Dirac Equation with Mixed Scalar–Vector–Pseudoscalar Linear Potential under Relativistic Symmetries

1 Introduction

The exact solutions play an important role in quantum mechanics. For example, the exact solutions to the Schrödinger equation for a hydrogen atom and for a harmonic oscillator in 3D [1–4] were important achievements at the beginning stage of quantum mechanics, which provided a strong evidence in favour of the theory being correct. Also, 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. One of the most important concepts for understanding the traditional magic numbers in stable nuclei is the spin symmetry breaking [5–8]. The potential having a linear form of position and a “Coulomb-like” potential play an important role in various branches of physics. The gluon condensation in high-energy physics can be studied with an effective linear potential [9]. The different properties of low-lying baryons have been searched by using a linear potential, and the corrections to the mass spectrum have also been calculated one-gluon exchange [10]. It has been suggested that the hadron spectrum can be studied by assuming that the quarks are bounded by a long-range potential plus a short-range potential (Coulomb-like) arising in gluon-photon exchange diagrams [11]. Also, a linear potential has often been used for the confinement potential in theories of quark confinement (known from lattice QCD) [12, 13]. The Dirac equation appears to be problematic for this potential. If linear potential is introduced as the time component of a Lorentz two-vector, no-bound state, solutions exist [14, 15]. If it is introduced as a Lorentz scalar, Bhalerao and Ram [16] have found only a very limited set of solutions. Some attempts to solve the Dirac equation with a linear potential have been made. For example, Abe and Fujita [17] found a simple and approximate analytical solution of the massless Dirac equation with a linear scalar potential. For more review of this, the reader can refer to references [18–22]. The spin and pseudospin symmetry concepts are introduced in nuclear theory [23, 24] so, one of the most important concepts for understanding the traditional magic numbers in stable nuclei is the spin symmetry breaking [5–8]. Conversely, The p-spin symmetry concepts observed in nuclear theory originally almost 40 years ago as a mechanism to explain different aspects of the nuclear structure is one of the most interesting phenomena in the relativistic quantum mechanics. Within the framework of the Dirac equation, p-spin symmetry used to feature deformed nuclei and superdeformation, as well as to establish an effective shell model [25–28], whereas spin symmetry is relevant for mesons [29]. Spin symmetry occurs when the scalar potential V_S is nearly equal to the vector potential V_V or equivalently (V_S ≈ V_V) and p-spin symmetry occurs when (V_S ≈ −V_V) [30, 31]. In recent years, by considering the importance of spin and p-spin symmetries, some authors have contributed many works in this field. For more review of this, readers can see references [32–36]. The purpose of this work is to solve the Dirac equation for the linear potential in scalar–vector–pseudoscalar under relativistic symmetries, i.e. spin and p-spin symmetries and their solutions are obtained systematically by using the Nikiforov–Uvarov (NU) method. To this end, in Section 2, we outline the NU method. In Section 3, we briefly introduce the Dirac equation with scalar–vector–pseudoscalar potential in view of spin and p-spin symmetries. In Section 4, we solve the Dirac equation with under spin and p-spin symmetries. Finally, Section 5 contains a summary and concluding remarks.

2 Basic Equations of the Nikiforov–Uvarov Method

In this section, we briefly describe the NU method. The method provides us with an exact solution of the nonrelativistic Schrödinger equation for certain kind of potentials [37]. The method is based on the solutions of general second-order linear differential equations with special orthogonal functions [38]. For a given potential, the Schrödinger equation in the spherical coordinates is reduced to a generalised equation of hypergeometric type with an appropriate s=s(r) coordinate transformation. Thus, it can be written in the following form:

(1)ψ″n(s) + τ˜(s)σ(s)ψ′n(s) + σ˜(s)σ2(s)ψn(s) = 0, (1)

where σ(s) and σ˜(s) are polynomials, at most of second degree, and τ˜(s) is a first-degree polynomial and ψ_n(s) is a function of the hypergeometric type. In order to find a particular solution of (1), let us decompose the wave function ψ_n(s) as follows:

(2)ψn(s) = ϕ(s)yn(s), (2)

Then we use

(3)[σ(s)ρ(s)]′ = τ(s)ρ(s), (3)

to reduce (1) to the form

(4)σ(s)y″n(s) + τ(s)y′n(s) + λyn(s) = 0. (4)

the coefficient of y′n(s) is taken in the form τ(s)/σ(s), where τ(s) is a polynomial of degree at most one, i.e.

(5)2ϕ′(s)ϕ(s) + τ˜(s)σ(s) = τ(s)σ(s), (5)

and hence the most regular form is obtained as follows,

(6)ϕ′(s)ϕ(s) = π(s)σ(s), (6)

where

(7)τ(s) = τ˜(s) + 2π(s), τ′(s) < 0, (7)

and the new parameter π(s) is a polynomial of degree at most one, and the prime denotes the differentiation with respect to s.

One is looking for a family of solutions corresponding to

(8)λ = λn = −nτ′(s) − n(n − 1)2σ″(s), (n = 0, 1, 2, …) (8)

The y_n(s) is the hypergeometric type function whose polynomial solutions are given by the Rodrigues relation

(9)yn(s) = Bnρ(s)dndsn[σn(s)ρ(s)], (9)

where B_n is the normalisation constant, and the weight function ρ(s) is the solution of the differential (3). By defining

(10)λ = K + π′(s), (10)

one obtains the polynomial π(s) as following equality

(11)π(s) = σ′(s) − τ˜(s)2 ± (σ′(s) − τ˜(s)2)2 − σ˜(s) + Kσ(s). (11)

In order to obtain the possible solutions according to the plus and minus signs of (11), the parameter K within the square root sign must be known explicitly. To provide this requirement, the expression under the square root sign has to be the square of a polynomial, because π(s) is a polynomial of degree at most one. In this case, an equation of the quadratic form is available for the constant K. Setting the discriminant of this quadratic equal to zero, the constant K is determined clearly. After determining K, the polynomial π(s) is obtained from (11), and then τ(s) and λ are also obtained by using (7) and (10), respectively.

3 The Dirac Equation in a 1+1 Dimensions

The two-dimensional Dirac equation can be obtained from the four-dimensional one with the mixture of spherically symmetric scalar, vector, and anomalous magnetic-like (tensor) interactions. If we limit the fermion motion to the x-direction (p_y=p_z=0) the four dimensional Dirac equation decomposes into two equivalent two-dimensional equations with two-component spinors and 2×2 matrices [39]. Then, results show that the scalar and vector interactions preserve their Lorentz structures, whereas the anomalous magnetic-like interaction turns out to be a pseudoscalar interaction. Furthermore, in the 1+1 world there is no angular momentum so that the spin is absent. Therefore, the 1+1 dimensional Dirac equation allows us to explore the physical consequences of the negative-energy states in a mathematically simpler and more physically transparent way. In this spirit the two-dimensional version of the anomalous magnetic-like interaction linear in the space coordinate has also received attention [40–45]. The 1+1 dimensional time-independent Dirac equation for a spin-1/2 particle of rest mass m under the action of vector V_V, scalar V_S and pseudoscalar V_P potentials can be written, in terms of the combinations (Σ=V_V + V_S) and (Δ=V_V − V_S), in units ℏ=c=1, as [46–50]

(12)Hψ = Eψ,H = σ1p + σ3m + 1 + σ32Σ + 1 − σ32Δ + σ2Vp (12)

where σ₁, σ₂ and σ₃ denote the Pauli matrices.

The connection to normal four-dimensional Dirac algebra is apparently found by the following replacement

(13)σ1→α→ = γ0γ→,σ2→γ5,σ3→β = γ0 (13)

where the γ^μ matrices are used in the Dirac representation and γ⁵=iγ⁰γ¹γ²γ³ [25]. If we suppose the wave function as ψ = (ψ+ψ−), we have two first-order equations for the upper, ψ₊ and the lower, ψ₋ components of the spinor as

(14a)−idψ−dx + mψ+ + Σψ+ − iVpψ− = Eψ+ (14a)

(14b)−idψ+dx − mψ− + Δψ− + iVpψ+ = Eψ−. (14b)

In terms of ψ₊ and ψ₋, the spinor is normalised as ∫−∞+∞(|ψ+|2 + |ψ−|2)dx = 1, so that ψ₊ and ψ₋ are square integrable. It is clear from the pair of coupled first-order differential equations (14) that both ψ₊ and ψ₋ must be discontinuous wherever the potential undergoes an infinite jump and have opposite parities if the Dirac equation is covariant under x → −x [45]. It is worth mentioning here that, solving (14) for ψ₋(x) and ψ₊(x) with (V_V=±V_S) gives one spinor component in terms of the other as

(15)ψ∓(x) = −i(ddx ∓ Vp)E ± mψ±(x) (15)

where E ≠ ±m. This equation is referred to as the “kinetic balance” relation. Because E=+m (E=−m) is an element of the positive (negative) energy spectrum of the Dirac Hamiltonian, then this relation with the top (bottom) sign is not valid for the negative (positive) energy solutions [51].

Substituting (15) into (14), with (V_V=±V_S) results in the following second-order differential equation:

(16)[−d2dx2 + 2VV(m ± E) ± dVpdx + m2 − E2 + Vp2]ψ±(x) = 0 (16)

Either for Δ=0 with E ≠ −m or Σ=0 with E ≠ +m the solution of the relativistic problem is mapped into a Sturm–Liouville problem in such a way that solution can be found by solving a Schrödinger-like problem.

At this stage we realise that the Dirac energy levels are symmetrical about E=0. This means that the pseudoscalar potential couples to the positive-energy component of the spinor in the same way it couples to the negative-energy component.

Giving ψ₊(ψ₋) as an element of the positive (negative) energy solutions. To obtain the other spinor component, we use the kinetic balance relation (15) with the top (bottom) sign. Therefore, the choice (V_V=+V_S)((V_V=−V_S)) dictates that the solution of (16) does not include the negative (positive) energy states and only makes up half of the solution space: the positive (E>0) (negative (E<0)) energy subspace. The union of these two subspaces constitutes the complete solution space of the Dirac equation.

This observation highlights the second critical property in this kind of problem that has to be considered carefully when presenting the physical interpretation. It associates with each choice of potential configuration in one sector of the energy spectrum, only the positive or the negative, but not both. This unsymmetrical treatment of the energy spectrum where half of the spectrum is missing is known to create problems such as particle–antiparticle interpretation of the relativistic theory [4, 52, 53].

4 Dirac Equation with Mixed Scalar–Vector–Pseudoscalar Linear Potential

4.1 Spin Symmetry Limit

In this subsection we solve the Dirac equation under spin symmetry limit with mixed scalar–vector–pseudoscalar linear potential. The exact spin symmetry is proved by Meng et al. [54, 55]. It occurs in the Dirac equation when (dΔdr = d(VV − VS)dr = 0) or Δ=C_s=constant. In this part, we are taking C_s=0 and scalar–vector–pseudoscalar potential as the linear potential

(17)Σ = asx (17)

(18)Vp = apx (18)

where a_s and a_p are constants. Therefore, (16) for ψ₊(x) becomes

(19)d2ψ+(x)dx2 + [−ap2 x2 + Asx + Ap]ψ+(x) = 0 (19)

where we have defined

(20a)As = −as(E + m) (20a)

(20b)Ap = E2 − m2 − ap (20b)

In mathematics, the parabolic cylinder functions are special functions defined as solutions to the differential equation (19) [56]. In order to solve (19) by means of the NU method, we should compare it with (1). The following values for the parameters are found as

(21)τ˜(x) = 0, σ(x) = 1, σ˜(x) = −ap2x2 + Asx + Ap (21)

Further, inserting these values into (11), we obtain

(22)π(x) = ±ap2x2 − Asx − Ap + K (22)

The expression under the square root of this equation must be the square of a polynomial of the first degree. This is possible only if its discriminant is zero and the constant parameter K can be determined from the condition that the expression under the square root has a double zero. Hence, K is obtained as K = As24ap2 + Ap. In that case, it can be written in the two possible forms of π(x):

(23)π(x) = ± (apx − As2ap), for K = As24ap2 + Ap (23)

According to the NU method, one of the two values of the polynomial π(x) is proper to obtain the bound states because τ(x) has a negative derivative. Therefore, the selected forms of π(x) and K take the following particular values:

(24)π(x) = −apx + As2ap, τ′(x) = −2ap < 0, ap > 0 (24)

In addition, after using (8) and (10) together with the assignments given in (21) and (24) , the following expressions for λ are obtained as

(25)λ = As24ap2 + Ap − ap (25)

(26)λn = λ = 2apn, (n = 0, 1, 2, …) (26)

Now, taking λ_n=λ, we can solve these equations to obtain the energy equation for the linear potential with spin symmetry in the Dirac theory and we obtain

(27)Ens± = −mas2 ± 22ap3(2m2ap + (4ap2 + a s2)(n + 1))4ap2 + as2 (27)

For a given value of n, this equation provides two distinct positive and negative energy spectra related with Ens− or Ens+, respectively. One of the distinct solutions is only valid to obtain the positive-energy Ens− bound states in the limit of the spin symmetry while Ens+ should be discarded. This could easily be seen by considering large n and imposing positivity of the energy.

Let us now find the corresponding eigenfunction for this system. Using (3), (6), and (7), we find

(28)ρ(x) = e−apx2 + Asapx (28)

(29)φ(x) = e−12(apx2 − Asapx) (29)

and further substituting (21) and (28) into (9), the relation (2) gives the radial upper-spinor wave function as follows:

(30)ψ+n(x) = N  2−n2Hn(apx − As2ap3/2)e−12(apx − As2ap3/2)2 (30)

where n=0, 1, …, H_n, is a n^th degree Hermite polynomial and the normalisation constant N is obtained from∫−∞+∞(|ψ+n(x)|2 + |ψ−n(x)|2)dx = 1. This upper-spinor component satisfies the restriction condition for the bound states, i.e. a_p>0. Also the lower spinor wave function, by using the recursion relations of Hermite polynomials (see [56]), can be obtained via

(31)ψ−n(x) = −iEn + m(ddx − Vp)ψ+n(x)=i Nap2−n2En + m(Hn + 1(apx − As2ap3/2) + As2ap3/2Hn(apx − As2ap3/2))e−12(apx − As2ap3/2)2 (31)

where E_n≠ −m.

Because the Hermite polynomial H_n has n distinct zeros we can conclude that ψ₊ has n nodes and if a_s=0, when there is a pure pseudoscalar potential linear in x, the expression for ψ₋ suggests that it has n + 1 nodes. Also, for x = As/4ap2, the lower component is proportional to a Hermite polynomial of degree n − 1.

4.2 P-Spin Symmetry Limit

Within the p-spin symmetry case, (d[V(r) + S(r)]dr = dΣ(r)dr = 0) or Σ(r)=C_ps=constant and p-spin symmetry is exact in the Dirac equation [54, 55, 57, 58]. In this part, we are taking C_ps=0 and scalar–vector–pseudoscalar potential as linear potential:

(32)Δ = apsx (32)

(33)Vp = apx (33)

Therefore, for the lower component in the p-spin symmetry we have

(34)d2ψ−(x)dx2 + [−ap2x2 + A˜psx + A˜p]ψ−(x) = 0 (34)

where we have defined

(35a)A˜ps = aps(m − E) (35a)

(36b)A˜p = E2 − m2 + ap (36b)

To avoid repetition, the functions required by the method for π(x), K, and τ(x) can be established as

(37)π(x) = apx − A˜ps2ap, for K = A˜ps24ap2 + A˜p (37)

(38)τ(x) = 2ap − A˜psap, τ′(x) = 2ap < 0, ap < 0 (38)

In the present case, because we took the negative sign of π(x) in the spin symmetric, we must choose the positive sign of π(x) for the p-spin symmetric limit with the following restriction a_p<0 required to obtain the negative-energy bound states.

As before the subsection, we obtain the explicit form of the energy equation as

(39)En˜ps± = m aps2 ± 2−2ap3(−2m2ap + (4ap2 + aps2)(n˜ + 1))4ap2 + aps2 (39)

For a given value of n˜, this equation provides two distinct energy spectra that En˜ps− solutions is only valid to obtain the negative-energy bound states in the limit of the p-spin symmetry. In our calculations for the p-spin symmetry wave functions, we firstly find the weight function as

(40)ρ(x) = eapx2 − A˜psapx (40)

Further, the first part of the wave function is being calculated as

(41)φ(x) = e12(apx2 − A˜psapx) (41)

By using (9) and (2) in the p-spin symmetry case, we may write down the lower-spinor wave function in the following fashion:

(42)ψ−n˜(x) = N˜  2−n˜2Hn˜(i(apx − A˜ps2ap3/2))e12(apx − A˜ps2ap3/2)2 (42)

Thus, the corresponding p-spin symmetric upper-component can be found as follows:

(43)ψ+n˜(x) = −iEn˜ − m(ddx + Vp) ψ−n˜(x)=i N˜ap2−n˜2En˜ − m(i Hn˜ + 1(i(apx − A˜ps2ap3/2))− A˜ps2ap3/2Hn˜(iapx − A˜ps2ap3/2))e12(apx − A˜ps2ap3/2)2 (43)

where En˜ ≠ m. In an analogy with the previous case, if a_ps=0, ψ₊ has n˜ + 1 nodes, while for x = A˜ps/4ap2, the upper component is proportional to a Hermite polynomial of degree n˜ − 1.

Let us finally remark that the negative energy solution for p-spin symmetry can be obtained directly from those of the positive energy solution for spin symmetry by performing the following replacements:

(44)Vp → −Vp, Σ → −Δ, E → −E(n˜ → n), ψ±→ ±ψ∓(the mass m is not changed). (44)

5 Concluding Remarks and Discussion

In this article, we have investigated the energy levels and wave functions of the Dirac equation with the mixed scalar–vector–pseudoscalar linear potential under spin and p-spin symmetries limits in 1+1 dimensions. We have found that there are only positive (negative) energy states for bound states in the case of spin (p-spin) symmetry based on the restriction condition a_p>0(a_p<0) for the wave functions. Some numerical results of (27) and (39) are given by Tables 1 and 2. Also, the variation of the energy spectrum (E_n) according to the pseudoscalar potential parameter a_p for some states is shown in Figures 1 and 2. In Figures 3 and 4 we plot the upper and lower components for the Dirac wave function in the case of spin symmetry for the three lowest bound states with n=0, 1, 2.

Figure 1:

The variation of the energy spectrum Ens− for linear potential in the case of spin symmetry with the potential parameter a_p>0; here we selected m=10, a_s=1.

Figure 2:

The variation of the energy spectrum En˜ps− for linear potential in the case of p-spin symmetry with the potential parameter a_p<0; here we selected m=10, a_s=1.

Figure 3:

The upper componet ψS+n of Dirac spinor for linear potential in the case of spin symmetry with the first three positive energy eigenvalues; here we selected m=1, a_p=1, a_s=1.

Figure 4:

The lower componet ψS−n of Dirac spinor for linear potential in the case of spin symmetry with the first three positive energy eigenvalues; here we selected m=1, a_p=1, a_s=1.

The solutions for Δ=0 with E=−m and Σ=0 with E=+m, excluded from the Sturm–Liouville problem, can be obtained directly from the original first-order equations (14a) and (14b) and are called isolated solutions.

However, the isolated solution has some distinctive characteristics when compared to the solutions of the Sturm–Liouville problem which led us to believe that, in fact, they belong to a different class of solutions. The isolated solution breaks the symmetry of the energy levels about E=0 exhibited by the solutions of the Sturm–Liouville problem, and the corresponding eigenspinor has only one component differing from zero. Furthermore, unlike the Sturm–Liouville solutions, the isolated solution is there even if V_p(x) is not so strong, i.e. there exists an isolated solution even if |a_p| ≤ 1.