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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) δ + δ ( ~α ~p + β M) δ δ Ψ = EΨ . (1) 1 1 µν ρσ µν − 2 2 ρσ − r µν ρσ νσ µρ   This equation holds in the CM system. The total energy of the system E can be expressed as

E = m2 q2 + M 2 q2. (2) − − p p The momentum operator in (1) can be written as

1 1 ∂ ~p = ~ = , (3) i ∇ i ∂~r 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 e2/r = α/r. ~α , β (j = 1, 2) are the usual 4 4 Dirac − − j j × matrices for particle 1 ( e, m) and 2 (+e, M), respectively. Ψ is 4 4 Dirac wave − αβ × function. We shall try to solve

1 1 α (~α ~ + β m) + ( ~α ~ + β M) Ψ= EΨ, (4) 1 i ∇ 1 − 2 i ∇ 2 − r   and find the eigen-value(s)

E = m2 q2 + M 2 q2, (2) − − p p which is specified by 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 angular parts of the large- large components Ψαβ (α, β = 1, 2) to be

1(l) , (l = 0, 1, 2, ) l · · · 3(l 1) +3 (l + 1) , (l = 1, 2, ) − l l · · · and 3(l) , (l = 1, 2, ) l · · · where l is the quantum number for the orbital angular momentum. Spin-angular parts of other (large-small, small-large, small-small) componentsofΨαβ 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 1 2.1. (l)l, general 1 We shall first discuss the simplest case of (l)l.Ψαβ can be taken as follows:

α β Ψαβ 1 1 F (r) 1(l) > 2 2 | l 1 3 i G(r) 3(l + 1) > +G˜(r) 3(l 1) > 2 4 { | l | − l } (5) 3 1 i H(r) 3(l + 1) > +H˜ (r) 3(l 1) > 4 2 { | l | − l } 3 3 K(r) 1(l) > 4 4 | l where 1(l) > and 3(l 1) > are normalized spin-angular wave functions (z-component of | l | ± l the total angular momentum has been omitted). The Breit equation (4) in 1 gives the fol- § lowing set of differential equations for the radial wave functions F (r), K(r), G(r), G˜(r),H(r) and H˜ (r):

α E + m M F r − −   l + 1 d l + 2 l d l 1 = + (G + H) − (G˜ H˜ ), 2l + 1 dr r − 2l + 1 dr − r − r   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 h i h i α l + 1 d l E + + m M H = (F + K) r − − 2l + 1 dr − r r    α = E + m + M G, r −   α l d l + 1 E + + m M H˜ = + (F + K) r − 2l + 1 dr r r    α = E + m + M G.˜ (6) 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 −     3/14 For later convenience, we introduce the dimensionless quantities: ρ = 2qr,

E M 2 q2 + m2 q2 y = = − − (> 0), 2αq p 2αqp M + m + E λ = (> 0), 2αq M + m E ν = − (> 0). (7) 2αq If the eigen-value E is a proper relativistic generalization of the Schr¨odinger case (as we shall see it is the case), y and λ are large quantities of the order of 1/α2, while ν is the order of unity. For ρ (r ) all radial wave function should behave like →∞ →∞ n ρ/2 F (ρ) ρ 1+ O(1/ρ) e− . (8) → { } ρ/2 qr 2 2 2 2 This is because e− = e− and E = m q + M q . Notice that if we consider 2 2 2 −2 − E, the total CM energy m + p + Mp + p , to bep larger than M + m, we would expect ipr that the radial wave functionsp shouldp contain a factor e . If we analitically continue p to qr imaginary values we get the factor e− and the energy expression (2) for bound states. The singlet case becomes very simple when two masses are equal M = m. First H = G, H˜ = G,˜  (9) 1 νρ K = − F  1+ λρ   Then, the combination  ρ/2 F + K = h˜(ρ)e− (10) obeys the differential equation 2 1 h˜′′ + h˜′ 1+ + − ρ ρ(1 + yρ)   α2y 1 α2 1 l(l + 1) 1 +h˜ 1 + = 0. 2 − ρ 4 ρ2 − ρ2 − 2ρ(1 + yρ)    1 2.2. S0, M = m 1 The S0, i.e., l = 0, case we have the equation for h˜ 2 1 h˜′′ + h˜′ 1+ + − ρ ρ(1 + yρ)   α2y 1 α2 1 1 +h˜ 1 + = 0. (11) 2 − ρ 4 ρ2 − 2ρ(1 + yρ)    If we ignore α2 and higher order terms in (11) and notice that α2y is of the order of unity, we find 2 α2y 1 h˜′′ + h˜′ 1+ + h˜ 1 = 0, (12) − ρ 2 − ρ    

4/14 which is precisely the Schr¨odinger equation for our issue and h˜ in this approximation is given 2 2 by F ( α y + 1, 2, ρ)= F (1 n, 2, ρ) apart from the normalization constant. α y 1 must − 2 − 2 − 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 = 0 h˜(ρ) can be expressed as

h˜(ρ)= ρsh(ρ),  (13) α2  where s = 1+ 1 .  − r − 4  2  Another solution s = 1 1 α is unacceptable from the square integrability of the − − − 4 wave function Ψαβ. h(ρ) obeysq the equation: 2 + 2s 1 h′′ + h′ 1+ + − ρ ρ(1 + yρ)   α2y 1 1 1 +h 1 s + sy = 0. (14) 2 − − ρ − 2 ρ(1 + yρ)      h(ρ) can be expanded in a Taylor series near ρ = 0:

∞ n h(ρ)= hnρ , (15) n=0 X 1 α2y 1 h = +1+ s + + sy h , 1 3 + 2s − 2 2 0   1 (n + 1)(n +3+2s) h y n+1 1 α2y 1 + n(n +1+2s) (n +1+ s + + sy) h − y − 2 2 n   α2y n 1 +1+ s hn 1 = 0. (16) − − − 2 −   It can be shown that h(ρ) can not be a polynomial:

N h(ρ)= h ρn (N < ) n ∞ n=0 X from the recurrence formulae (16). So the sum in equation (15) must extend to n = , a ∞ sharp difference from the Schr¨odinger case. Note that the expansion (15) holds only the tiny region ρ< 1/y O(α2). ∼ Next we discuss on ρ = , which is an irregular singular point. Assuming the form ∞ c c h(ρ)= eλρρk c + 1 + 2 + at ρ (17) 0 ρ ρ · · · →∞  2  where λ, k, cn 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< . ∞

5/14 k turns out to be α2y k = 1 s (18) 2 − − appeared in (14), whose value is not fixed from the situation at ρ . Further discussion →∞ will be given in 2.2.4. § 1 2.2.1. Detailed discussion on S0, m = M. Note again ρ/2 s F + K = e− ρ h(ρ). h(ρ) obeys the diffrential equation, (14), d2h dh γ N 1 d + 1+ + h = δ h, dρ2 dρ − ρ ρ −ρ(1 + yρ) dρ −           α2  where γ =2+2s, s = 1+ 1 4 ,  − −  q   1  δ = 2 + sy,   (19)  2  N = α y 1 s, 2 − −   ρ = 2qr,     2√m2 q2 E  y = − = ,  2αq 2αq   2  α 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(α2) 10 4. As the first step we shall ≃ − ignore the RHS of (19). Then h can immediately solved: . h =. F ( N, γ, ρ). (20) − Normalizability of the radial wave function requires N equals to positive integer: α2y N = 1 s = n 1, n = 1, 2, 3, . (21) 2 − − − · · · n is the pricipal quantum number. (20) is the relativistic extension form of the non- relativisitic Coulomb wave function 2 N = α y 1 s n 1, 2 − − → − F ( N, γ, ρ) (22) −   γ =2+2s 2. → The energy eigenvalues derived from the condition (21) are

2 2 1 En = 2 m q = 2m 1 − s − 1 + (αy)2 p α2 = 2m 1 s − (α2y)2 + α2

α2 = 2m 1 . (23) − 2(n + s) 2 + α2 s { }

6/14 When n = 1,

α2 E1 = 2m 1 . (24) r − 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 the eq. (19), then we find

d2f 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. 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(α2), so that f in the RHS of (27) can be ignored. This equation ≃ gives immediately the solution

ρ ρ ρ e dρ σ γ 1 d f(ρ)= e− F ( N,γ,σ)σ dσ δ F . (F )2ργ − −σ(1 + yσ) dσ − Z0 Z     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 ) →∞ § c h(ρ)= ρβ c + 1 + , β = N 0 ρ · · ·   as will be described in 2.2.4 F (ρ) contributed to ρβ, while f(ρ) does not, i.e., § f(ρ) 1 O , when ρ . ρβ ≃ yρ →∞  

Therefore (21) gives the exact eigenvalues. To find f(ρ), we may use 1/y expansion, or α2 expansion, since y is of the order of 1/α2.

7/14 Introducing

∞ h(ρ) = f (ν)(ρ), ν=0 X f (0)(ρ) = F (ρ)= F ( N, γ; ρ),N = n 1, − − ∞ f(ρ) = f (ν)(ρ), ν=1 X where f ν(ρ) is of the order of (1/y)ν O(α2ν ), and the differential operators ∼ d2 γ d N D(0) = + 1+ + , dρ2 − ρ dρ ρ   1 d D(1) = δ , −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 serries of equations

D(0)f (0) = 0, (0) (ν) (1) (ν 1) D f (ρ) = D f − (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. 6 1 2.2.2. f(ρ) for the ground state of S0, m = M, near ρ = 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), (28) f(0) = 0. The Taylor expansion of f(ρ) near ρ = 0 is given by

∞ n f(ρ)= fn ρ n=1 X f ’s are equal to h (n 1) and given by (16). Noting that n n ≥ . 2 . s =. α , α2y =. 2, − 8 . . sy =. 1 , δ =. 1 , − 4 4 . . N =. 0, γ =. 2, one finds . 1 . 1 1 f0 = 0, f1 =. , f2 =. yf1 = y, 12 − 4 −48

8/14 and in general

. ( y)n/2 fn+1 =. − . (n + 1)(n + 2)(n + 3)

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

2.2.3. h(ρ) for the ground state of 1S , m = M, near ρ = 1/y, mathematical curiosity. 0 − The eq. (14) has the regular singular point at ρ = 1/y. The regular solution near ρ = 1/y − − is given by

∞ n h(x)= gnx , (29) n=2 X where x = ρ 1 . − y Another solution near x =0= ρ 1 contains log of the solution (29). − y The recurrence formulae for gn’s are

3g = 2(1 + γ)y + 2 γ g , (30) 3 { − } 2

(n + 1)(n 1)gn+1 + n(n + γy)δ gn y(n 1 N)gn 1. (31) − − { } − − − − Ignoring O(1/y) terms, one finds

. . 3 . 2 g3 =. 2yg2, g4 =. yg3 =. 3y g2. 2

Taking g2 = 1,

. ny . n 1 gn+1 =. gn =. ny − , (n> 1). (n 1) − So that h(ρ) turn out to be

∞ n n 1 . ∞ n+1 n h(ρ) = gnx y − =. nx y n=2 n=1 X X x2y (1 + yρ)2 = = . 1 yx y2ρ − This expression is valid at the unphysical range 2/y<ρ< 0 and good within α2. −

9/14 2.2.4. Behavier of h(ρ) at ρ . ρ = is the irreguler singular point, so that h(ρ) can →∞ ∞ be expressed in the form (17) as described in 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)yc + (n β) (n + 1 γ β)y n + δ c n+1 − { − − − } n + (n 1 β)(n γ 1 β)cn 1 = 0, − − − − − − where

γ = 2+2s, 1 δ = , 2 α2y β = N = 1 s = n 1. 2 − − − N has already been fixed in Eq. (21). We repeat that ρβ is the contribution from F (ρ) and f(ρ)= h(ρ) F (ρ) does not − contribute to ρβ term. Furthermore it is easy to show f(ρ) 1 O when ρ . ρβ ∼ yρ →∞   This shows (21) is in fact the correct (or exact) eigenvalues.

2.2.5. Comparison of Breit case with Dirac Coulomb solution. For simplicity we shall 2 1 discuss only ground states of S1/2 (Dirac) and S0 (Breit). with Coulomb potential[4] reads

1 ~α~ + βm α Ψ= EΨ, i ∇ − r n o (32) E = m2 q2. − 2 Ψ for S1/2, spin up, can be written p

F (r)Y σ3+1 s spin up Ψ= 00 2 1/2 (33) ˆ σ3+1 K(r)(~σ~r) 2 ! p1/2 spin up,

1 0 1 σ3 = , Y00(θ, φ)= , 0 1 √2π − ! 1 ~σ~rˆ = (~σ~r). r The differential equation for F is given by

2 2 d F 1 α 2 dF α 2 2 + α 2 + + E + m F = 0. (34) dr E + r + m · r r dr r −     

10/14 Introducing dimensionless quantities E + m ρ = 2qr, y = , 2αq and write F to be

ρ/2 s F (ρ)= e− ρ h(ρ), where s is introduced to remove 1/ρ2 terms as usual. Then the differential equation for h(ρ) is given by 2 + 2s 1 h′′ + h′ 1+ + − ρ ρ(1 + yρ)   αE 1 1 1 + h 1 s + sy = 0, (35) q − − ρ − 2 ρ(1 + yρ)      where E = m2 q2, binding energy B = m m2 q2, y = (E + m)/2αq. However s is − − − different fromp Breit case p s = 1+ 1 α2. − − p The form (35) in the Dirac case looks exactly the same as the Breit case (14). Only differences are in below.

Dirac Breit α2 s s = 1+ 1 α2 s = 1+ 1 − − − r − 4 αE p α2y 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)

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

q = αm.

Therefore

E = m2(1 α2) − p B = m E = 1 1 α2 m. − − − n p o As is well-known, Dirac equation with Coulomb force can be solved for all cases of n and s ρ/2 l in terms of two confluent hypergeomotric functions (multiplied by ρ e− ).

11/14 However one cannot do

N =0 and δ = 0 simultaneously in the Breit case.

α2y N =0 means 1 s = 0. 2 − − then . 1 . 1 δ =. + sy =. . 2 4 While δ = 0 demands

1 α2 . y =. 0 2 − 8 but

α2y . =. 2 1st exiced state! 2 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 conflu- ent 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).

2.3. 1S , M = m 0 6 Unequal mass case M = m can be treated in exactly the same way as in 2.2 for the case of 6 § M = m. The equation for the quantity h˜(ρ):

ρ/2 F + K = h˜(ρ)e− 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 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. §

12/14 To remove the 1/ρ2 terms, put

h˜(ρ)= ρsh(ρ),

finding again s = 1+ 1 α2/4. The differential equation for h(ρ) reads − − p γ 1 (m ¯ M¯ )2ρ 1+ yρ h +h′ 1+ + − + ′′ − ρ (1 + yρ)2 (m ¯ M¯ )2ρ2 1+ yρ ρ  − −   α2y 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  − −   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  We should notice here that (A)+(B) and (C) reduce δ 1 (A) + (B) − , δ = + sy, → ρ(1 + yρ) 2

(C) 0, and (D) = 0, → 1 when the equal masses case, M = m. (39) reduces to (14) for the case, S0,m = M. We shall rewrite the term (D) as follows:

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   "  # α2y (m ¯ 2 M¯ 2) 2 1 (F ) − . ··· − 2 y2 ρ   The last term (F ) shall be added to the first line of th coefficient h(ρ) in (19): γ h′′ + h′ 1+ 2(B) − ρ −   α2y m¯ 2 M¯ 2 2 1 + h 1 − 1 s 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 same technique described in 2.3, 1S, m = M. § h(ρ) = F ( N, γ, ρ)+ f(ρ), (41) −

13/14 α2y m¯ 2 M¯ 2 2 N = 1 − 1 s, 2 − y2 − − (   ) N = n 1, n = 1, 2, . (42) − · · · n is the principal quantun 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, terms of the order of α4 are ignored. Other radial wave functions, F,K,G,H, G˜ and H˜ can now be writen down. The energy eigenvalues from (42)

α2y m¯ 2 M¯ 2 2 αE m2 M 2 2 1 − = 1 − = n + s n,¯ 2 − y2 4q − E2 ≡ (   ) (   )

E = M 2 q2 + m2 q2 (43) − − is easily calculated. Therefor the binding energyp B = Mp+ m E is − ∞ α 2m B = M + m E = e , − m 2¯n m=1 X   2mM e = , 1 (M + m)   2mM 3mM  e = 1 ,  2 2  (44) (M + m) − (M + m)     2 4mM 5mM 5(mM)  e3 = 1 +  (M + m) − (M + m)2 (M + m)4      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.

ρ/2 s Knowing F + K = e− ρ h(ρ), all relevant radial wave functions, F,K,G,H, G˜ and H˜ can easily be derived. References [1] G. Breit Phys. Rev. 34 (1929) 553, 36 (1930) 383, 39 (1932) 616. [2] H.A. Bethe and E.E. Salpeter, of One- and Two- Systems, p. 88-436, Bd XXXV Atom I, Handbuch der Physik, ed. by S. Fl¨ugge (1957), esp. p. 175-203 and p. 256-290. [3] J.R. Sapirstein and D. R. Yennie, Theory of the Hydrogentic Bound States, p. 560-672, and V.W. Hughes and G. Zu Putlitz. Muonium, p. 827-904, esp. II. 2 Breit equation, p. 826. Both in , ed. by T. Kinoshita, (Advanced Series on Direcctions in High Energy Physics, vol. 7), 1990 World Scientific. [4] C. G. Darwin, Proc. Roy. Soc. London(A) 115 (1928) 654. W. Gordon, Zs. f. Phys. 48 (1928) 11. [5] Prof. T. Rijken was kindly informed me of this fact (according to his son).

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