A . Pick R > 0 such that 1 BR(0) f i = f i 1 A λ ν almost everywhere for all i ∈ {1, . . . , n} and note that 1 BR(0) n f 1 ⊗ s · · · ⊗ s f n = f 1 ⊗ s · · · ⊗ s f n 1 A n λ ⊗n ν almost everywhere. Let C = sup k∈BR(0) ω(k). Using the fundamental theorem of calculus we find the following point wise inequality for k ∈ B R (0) n : We therefore find the following point wise estimates on B R (0) n : Multiplying by 1 p! and summing over p yields a finitie number so f 1 ⊗ s · · · ⊗ s f n is analytic for H 0 (ξ). Now, Ω is an eigenvector for H 0 (ξ) and therefore analytic we see J (D) is a total set of analytic vectors for H 0 (ξ, A) and therefore it spans a core for H 0 (ξ, A) by Nelson analytic vector theorem.
This also holds for n = 0 as we in this case obtain H 0 (ξ, A) p Ω 2 = K(ξ) 4p ≤ ( C 1/2 ) 4p . Multiplying by 1 n! and summing over n yields a finitie number so Thus ε(f 1 ) is semi analytic for H 0 (ξ). This implies {ǫ(f ) | f ∈ D} spans a dense subspace of semi analytic vectors for H 0 (ξ, A), which is a core by the Masson-McClary theorem.

Hypothesis 2:
We assume exists and is strictly positive.
It is well known that Hypothesis 1 and 2 are fulfilled in this case. We can now state the main theorem of this paper: Theorem 3.3. Assume Hypothesis 1 and 2 along with ν ≥ 3. Then H µ (ξ) has no ground states for any ξ and µ = 0.

Proof of Theorem 3.3
We start with proving series of lemmas which we shall need. We work under Hypothesis 1 and 2. The first Lemma is known and we only sketch the proof.
One now easily calculates using Lemma 2.2 For any x ∈ R n \{0} we write x = x −1 x. The next small lemma is basically spherical coordinates.
where e 1 is the first standard basis vector. If U is open then λ ν (U ∩ B 1 (0)) = 0.
For every ξ ∈ R n and 0 < ε < 1 we define where k = k/ k . The following Lemma is essential: Proof. We start by proving (1). Assume ξ = 0 and k = 0. If ω(k) = 0 then by Hypothesis 2 we have ω(k ′ ) < 0 for all k ′ ∈ B |k| (0) which contradicts Hypothesis Assume now ξ = 0 and let k / ∈ Rξ. By rotation invariance of Σ (Lemma 4.1) we may calculate Page 9] either k and ξ − k are linearly dependent or k and ξ are linearly dependent. In any case ξ and k are linearly independent which ( as ξ = 0) implies k = aξ for some a ∈ R.
has a ground state for some µ = 0 and ξ ∈ R ν . Then the corresponding eigenspace is non degenerate.
We will now sharpen this result.
Then ψ (n) is an eigenvector for H n,A (ξ) corresponding to the eigenvalue Σ(ξ).
on a set of positive λ ⊗n ν measure. Sub-additivity of ω along with Lemmas A.3 and A.5 gives most hold on a set of positive λ ⊗n ν measure. By Lemma 4.4 we se that this can only hold for k ∈ (R ν ) n with k 1 + · · · + k n ∈ Span(ξ). But the rank theorem implies that the set of k satisfying this is a subspace of (R ν ) n of dimension νn − (ν − 1) < νn. However such a subspace must have λ ν measure 0 which is a contradiction.
The next two Lemmas are an adapted version of the corresponding ones found in [8]. For ξ ∈ R ν and k = 0 we define Lemma 4.7. Fix ξ ∈ R ν and R > 0. Then k·∇K(ξ−dΓ (k))Q 0 (k, ξ) is uniformly bounded for k in B(0, R)\{0}. We also have Proof. Note k · ∇K(ξ − dΓ (k))Q 0 (k) is bounded for k = 0 by the closed graph theorem and Lemma 3.2. For ψ ∈ F (H) we find by equation (2.2) that for any R > 0 and converges strongly to ∂ i K(ξ − dΓ (k))P 0 (ξ). We have Now ω is continous and goes to 0 as k tends to 0 so Q 0 (k, ξ) goes strongly to P 0 (ξ). Hence it is enough to see But this is obvious from the equality because the first term is constant and the other term is uniformly bounded and goes to 0.
For ξ ∈ R ν and k / ∈ Rξ we may by Lemma 4.4 define If ν ≥ 3 then S ε is open, non-empty and invariant under positive scalings. Furthermore, Proof. As ξ is fixed in this proof it will be omitted from the nation of Q, Q 0 , B and P 0 . If P 0 = 0 we can pick v(ξ) = 0. If P 0 (ξ) = 0 then is has dimension 1 by Lemma 4.6 and is spanned by a vector ψ ∈ D(H(ξ)). Using P 0 = |ψ ψ| we find that v(ξ) i = ψ, ∂ i K(ξ − dΓ (k))ψ does the trick. Furthermore, S ε is obviously open and invariant under positive scaling since this holds for S ε (ξ) and S ε (C ω v(ξ)). Furthermore any non-zero vector which is orthogonal to ξ and v(ξ) is in S ε and such vector will always exist if ν ≥ 3. It remains only to prove equation 4.7. By Lemma 4.4 we may pick R(ξ, ε) > 0 such that for k ∈ S ε (ξ) ∩ B R(ξ,ε) (0) we have with D(ξ, ε) < 1. Hence we find Using Lemma 3.2 we may calculate for k ∈ S ε (ξ): where o 2 (k) := −Q 0 (k)ω(k) −1 E ξ (−k)Q(k). Note o i (k) goes to 0 in norm for k tending to 0 in S ε (ξ) by equation (4.8), Lemma 3.2 and the uniform bound Q 0 (k) ≤ 1. Inserting equation (4.13) into equation (4.11) we find Q(k) = Q 0 (k) + |k| ω(k) Q 0 (k)( k · ∇K(ξ − dΓ (ω))Q 0 (k) (4.14) Note o(k) goes to 0 in norm for k tending to 0 in S ε (ξ) by equation (4.8), Lemmas 3.2 and 4.7, the uniform bound Q 0 (k) ≤ 1 and the fact that |k|ω(k) −1 has a limit for k tending to 0. Using equation (4.14) and appealing to the limit found in Lemma 4.7 along with the uniform bounds in Lemma 4.7 and equation (4.8) we now see (1 − P 0 )Q(k) and Q(k)(1 − P 0 ) goes to 0 weakly for k tending to 0 inside S ε (ξ). Hence we find From equation (4.11) we find The second and third term converges to 0 in norm since for k tending to 0 inside S ε (ξ) since D k and k · v(ξ)P 0 Q(k)P 0 are uniformly bounded by equation (4.8) and o 1 (k) converges to 0 in norm since for k tending to 0 inside S ε (ξ). Sandwiching the first term with two vectors φ, ψ ∈ F (H) we find Now ∂ i K(ξ −dΓ (k))P 0 ψ, (1−P 0 )Q(k)P 0 φ converges to 0 for k going to 0 inside S ε (ξ) by equation (4.15) and |k| ω(k) k i remains bounded as k goes to 0. Therefore first term goes weakly to 0 for k going to 0 inside S ε (ξ).
Proof (Theorem 3.3). Fix notation from Lemma 4.8. Assume that a ground state ψ exist and pick η ∈ D(N 1/2 ) such that ψ, η > 1 2 . Then by Lemma B.14 in Appendix B we have the pull through formula and since (1 − C ω k · v(ξ)) −1 η, ψ is uniformly bounded from below in S ε (ξ) by 1 2 we find that there is R > 0 such that for all k ∈ S ε (ξ) ∩ B R (0). Using Hypothesis 1 and 2 we see ω(Re 1 ) 2 > 0 because if that was not true then ω ≤ 0 on B R (0) which is a contradiction. Hence we find as v ∈ H we find that the integral of ω(k) −2 |v(k)| 2 over B R (0) must be infinite. Using Lemma 4.2 we find as λ ν (B 1 (0)) we see that the latter integral must be infinite. Furthermore since S ε (ξ) is open and not empty we have which is integrable with integral (N + 1) 1/2 η 2 ψ 2 by definition of the Fock space norm. This is the desired contradiction.
A. Partitions of unity and the essential spectrum.
In this section we prove a few technical ingredients. Hypothesis 1 will be assumed throughout this section. Define The following Lemma can be found in e.g. [9]: The following Lemma is obvious Lemma A.2. There is a unique isomorphism Note that we may identify (1) , . . . , k σ(n) ).

Now we define
which is strongly resolvent measurable in (k 1 , . . . , k n ) ∈ (A c ) n since ξ → H(ξ) is strong resolvent measurable by Lemma 3.2. In particular defines a selfadjoint operator on L 2 (R nν , B(R nν ), λ nν , F (H A )) and it is reduced by the projection 1 ⊗ S n . To see this we note that 1 ⊗ S n commutes with the unitary group of H n,A (ξ) since H
Let g 1 , . . . , g n ∈ H A c and let K ⊂ CS A be a subspace. Define If ψ ∈ Span(J (K)) we have where λ ∈ C.
Proof. Pick ε > 0 such that the balls B ǫ (k 1 ), . . . , B ǫ (k ℓ ) are pairwise disjoint and we have (B ǫ (k 1 )∪· · ·∪B ǫ (k ℓ ))∩{v = 0} is a λ ν 0-set. Let ε n = ε n , B Note that CS ⊂ A ∞ so A ∞ is a dense subspace of H. In particular, J (A ∞ ) spans a core for H µ (ξ − k 0 ) by Lemma 3.2. For each p ∈ N we may thus pick for all n ≥ u 1 (p). Note now that ψ p may we written as for some a(p), b(p), c(p), α i,j (p) constants and f j i (p) ∈ A ∞ . Note that each f j i (p) is in fact contained in some A l(i,j,p) by definition so defining u 2 (p) = max i,j {l(i, j, p)} we see that ψ p ∈ Span(J (A l )) for any l ≥ u 2 (p). Define now u p inductively by u 1 = max{u 1 (p), u 2 (p)} and u p+1 = max{u 1 (p), u 2 (p), u p−1 } + 1.
To summarise we have found vectors ψ p ∈ D(H(ξ)) and a strictly increasing sequence of numbers {u p } ∞ p=1 ⊂ N such that For each n ∈ N and A ∈ {B c n , B n } define V n = V Bε n (k0) c and j n,A : H i → H B c n ⊕H Bn by j n,B c n (f ) = (f, 0) and j n,Bn f = (0, f ). Furthermore we set Q n,A = V * n j n,A and let U n be the unitary map from Lemma A.4 corresponding to B c n . Fix f ∈ H. Then the following equalities holds λ ν almost everywhere: since P Bn (f ) = f 1 Bn λ ν -almost everywhere and P B c n (f ) = f 1 B c n λ ν -almost everywhere. For f ∈ A n we have 1 B c n f = f and so we obtain the two equalities Q n,Bn P Bn g (i) n = 1 Bn g (i) n = g (i) for all i ∈ {1, . . . , ℓ}. We now define the Weyl sequence as follows: We will now prove (1) φ p ∈ D(F 1 ).
(2) φ p is orthogonal to φ r for p = r.
(1): Define for all p ∈ N the set (2): Let r < p. Then φ r ∈ Span(C r ) and φ p ∈ Span(C p ), so we just need to see that every element in C p and C r are orthogonal. Let ψ 1 ∈ C p and ψ 2 ∈ C r . Note every tensor in C p has a factor g (1) up and that this factor is orthogonal to g (i) ur for all i by construction. Furthermore for any h ∈ A ur we see that h is supported in B c ur ⊂ B c up and hence g up h = 0, so g (1) up is orthogonal to any element in A ur . This implies ψ 1 contains a factor orthogonal to all factors in ψ 2 and thus ψ 1 is orthogonal to ψ 2 .
(3): Q up,B c up and Q up,Bn p are isometrics and which implies Γ (Q np,B c up ) and Γ (Q np,Bu p ) are isometrics. Using equations (A.7) and (A.8) we calculate where we used g up are normalised and orthogonal if i = j and (4): Define the function g up = g (1) up ⊗ s · · · ⊗ s g (ℓ) up . Using Lemma A.3 we see that (H(ξ) − Σ(ξ − k 0 ) − ω n (k 1 , . . . , k n ))φ p is given by Using the triangle inequality, Γ (P B c np )ψ p = 1, Γ (Q np,B c up )Γ (P B c np )ψ p = ψ p and Lemma A.3 we find γ ≤ C 1 + C 2 + C 3 where Let f : (R ν ) n → R + be non negative and symmetric. Using that the g up have disjoint support one finds Thus using permutation invariance of f we find By continuity of ω we now see √ ℓ!γ goes to 0 for p tending to ∞.

B. Proof of pull though formula
This appendix is devoted to proving the pull through formula. The in case K(k) = |k| 2 one could compute everything directly using tools as in [8]. However the other possible choices of K require a more sophisticated approach ao we use the formalised developed in [3] and the reader should consult this paper for the proofs. Let H = L 2 (M, E, µ), where (M, E, µ) is assumed to be σ-finite. We start by defining where · is the Fock space norm. This makes sense since P n (F + (H)) ⊂ F (H). We now have Lemma B.1. The map d defines a metric on F + (H) and turns this space into a complete separable metric space and a topological vector space. The topology and Borel σ-algebra is generated by the projections P n . If a sequence {ψ n } ∞ n=1 ⊂ F (H) is convergent/Cauchy then it is also convergent/Cauchy with respect to d. Also any total/dense set in F b (H) will be total/dense in F + (H) as well.
For each a ∈ R we define · a,+ = lim Note · a,+ restricts to a norm on F a,+ (H) that comes from an inner product. In particular F a,+ (H) is a Hilbert space and for a ≥ 0 we have F a,+ (H) = D((N + 1) a ). We summarise as follows Lemma B.2. · a,+ defines measurable map from F + (H) to [0, ∞], and restricts to a norm on the spaces F a,+ (H) that comes from an inner product turning F a,+ (H) into a Hilbert space.
The point of defining a metric on F + (H) and finding a dense set is that most of the operations we will encounter in this chapter are continous on F + (H). Therefore many operator identities only needs to be proven on well behaved vectors. Fix now v ∈ H. We now define the following maps on F + (H) Where a n (v) is annihilation from H ⊗s(n+1) to H ⊗sn and a † n (f ) is creation from H ⊗sn to H ⊗(n+1) .
Lemma B.4. The map Γ + (U ) is an isometry on F + (H) and is thus continuous. Furthermore we have We will now consider a class of linear functionals on F + (H). For each n ∈ N we let Q n : F + (H) → N denote the linear projection which preserves the first n entries of (ψ (n) ) and projects the rest of them to 0. For ψ ∈ N there is K ∈ N such that for n ≥ K we have Q n ψ = ψ. For φ ∈ F + (H) we may thus define the pairing where n ≥ K.
Lemma B.5. The map Q n above is linear and continuous into F (H). The paring ·, · + is sesquilinear, and continuous in the second entry. If φ ∈ F a,+ (H) then ψ → ψ, φ + is continous with respect to · −a,+ . Furthermore, the collection of maps of the form ψ, · + will separate points of F + (H).
Lemma B.2 shows that x → f (x) a,+ is measurable for functions f ∈ C(X, X , ν) and so the integral always makes sense. If a = 0 then it is finite if and only if f ∈ L 2 (X, X , ν, F b (H)). We write f ∈ C(X, X , ν) as (f (n) ) where f (n) = x → P n f (x). For f, g ∈ C(X, X , ν) we define We can now summarise.
Lemma B.8. d is a complete metric on C(X, X , ν) such that C(X, X , ν) becomes separable topological vector space. The topology is generated by the maps f → (x → P n f (x)). Furthermore L 2 (X, X , ν, F b (H)) ⊂ C(X, X , ν) and convergence in L 2 (X, X , ν, F b (H)) implies convergence in C(X, X , ν). Also the map x → f (x) a,+ is measurable for any f in C(X, X , ν) and a ∈ R.
We now move on to discuss some actions on this space. This is strongly related to the direct integral and readers should look up the results in [12]. Let n ≥ 1, v ∈ H, U be unitary on H, ω = (ω 1 , . . . , ω p ) a tuple of selfadjoint multiplication operators on H, m : M n → R p measurable and g : R p → R a measurable map. Then we wish to define operators on C(M ℓ , E ⊗ℓ , µ ⊗ℓ ) for ℓ ≥ 1 by and Γ ⊕,0 = Γ + (U ). We have the following lemma.
We can now prove the pull-trough formula.