Eigenvalues of the Breit Equation

Eigenvalues of the Breit Equation {eqnarray*} [(\vec{\alpha}_{1} \vec{p} + \beta_{1}m)_{\alpha \alpha^{\prime}} \delta_{\beta \beta^{\prime}} + \delta_{\alpha \alpha^{\prime}} (-\vec{\alpha}_{2} \vec{p} + \beta_{2}M)_{\beta \beta^{\prime}} - \frac{e^{2}}{r} \delta_{\alpha \alpha^{\prime}} \delta_{\beta \beta^{\prime}}] \Psi_{\alpha^{\prime} \beta^{\prime}} = E \Psi_{\alpha \beta}, {eqnarray*} in which only the static Coulomb potential is considered, have been found. Here the detailed discussion on the simple caces, $^{1}S_{0},\ m=M$ and $m \neq M$ is given deriving the exact energy eigenvalues. The $\alpha^2$ expansion is used to find radial wave functions. The leading term is given by classical Coulomb wave function. The technique used here can be applied to other cases.


Introduction and the Breit equation 1.Introduction
The Breit equation [1] has traditionally [2] considered to be singular and has never been tried to solve itself. Instead, the Pauli approximation or a generalized Foldy-Wouthysen transformation has been applied to derive the effective Hamiltonian H eff , consisting of the Schrödinger Hamiltonian and many other relativistic correction terms [2] [3]. Then perturbation method was used to evaluate the eigen-values (binding energies) in power series in α as high as possible (or as much as needed for experimental verification of QED).
Here we shall try to directly solve the Breit equation. The radial wave fundtion for the simplest case, 1 S leptonium states with equal masses, is given by the confluent Heun function. The auther is not familiar to handle this function. Instead, the approximate method using α 2 ∼ 10 −4 expansion shall be used to find wave funcitons. Its leading term is given by Coulomb wave function.
However, the exact eigenvalues are found in singlet-states with equal m = M and unequal masses m = M cases. We shall study the simplest case, 1 S, equal lepton masses, m = M , in greater details here. The technique used here shall be applied to other cases, 1 (l) l , 3 (l) l and 3 P 0 including m = M , to find exact eigenvalues.
More complicated cases, m = M described in §2.3 below and triplet cases, shall be described in due course.

Basic equation
The Breit equation for two leptons (Dirac particles), with charge and mass, (−e, m) and (+e, M ), interacting through the (static) Coulomb potential −e 2 /r = −α/r (α . = . 1/137) is given by This equation holds in the CM system. The total energy of the system E can be expressed as The momentum operator in (1) can be written as introducing the operator r canonically conjugate to p. Then the (static) Coulomb potential (contribution form one longitudinal photon exchange between two oppositely charged point Dirac particles) can be written as −e 2 /r = −α/r. α j , β j (j = 1, 2) are the usual 4 × 4 Dirac matrices for particle 1 (−e, m) and 2 (+e, M ), respectively. Ψ αβ is 4 × 4 Dirac spinor wave function.
We shall try to solve and find the eigen-value(s) which is specified by discrete real value(s) q n . The two particle system can be classified by the total angular momentum and its zcomponent and parity. In so doing it is sufficient to specify spin angular parts of the largelarge components Ψ αβ (α, β = 1, 2) to be 1 (l) l , (l = 0, 1, 2, · · · ) 3 (l − 1) l + 3 (l + 1) l , (l = 1, 2, · · · ) and 3 (l) l , where l is the quantum number for the orbital angular momentum. Spin-angular parts of other (large-small, small-large, small-small) components of Ψ αβ are completely fixed by those of the large-large components. We shall describe the singlet case in §2 in detail. Other cases can be handled similarly.
2/14 2. The Singlet Cases 2.1. 1 (l) l , general We shall first discuss the simplest case of 1 (l) l . Ψ αβ can be taken as follows: where | 1 (l) l > and | 3 (l ± 1) l > are normalized spin-angular wave functions (z-component of the total angular momentum has been omitted). The Breit equation (4) in §1 gives the following set of differential equations for the radial wave functions F (r), K(r), G(r),G(r), H(r) andH(r): Therefore, we have the relations: For later convenience, we introduce the dimensionless quantities: If the eigen-value E is a proper relativistic generalization of the Schrödinger case (as we shall see it is the case), y and λ are large quantities of the order of 1/α 2 , while ν is the order of unity.
For ρ → ∞(r → ∞) all radial wave function should behave like This is because e −ρ/2 = e −qr and E = m 2 − q 2 + M 2 − q 2 . Notice that if we consider E, the total CM energy m 2 + p 2 + M 2 + p 2 , to be larger than M + m, we would expect that the radial wave functions should contain a factor e ipr . If we analitically continue p to imaginary values we get the factor e −qr and the energy expression (2) for bound states. The singlet case becomes very simple when two masses are equal M = m. First Then, the combination obeys the differential equatioñ If we ignore α 2 and higher order terms in (11) and notice that α 2 y is of the order of unity, we findh 4/14 which is precisely the Schrödinger equation for our issue andh in this approximation is given by F (− α 2 y 2 + 1, 2, ρ) = F (1 − n, 2, ρ) apart from the normalization constant. α 2 y 2 − 1 must be equal to an integer n − 1 = 0, 1, 2, · · · .
(11) shows that ρ = 0 is the regular singular point while ρ = ∞ is an irregular singular point.
Near ρ = 0h(ρ) can be expressed ash Another solution s = −1 − 1 − α 2 4 is unacceptable from the square integrability of the wave function Ψ αβ . h(ρ) obeys the equation: h(ρ) can be expanded in a Taylor series near ρ = 0: It can be shown that h(ρ) can not be a polynomial: from the recurrence formulae (16). So the sum in equation (15) must extend to n = ∞, a sharp difference from the Schrödinger case. Note that the expansion (15) holds only the tiny region ρ < 1/y ∼ O(α 2 ). Next we discuss on ρ = ∞, which is an irregular singular point. Assuming the form where λ, k, c n are constant. Introducing (17) into the differential equation (14), we find λ is either 0 or 1 while k is undermined. From the square-integrability of the wave function we must choose λ to be 0 and k < ∞.

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k turns out to be (14), whose value is not fixed from the situation at ρ → ∞. Further discussion will be given in §2.2.4.
h(ρ) obeys the diffrential equation, (14), α 2 y and sy are of the order of unity O(1). The RHS of equation (19) is of the order of 1/y or O(α 2 ) ≃ 10 −4 . As the first step we shall ignore the RHS of (19). Then h can immediately solved: Normalizability of the radial wave function requires N equals to positive integer: n is the pricipal quantum number. (20) is the relativistic extension form of the nonrelativisitic Coulomb wave function The energy eigenvalues derived from the condition (21) are 6/14 When n = 1, The approximate solution F (−N, γ, ρ) is correct ignoring terms of the order O(1/y) ≃ α 2 ≃ 10 −4 . We write When n = 1, F (0, γ, ρ) = 1, We shall introduce into the eq. (19), then we find whose solution gives h = F + f . Or, alternatively, we may use an approximate method: F is the order of unity, while f (ρ) is the order of O(1/y) ≃ O(α 2 ), so that f in the RHS of (27) can be ignored. This equation gives immediately the solution It is difficult to perform integretion here, but computer shall easily do the jobs. The equation (21) give the exact eigenvalues for our problem. The reason is as follows. When ρ → ∞, the function h(ρ) = F (ρ) + f (ρ) behaves like (see §2.2.4 ) h(ρ) = ρ β c 0 + c 1 ρ + · · · , β = N as will be described in §2.2.4 F (ρ) contributed to ρ β , while f (ρ) does not, i.e., Therefore (21) gives the exact eigenvalues. To find f (ρ), we may use 1/y expansion, or α 2 expansion, since y is of the order of 1/α 2 .

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Introducing where f ν (ρ) is of the order of (1/y) ν ∼ O(α 2ν ), and the differential operators where D (ν) is of the order of (1/y) ν (here ν = 0, and 1), the differential equation (19) will give the following serries of equations It is easy to see that shall be good if the terms of the order of α 4 ∼ 10 −8 are ignored. These techniques can be applied to other singlet and triplet cases with m = M .
We shall give the recurrence formulae here for the case of 1 S, m = M N has already been fixed in Eq. (21). We repeat that ρ β is the contribution from F (ρ) and f (ρ) = h(ρ) − F (ρ) does not contribute to ρ β term. Furthermore it is easy to show This shows (21) is in fact the correct (or exact) eigenvalues.

Comparison of Breit case with Dirac Coulomb solution.
For simplicity we shall discuss only ground states of 2 S 1/2 (Dirac) and 1 S 0 (Breit).

Dirac equation with Coulomb potential[4] reads
Ψ for 2 S 1/2 , spin up, can be written The differential equation for F is given by 10/14 Introducing dimensionless quantities and write F to be where s is introduced to remove 1/ρ 2 terms as usual. Then the differential equation for h(ρ) is given by where E = m 2 − q 2 , binding energy B = m − m 2 − q 2 , y = (E + m)/2αq. However s is different from Breit case The form (35) in the Dirac case looks exactly the same as the Breit case (14). Only differences are in below.
And both αE q − 1 − s = 0 and 1 2 + sy = 0 (37) holds and they give the same value of q, As is well-known, Dirac equation with Coulomb force can be solved for all cases of n and l in terms of two confluent hypergeomotric functions (multiplied by ρ s e −ρ/2 ).

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However one cannot do N = 0 and δ = 0 simultaneously in the Breit case.
While δ = 0 demands and N cannot be zero. From these situations, the Breit case is more complicated than the Dirac case, as described in this article.

Confluent Heun function.
The solution h(ρ) of differential equation (14) is confluent Heun function [5], which has 5 parameters. Unfortunately the author is not familier to this function, so that the auther has used physical ground and 1/y expansion of the wave function h(ρ). The equation for the quantityh(ρ): whereM = M/2αq andm = m/2αq. As in §2.2, ρ = 0 and ρ = ∞ in eq. (38) are the regular and irregular singular points, respectively. Therefore, we can adopt the same procedure as in §2.2.

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To remove the 1/ρ 2 terms, puth (ρ) = ρ s h(ρ), finding again s = −1 + 1 − α 2 /4. The differential equation for h(ρ) reads We should notice here that (A)+(B) and (C) reduce We shall rewrite the term (D) as follows: The last term (F ) shall be added to the first line of th coefficient h(ρ) in (19): n is the principal quantun number. The background part f is O(α 2 ) and can be found by solving N, γ, ρ).
Thus the accuracy of the solution F + f is quite good, terms of the order of α 4 are ignored.
Other radial wave functions, F, K, G, H,G andH can now be writen down. The energy eigenvalues from (42) and so on. The first term is exactly the well-known form of the reduced mass.
It is easy to see that (43) reduces to (23) when M = m.