# Exact Solution of the Dirac Equation for the Yukawa Potential with Scalar and Vector Potentials and Tensor Interaction

Authors: Mona Azizi; Nasrin Salehi; Ali Akbar Rajabi
Journal: ISRN High Energy Physics
 (2013)

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## Abstract

We present exact solutions of the Dirac equation with Yukawa potential in the presence of a Coulomb-like tensor potential. For this goal we expand the Yukawa form of the nuclear potential in its mesonic clouds by using Taylor extension to the power of seventh and bring out its simple form. In order to obtain the energy eigenvalue and the corresponding wave functions in closed forms for this potential (with great powers and inverse exponent), we use ansatz method. We also regard the effects of spin-spin, spin-isospin, and isospin-isospin interactions on the relativistic energy spectra of nucleon. By using the obtained results, we have calculated the deuteron mass. The results of our model show that the deuteron spectrum is very close to the ones obtained in experiments.


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## Body

## 1. Introduction

It is well known that the exact energy eigenvalues of the bound state play an important role in quantum mechanics. In particular, the Dirac equation which describes the motion of a spin-1/2 particle has been used in solving many problems of nuclear and high-energy physics. Recently, there has been an increase in searching for analytic solution of the Dirac equation [1–11]. Recently, tensor couplings have been used widely in the studies of nuclear properties [12–22], and they were introduced into the Dirac equation by substitution P→→P→-imωβ·x^U(r) [16, 23], where m is one of the particles, mass and ω refers to harmonic oscillator. In this work, we are going to solve the relativistic Dirac equation for Yukawa potential in fresh approaches. We first discuss the Dirac equation with a tensor potential for Yukawa potential. Then, we solve the resulting equations for deuteron nuclei and obtain the parameters for it. We are also able to number the mass of Deuteron by using obtained parameters, and the equation is in conformity with experimental measure.


## 2. Yukawa Potential

The screened Coulomb potential, also known as the Yukawa potential in atomic physics and the Debye-Huckel potential in plasma physics, is of interest in many areas of physics. It was originally used to model strong nucleon-nucleon interactions due to meson exchange in nuclear physics by Yukawa [19, 20]. It is also used to represent a screened Coulomb potential due to the cloud of electronic charges around the nucleus in atomic physics or to account for the shielding by outer charges of the Coulomb field experienced by an atomic electron in hydrogen plasma. However, the Schrödinger equation for this potential cannot be solved exactly; hence, various numerical and pertubative methods have been devised to obtain the energy levels and related physical quantities [24]. The generic form of this potential is given by
(1)U(x)=-gxV(x)g>0,
that g is the potential strength, and
(2)V(x)=e-kx.In this part, we expand the Yukawa potential in its meson clouds,x=a, by using Taylor extension to the power of seventh:
(3)U(x)=-gx[e-ka-ke-ka(x-a)+k22!e-ka(x-a)2-k33!e-ka(x-a)3+k44!e-ka(x-a)4-k55!e-ka(x-a)5+k66!e-ka(x-a)6-k77!e-ka(x-a)7+⋯].
We can reduce this equation to
(4)U(x)=ax6-bx5+cx4-dx3+ex2-fx+h-Lx.
Now, we have the general Yukawa equation, that a, b, c, d, e, f, h, and L are
(5)a=k77!,b=-(77!k6+77!k6),c=(217!k5+66!k5+15!k5),d=-(357!k4+156!k4+55!k4+13!k4),e=(357!k3+206!k3+105!k3+64!k3+33!k3+12!k3),f=-(217!k2+156!k2+105!k2+64!k2+33!k2+12!k2),h=(77!k+66!k+55!k+44!k+33!k+12!k+k),L=-(77!+66!+55!+44!+33!+12!+2).


## 3.     Exact Analytical Solution of the Dirac Equation for Yukawa Potential

We are interested in the particular solution of the Dirac equation with the Yukawa interaction potential (4). In this section, we anticipate the answer and then obtain the parameters for highest nuclei, deuteron. A deuteron consists of a neutron and a proton. Let us assume one nucleon is fixed and another nucleon is moving around the center of mass. At last, by using single particle model, we investigate the effects of one pion exchange for a single nucleon. The Dirac Hamiltonian is
(6)H=α→·P→+β(m+S)+V-iβα→·x^U,
where S is the scalar potential, V is the vector potential, and U is the tensor part. We have  Δ=V-S and Σ=V+S. The Dirac equation can be solved exactly for the cases Δ=0 and Σ=0·α→=(0σ→σ→0) and β=(I00-I) which are the usual Dirac matrices. We will indicate upper and lower components of Dirac spinor with ΨA and ΨB. Expressing the Dirac matrices in terms of the Pauli matrices: we obtain the following coupled equations, [25, 26](7a)σ→·P→ΨA+(V-S-m)ΨB+iσ→·xU(x)ΨA=EΨB,(7b)σ→·P→ΨB+(V+S+m)ΨA-iσ→·xU(x)ΨB=EΨA.We assume that V, S, and U are radial potentials. Using the identity
(8)σ→·P→=(σ→·x→)x2(σ→·x→)(σ→·P→)=(σ→·x→)x2(-ix∂∂x+iσ→·L→).
With (7a) and (7b), we find the following second-order differential equations for ΨA and ΨB: (9a)P→2ΨA+(U2+dudx+2ux+(dΔ/dx)(E+m-Δ)U)ΨA+(4U+2(dΔ/dx)(E+m-Δ))(S→·L→2)ΨA-(dΔ/dx)(E+m-Δ)∂ΨA∂x=(E+m-Δ)(E+m-Σ)ΨA,(9b)P→2ΨB+(U2-dudx-2ux-(dΣ/dx)(E-m-Σ)U)ΨB+(-4U+2(dΣ/dx)(E-m-Σ))(S→·L→2)ΨB+(dΣ/dx)(E-m-Δ)∂ΨB∂x=(E+m-Δ)(E-m-Σ)ΨB.Here S→ stands for the (1/2)σ→ spin operator and L→ for orbital angular momentum operator. The Hamiltonian operator commutes with the total angular momentum operator which is given by J→=L→+S→ and with the parity operator. Therefore, the eigenfunctions defined by Hamiltonian can be expressed as
(10)Ψnjmk=(ignk(x)xϕjmkl(x^)fnk(x)xϕjm(-k)l-(x^)),
where ϕjmkl(x^) denotes the spin spherical harmonics. The quantum number k is related to ℓ and j as follows:
(11)k={-(ℓ+1)=-(j+12)j=ℓ+12,ℓ=+(j+12)j=ℓ-12.
For spin spherical harmonics we have
(12)σ→·x^ϕjmkl=ϕjm(-k)l-.
Equations (9a) and (9b) reduced to a set of coupled equations for the radial wave functions gk and fκ:(13a)(d2dx2-k(k+1)x2+2kxU-dUdx-U2)gk(x)+dΔ/dx(E+m-Δ)(ddx+kx-U)gk(x)=-(E+m-Δ)(E-m-Σ)gk(x),(13b)(d2dx2-k(k-1)x2+2kxU+dUdx-U2)fk(x)+dΣ/dx(E-m-Δ)(ddx-kx+U)fk(x)=-(E+m-Δ)(E-m-Σ)fk(x).Using the following relations [13]:
(14)2S→·L→=J→2-L→2-S→2,σ→·L→ϕjmkl=-(k+1)ϕjmkl,σ→·L→ϕjm(-k)l-=-(k+1)ϕjm(-k)l-,
for particle of m mass and E energy, we have
(15)S=V=12Σ,Σ(x)=ax6-bx5+cx4-dx3+ex2-fx+h-Lx,Δ(x)=0,U(x)=-1x.
We assume the case S(x)=V(x) [27–31] and consider k(k+1)=ℓ(ℓ+1), then the upper component in (13a) is as follows:
(16)(d2dx2-ℓ(ℓ+1)+2+2kx2-Σ(r)(E+m))gk(x)=(m2-E2)gk(x).
As we know in deuteron, two nucleons have in D-state, so
(17)[d2dx2-ℓ(ℓ+1)+2-2(ℓ+1)x2]gk(x)=((a1x6-b1x5+c1x4-d1x3+e1x2-f1x+h1-L1x)-ℰa1x6-b1x5+c1x4)gk(x).
Therefore, we have a Schrödinger-like equation [31]
(18)g1′′(x)=[(a1x6-b1x5+c1x4-d1x3+e1x2-f1x+h1-L1x)-ℰ+ℓ(ℓ-1)x2].
By setting
(19)a(E1+m)=a1,b(E1+m)=b1,c(E1+m)=c1,d(E1+m)=d1,(20)e(E1+m)=e1,f(E1+m)=f1,h(E1+m)=h1,L(E1+m)=L1(E12-m2)=ℰ.
In order to solve (18), we suppose the following form for the wave function:
(21)g(x)=M(x)eZ(x).
Now, for the functions M(x) and Z(x), respectively, we make use of the following ansatz [27, 32, 33]:
(22)M(x)={1ifgroundstate∏i=1v(x-aiv)ifv>0,Z(x)=-14αx4-13βx3-12ηx2-τx+δlnx.For a particular grand angular quantum numberγ, there are different solutions which are labeled by υ (υ determines the number of the nodes of the wave function). By substitution of M(x) and Z(x) into (21) and then taking the second-order derivative of the obtained equation, we can get
(23)g′′(x)=[Z′′+Z′2+M′′+2M′Z′M].
We consider the ground state, which is called the 0th node solution of the above differential equation. By equating (18) and (23) for υ=0, it can be found that
(24)α=a1,β=-b12a1,η=c1-β12α,τ=-d1-2βη2α,δ=ℓ-1,ℰ=η(1-2δ)-τ2-h1.
The upper component is as follows:
(25)g1(x)=N0xℓ-1exp(-14αx4-13βx3-12ηx2-τx)
and another component of wave function is
(26)f1(x)=(σ→·P→+iσ→·r^U)(E+m)g1(x).
By substituting (25) into (26), we obtained the following equation:
(27)f1(x)=-iσ→·x^(E+m)[ddx-iσ→·L→-U]N0xℓ-1×exp(-14αx4-13βx3-12ηx2-τx).
Therefore, the total wave function is
(28)Ψ=N0(1-iσ→·x^(E+m)[αx3+βx+ηx+τ+1x])×exp(-14αx4-13βx3-12η-τx).
Now we have to obtain parameters a, b, c, d, e, f, h, and L, by using of (5):
(29)b=14ka,c=105k2a,d=440k3a,e=2555k4a,f=6846k5a,h=8659k6a.
We know that k=1/x=0.714(fm-1) [34], by using of (24), we can get amount of parameters α, β, η, and τ:(30)α=a1,β=-9.8a1,η=83.08a1,τ=83.08a1.
The bonding energy for deuteron is reported 2.224 MeV [35]. This is useful for calculating the numeric values of parameters in (30):
(31)ε=E2-m2=η(1-2δ)-τ2-h1.
By substituting the amounts of (31) and ℓ=1, we obtained the numeric values of parameters in (30). The parameters are reported in Table 1, while the parameters of Yukawa potential are reported in Table 2.Table 1
The best adaption with experimental value isa1=3.06136,α=1.74967,β=-17.146815,η=145.362999, τ=-141.02380, and  E  (MeV) = 2.19020.
a
1
α
β
η
τ
E (MeV)
3.06131
1.74965
−17.146628
145.361420
−141.02227
2.89658
3.06134
1.74967
−17.146785
145.362749
−141.02356
2.49699
3.06135
1.74967
−17.146805
145.362915
−141.02372
2.34861
3.06136
1.74967
−17.146815
145.362999
−141.0238
2.19020
3.0638
1.74967
−17.147648
145.377006
−141.0234
1.83278Table 2
The value of the Yukawa potential parameters.
a (fm−7)
b (fm−6)
c (fm−5)
d (fm−4)
e (fm−3)
f (fm−2)
h (fm−1)
0.21961
4.30589
45.24045
265.53556
2159.7519
8109.2235
14362.5603


## 4. The Hyperfine Interaction

As the Baryons have spin and isospin, the deuteron neutron and proton can interact together. We introduce the hyperfine interaction potential. The complete potential is(32)〈Hint(x)〉=V(x)+〈HS(x)〉+〈HI(x)〉+〈HSI(x)〉.In this work, we have added the hyperfine interaction potentials (HS(x),  HI(x), and HSI(x)) which yield properties very close to the experimental results. By regarding V(x) as the nonpertubative potential and the other terms in (32) as pertubative ones according to this explanation. The spin and isospin potential contains a δ-like term. We have modified it by a Gaussian function of the nucleons pair relative distance [36]
(33)HS=AS1(πσS)3e-x2/σS2∑(S→i·S→j),
where Si is the spin operator of the ith nucleon  and x is the relative nucleon pair coordinate.As and σS  are constants: σS=2.87fm and AS=67.4fm2 [34, 37]. We know that the deuteron consists of two nucleons. Where 1 indicate the first nucleon, and 2 indicate the second nucleon. We have
(34)HS=AS1(πσS)3e-x2/σS2(12)[S2-S12-S22].
Furthermore, we add two hyperfine interaction terms to the Hamiltonian nucleon pairs similar to (33) [38]
(35)HI=AS1(πσI)3e-x2/σI2∑(I→i·I→j),
where Ii is the isospin operator of the ith nucleon, and the constants are described as σI=3.45fm  and AI=51.7fm2 [37, 39]. The another one is a spin-isospin interaction, given by similar (33) [38, 40]
(36)HSI=ASI1(πσSI)3e-x2/σSI2∑(S→i·S→j)(I→i·I→j).
The fitted parameters are σSI=2.31fm and ASI=-106.2fm2 [35, 41], where ψγ is the perturbed wave function, and we write it as
(37)ψγ=ψγ′+∑γ′=γ〈ψγ′0∣Hint∣ψγ′0〉ψγ′′0Eγ0-Eγ′0.
The deuteron mass are given by two nucleons masses, and the eigenenergies of the Dirac equation E, (E is a function of  η, δ, τ, h, m) with the first-order energy correction from potential Hint can be obtained by using the unperturbed wavefunction (33), (35), and (36). The total potential for the ground state as well as the other states can be written as [40, 42]
(38)〈Hint〉=∫ψγ*Hinψγx2dxdΩ∫ψγ*ψγx2dxdΩ.
We first assume γ=0. The potentials can be extracted from (33), (35), and (36):
(39)HS=0.10059(MeV),HI=0.04782(MeV),HSI=-0.50053(MeV).
Now, we can calculate the deuteron mass according to the following formula:
(40)MD=mn+mp+Eυγ+〈Hint〉.
By substituting mn=mp=938(MeV)=4.65(fm-1), energy eigenvalue (Eυγ), and the expectation values of Hint in (40) we have
(41)MD=1877.83808(MeV/c2).By comparing the experimental amount of deuteron mass (MD=1875.612(MeV/c2)=9.29(fm-1)) with our calculated mass for deuteron, we found that a good agreement has been obtained by our model [35].


## 5.  Conclusion

In this work, we offer a new approach for the solving of Yukawa potential. We expand this potential around of its mesonic cloud that gets a new form with great powers and inverse exponent. We calculate wave function of the Dirac equation for a new form of Yukawa potential, including a coulomb-like tensor potential. By considering the effects of pertubative interaction potential, we see that the theoretical and experimental masses are in complete agreement. These improvements in reproduction of deuteron mass is obtained by using a suitable form for confinement potential and exact analytical solution of the Dirac equation for our proposed potential. Finally, one can use this model for another nuclei, by relativistic or nonrelativistic equation and get the properties of them and the amount of “g”, the strength of nuclear force.



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