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 contribution 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.

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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.

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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. ρ ρ(1 + yρ) 2 ρ 2 ρ(1 + yρ) (14)

4/13 PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari h(ρ) can be expanded in a Taylor series near ρ = 0: ∞ n h(ρ) = hnρ , (15) n=0 1 α2 y 1 1 h = − + 1 + s + + sy h ,(n + 1)(n + 3 + 2s) hn+ 1 3 + 2s 2 2 0 y 1 1 α2 y 1 + n(n + 1 + 2s) − (n − + 1 + s + + sy) hn y 2 2 α2 y − n − 1 − + 1 + s hn− = 0. (16) 2 1 It can be shown that h(ρ) cannot be a polynomial: N

n Downloaded from h(ρ) = hnρ (N < ∞) n=0 from the recurrence formulae (16). So the sum in Eq. (15)mustextendton =∞, a sharp difference from the Schrödinger case. Note that the expansion (15) holds only in the tiny region ρ<1/y ∼

O(α2). http://ptep.oxfordjournals.org/ Next we discuss ρ =∞, which is an irregular singular point, assuming the form λρ k c1 c2 h(ρ) = e ρ c0 + + +··· at ρ →∞, (17) ρ ρ2 where λ, k, cn are constant. Introducing (17) into the differential equation (14), we ﬁnd that λ is either 0 or 1 while k is undetermined. From the square-integrability of the wave function, we must choose

λ to be 0 and k < ∞. at CERN LIBRARY on June 21, 2013 k turns out to be α2 y k = − 1 − s, (18) 2 which appeared in (14); its value is not ﬁxed from the situation at ρ →∞. Further discussion will be given in Sect. 2.2.4.

1 2.2.1. Detailed discussion of S0, m = M. Note again F + K = e−ρ/2ρsh(ρ). h(ρ) obeys the differential equation (14), ⎫ 2 γ d h dh N 1 d ⎪ + −1 + + h = − − δ h, ⎪ dρ2 dρ ρ ρ ρ(1 + yρ) dρ ⎪ ⎪ ⎪ α2 ⎪ where γ = 2 + 2s, s =−1 + 1 − , ⎪ 4 ⎪ ⎬⎪ δ = 1 + , sy , 2 ⎪ (19) α2 y ⎪ N = − 1 − s, ⎪ ⎪ 2 ⎪ ρ = 2qr, ⎪ ⎪ ⎪ 2 m2 − q2 E ⎪ y = = ⎭ 2αq 2αq where α2 y and sy are of the order of unity O(1).

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The r.h.s. of Eq. (19) is of the order of 1/y or O(α2) 10−4. As a ﬁrst step we shall ignore the r.h.s. of Eq. (19). Then h can be solved immediately: . h =. F(−N,γ,ρ). (20)

Normalizability of the radial wave function requires N equal to a positive integer: α2 y N = − 1 − s = n − 1, n = 1, 2, 3,.... (21) 2 n is the principal quantum number. Equation (20) is the relativistic extension form of the non- relativistic Coulomb wave function: ⎧ ⎨ α2 = y − − → − , (− ,γ,ρ) N 1 s n 1 F N ⎩ 2 (22) γ = 2 + 2s → 2. Downloaded from The energy eigenvalues derived from condition (21)are 2 2 1 En = 2 m − q = 2m 1 − 1 + (αy)2

http://ptep.oxfordjournals.org/ α2 = 2m 1 − (α2 y)2 + α2 α2 = 2m 1 − . (23) {2(n + s)}2 + α2 = When n 1, at CERN LIBRARY on June 21, 2013 α2 E = 2m 1 − . (24) 1 4 The approximate solution F(−N,γ,ρ) is correct, ignoring terms of the order O(1/y) α2 10−4. We write h(ρ) = F(−N,γ,ρ)+ f (ρ). (25) When n = 1, F(0,γ,ρ)= 1, f (0) = 0. (26) We shall introduce h = F + f, into Eq. (19), then we ﬁnd d2 f df γ N 1 d + −1 + + f = − − δ (F + f ), (27) dρ2 dρ ρ ρ ρ(1 + yρ) dρ whose solution gives h = F + f . Alternatively, we may use an approximate method: F is the order of unity, while f (ρ) is the order of O(1/y) O(α2),sothat f in the r.h.s. of Eq. (27) can be ignored. This equation immediately gives the solution ρ eρdρ ρ 1 d f (ρ) = e−σ F(−N,γ,σ)σγ dσ − − δ F . 2 γ 0 (F) ρ σ(1 + yσ) dσ It is difﬁcult to perform integration here, but a computer shall easily do the job.

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Equation (21) give the exact eigenvalues for our problem. The reason for this is as follows. When ρ →∞, the function h(ρ) = F(ρ) + f (ρ) behaves like (see Sect. 2.2.4) c h(ρ) = ρβ c + 1 +··· ,β= N. 0 ρ

As will be described in Sect. 2.2.4, F(ρ) contributes to ρβ, while f (ρ) does not, i.e., f (ρ) 1 O , when ρ →∞. ρβ yρ Therefore, Eq. (21) gives the exact eigenvalues. To ﬁnd f (ρ),wemayuse1/y expansion, or α2 expansion, since y is of the order of 1/α2. Introducing ∞ h(ρ) = f (ν)(ρ), Downloaded from ν=0 ( ) f 0 (ρ) = F(ρ) = F(−N,γ; ρ), N = n − 1, ∞ f (ρ) = f (ν)(ρ), http://ptep.oxfordjournals.org/ ν=1 where f ν(ρ) is of the order of (1/y)ν ∼ O(α2ν), and the differential operators 2 ( ) d γ d N D 0 = + −1 + + , dρ2 ρ dρ ρ ( ) 1 d D 1 =− − δ , at CERN LIBRARY on June 21, 2013 ρ(1 + yρ) dρ where D(ν) is of the order of (1/y)ν (here ν = 0 and 1), the differential equation (19) will give the following series of equations: ( ) ( ) D 0 f 0 = 0, ( ) (ν) ( ) (ν− ) D 0 f (ρ) = D 1 f 1 (ρ), (ν = 1, 2,...).

It is easy to see that f (0)(ρ) + f (1)(ρ) = F(ρ) + f (1)(ρ) shall be good if the terms of the order of α4 ∼ 10−8 are ignored. These techniques can be applied to other singlet and triplet cases with m = M.

1 2.2.2. f (ρ) for the ground state of S0, m = M,nearρ = 0 (ρ < 1/y). The function h(ρ) for n = 1 is given by

h(ρ) = F(0,γ,ρ)+ f (ρ) = 1 + f (ρ), f (0) = 0. (28)

The Taylor expansion of f (ρ) near ρ = 0 is given by ∞ n f (ρ) = fn ρ ; n=1

7/13 PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari the fn are equal to hn(n ≥ 1) andaregivenby(16). Noting that . α2 . s =. − ,α2 y =. 2, 8 . 1 . 1 sy =. − ,δ=. , 4 4 . . N =. 0,γ=. 2, one ﬁnds . 1 . 1 1 f = 0, f =. , f =. − yf =− y, 0 1 12 2 4 1 48 and in general . (− )n/ = y 2 fn+ . . 1 (n + 1)(n + 2)(n + 3)

∞ ∞ Downloaded from . (− )n−1 = n y 1 n f (ρ) . fnρ = ρ n(n + 1)(n + 2) 2 n=1 n=1 1 (1 + yρ)2 1 3 1 =− log + + , ( ρ)2 + ρ ρ 2y 2 y 1 y 4 2y http://ptep.oxfordjournals.org/ − 1 <ρ< 1 (ρ) ( / ) = (α2) (ρ) which holds at y y . f is of the order of O 1 y O and this yf is good within the errors of O(1/y) = O(α2). General cases n > 1 can be treated analogously.

1 2.2.3. h(ρ) for the ground state of S0, m = M,nearρ =−1/y, mathematical curiosity. Equation (14) has a regular singular point at ρ =−1/y. The regular solution near ρ =−1/y is given by ∞ at CERN LIBRARY on June 21, 2013 n h(x) = gn x , (29) n=2 = ρ − 1 where x y . = = ρ − 1 Another solution near x 0 y contains the log of solution (29). The recurrence formulae for gn are

3g3 ={2(1 + γ)y + 2 − γ }g2, (30)

−(n + 1)(n − 1)gn+1 +{n(n + γ y)δ}gn − y(n − 1 − N)gn−1. (31) Ignoring the O(1/y) terms, one ﬁnds . . 3 . g =. 2yg , g =. yg =. 3y2g . 3 2 4 2 3 2 Taking g2 = 1, . . = ny = n−1 gn+ . gn . ny ,(n > 1), 1 (n − 1) so h(ρ) turns out to be ∞ ∞ . n n−1 = n+1 n h(ρ) = gn x y . nx y n=2 n=1 2 ( + ρ)2 = x y = 1 y . 1 − yx y2ρ This expression is valid over the unphysical range −2/y <ρ<0 and good within α2.

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2.2.4. Behavior of h(ρ) at ρ →∞. ρ =∞is the irregular singular point, so that h(ρ) can be expressed in the form (17), as described in Sect. 2.2. We shall give the recurrence formulae here for the case of 1S, m = M: c c h(ρ) = ρβ c + 1 + 2 +··· , 0 ρ ρ2 δ c = −β(β + γ − 1) + c , 1 y 0

(n + 1)ycn+1 + (n − β){(n + 1 − γ − β)y − n + δ}cn + (n − 1 − β)(n − γ − 1 − β)cn−1 = 0, where

γ = 2 + 2s,

δ = 1, Downloaded from 2 α2 y β = N = − 1 − s = n − 1. 2

N has already been ﬁxed in Eq. (21). http://ptep.oxfordjournals.org/ We repeat that ρβ is the contribution from F(ρ),and f (ρ) = h(ρ) − F(ρ) does not contribute to the ρβ term. Furthermore, it is easy to show that f (ρ) 1 ∼ O when ρ →∞. ρβ yρ This shows that Eq. (21) does in fact give the correct (or exact) eigenvalues. at CERN LIBRARY on June 21, 2013

2.2.5. Comparison of the Breit case with the Dirac Coulomb solution. For simplicity we shall 2 1 only discuss the ground states of S1/2 (Dirac) and S0 (Breit). The Dirac equation with Coulomb potential [7,8] reads 1 α α∇+ βm − = E, i r (32) E = m2 − q2.

2 for S1/2, spin up, can be written as ⎛ ⎞ σ + F(r)Y 3 1 ⎜ 00 ⎟ s / spin up = ⎝ 2 ⎠ 1 2 σ + , K (r)(σrˆ) 3 1 p1/2 spin up 2 (33) 10 1 σ3 = , Y00(θ, φ) = √ , 0 −1 2π

1 σrˆ = (σr). r The differential equation for F is given by d2 F 1 α 2 dF α 2 + · + + E + − m2 F = 0. (34) 2 + α + 2 dr E r m r r dr r

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We introduce the dimensionless quantities

E + m ρ = 2qr, y = , 2αq and write F as

F(ρ) = e−ρ/2ρsh(ρ), where s is introduced to remove the 1/ρ2 terms as usual. Then the differential equation for h(ρ) is given by 2 + 2s 1 αE 1 1 1 h + h −1 + + + h − 1 − s − + sy = 0, ρ ρ(1 + yρ) q ρ 2 ρ(1 + yρ) (35) Downloaded from where E = m2 − q2, binding energy B = m − m2 − q2,andy = (E + m)/2αq.However,s is different from the Breit case: s =−1 + 1 − α2. http://ptep.oxfordjournals.org/

The form (35) in the Dirac case looks exactly the same as the Breit case (14). The only differences are given below.

Dirac Breit α2 s s =−1 + 1 − α2 s =−1 + 1 −

4 at CERN LIBRARY on June 21, 2013 αE α2 y N − 1 − s − 1 − s q 2

However, the ground-state solution of (35) in the Dirac case is

h(ρ) = F(0, 2 + 2s,ρ)= 1, for n = 1. (36)

Both αE 1 − 1 − s = 0 and + sy = 0 (37) q 2 hold, and they give the same value of q,

q = αm.

Therefore, E = m2(1 − α2) B = m − E = 1 − 1 − α2 m.

As is well known, the Dirac equation with Coulomb force can be solved for all cases of n and l in terms of two conﬂuent hypergeometric functions (multiplied by ρse−ρ/2).

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However, one cannot do N = 0 and δ = 0 simultaneously in the Breit case:

α2 y N = 0 means − 1 − s = 0, 2 then . 1 . 1 δ =. + sy =. , 2 4 while δ = 0 demands 1 α2 . − = y . 0 Downloaded from 2 8 but α2 y . =. 2 1st excited state! 2 http://ptep.oxfordjournals.org/ and N cannot be zero. From these situations, the Breit case is more complicated than the Dirac case, as described in this article.

2.2.6. Conﬂuent Heun function. The solution h(ρ) of differential equation (14) is the conﬂuent Heun function (Prof. T. Rijken kindly informed me of this fact (via his son)); this function has 5 parameters. Unfortunately, the author is not familiar with this function, so that here the physical at CERN LIBRARY on June 21, 2013 ground and 1/y expansion of the wave function h(ρ) are used.

1 2.3. S0, M = m The unequal mass case M = m can be treated in exactly the same way as in Sect. 2.2 for the case of M = m. The equation for the quantity h˜(ρ),

F + K = h˜(ρ)e−ρ/2, is given by 2 1 2(1 + yρ) h˜ + h˜ −1 + − + ρ ρ(1 + yρ) ρ{(1 + yρ)2 − (M¯ −¯m)2ρ2} 1 1 1 + yρ + h˜ − + − ρ 2ρ(1 + yρ) ρ{(1 + yρ)2 − (M¯ −¯m)2ρ2} 1 α2 (M¯ +¯m)2ρ2 (1 + yρ)2 + h˜ + 1 − − (M¯ −¯m)2 = 0, (38) 4 4 (1 + yρ)2 ρ2 where M¯ = M/2αq and m¯ = m/2αq. As in Sect. 2.2, ρ = 0 and ρ =∞in Eq. (38) are regular and irregular singular points, respectively. Therefore, we can adopt the same procedure as in Sect. 2.2.

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To remove the 1/ρ2 terms, we put h˜(ρ) = ρsh(ρ), again ﬁnding s =−1 + 1 − α2/4. The differential equation for h(ρ) reads ¯ 2 γ 1 (m¯ − M) ρ 1 + yρ h + h −1 + + + ρ (1 + yρ)2 − (m¯ − M¯ )2ρ2 1 + yρ ρ α2 y 1 + h − 1 − s 2 ρ 1 (m¯ − M¯ )2ρ 1 + yρ s s (A) ··· + + − (1 + yρ)2 − (m¯ − M¯ )2ρ2 1 + yρ ρ ρ ρ2 1 1 (m¯ − M¯ )2ρ 1 + yρ (B) ··· − + 2 (1 + yρ)2 − (m¯ − M¯ )2ρ2 1 + yρ ρ Downloaded from 1 α2 (C) ··· + + y2 − 2(m¯ 2 + M¯ 2) 4 4 α2 (m¯ 2 − M¯ 2)2 (D) ··· + ρ2 = 0. (39) 4 (1 + yρ)2 http://ptep.oxfordjournals.org/ We should notice here that (A) + (B) and (C) reduce as follows: −δ 1 (A) + (B) → ,δ= + sy, ρ(1 + yρ) 2 (C) → 0, and (D) = 0,

1 for the equal masses case, M = m. Equation (39) reduces to (14)forthe S0, m = M case. We shall rewrite the term (D) as follows: at CERN LIBRARY on June 21, 2013 α2 (m¯ 2 − M¯ 2)2 yρ 2 (D) = 4 y2 1 + yρ α2 (m¯ 2 − M¯ 2)2 yρ 2 2y (E) ··· = + 4 y2 1 + yρ ρ

2 α2 y (m¯ 2 − M¯ 2) 1 (F) ··· − . 2 y2 ρ The last term (F) shall be added to the ﬁrst line of the coefﬁcient h(ρ) in (19): 2 2 ¯ 2 2 γ α y m¯ − M 1 h + h −1 + − 2(B) + h 1 − − 1 − s ρ 2 y2 ρ

+ h[(A) + (B) + (C) + (E)] = 0. (40) (A), (B), (C),and(E) are of the order of 1/y or α2. Now we can use the technique described in Sect. 2.3 for 1S, m = M: h(ρ) = F(−N,γ,ρ)+ f (ρ), (41) 2 α2 y m¯ 2 − M¯ 2 N = 1 − − 1 − s, 2 y2

N = n − 1, n = 1, 2,..., (42)

12/13 PTEP 2013, 063B03 Y. Yamaguchi and H. Kasari where n is the principal quantum number. The background part f is O(α2) and can be found by solving γ N d f + f −1 + + f = +2(B) − (A) − (B) − (C) − (E) F(−N,γ,ρ). ρ ρ dρ Thus the accuracy of the solution F + f is quite good, and terms of the order of α4 are ignored. The other radial wave functions, F, K , G, H, G˜ ,andH˜ , can now be written down. The energy eigenvalues from Eq. (42), 2 α2 y m¯ 2 − M¯ 2 αE m2 − M2 2 1 − = 1 − = n + s ≡¯n, 2 y2 4q E2 E = M2 − q2 + m2 − q2, (43) are easily calculated. Therefore, the binding energy B = M + m − E is Downloaded from ∞ α 2m B = M + m − E = em , 2n¯ m=1

⎫ http://ptep.oxfordjournals.org/ 2mM ⎪ e = , ⎪ 1 ( + ) ⎪ M m ⎪ ⎬⎪ 2mM 3mM e = 1 − , 2 ( + ) ( + )2 ⎪ (44) M m M m ⎪ ⎪ 4mM 5mM 5(mM)2 ⎪ e = 1 − + ⎭⎪ 3 (M + m) (M + m)2 (M + m)4 at CERN LIBRARY on June 21, 2013 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)whenM = m. Knowing that F + K = e−ρ/2ρsh(ρ), all relevant radial wave functions, F, K , G, H, G˜ ,andH˜ , can easily be derived.

References [1] G. Breit, Phys. Rev. 34, 553 (1929). [2] G. Breit, Phys. Rev. 36, 383 (1930). [3] G. Breit, Phys. Rev. 39, 616 (1932). [4] H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Systems,InAtoms I/Atome I (Encyclopedia of Physics/Handbuch der Physik), ed. S. Flügge (Springer, Berlin, 1957), Vol. 7, pp. 88–436. [5] J. R. Sapirstein and D. R. Yennie, Theory of Hydrogenic Bound States. In Quantum Electrodynamics (Advanced Series on Directions in High Energy Physics), ed. T. Kinoshita (World Scientiﬁc, 1990), Vol. 7, pp. 560–672. [6] V.W. Hughes and G. Zu Putlitz, Muonium. In Quantum Electrodynamics (Advanced Series on Directions in High Energy Physics), ed. T. Kinoshita (World Scientiﬁc, 1990), Vol. 7, pp. 826–904. [7] C. G. Darwin, Proc. R. Soc. London, Ser. A 115, 654 (1928). [8] W. Gordon, Z. Phys. 48, 11 (1928).

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