# A Novel Approach to Solve Quasiexactly Solvable Pauli Equation

Authors: Ramazan Koç; Eser Olğar; Haydar Mutaf
Journal: Advances in Mathematical Physics
 (2015)

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## Abstract

The spectra for some specific forms of external magnetic fields in Pauli equation are obtained in the framework of the asymptotic iteration method (AIM). AIM is applied to find the solution of Pauli equation. When the method is applied to quasiexactly solvable systems, it not only easily gives the corresponding spectrum, but also produces accurate results for the eigenvalues of the system havingsl(2) symmetry.


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## Body

## 1. Introduction

It is well known that motion of the charged particles in electromagnetic fields described by Pauli equation is one of the fascinating phenomena in modern physics, providing a general approach to understanding the properties of electrons. The Hamiltonian for this system whenħ=e=2me=c=1 is
(1)H=px+Ax2+py+Ay2+g2∇×Azσz,
where px, py are the momentum operators, g=2 is the gyromagnetic ratio, and A and σz are the vector potential of the electromagnetic field and the Pauli matrix, respectively. The Pauli equation has been solved exactly for uniform magnetic field Bx=By=0, Bz=B0, leading the Landau levels. Moreover, it was proved in [1] that for any general magnetic field Bz=B(x,y) perpendicular to the xy plane, the ground state is solved exactly as determined in [2–4]. The general result of [1] can be viewed as a very special case of the recently discovered QES systems that only a finite spectrum to be solved with algebraic methods [5–9]. The Pauli equation has been solved in a quasiexactly solvable model [10] for possible (nonuniform) magnetic fields [11]. If not impossible, it seems difficult to find other forms of the magnetic field such that the Pauli equation could be exactly solved. Therefore, the forms of magnetic fields, for which Pauli equation could be exactly solved, are indeed rather rare.Therefore, a natural question arising at this point is can the asymptotic iteration method be applied to these systems? The answer to this question is the main topic of this paper. An asymptotic iteration technique which improves both analytical and numerical solutions of the problems has been introduced by Çiftçi et al. [12]. Afterwards, it has been developed for many physical systems in order to obtain the whole spectra [13, 14]. It has been also suggested to solve the Schrödinger equation and developed for some quantum optical systems [15]. Later, a considerable attention has been paid to AIM [12, 16–19]. The results obtained by this method fit with the solutions of many exactly solvable differential equations in the form of the Schrödinger equation. It also gives accurate result for the nonexactly solvable Schrödinger equation such as sextic oscillator, cubic oscillator, and deformed coulomb potential, which are important in applications to many problems in physics. Although AIM has been applied to solve the Schrödinger equation, its application to the solution of the Pauli equation is still missing. It is well known that the AIM is very powerful in describing many physical problems. It improves both analytical and numerical solutions and also it is useful for understanding the nature of physical structures.One can show that the termination relationΔn(x)=0 in the corresponding method reflects effectively the peculiar properties of the functions λ0 and s0. After appropriate change of the variable, their hidden polynomial nature can be revealed. On these polynomials of finite order, the finite-dimensional representations of the hidden sl(2,R) algebraic structure underlying the quasiexact solvability are realized. Also with this hidden polynomial structure, the nonlinear supersymmetry discussed in [10] can be related.


## 2. Formalism of the Asymptotic Iteration Method

The AIM is proposed to solve the second-order differential equations and details can be found in [12]. In this section we systematically apply the method to obtain eigenvalues of the Pauli equation for various nonuniform magnetic fields.Consider the following second-order differential equation:(2)y′′=λ0y′+s0y,
where λ0≠0 and λ0, s0 are functions defined in C∞(a,b) and they have sufficiently many continuous derivatives. y′ and y′′ denote derivatives of y with respect to x. It is easy to show that (n+2)th derivative of the function y can be written as
(3)y(n+2)=λny′+sny,
where λn and sn are given by the recurrence relations
(4)λn=λn-1′+sn-1+λn-1λ0,sn=sn-1′+λn-1s0.
If we have, for sufficiently large n,
(5)snλn=sn-1λn-1=αnx,
then the solution of (2) can be written as follows:
(6)y=exp⁡-∫xαdt·C1+C2∫xexp⁡∫sλ0+2αdtds,
where C1 and C2 are the integration constants. In calculating the parameters in (4), when n=0, we take the initial conditions as λ-1=1 and s-1=0 [14] and the termination condition Δn(x) is Δn(x)=0 for
(7)Δn(x)=λn(x)sn-1(x)-λn-1(x)sn(x).Before going further we note that, in this paper, we will determine magnetic fields that lead to QES solvable potential forms in Pauli equation. We consider the magnetic fields only in asymmetric gauge. Solution of the Pauli equation including nonuniform magnetic fields has recently been studied in the framework of the QES systems [10, 11]. Ho and Roy presented a general procedure for determining possible (nonuniform) magnetic fields such that the Pauli equation becomes QES with an underlying sl(2) symmetry. But our paper includes a part of the motivation those QES systems which can be solved by using AIM and it gives the whole spectrum with accurate results.The following section includes the solution of Pauli Equation within the framework of the AIM.


## 3. Magnetic Field

In this paper, we deal with magnetic field in asymmetric gauge with vector potentials as in [11]
(8)Ax(x,y)=0,Ay(x,y)=-W¯(x).
In this case components of the magnetic field B with asymmetric gauge are given by
(9)Bx=By=0,Bz=B(x)=-W¯′(x),
where W¯(x) is an arbitrary function of x and W¯′(x)=dW¯(x)/dx. The Pauli Hamiltonian then takes the form
(10)H=px2+py-W¯x2-W¯′(x)σz.
The variables x and y can be separated by introducing the wavefunction
(11)ψ(x,y)=e-ikyϕ(x).
Here k (-∞<k<∞) are the eigenvalues of py, and ϕ(x)=ϕ+(x)ϕ-(x) is a two-component function of x. Then the Hamiltonian, H, can be written as
(12)H∓=-d2dx2+W¯(x)+k2∓W¯′(x).It is obvious that the Hamiltonian has been written in the one-dimensional form of the Schrödinger equation with the potential. This final form is the SUSY form of the Pauli equation with the superpotentialW(x)=W¯(x)+k.Now, our aim is to determine the magnetic fields that lead to the QES potential forms. By defining the magnetic fields in appropriate forms that construct QES potentials, we will try to apply AIM to the resultant Hamiltonian.


### 3.1. Class I

In order to obtain the Class I potential, the corresponding magnetic field can be defined as(13)B=-W¯′x=-ae-αx-beαx.
In this case we can write the following eigenvalue problem Hψ(x)=Eψ(x) form (12) for H- part that satisfies the Schrödinger equation under the normalizable zero energy state condition
(14)-d2dx2+Vxψx=Eψx,
where
(15)Vx=1α2a2exp⁡-2αx+b2exp⁡⁡2αxhhhhii-a2kα+α2exp⁡-αxhhhhii+b2kα-α2exp⁡αx+k2α2-2ab
is the Class I potential according to Turbiner’s classification in [6]. After redefining the variable as z=exp⁡(αx) and using ψ=z-1/2χ(z) in (14), we get the following differential equation:
(16)-d2dz2+A2z4+B2-Cz3-Dz+γγ+1z2χ(z)=0,
where A=a/α2, B=b/α2,  C=a(2k+α)/α3,  D=b(-2k+α)/α3, and γ(γ+1)=((α2(k2-E)-2ab)/α4)-1/4. Before applying AIM to (16), we have to obtain the asymptotic form for χ(z). It is easy to see that when z→0 and z→∞, χ(z)~zs+1exp⁡-A/z-Bz, and after substituting χ(z)=zs+1exp⁡-A/z-Bzf(z) into (16), we obtain
(17)f′′=2B-s+1z-Az2f′+uz-vz3f,
where u=2s+1B-D, v=2sA+C, and s(s+1)=γ(γ+1)+2BA. Now, we can apply AIM to this final equation with λ0=2B-(s+1)/z-A/z2 and s0=(u/z)-((2sA+C)/z3) and we will show that (17) has exact solutions under some constraints on the parameters A, B, C, and D.If we apply AIM quantization condition, that is, from (4) by substituting λ0 and s0, in order to obtain polynomial solutions we see that the parameters must be equal to u=-2nB, v=-2A and s values are given in Table 1; we will show the results for type 1 and a=b=α=1 for simplicity. Note that there are two types of solutions that give us the polynomials, for example, for n=2, v=-2A (as given above), and v=(6/B)(s+2)(2s+3)-AB. If we use the first value of v we get the results given in Table 1, but one can use the second value of s and get different results which give us the polynomials. We will not give this second situation here because of complication on the calculations. In this paper, we use only v=-2A type solution. If we use u=2s+1B-D,  v=2sA+C,  s(s+1)=γ(γ+1)+2BA,  u=-2nB, and v=-2A, we have D=2(n+s+1)B and C=-2(s+1)A. If we go back to definitions of A, B, C, and D, we find the following expression for the energy as follows:
(18)Ens=n(1+n+2s)α4+2ab(1+α2)α2.
If the parameters in (18) are mapped to the parameters in energy values in [10, 11], the energy values can be in accord. In Table 1, we will show the results for type 1 and a=b=α=1 for simplicity. Similarly, in Table 2, we will show the results of different values of s for A=B=1 for type 2 solution.Table 1
Type 1 energy values whenA=B=1.
n
u
=
-
2
n
B
v
=
-
2
A
s
E
n
s
E
n
s
k
0
0
-
2
-
1
4
Infinity
1
-
2
-
2
-
2
-
0.5
2
5
0
-
8
2
-
4
-
2
-
2.63571
-
0.54284
-
51.3481
-
1.71698
3.13208
-
0.790868
-
0.14731
9.41074
0.997333
3
-
6
-
2
-
3.12512
-
2.75072
-
11.7596
-
5
/
2
1.0
-
1.77778
-
1.51563
6.90622
0.627819
0.14075
16.8445
0.998845
4
-
8
-
2
-
3.54136
-
4.33088
-
4.89446
-
3.10868
-
0.86944
-
1.70161
-
2.39350
4.852
0.15676
-
1.34675
13.226
0.864101
0.390287
27.1223
0.994757
5
-
10
-
2
-
3.92981
-
5.2981
-
2.60348
-
3.61617
-
2.1617
-
1.30116
-
3.07259
3.2741
-
0.101646
-
2.29765
11.0235
0.614865
-
1.1973
22.027
0.937851
0.613514
0.1351
0.991394Table 2
Type 2 energy values whenA=B=1.
n
u
=
-
2
n
B
v
s
0
0
-
2
-
1.0
1
-
2
-
2
-
2.0
-
0.5
2
-
4
6
(
s
+
2
)
(
2
s
+
3
)
-
2.5
-
2.22871
-
2.0
-
1.27129
-
1.0
3
-
6
4
(
12
s
3
+
72
s
2
+
125
s
+
54
)
10
s
+
23
3.13571
-
2.80396
-
2.63188
-
2.5
-
2.21698
-
2.11304
-
1.86812
-
1.08301
-
0.647315


### 3.2.  Class IX

Let us assume that the magnetic field takes the form of(19)B=-W¯′(x)=ax-3+bx-2.
With this magnetic field, we can write the following eigenvalue problem for H- after replacing x by z:
(20)-d2dz2+A2z4+B2-Cz3-Dz+Fz2χ(z)=0,
where A=a, B=k2-E, C=2a(1+b), D=2bk, and F=b(b+1)+2ak. This potential form is the Class IX potential form. By applying the same procedure performed in the previous potential form, when z→0 and z→∞, χ(z)~zs+1exp⁡-A/z-Bz, and after substituting χ(z)=zs+1exp⁡-A/z-Bzf(y) into (20), we have
(21)f′′=2B-s+1z-Az2f′+uz-vz3f.
Here u=2s+1B-D, v=2sA+C, and s(s+1)=F+2BA. In this form of differential equation, if we apply AIM to this final equation with λ0=2B-(s+1)/z-A/z2 and s0=u/z-(2sA+C)/z3, we will obtain the exact solutions of (20) under some constraints on the parameters A, B, C, and D. It can be easily seen that the solutions of this form of potential are equal to the solution of the first example except for definitions of A, B, C, and D. In this type of potential, the definitions of Ensk and b for the type 1 solution can be obtained as below:
(22)b=-s,Ensk=k2--ksn+s+12.


## 3.1. Class I

In order to obtain the Class I potential, the corresponding magnetic field can be defined as(13)B=-W¯′x=-ae-αx-beαx.
In this case we can write the following eigenvalue problem Hψ(x)=Eψ(x) form (12) for H- part that satisfies the Schrödinger equation under the normalizable zero energy state condition
(14)-d2dx2+Vxψx=Eψx,
where
(15)Vx=1α2a2exp⁡-2αx+b2exp⁡⁡2αxhhhhii-a2kα+α2exp⁡-αxhhhhii+b2kα-α2exp⁡αx+k2α2-2ab
is the Class I potential according to Turbiner’s classification in [6]. After redefining the variable as z=exp⁡(αx) and using ψ=z-1/2χ(z) in (14), we get the following differential equation:
(16)-d2dz2+A2z4+B2-Cz3-Dz+γγ+1z2χ(z)=0,
where A=a/α2, B=b/α2,  C=a(2k+α)/α3,  D=b(-2k+α)/α3, and γ(γ+1)=((α2(k2-E)-2ab)/α4)-1/4. Before applying AIM to (16), we have to obtain the asymptotic form for χ(z). It is easy to see that when z→0 and z→∞, χ(z)~zs+1exp⁡-A/z-Bz, and after substituting χ(z)=zs+1exp⁡-A/z-Bzf(z) into (16), we obtain
(17)f′′=2B-s+1z-Az2f′+uz-vz3f,
where u=2s+1B-D, v=2sA+C, and s(s+1)=γ(γ+1)+2BA. Now, we can apply AIM to this final equation with λ0=2B-(s+1)/z-A/z2 and s0=(u/z)-((2sA+C)/z3) and we will show that (17) has exact solutions under some constraints on the parameters A, B, C, and D.If we apply AIM quantization condition, that is, from (4) by substituting λ0 and s0, in order to obtain polynomial solutions we see that the parameters must be equal to u=-2nB, v=-2A and s values are given in Table 1; we will show the results for type 1 and a=b=α=1 for simplicity. Note that there are two types of solutions that give us the polynomials, for example, for n=2, v=-2A (as given above), and v=(6/B)(s+2)(2s+3)-AB. If we use the first value of v we get the results given in Table 1, but one can use the second value of s and get different results which give us the polynomials. We will not give this second situation here because of complication on the calculations. In this paper, we use only v=-2A type solution. If we use u=2s+1B-D,  v=2sA+C,  s(s+1)=γ(γ+1)+2BA,  u=-2nB, and v=-2A, we have D=2(n+s+1)B and C=-2(s+1)A. If we go back to definitions of A, B, C, and D, we find the following expression for the energy as follows:
(18)Ens=n(1+n+2s)α4+2ab(1+α2)α2.
If the parameters in (18) are mapped to the parameters in energy values in [10, 11], the energy values can be in accord. In Table 1, we will show the results for type 1 and a=b=α=1 for simplicity. Similarly, in Table 2, we will show the results of different values of s for A=B=1 for type 2 solution.Table 1
Type 1 energy values whenA=B=1.
n
u
=
-
2
n
B
v
=
-
2
A
s
E
n
s
E
n
s
k
0
0
-
2
-
1
4
Infinity
1
-
2
-
2
-
2
-
0.5
2
5
0
-
8
2
-
4
-
2
-
2.63571
-
0.54284
-
51.3481
-
1.71698
3.13208
-
0.790868
-
0.14731
9.41074
0.997333
3
-
6
-
2
-
3.12512
-
2.75072
-
11.7596
-
5
/
2
1.0
-
1.77778
-
1.51563
6.90622
0.627819
0.14075
16.8445
0.998845
4
-
8
-
2
-
3.54136
-
4.33088
-
4.89446
-
3.10868
-
0.86944
-
1.70161
-
2.39350
4.852
0.15676
-
1.34675
13.226
0.864101
0.390287
27.1223
0.994757
5
-
10
-
2
-
3.92981
-
5.2981
-
2.60348
-
3.61617
-
2.1617
-
1.30116
-
3.07259
3.2741
-
0.101646
-
2.29765
11.0235
0.614865
-
1.1973
22.027
0.937851
0.613514
0.1351
0.991394Table 2
Type 2 energy values whenA=B=1.
n
u
=
-
2
n
B
v
s
0
0
-
2
-
1.0
1
-
2
-
2
-
2.0
-
0.5
2
-
4
6
(
s
+
2
)
(
2
s
+
3
)
-
2.5
-
2.22871
-
2.0
-
1.27129
-
1.0
3
-
6
4
(
12
s
3
+
72
s
2
+
125
s
+
54
)
10
s
+
23
3.13571
-
2.80396
-
2.63188
-
2.5
-
2.21698
-
2.11304
-
1.86812
-
1.08301
-
0.647315


## 3.2.  Class IX

Let us assume that the magnetic field takes the form of(19)B=-W¯′(x)=ax-3+bx-2.
With this magnetic field, we can write the following eigenvalue problem for H- after replacing x by z:
(20)-d2dz2+A2z4+B2-Cz3-Dz+Fz2χ(z)=0,
where A=a, B=k2-E, C=2a(1+b), D=2bk, and F=b(b+1)+2ak. This potential form is the Class IX potential form. By applying the same procedure performed in the previous potential form, when z→0 and z→∞, χ(z)~zs+1exp⁡-A/z-Bz, and after substituting χ(z)=zs+1exp⁡-A/z-Bzf(y) into (20), we have
(21)f′′=2B-s+1z-Az2f′+uz-vz3f.
Here u=2s+1B-D, v=2sA+C, and s(s+1)=F+2BA. In this form of differential equation, if we apply AIM to this final equation with λ0=2B-(s+1)/z-A/z2 and s0=u/z-(2sA+C)/z3, we will obtain the exact solutions of (20) under some constraints on the parameters A, B, C, and D. It can be easily seen that the solutions of this form of potential are equal to the solution of the first example except for definitions of A, B, C, and D. In this type of potential, the definitions of Ensk and b for the type 1 solution can be obtained as below:
(22)b=-s,Ensk=k2--ksn+s+12.


## 4.  Results and Discussion

The energy obtained for Class IX potential in the form of parametersA, B, C, D, and F agrees with those in [20, 21]. Similar to the Class I potential, the values obtained for the second type solution in Table 2 are valid for this potential. The energy in (22) depends only on k, s, and n. For simplicity, the energy values for k=1 are also tabulated in Table 1.Similar to the procedure presented in this study, it is possible to obtain the same solutions for other class potentials described by Turbiner [6] except for Class IV and Class V potentials. For these two potentials, an appropriate superpotential to construct the potentials has not been written because of the constant k in the W¯(x)+k. The other six classes satisfy the requirements of using AIM which is the differential equations conditions being transformable to a specific form in (2), with quite restrictive asymptotic conditions on the coefficient functions of α(x) in (5).


## 5. Conclusions

In this work, we have shown that the AIM method can also be used to obtain the QES energies obtained in [10, 11] by using the known forms of the magnetic field B with asymmetric gauge obtained in the same paper. We follow these corresponding studies and consider the magnetic field in that form. The procedure presented here makes full use of the close connection between QES systems and SUSY-like form. Additionally, an attempt to give the QES solution of the Pauli equation by using asymptotic iteration technique is presented. The application of AIM is considered for Class I and Class IX QES systems. By the same procedure, AIM can be applied to the other QES potentials in one dimension except for Class IV and Class V. The whole spectrum of the QES systems is presented.Finally, we will try to apply this method to other QES potentials in all branches of physics.



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*Source: 304093-2015-01-29.xml*