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Spectral problems about many-body Dirac operators mentioned by Derezi\'{n}ski

We consider spectral problems for many-body Dirac operators mentioned by Derezi\'{n}ski in the IAMP News Bulletin of January 2012. In particular, we derive a representation of the Dirac Coulomb operator for a helium-like ion as a matrix operator of order sixteen. We show that it is essentially self-adjoint (under natural restrictions on the coupling constants), that the essential spectrum of its closure is the whole real line and that it has no eigenvalues.


Introduction
In [2], Dereziński mentioned open problems about many-body Dirac operators. These problems were originally formulated by B. Jeziorski who is a chemist from the University of Warsaw (cf. J. Sucher [10] p.6). Among them there are spectral problems on Dirac-Coulomb operator H DC for a helium-like ion, which has the form H DC = H(1, Z) + H(2, Z) + 1 where is the usual Dirac operator for an electron i in the hydrogen-like ion of charge Z and of mass m. In the above notation, r i and p i , i = 1, 2 are a position vector and the momentum operator, respectively, of the i-th electron, The vector α is a vector operator whose components α 1 , α 2 , α 3 , together with the operator β ≡ α 4 are Hermitian matrices of order four satisfying the anti-commuting relations α j α k + α k α j = 2δ jk (j, k = 1, 2, 3, 4).
Since the domain of the Dirac operator H(i, Z) is a subspace of four-component wave functions depending on the three coordinates of the i-th electron, we may infer that H DC acts on sixteen-component wave functions which depend on the six coordinates of two electrons and have the anti-symmetric property due to the Pauli principle. Mathematically, the operator should be written as where is an operator in H = L 2 (R 3 ; C 4 ). The domain of H DC is a subspace of the antisymmetric tensor product H ⊗ A H.
In this paper we shall rigorously derive a representation of H DC as a matrix operator of order sixteen and give an answer to its spectral problems, especially essential self-adjointness, continuous (essential) spectrum and absence of eigenvalues. Our method was inspired by a simpler model operator (see (4.5)).

Two-electron problems
As far as we know, there is no systematic derivation of relativistic systems in the physics and quantum chemistry literature, so that two-body relativistic systems with which we are concerned seem to be less familiar than nonrelativistic ones. The first relativistic equation for two particles which was extensively used in the past was the Breit equation; it is a differential equation for a relativistic wave function for two electrons, interacting with each other and with an external electromagnetic field. It is not fully Lorentz invariant and is only an approximation. It reads (see [1]) where r 12 = r 1 − r 2 , r 12 = |r 12 | and is the Dirac Hamiltonian and the Dirac matrices α j and β j operate on the spinor of U (for electron j). The wave function U depends on the positions r 1 and r 2 and has sixteen spinor components. If we neglect the right hand side of the equation (2.1) with A = 0, then we get the Dirac-Coulomb equation where H D [j] is the usual Dirac operator acting on the j-th electron: Here, σ = (σ 1 , σ 2 , σ 3 ) are the Pauli matrices The letters ℓ and s refer to the large and the small part of the wave function.
In the relativistic theory, we have to handle a two-fold tensor product space of C 4 -valued functions because the usual Dirac operator acts on four-vectors belonging to L 2 (R 3 ; C 4 ). For H = L 2 (R 3 ; C 4 ), the two-fold tensor product H ⊗2 = H⊗H can be identified with Here we have identified L 2 (R 6 ; C 4 ) with L 2 (R 6 ; C 2 ) ⊗ C 2 in the following way.
In this notation, we see that any product function and t M stands for the transposed matrix of the 2 × 2 matrix M . , where a slightly different notation from ours is adopted.
Moreover, it should be pointed out that the four components Ψ ℓℓ , Ψ ℓs , Ψ sℓ , Ψ ss in (2.7) are not functions from R 3 ⊗ R 3 to C 4 , but they are functions from In a similar way, we can define the symmetric tensor product space.
In the matrix formulation, we define the standard inner product of H 2 A or H 2 S by (2.8) The notation F ij , G ij will be used as an abbreviation for the integral over the trace.

The two-electron Dirac-Coulomb Hamiltonian
As usual, the free and the one-particle Dirac operator are denoted by Here p = t (p 1 , p 2 , p 3 ), p j = −i∂ r j ; m is a nonnegative number and k ∈ R. Theorem 3.2 prepares for a representation of Then it holds that Proof: In general, the Kronecker product of two matrices X = (x ij ) and Y = (y ij ) is defined by (3.6) Q.E.D.
Then it holds that

7)
where Proof: Denote f ℓ and f s by i and j, respectively. Let us consider two elements where for k = 1, 2, . (3.10) We can regard Ψ k (k = 1, 2) as functions of x = (r, ω) as follows.
where χ ± are two orthonormal functions describing the spin of electrons. Then we have We see that Similarly, the identity
3. If |k| ≤ √ 3/2, then H has no eigenvalues in R\(−m, m) and there are countably many eigenvalues in (−m, m) whose only accumulating points are ±m.
When the scalar potential k/|r| is replaced by any symmetric matrix potential V (r) satisfying |V (r)| ≤ k/|r|, we have

Now we shall state our main results for the Hamiltonian
These are just the first attempts to answer the spectral problem. with |k ′ | < 1/2.
The unique self-adjoint extension is denoted by the same symbol H DC again. We also consider the following simple model operator H in L 2 (R 6 ) 4 . It is easier to handle than H DC because an orthogonal change of variables reduces it to a Dirac operator in L 2 (R 3 ) 4 with a double-well potential.

The canonical form of H DC on the anti-symmetric space
We return to the familiar notation x j instead of r j . We may represent H DC as follows. Recall Then We shall check the first component of H DC Ψ. In view of (5.5) and we find As for the second component, we see As for the third and fourth components, similar computations yield The anti-symmetric property implies the following identities.

Lemma 5.3 Suppose that F, G ∈ H 2
A . Then for any quadruple of indices i, j, k, ℓ, if V is a diagonal matrix satisfying for any i, j = 1, 2.
Hence for any ψ, ϕ ∈ H 2 S , Here, we have used Similarly, we see that for any ψ, ϕ ∈ H 2 A , Q.E.D.
4×4 be an Hermitian matrix which commutes with I 2 ⊗ ( σ · y 1 ) ( y 1 ∈ R 3 ), and which satisfies Then for all v ∈ H 1 (R 6 ; C 4 ) we have These inequalities can be proved by introducing polar coordinates in the variable y 1 just as in the case of the usual Dirac operator (Schmincke [8,9]). Without loss of generality, we may assume that m = 0.
Proof: It suffices to show that (H n ± iI 4 )(C ∞ 0 (R 6 ; C 4 )) = L 2 (R 6 ; C 4 ). (6.18) Due to the cutoff function, the scalar potential V n has singularities which do not coincide, so that we are able to employ a technique similar to the one developed by Vogelsang ([13]) in view of the following lemma which can be shown to be based on Lemma 6.4.
Q.E.D. Proof of Lemma 6.2 From Proposition 6.6, it follows that Q.E.D.
It follows that From Lemma 6.7, it follows that {w n } becomes a singular sequence of H DC − λ + µ. In fact, it satisfies that as n → ∞ w n ⇀ 0, w n L 2 = 1. (6.33) The last assertion w n L 2 = 1 follows from the fact that the eigenvectors u(ξ), v(ξ) corresponding to the different eigenvalues are orthogonal to each other. Since

Absence of eigenvalues
Proof of Theorem 4. 3 We first consider the case where m > 0. Let u ∈ H 2 A be a solution to Then we see that for any function ϕ(y 1 , Then we see that for   Therefore, we note that each eigenvalue E n (κ) of H(κ) is an analytic function of κ > 0. Now we claim Lemma 6.8 Let u ∈ H 2 A be a solution to the equation (6.34). Then u = 0 almost everywhere in R 6 .