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Eigenvalues of the Breit Equation Prog. Theor. Exp. Phys. 2013, 063B03 (13 pages) DOI: 10.1093/ptep/ptt041 Eigenvalues of the Breit equation Yoshio Yamaguchi1,† and Hikoya Kasari2,∗ 1Nishina Center, RIKEN, 350-0198 2-1 Hirosawa, Wako, Saitama, Japan 2Department of Physics, School of Science, Tokai University 259-1292 4-1-1 Kitakaname, Hiratsuka, Kanagawa, Japan ∗E-mail: [email protected] Received January 9, 2013; Accepted April 12, 2013; Published June 1, 2013 ............................................................................... Eigenvalues of the Breit equation Downloaded from e2 (α p + β m)αα δββ + δαα (−α p + β M)ββ − δαα δββ αβ = Eαβ , 1 1 2 2 r in which only the static Coulomb potential is considered, have been found. Here a detailed 1 discussion of the simple cases, S0, m = M, and m = M, is given, deriving the exact energy eigenvalues. An α2 expansion is used to find radial wave functions. The leading term is given by http://ptep.oxfordjournals.org/ the classical Coulomb wave function. The technique used here can be applied to other cases. ............................................................................... Subject Index B87 1. Introduction and the Breit equation at CERN LIBRARY on June 21, 2013 1.1. Introduction The Breit equation [1–3] has traditionally [4] been considered to be singular, and no attempt has ever been made to solve it. Instead, the Pauli approximation or a generalized Foldy–Wouthuysen trans- formation has been applied to derive the effective Hamiltonian Heff, consisting of the Schrödinger Hamiltonian 1 1 1 e2 H =− + p2 − 0 2 m M r and many other relativistic correction terms [4–6]. The perturbation method was then used to evaluate the eigenvalues (binding energies) in a power series with α as high as possible (or as high as necessary for experimental verification of quantum electrodynamics). Here we shall try to solve the Breit equation directly. The radial wave function for the simplest case, 1S leptonium states with equal masses, is given by the confluent Heun function. The author is not familiar with handling this function. Instead, an approximate method using an α2 ∼ 10−4 expansion shall be used to find the wave functions. Its leading term is given by the Coulomb wave function. However, the exact eigenvalues are found in singlet states with equal, m = M, and unequal, m = M, mass cases. We shall study the simplest case, 1S, equal lepton masses, m = M, in greater detail 1 3 3 here. The technique used here shall be applied to other cases, (l)l, (l)l ,and P0 including m = M, to find the exact eigenvalues. † This author contributed mainly to this work. © The Author(s) 2013. Published by Oxford University Press on behalf of the Physical Society of Japan. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari More complicated cases, i.e., m = M, described in Sect. 2.3 below, and triplet cases, shall be described in due course. 1.2. 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 −e2/r =−α/r (α=. 1/137) is given by e2 (α p + β m)μνδρσ + δμν(−α p + β M)ρσ − δμνδρσ νσ = Eμρ . (1) 1 1 2 2 r This equation holds in the center of mass (CM) system. The total energy of the system E can be expressed as Downloaded from E = m2 − q2 + M2 − q2. (2) The momentum operator in (1) can be written as http://ptep.oxfordjournals.org/ 1 1 ∂ p = ∇= , (3) i i ∂r introducing the operator r canonically conjugate to p. Then the (static) Coulomb potential (the contri- bution from one longitudinal photon exchange between two oppositely charged point Dirac particles) 2 can be written as −e /r =−α/r. α j ,βj ( j = 1, 2) are the usual 4 × 4 Dirac matrices for particles 1 (−e, m) and 2 (+e, M), respectively. αβ is a 4 × 4 Dirac spinor wave function. at CERN LIBRARY on June 21, 2013 We shall try to solve 1 1 α α ∇+ β m + −α ∇+ β M − = E, (4) 1 i 1 2 i 2 r and find the eigenvalue(s) E = m2 − q2 + M2 − q2, (2) which are specified by the discrete real value(s) qn. The two-particle system can be classified by the total angular momentum and its z-component and parity. In so doing, it is sufficient to specify the spin-angular parts of the large–large components αβ (α, β = 1, 2) as 1 (l)l ,(l = 0, 1, 2,...) 3 3 (l − 1)l + (l + 1)l ,(l = 1, 2,...) 3 (l)l ,(l = 1, 2,...) where l is the quantum number for the orbital angular momentum. The spin-angular parts of the other (large–small, small–large, small–small) components of αβ are completely fixed by those of the large–large components. We shall describe the singlet case in Sect. 2 in detail. Other cases can be handled similarly. 2/13 PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari 2. The singlet cases 1 2.1. (l)l, general 1 We shall first discuss the simplest case of (l)l . αβ can be taken as follows: αβ αβ 11 F(r)|1(l) > 22 l 13 i{G(r)|3(l + 1) > +G˜ (r)|3(l − 1) >} 24 l l (5) 31 i{H(r)|3(l + 1) > +H˜ (r)|3(l − 1) >} 42 l l 33 K (r)|1(l) > 44 l Downloaded from 1 3 where | (l)l > and | (l ± 1)l > are normalized spin-angular wave functions (the z-component of the total angular momentum has been omitted). The Breit equation (4) in Sect. 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): http://ptep.oxfordjournals.org/ α l + 1 d l + 2 l d l − 1 E + − m − M F = + (G + H) − − (G˜ − H˜ ), r 2l + 1 dr r 2l + 1 dr r √ d l + 2 √ d l − 1 l + (G − H) + l + 1 − (G˜ − H˜ ) = 0, dr r dr r α α E + − (m + M) F = E + + (m + M) K, r r at CERN LIBRARY on June 21, 2013 α l + 1 d l α E + + m − M H =− − (F + K ) = E + − m + M G, r 2l + 1 dr r r α l d l + 1 α E + + m − M H˜ = + (F + K ) = E + − m + M G˜ . (6) r 2l + 1 dr r r Therefore, we have the relations: α α E + + m − M H = E + − m + M G, r r α α E + + m − M H˜ = E + − m + M G˜ . r r For later convenience, we introduce the dimensionless quantities: ρ = 2qr, E M2 − q2 + m2 − q2 y = = (> 0), 2αq 2αq M + m + E λ = (> 0), 2αq M + m − E ν = (> 0). (7) 2αq If the eigenvalue E is a proper relativistic generalization of the Schrödinger case (as we shall see is the case), y and λ are large quantities of the order of 1/α2, while ν is of the order of unity. 3/13 PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari For ρ →∞(r →∞), all radial wave functions should behave like −ρ/ F(ρ) → ρn{1 + O(1/ρ)}e 2. (8) −ρ/2 −qr This is because e = e and E = m2 − q2 + M2 − q2. Notice that, if we consider E,the total CM energy m2 + p2 + M2 + p2, to be larger than M + m, we would expect that the radial wave functions should contain a factor eipr. If we analytically 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 the two masses are equal, M = m.First, ⎫ H = G, H˜ = G˜ , ⎬ 1 − νρ . (9) K = F ⎭ 1 + λρ Then, the combination Downloaded from −ρ/ F + K = h˜(ρ)e 2 (10) obeys the differential equation 2 2 2 1 α y 1 α 1 l(l + 1) 1 http://ptep.oxfordjournals.org/ h˜ + h˜ −1 + + + h˜ − 1 + − − = 0. ρ ρ(1 + yρ) 2 ρ 4 ρ2 ρ2 2ρ(1 + yρ) 1 2.2. S0, M = m 1 ˜ For the S0, i.e., l = 0, case we have the following equation for h: 2 2 2 1 α y 1 α 1 1 h˜ + h˜ − + + + h˜ − + − = . (11) 1 1 0 at CERN LIBRARY on June 21, 2013 ρ ρ(1 + yρ) 2 ρ 4 ρ2 2ρ(1 + yρ) If we ignore α2 and higher-order terms in (11) and notice that α2 y is of the order of unity, we find 2 2 α y 1 h˜ + h˜ −1 + + h˜ − 1 = 0, (12) ρ 2 ρ which is precisely the Schrödinger equation for our issue and h˜ in this approximation is given by (− α2 y + , ,ρ)= ( − , ,ρ) α2 y − F 2 1 2 F 1 n 2 , apart from the normalization constant. 2 1 must be equal to an integer n − 1 = 0, 1, 2,.... Equation (11)showsthatρ = 0 is the regular singular point while ρ =∞is an irregular singular point. Near ρ = 0, h˜(ρ) can be expressed as ⎫ ˜(ρ) = ρs (ρ), h h ⎬⎪ . α2 ⎪ (13) where s =−1 + 1 − ⎭ 4 =− − − α2 Another solution s 1 1 4 is unacceptable from the square-integrability of the wave function αβ . h(ρ) obeys the equation: 2 2 + 2s 1 α y 1 1 1 h + h −1 + + + h − 1 − s − + sy = 0.
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