arXiv:1008.4224v4 [math.AP] 5 Nov 2010 h ytm stebnigeeg sqatzd h u fi n the and it bin of the sum gives the light quantized, of is speed energy the binding of the square As syste the system. the and the of ever defect of defect mass mass mass rest the the the gives of a of difference sum of the the and mass than system, less the is the state that, bound indicate a in experiments system and relativity special Both Introduction 1 olg fPyisadEetoi nomto,Yna Univ Yan’an Information, Electronic and Physics of College PACS: Keywords: colo lcrncadIfrainEgneig UA Be BUAA, Engineering, Information and Electronic of School define odneuto ntesm oeta edcnas esolved be also can method. field ar same potential potentials the same vector the sol and in scalar be equation the Gordon can pr that field we condition potential the potentials, under certain vector and a scalar in mas Schr¨odinger system the equation of with concept equation the h introducing Gordon we By comparison, equation. For Dirac the significance. probability eq the Klein-Gordon have the field of solutions transpl stationary the to w compa that relativistic proves us when the allows into form Schr¨odinger equation the only better of proach not a which in limit, expressed non-relativistic be clearly can equation r tt aefntosa pca ouin ieΨ( defi like we solutions If special reference. as combines functions of which wave frames strictly, state inertial laws ary two relevant in of functions set wave a and equation ese oitoueamteaia ehdt eieteKlei the derive to method mathematical a introduce to seek We li-odnEuto na in Equation Klein-Gordon ttoaySltoso the of Solutions Stationary m 36.m 03.65.Ge 03.65.Pm, = E/c eaiitcwv qain;KenGro qain on state Bound equation; Klein-Gordon equations; wave Relativistic 2 hc scle h aso h ytm hnteKlein-Gordo the then system, the of mass the called is which , a’n hax,Cia 716000 China, Shaanxi, Yan’an, oeta Field Potential [email protected] [email protected] unqn Bi Guangqing hn,100191 China, ukiBi Yuekai Abstract 1 r t , v qain,btalso but equations, ave = ) aini potential a in uation n h ovn ap- solving the ant qa,teKlein- the equal, e v lodiscussed also ave e xcl,then exactly, ved ψ ,wieteproduct the while m, noteKlein- the into s xcl yusing by exactly ( etestation- the ne v hti the if that ove h relativistic the r ) e ihthe with red atceforming particle y e − iEt/ n-Gordon igeeg of energy ding many-particle etms of mass rest h ¯ ersity, ijing, and , n every particle forming the system is the energy level of the system. For instance, the mass of an atomic nucleus is obviously less than the sum of the rest mass of every nucleon forming the atomic nucleus. Therefore, in order to express the mass defect explicitly, there is a necessity to introduce the concept of system mass, which differs from the sum of the rest mass of every particle forming the system. By introducing this concept, we can express relativistic wave equations in a better form when compared with the non-relativistic limit. By means of this method, we are able to solve the Klein-Gordon equation just like solving the Schr¨odinger equation, and the solutions can also have the probability significance just like the solutions of the Schr¨odinger equation. As the Schr¨odinger equation has been thoroughly studied, it surely creates conditions for studying the Klein-Gordon equation thoroughly.
2 Relativistic Wave Equations and the Probability Significance of Their Stationary Solutions
As we know, for a free particle, its energy and momentum are both constant, therefore, it is quite natural to assume that its matter wave is a plane wave. According to the de Broglie relation ¯hk = p, E =¯hω. We have the wave function of a free particle:
i − h (Et−p r) Ψp(r,t)= Ae ¯ . (1)
2 Where E is the total energy of the particle containing the intrinsic energy m0c , and p is the momentum of the particle. The quantization method of quantum mechanics is assuming E and p are corre- spondingly equivalent to the following two differential operators: ∂ E → i¯h , p →−i¯h∇. ∂t This relation is obviously tenable for the wave function of a free particle, while ar- bitrary wave function is equal to the linear superposition of the plane waves of free particles with all possible momentum, i.e. ∞ 1 − i (Et−p r) Ψ(r,t) = c(p,t)e h¯ dp dp dp , (2) (2π¯h)3/2 x y z −∞ ∞ 1 i (Et−p r) c(p,t) = Ψ(r,t)e h¯ dx dy dz. (3) (2π¯h)3/2 −∞ Therefore, this quantization method is also tenable for arbitrary wave functions. Later, we will prove the right-hand side of (1) is relativistic, thus this quantiza- tion method itself is relativistic. Then why the Schr¨odinger equation derived from
2 this method is non-relativistic? That is because the relation between E and p is non-relativistic, which is due to the fact that the kinetic energy is non-relativistic. Therefore, if we want to establish the relativistic wave equations, we need to introduce the relativistic kinetic expression.
Assuming that any particle with the rest mass m0, no matter how high the speed is, no matter it is in a potential field or in free space, and no matter how it interacts with other particles, its kinetic energy is:
2 2 2 4 1/2 2 Ek = (c p + m0 c ) − m0c . (4)
2 If Ek + m0c is denoted by E, then (4) can be expressed as
2 2 2 2 4 E = c p + m0 c .
Thus, it leaves us a matter of mathematical techniques. Therefore, we introduce the mathematical method in reference [1]-[4]. Power functions and exponential functions play special roles in this method, which are called the base functions as we can establish mapping relations between them and arbitrary functions in a certain range. For instance, in quantum mechanics, (2) determines the mapping relations between wave functions of free particles and arbitrary wave functions. Similarly to reference [1]- [4], we can introduce the base function and relevant concepts in quantum mechanics: Definition 1 The right-hand side of (1) is defined as the base function of quantum mechanics, where E and p are called the characters of base functions, while E and p are not only suitable for free particles, but also suitable for any system, and the relation between E and p is called the characteristic equation of wave equations. Different system has different characteristic equations. For instance, E2 = c2p2 + 2 4 m0 c is the characteristic equation for free-particle system. According to differential laws, we have ∂ i¯h Ψ = EΨ , −i¯h∇Ψ = pΨ . (5) ∂t p p p p
Definition 2 Let m0 = m01 + m02 + + m0N be the total rest mass of an N-particle system, E′ be the sum of the kinetic energy and potential energy of all the N particles, then the actual mass of the system, which is called the system mass, is defined as 1 m = m + E′. (6) 0 c2 Definition 3 If the system is in a bound state(E′ < 0), then the absolute value of E′ is ′ 2 2 2 |E | = m0c − mc = △mc ,
which is called the binding energy of the system, where △m = m0 − m is the mass defect of the system. Definition 4 The total energy of the system E is defined as the sum of the rest energy, kinetic energy and potential energy of all the particles forming the system, 2 ′ i.e. E = m0c + E .
3 According to Definition 2 and 4, the total energy of the system is equal to the product of the system mass and the square of the speed of light, i.e. E = mc2, thus the system mass is uniquely determined by the energy level of the system. Definition 5 In relativistic quantum mechanics, the stationary state wave func- tion is defined as the following special solution, i.e.
Ψ(r,t)= ψ(r)e−iEt/h¯ . (7)
Let U(r) be the potential energy of a particle moving in an potential field, then ′ according to Definition 2, we have Ek = E − U. Combining it with (4), we have
′ 2 2 2 4 1/2 2 E − U = (c p + m0 c ) − m0c . (8)
This is the characteristic equation of relativistic wave equations for a single-particle system. In order to make it easier to solve the corresponding relativistic wave equa- tion, the characteristic equation (8) should be transformed to remove the fractional power, then we have ′ 2 2 2 2 2 4 (E − U + m0c ) = c p + m0 c . (9) Expanding the left-hand side of (9) and by using (6), we have
2 2 ′ p 2m U E = + U − 2 , m0 + m m0 + m (m0 + m)c 2 2 p 2m U 2 E = + U − 2 + m0c . (10) m0 + m m0 + m (m0 + m)c
The essence of this expression is the relativistic Hamiltonian. Therefore, taking (10) as the characteristic equation, multiplying both sides of the equation by the base function Ψp(r,t), and by using (5), we have
2 2 ∂Ψp ¯h 2 2m U 2 i¯h = − ∇ Ψp + UΨp − 2 Ψp + m0c Ψp. ∂t m0 + m m0 + m (m0 + m)c
According to (2), in the operator equation which is tenable for the base function Ψp, as long as each operator in the operator equation is a linear operator and each linear
operator does not explicitly contain the characters E and p of Ψp, then this operator equation is also tenable for an arbitrary wave function Ψ(r,t). Whereas, considering that the system mass m is equivalent to the character E of Ψp, this operator equation is not tenable for arbitrary wave functions, but tenable for an stationary state wave function like (7). Thus we have:
A particle with the mass m0 moving in the potential field U(r) can be described by the wave function Ψ(r,t), an arbitrary stationary state wave function Ψ satisfies the following relativistic wave equation
2 2 ∂Ψ ¯h 2 2m U 2 i¯h = − ∇ Ψ+ UΨ − 2 Ψ+ m0c Ψ. (11) ∂t m0 + m m0 + m (m0 + m)c
4 The corresponding ψ(r) satisfies the following relativistic wave equation
2 2 ¯h 2 2m U 2 Eψ = − ∇ ψ + Uψ − 2 ψ + m0c ψ. (12) m0 + m m0 + m (m0 + m)c
Or, if the energy corresponding to the energy operator is interpreted as E′ in Definition 2, then the relativistic wave equation is
2 2 ∂Ψ ¯h 2 2m U i¯h = − ∇ Ψ+ UΨ − 2 Ψ. (13) ∂t m0 + m m0 + m (m0 + m)c The corresponding stationary form is
2 2 ′ ¯h 2 2m U E ψ = − ∇ ψ + Uψ − 2 ψ. (14) m0 + m m0 + m (m0 + m)c
2 ′ As E − m0c = E , (11) and (13) are equivalent. In the non-relativistic limit, we have
U 2 m → m0, 2 → 0, (m0 + m)c
(13) approaches the Schr¨odinger equation. Let w(r,t)=Ψ∗(r,t)Ψ(r,t), differentiating both sides of the equation with respect to time t, and considering
2 ∂Ψ i¯h 2 1 2m 1 U 1 2 = ∇ Ψ+ UΨ − 2 Ψ+ m0c Ψ, ∂t m0 + m i¯h m0 + m i¯h (m0 + m)c i¯h
∗ 2 ∂Ψ i¯h 2 ∗ 1 2m ∗ 1 U ∗ 1 2 ∗ = − ∇ Ψ − UΨ + 2 Ψ − m0c Ψ . ∂t m0 + m i¯h m0 + m i¯h (m0 + m)c i¯h We have the probability conservation equation, i.e. ∂w i¯h i¯h = (Ψ∗∇2Ψ − Ψ∇2Ψ∗)= ∇ (Ψ∗∇Ψ − Ψ∇Ψ∗)= −∇ J. ∂t m0 + m m0 + m In other words, w(r,t)=Ψ∗(r,t)Ψ(r,t) can be interpreted as the probability density. Thus we have Let w(r,t)=Ψ∗(r,t)Ψ(r,t) be the probability density, then the corresponding probability current density vector is i¯h J = (Ψ∇Ψ∗ − Ψ∗∇Ψ), (15) m0 + m making the following equation tenable, i.e. ∂w + ∇ J =0. (16) ∂t Here Ψ(r,t) is a stationary state wave function like (7), determined by wave function (11) or (13) and natural boundary conditions.
5 Now, Let us introduce the concept of system mass into the Dirac equation, so as to compare it with (14). According to reference [5], we can express the Dirac equation as the following two equations:
2 cσ pψ2 + µ0c ψ1 + Uψ1 = Eψ1,
p 2 cσ ψ1 − µ0c ψ2 + Uψ2 = Eψ2. 2 Considering E = µc , solve the second equation for ψ2, then substitute the result into the first equation, thus we have c2 p p 2 (σ ) 2 (σ )ψ1 + µ0c ψ1 + Uψ1 = Eψ1. (µ0 + µ)c − U Where c2 p p (σ ) 2 (σ )ψ1 (µ0 + µ)c − U c2 c2 = (σ p)(σ p)ψ + (σ p) (σ p)ψ . (µ + µ)c2 − U 1 (µ + µ)c2 − U 1 0 0 2 ′ 2 As (σ p)(σ p) = p , E = E + µ0c , the equation can be transformed into the following form: c2p2 c2 E′ψ = ψ + Uψ + (σ p) (σ p)ψ . 1 (µ + µ)c2 − U 1 1 (µ + µ)c2 − U 1 0 0 As it is known, for any scalar function U, we have
(σ pU)(σ p)ψ1 = {−i¯h(∇U p)+¯hσ[(∇U) × p]} ψ1.
If U is simply a function of r, then 1 dU ∇U = r, r dr dU ∂ψ −i¯h(∇U p)ψ = −¯h2∇U ∇ψ = −¯h2 1 , 1 1 dr ∂r 1 dU 1 dU (∇U) × p = (r × p)= L. r dr r dr Thus the equation can be expressed as
2 2 ′ c p E ψ1 = 2 ψ1 + Uψ1 (µ0 + µ)c − U c2 dU 2 ∂ψ + S Lψ − ¯h2 1 . [(µ + µ)c2 − U]2 dr r 1 ∂r 0 Where S = (1/2)¯hσ is the spin angular momentum operator. Multiplying both sides 2 ′ 2 2 2 of the equation by (µ0 + µ)c − U, noting that E = E − µ0c = µc − µ0c , we have ′ 2 2 2 2 2 E (µ0 + µ)c ψ1 = c p ψ1 +2µc Uψ1 − U ψ1 c2 dU 2 ∂ψ + S Lψ − ¯h2 1 . (µ + µ)c2 − U dr r 1 ∂r 0
6 Therefore, the Dirac equation can be expressed as the following form:
2 2 ′ ¯h 2 2µ U E ψ = − ∇ ψ + Uψ − 2 ψ µ0 + µ µ0 + µ (µ0 + µ)c 1 dU 2 ∂ψ (17) + S Lψ − ¯h2 . (µ + µ)[(µ + µ)c2 − U] dr r ∂r 0 0
Where µ is the system mass corresponding to µ0, S is the spin angular momentum operator, L is the orbital angular momentum operator, and U is simply a function of r.
3 The System Mass of Pionic Hydrogen Atoms and Relativistic Wave Functions
A pionic hydrogen atom is a system formed by a negative pion and an atomic nucleus. Reference [6] expounded the relativistic energy levels of such a system. Let us take the pionic hydrogen atom for example. We can transplant the mathematical methods of the non-relativistic quantum mechanics in reference [7] into the relativistic quantum mechanics, thus determine the system mass and relativistic wave functions of such a system. − Considering a negative pion π with the rest mass m0 moving in an nuclear electric field, taking the atomic nucleus as the origin of coordinates, then the potential energy 2 −1/2 of the system is U = −Zes/r, es = e(4πε0) , according to (14), the relativistic stationary state wave equation of the system is
2 2 2 4 ′ ¯h 2 2m Zes 1 Z es E ψ = − ∇ ψ − ψ − 2 2 ψ. m0 + m m0 + m r (m0 + m)c r The equation expressed in spherical polar coordinates is
2 2 ′ ¯h ∂ 2 ∂ 1 ∂ ∂ 1 ∂ E ψ = − r + sin θ + 2 2 ψ m0 + m ∂r ∂r sin θ ∂θ ∂θ sin θ ∂ϕ 2 2 4 2m Zes 1 Z es − ψ − 2 2 ψ. m0 + m r (m0 + m)c r By following the solving procedure of the Schr¨odinger equation in reference [7], we use the method of separation of variables to solve this equation. Let ψ(r, θ, ϕ)= R(r)Y (θ, ϕ), by substituting it into the equation, we have
(m + m)E′r2 1 ∂ ∂R Ze 2 Z2e 4 0 + r2 +2m s r + s ¯h2 R ∂r ∂r ¯h2 ¯h2c2 1 1 ∂ ∂Y 1 ∂2Y = − sin θ + = λ. Y sin θ ∂θ ∂θ sin2 θ ∂ϕ2 1 ∂ ∂R (m + m)E′ 2m Ze 2 Z2e 4 1 r2 + 0 R + s + s − λ R =0. (18) r2 ∂r ∂r ¯h2 ¯h2 r ¯h2c2 r2
7 1 ∂ ∂Y 1 ∂2Y sin θ + + λY =0. (19) sin θ ∂θ ∂θ sin2 θ ∂ϕ2 According to (19), denoting λ = l(l + 1), l =0, 1, 2,..., obviously the solution of the
equation is the spherical harmonics Ylm(θ, ϕ). Now let us solve the radial equation (18), discussing the situation of the bound state (E′ < 0). Let
4(m + m)|E′| 1/2 2mZe 2 α′ = 0 , β = s . (20) ¯h2 α′¯h2 Let R(r)= u(r)/r, considering
1 d dR 1 d2 r2 = (rR), r2 dr dr r dr2 and by using the variable substitution ρ = α′r, then the equation (18) can be expressed as (α is the fine structure constant, which is absolutely different from α′):
d2u β 1 l(l + 1) − Z2α2 + − − u =0. (21) dρ2 ρ 4 ρ2 Firstly, let us study the asymptotic behaviour of this equation, when ρ → ∞, the equation can be transformed into the following form:
d2u 1 − u =0, u(ρ)= e±ρ/2. dρ2 4
As eρ/2 is in conflict with the finite conditions of wave functions, we substitute u(ρ)= e−ρ/2f(ρ) into the equation, then we have the equation satisfied by f(ρ):
d2f df β l(l + 1) − Z2α2 − + − f =0. (22) dρ2 dρ ρ ρ2 Now let us solve this equation for series solutions. Let
∞ s+ν f(ρ)= bν ρ , b0 =0. (23) ν=0 In order to guarantee the finiteness of R = u/r at r = 0, s should be no less than 1. By substituting (23) into (22), as the coefficient of ρs+ν−1 is equal to zero, we have the relation satisfied by bν : s + ν − β b = b . (24) ν+1 (s + ν)(s + ν + 1) − l(l +1)+ Z2α2 ν
If the series are infinite series, then when ν → ∞ we have bν+1/bν → 1/ν. Therefore, when ρ → ∞, the behaviour of the series is the same as that of eρ, then we have
α′ α′ R = u(ρ)= e−ρ/2f(ρ). ρ ρ
8 Where f(ρ) tends to infinity when ρ → ∞, which is in conflict with the finite con- ditions of wave functions. Therefore, the series should only have finite terms. Let s+nr bnr ρ be the highest-order term, then bnr+1 = 0, by substituting ν = nr into (24)
we have β = nr + s. On the other hand, the series starts from ν = 0, and doesn’t
have the term ν = −1, therefore, b−1 = 0. Substituting ν = −1 into (24), considering 2 2 b0 = 0, we have s(s−1) = l(l +1)−Z α . Denoting n = nr +l +1, then the following set of equations can be solved for s and β:
s(s − 1) = l(l + 1) − Z2α2 β = nr + s (25) n = nr + l +1.
We have s =1/2 ± (l +1/2)2 − Z2α2, taking s =1/2+ (l +1/2)2 − Z2α2, then