Annals of Physics 333 (2013) 323–334

Contents lists available at SciVerse ScienceDirect

Annals of Physics

journal homepage: www.elsevier.com/locate/aop

A position-dependent mass model for theThomas–Fermi potential: Exact solvability andrelation to δ-doped semiconductorsAxel Schulze-Halberg a,∗, Jesús García-Ravelo b, Christian Pacheco-García b,José Juan Peña Gil ca Department of Mathematics and Actuarial Science, Indiana University Northwest, 3400 Broadway, Gary IN 46408, USAb Escuela Superior de Física y Matemáticas, Instituto Politécnico Nacional, Unidad Profesional Adolfo López Mateos, Zacatenco,07738 México D.F., Mexicoc Universidad Autónoma Metropolitana - Azcapotzalco, CBI - Area de Física Atómica Molecular Aplicada, Av. San Pablo 180, ReynosaAzcapotzalco, 02200 México D.F., Mexico

a r t i c l e i n f o

Article history:Received 16 October 2012Accepted 7 December 2012Available online 21 December 2012

Keywords:Position-dependent massSchrödinger equationThomas–Fermi potentialδ-doped semiconductor

a b s t r a c t

We consider the Schrödinger equation in the Thomas–Fermi field,a model that has been used for describing electron systems inδ-doped semiconductors. It is shown that the problem becomesexactly-solvable if a particular effective (position-dependent)massdistribution is incorporated. Orthogonal sets of normalizablebound state solutions are constructed in explicit form, and theassociated energies are determined. We compare our resultswith the corresponding findings on the constant-mass problemdiscussed by Ioriatti (1990) [13].

© 2012 Elsevier Inc. All rights reserved.

1. Introduction

Effective-mass theory has been widely used in solid-state physics, in particular for studyingimpurity states in semiconductors [1–4], allowing to reformulate various existing models to interpretexperimental data [5–7]. As an example let us mention the study of δ-doped semiconductors, whichprovides an important application of effective-mass theory. In particular, δ-doped quantumwells canhave two-dimensional conductive channels with a high density of charge carriers. In the special caseof GaAs-based systems, the n-type δ-doped quantum wells have been fabricated using Si atoms as

∗ Corresponding author. Tel.: +1 219 980 6780.E-mail addresses: xbataxel@gmail.com, axgeschu@iun.edu (A. Schulze-Halberg).

0003-4916/$ – see front matter© 2012 Elsevier Inc. All rights reserved.doi:10.1016/j.aop.2012.12.002

324 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

donor impurities [8,9]. In the simplest case, electron quantum states of semiconductor structurescan be analyzed by resolving the corresponding one-dimensional Schrödinger and Poisson equationsin a self-consistent manner [10–12]. An alternative approach to analyze the sub-band structureof the above systems is to employ the Thomas–Fermi local-density-functional approximation inone dimension, leading to an effective-mass Schrödinger equation [13,14]. The latter approach hasbeen used to study the influence of hydrostatic pressure on the electron states of δ-doped GaAsstructures [15]. Furthermore, the effect of a position-dependent mass on the electron states inquantumwells has been discussed and the results have been compared with those where the systemfor a constantmass is considered [16–22].While the vastmajority of the previouslymentioned resultsare based on approximations, in this work we present the analysis of an exactly-solvable modelwithin the framework of position-dependentmass formulation and Thomas–Fermi theory of δ-dopedsemiconductor structures. In particular, we construct bound-state energies and closed-form solutionsof a position-dependent mass Schrödinger equation for a Thomas–Fermi potential, comparing ourresults with their counterparts for constant mass [13]. The remainder of this paper is organized asfollows. Section 2 introduces our effective-mass equation and its general solution, which is thenadjusted to the boundary conditions dictated by our physical model (Section 3). Depending on theparity of our solutions, the latter boundary conditions are either of Dirichlet or of Neumann type.

2. The position-dependent mass model

We consider the following one-dimensional Schrödinger equation on the whole real line,

12m

Ψ ′′−

m′

2m2Ψ ′

+ (E − V ) Ψ = 0, (1)

where Ψ = Ψ (x) stands for the solution,m = m(x) denotes the nonnegative effective mass function,V = V (x) represents the potential and the real-valued constant E is the stationary energy. In thepresent work we focus on the Thomas–Fermi potential

V = −V0

(β|x| + c)4, (2)

for positive constantsβ, V0 and c .When dealingwith explicitmodels in the next section, Eq. (1)will beequippedwith boundary conditions of Dirichlet or Neumann type. Now, the solutionΨ of our effectivemass equation (1) will undergo the point transformation

Ψ (x) =

m2

14Φ(u(x)) u(x) =

x2 m(s) ds. (3)

Assuming the coordinate change u to be invertible with inverse x = u−1, the transformation (3)converts our initial equation (1) into the form of a conventional Schrödinger equation:

Φ ′′+

E +

V0

(β |x| + c)4+

x′′

2 x′

2

−x′′′

2 x′

Φ = 0. (4)

This equation can be employed as follows [23]. First, a coordinate change x = x(u) is chosen, such thatthe transformed equation (4) can be solved forΦ . The solution of our initial equation (1) is then givenaccording to the rule given by the point transformation (3). At the same time, the change of coordinatex = x(u) determines the effective mass m in (1) by differentiating the coordinate change in (3) andmaking use of the inverse function theorem u′

= 1/x′:

m(x) =1

2 x′[u(x)]. (5)

Clearly, the principal idea behind our point transformation formalism is the fact that solvable casesof our conventional Schrödinger equation like (4) are much easier to find than for the effective mass

A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334 325

equation (1). We now inspect the coefficient of Φ in the transformed equation (4). Clearly, the lasttwo terms of that coefficient turn constant if x is exponential in u. We will therefore look for a changeof coordinates x = x(u) that fulfills the following equation:

β |x| + c = c exp(β u), (6)

where u ≥ 0 and the particular choice of coefficients was made in order to simplify subsequentcalculations. It is immediate to see that the following function

u =1β

logβ

c|x| + 1

, (7)

solves Eq. (6), complies with (3) and can be inverted on both intervals (−∞, 0) and (0,∞), wherethe respective inverse function x = u−1 on each of these interval satisfies

|x| =cβ

[exp(βu)− 1]. (8)

Wewill nowdetermine the transformed equation (4) by substituting (8). The derivative terms becomex′′

2x′

2

−x′′′

2x′= −

β2

4,

note that the sign of x does not play a role, since we are dealing with ratios. Thus, the coordinatechange (6) renders our transformed equation (4) in the form

Φ ′′+

E +

V0

c4exp(−4 β u)−

β2

4

Φ = 0. (9)

Note that despite the piecewise definition of our coordinate change (6), the transformed equation (9)is given by a single expression. This is due to the fact that the signs, which differ in the two lines of (6),cancel out when the latter definition is plugged into (4). Now, the Schrödinger equation (9) admits itsgeneral solution to be in closed form:

Φ = c1 J√β2−4E4β

√V0

2 β c2exp(−2 β u)

+ c2 Y√

β2−4E4β

√V0

2 β c2exp(−2 β u)

. (10)

Here, J andY stand for the Bessel functions of the first kind [19],while c1, c2 denote arbitrary constants.While the function (10) is a solution of the transformed equation (10), we must reverse our pointtransformation (3) in order to obtain the sought solutionΨ of our initial equation (1). Beforewe can doso, we need to construct the effective mass distributionm that is needed in our point transformation.Since we know that u is given by (7), we obtain the mass m by differentiating the coordinate changein (3) and solving form:

m =12(u′)2 =

12 (β|x| + c)2

. (11)

The graph of this function can be inspected in Fig. 1. Note that we are using reduced units, that is,distances and masses are measured in terms of Bohr’s radius a0 and twice the electron mass 2me.In the strict sense, m is not defined at x = 0, as the coordinate change u cannot be differentiatedthere. It is therefore understood that we definem(0) = 1/(2c2). Now that we have the effective massdistribution, we can insert it, together with (7) and (10), into (3), which gives the general solution Ψof our initial equation (1):

Ψ = c1ψ1 + c2ψ2, (12)

326 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

Fig. 1. The position-dependentmass function as defined in (11) for the settingsβ =2

15π , c = (5π)−45 23

35 , andV0 =

215π

2.

where the two linearly independent solutions ψ1 and ψ2 are given by

ψ1 =

1

β|x| + cJ√

β2−4E4β

√V0

2β(β|x| + c)2

(13)

ψ2 =

1

β|x| + cY√

β2−4E4β

√V0

2β(β|x| + c)2

,

note that some irrelevant factors have been absorbed into the constants c1 and c2 of (12). The initialequation (1) reads after insertion of our effective mass distribution (11)

(β|x| + c)2Ψ ′′(x)+ 2β(β|x| + c)[θ(x)− θ(−x)]Ψ ′+

E +

V0

(β|x| + c)4

Ψ = 0. (14)

Here, θ stands for the Heaviside function, which appears due to the fact that the mass m is notdifferentiable at x = 0. As a consequence, the coefficient of Ψ ′ in Eq. (14) is not continuous there,but exhibits a step from negative to positive values.

3. Bound state solutions

The next step consists in satisfying physical boundary conditions, that is, finding particular casesof the general solution (12), which resemble the behavior of quantum-mechanical bound states fornegative energy values. Since we want these states to be defined on the whole real line, the soughtsolutions must vanish at the infinities. As we will show now, this requirement forces c2 = 0 in (12):consider the series expansion of the particular solution ψ2 around x = ∞. The leading term is givenby

1β|x| + c

Y√β2−4E4β

√V0

2β (β|x| + c)2

= a x

−12 +

14 −

Eβ2

+ b x−

12 −

14 −

Eβ2

+ Ox−

32

, (15)

where a and b stand for nonzero constants that involve the parameters β, c and V0. Since we assumeE to be a negative number, the first exponent on the right hand side of (15) satisfies the estimate

−12

+

14

−Eβ2

= −12

+

14

+|E|

β2> 0.

Thus, the first term on the right hand side of (15) is always unbounded at x = ∞, such that thissolution must be discarded using c2 = 0. Now, in order to find out how the remaining solution ψ1

A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334 327

behaves at x = ∞, we find the leading term of its series expansion there:1

β |x| + cJ√

β2−4E4β

√V0

2 β (β |x| + c)2

= c x

−12 −

14 −

Eβ2

+ Ox−

32

, (16)

where the constant c depends on the parameters β, c and V0. Since the exponent on the right handside of (16) fulfills the estimate (recall E < 0)

−12

−

14

−Eβ2

= −12

−

14

+|E|

β2< 1, (17)

the solution ψ1 stays bounded at x = ∞, behaving there like x−α for a constant α > 1. Therefore, ψ1is integrable at infinity, which together with the fact that ψ1 is continuous on the positive half-lineand bounded at x = 0, gives ψ1 ∈ L2(0,∞). Following a common approach, we will now require ourmodel to exhibit parity, see [13] and references therein. Besides being a critical step in the presentconsiderations, parity of the solution ψ1 will automatically ensure the L2-norm to be finite over thewhole real line. Finally, let us rename our solution ψ1 to Ψ , since we will only be dealing with thissolution and not consider ψ2 anymore.

3.1. Odd-parity states

We can now start by imposing odd parity, implying the condition Ψ (0) = 0 that results in thefollowing explicit condition:

J 14 −

Eβ2

√V0

2 β c2

= 0. (18)

The parametersβ, c andV0 are fixed, thereforewewill use our energy E to complywith condition (18),which has a finite number of pairwise different solutions in E, as is well-known [19]. These solutions,as well as their number, and their dependence on the Bessel function’s argument cannot be givenin closed form. Instead, one has to rely on numerical approximations. Let us therefore assume that(18) has n pairwise different solutions for E, which we denote by E1, E3, E5, . . . , E2n−1. The reasonfor choosing odd indices is to denote that these energies belong to solutions with odd parity, whichwill be established below. Without restriction we can further assume that the energies are labeled inascending order, such that Ej < Ej+1, j = 1, 3, . . . , 2n − 1. In case (18) has no solutions, obviouslyno energy needs to be introduced at all. Now, to each of the energies Ej, j = 1, 3, . . . , 2n − 1, thereis a corresponding solution Ψj of our effective position-dependent mass equation (14) that can beextracted from (13):

Ψj =

1

β |x| + cJ

14 −

Ejβ2

√V0

2 β (β |x| + c)2

, j = 1, 3, . . . , 2n − 1. (19)

Observe that our energy on the right hand side is now given by the value Ej. Although the solution(19) is formally defined on the whole real line, at this point wewill keep focusing on the positive half-line and zero until all necessary properties of our solutions have been found. One of these propertiesconcerns the number of nodes that a solution Ψj exhibits. While usually this can be found out bymeans of standard results from Sturm–Liouville theory, the present problem is irregular due to thediscontinuous coefficient in (14) and due to the unbounded domain. Therefore, we cannot employthe Sturm–Liouville formalism, but must proceed by hand. First, it is immediate to see that in (19)only the Bessel function itself contributes nodes, as the factor in front does not vanish for any x. Next,we observe that the Bessel function’s argument is a strictly decreasing function on the positive half-line, taking its maximum at x = 0. Since for any real j the Bessel function x → Jj(x) has only a finitenumber of nodes in any interval of the form [0, k], k > 0, we conclude that each solutionΨj must have

328 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

a finite number of nodes on the positive half-line, plus one at x = 0 due to our condition (18). Let usnow consider the solution Ψ1, that is, the solution (19) corresponding to the energy value E1 < 0. Byconstruction this is the lowest value for E, such that the function Ψ1 vanishes at x = 0. At the sametime, this lowest value E1 generates the highest order of the Bessel function J in (19) among all otherenergies Ej, j = 3, 5, . . . , 2n − 1, that fulfill the condition (18). Now, it is well-known that increasingthe index of the Bessel function x → Jj(x) for any real j moves its nodes away from x = 0 [19].Consequently, increasing the index of the Bessel function in our solution (19)moves thenodes towardsx = 0, as the Bessel function’s argument is strictly decreasing in x. Consequently, if E1 generatesalready the highest order of the Bessel function J , then the solution Ψ1 cannot have any nodes exceptthe one at x = 0. Now, the same argumentation can be applied to the energy E3 with the result thatthe corresponding solutionΨ3 has two nodes, one of which is at x = 0. Straightforward generalizationof this argument yields that the solutionΨj, j = 1, 3, . . . , 2n−1, as given in (19), has exactly (j+1)/2nodes, one of which is at x = 0. Another property of the solutions Ψj, j = 1, 3, . . . , 2n − 1, is theirorthogonality on the positive half-axis with respect to the standard internal product on L2(0,∞),that is,

∞

0Ψj Ψk dx = 0, j, k = 1, 3, . . . , 2 n − 1, j = k. (20)

We will now prove this relation through an explicit calculation. To this end, let us take the explicitform of our solutions as given in (19) and substitute it into the left hand side of (20):

∞

0Ψj Ψk dx =

∞

0

1β |x| + c

J14 −

Ejβ2

√V0

2 β (β |x| + c)2

J

14 −

Ekβ2

√V0

2 β (β |x| + c)2

dx. (21)

Wewill proceedmanually, applying the following change of variable t = t(x) and its inverse x = t−1,given by

t =

√V0

2β(β|x| + c)2x =

V140

2β3t−

cβ, (22)

recall that we are still operating solely on the positive half-axis at this point. Clearly, out variablechange is defined for x ∈ (0,∞) and t ∈ (0, 1), where x = 0 (this and subsequent statements in thelimit sense) implies t = 0 and x = ∞ implies t = 0. These relations will be important once we adjustthe integration limits in (21) to the new variable t . Before we do so, let us observe that (22) gives

dx = −V

140

(2βt)32dt. (23)

Now, insertion of this relation, together with (22), into (21) gives the result∞

0Ψj Ψk dx =

12β

1

0

1tJ

14 −

Ejβ2

(t)J14 −

Ekβ2

(t)dt, (24)

where in the last step we have compensated the negative sign in (23) by exchanging the upper andlower limit of integration. The integral (24) can be evaluated explicitly [20], giving the following,seemingly complicated, result:

∞

0Ψj Ψk dx = 2c2(Ek − Ej)J 1

4 −Ejβ2

(0)J14 −

Ekβ2

(0) (25)

+2

β2 − 4Ej +β2 − 4Ek

J14 −

Ejβ2

√V0

2βc2

J

14 −

Ekβ2

√V0

2βc2

A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334 329

+

√V0

c2(Ej − Ek)J1+

14 −

Ejβ2

√V0

2βc2

J

14 −

Ekβ2

√V0

2βc2

−

√V0

c2(Ej − Ek)J

14 −

Ejβ2

√V0

2βc2

J1+

14 −

Ekβ2

√V0

2βc2

. (26)

Each term of the latter expression vanishes identically, simply because of our boundary condition(18) and since the Bessel function J for positive order becomes zero at x = 0. Note that the result(26) is only valid for j = k, as otherwise several denominators would become zero. We have thereforeproved that the solutions given in (19) form a finite, orthogonal set in the space L2(0,∞). Now weare ready to retake the issue of extending our problem’s domain to the whole real line. As far as ourposition-dependent mass equation (14) is concerned, we can simply consider it for all x ∈ R, since allcoefficients remain nonsingular. The solutions (19) will be adjusted as follows:

Ψj = [θ(x)− θ(−x)]

1

β|x| + cJ

14 −

Ejβ2

√V0

2β(β |x| + c)2

, j = 1, 3, . . . , 2n − 1, (27)

recall that θ denotes the Heaviside function, the effect of which is simply to establish odd parity. Sinceall solutions vanish at x = 0, the continuation (27) through there is smooth, such that these functionssolve our position-dependent mass equation (14) on the whole real line. Note that orthogonality isnot affected by the adjusted definition (27), as the integral (24) on the x-interval (0,∞) yields thesame result as on the x-interval (−∞, 0) except for the sign. A property that does change with (27),is the number of nodes that each solution exhibits. This number increases, because parity causes thenodes on the positive half-line to be reflected onto the negative half-line. If we assume to have n − 1nodes on the positive half-line and one node at x = 0, we get n − 1 nodes more on the negative half-line. Hence, the solution Ψ1, redefined according to (27), would still have only one node at x = 0. Thesolution Ψ3 has now one node at x = 0, one node for x > 0 and one node for x < 0, which makes atotal of three nodes. Straightforward generalization shows that the solutionΨj, j = 1, 3, . . . , (2n−1),has a total of j nodes on the real line, one of which is at x = 0. In particular, each solution in (27) hasan odd number of nodes. Therefore, we will now proceed with the construction of solutions to ourposition-dependent mass equation (14) that have even parity and thus exhibit an even number ofnodes.

3.2. Even-parity states

The general way of obtaining even-parity solutions is very similar to the case of odd parity, in thefollowingwewill therefore shorten our argumentswhere convenient. First, we go back to the solutionΨ in its form (13), defined on the positive half-line. This time, instead of imposing (18), we require

Ψ ′(0) = 0, (28)

where the derivative is understood to be from the right, since at this point we are considering thepositive half-line only. If (28) is fulfilled, we can continue a solution Ψ smoothly through x = 0, suchthat it becomes defined on the whole real line. Before we do so, we must comply with condition (28)that reads explicitly after substitution of (13)

ddx

1

β |x| + cJ√

β2−4E4β

√V0

2 β (β |x| + c)2

x=0

= 0 (29)

❀2

√V0

c2J 1

4 −Eβ2

√V0

2 β c2

− (β +

β2 − 4 E) J

1+

14 −

Eβ2

√V0

2 β c2

= 0. (30)

Since the parameters β, c and V0 are fixed, we must solve this condition for our energy E. As we haveseen in the context of (18), our function Ψ to be differentiated in (29) is a smooth function that has

330 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

a finite number of nodes and as such a finite number of extrema. Since (28) can be interpreted as acondition for Ψ having an extremum in x = 0, it must have a finite number of solutions for E. Thisis so, because varying the order of our Bessel function J makes its nodes and extrema shift. We cantherefore assume that Eq. (28) has n pairwise different solutions thatwe label as E0, E2, E4, . . . , E2n−2.Without restriction we can further suppose that E0 < E1 < · · · < E2n−2 < 0. Clearly, these solutionscannot be obtained in closed form, but must be approximated numerically. To each of the energies Ejthere is a corresponding solution Ψj, j = 0, 2, . . . , 2n − 2, that can be extracted from (13) as

Ψj =

1

β |x| + cJ

14 −

Ejβ2

√V0

2 β (β |x| + c)2

, j = 0, 2, . . . , 2n − 2. (31)

Note that the energy on the right hand side is now given by a solution Ej of our condition (28). Sincethe solution (31) has the same form as its counterpart in (19), its number of nodes can be determinedby the samemethod. The only difference between the two cases lies in the fact that the solutions (19)vanish at zero due to (18), while their counterparts (31) do not. Therefore, by the argumentation givenin the context of (19), we conclude that each functionΨj, j = 0, 2, . . . , 2n−2, has exactly j/2 nodes onthe positive half-line. Next, we will establish orthogonality of the functions in (31) with respect to thecommon internal product in L2(0,∞). For two functions Ψj and Ψk, j, k = 0, 2, . . . , (2n − 2), j = k,wemust evaluate the integral in (26), the form of which remains exactly the same as for the functions(19). The only difference is that the energies on the right hand side of (26) are now given by solutionsof our present condition (28). Consequently, since (18) is not valid anymore, we must find a new wayto show that (26) equals zero. As a first step we observe that the expressions in line (25) on the righthand side of (20) vanish, because the Bessel function’s order is positive (since Ej, Ek < 0), such that itevaluates to zero at vanishing argument. In order to deal with the remaining terms on the right handside of (26), we rewrite our condition (30) as follows:

J14 −

Ejβ2

√V0

2 β c2

=

1

2 c2√V0 (β +

β2 − 4 Ej)

J1+

14 −

Ejβ2

√V0

2 β c2

. (32)

If we now substitute this into the result of our integral (26) without simplifying, we obtain thefollowing lengthy expression (recall that the terms on the right hand side of line (25) are zero):

∞

0Ψj Ψk dx

=2

β2 − 4 Ej +β2 − 4 Ek

1

2 c2√V0 (β +

β2 − 4 Ej)

J1+

14 −

Ejβ2

√V0

2 β c2

×

1

2 c2√V0 (β +

β2 − 4 Ek)

J1+

14 −

Ekβ2

√V0

2 β c2

+

√V0

c2 (Ej − Ek)

× J1+

14 −

Ejβ2

√V0

2 β c2

1

2 c2√V0 (β +

β2 − 4 Ek)

J1+

14 −

Ekβ2

√V0

2 β c2

−

√V0

c2 (Ej − Ek)

1

2 c2√V0 (β +

β2 − 4 Ej)

J1+

14 −

Ejβ2

√V0

2 β c2

× J1+

14 −

Ekβ2

√V0

2 β c2

. (33)

The latter expression contains only two different Bessel functions instead of the four in (24). Aftercollecting coefficients, we get the more concise form

A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334 331∞

0Ψj Ψk dx =

2 V0

c4 (Ej − Ek)β +

β2 − 4 Ek

−2 V0

c4 (Ej − Ek)β +

β2 − 4 Ej

+

8 V0

c4β +

β2 − 4 Ej

β +

β2 − 4 Ek

β2 − 4 Ej +

β2 − 4 Ek

× J1+

14 −

Ejβ2

√V0

2 β c2

J1+

14 −

Ekβ2

√V0

2 β c2

= 0, (34)

where in the last step we simply joined terms of the coefficient in square brackets. We have thereforeshown that the solutions in (31) form an orthogonal set with respect to the usual inner product inL2(0,∞). We can now extend the domain of our solutions to the whole real line by simply admittingall real values for x in (31). Since Ψ ′(0) vanishes, the solutions will have even parity and continuesmoothly through x = 0. Orthogonality is clearly maintained on the whole real line. The numberof nodes that a solution exhibits, will double once the solution is defined for all x, such that Ψj, j =

2, 4, . . . , 2n − 2, has exactly j nodes.

3.3. Summary

We have now completed our analysis of the position-dependent mass equation (14) under theboundary conditions that the solutions must vanish at the infinities and exhibit parity. We find from(27) and (31) that the entire set of solutions Ψj ∈ L2(R), j = 0, 1, 2, . . . , of Eq. (14) reads

Ψj =

1

β |x| + cJ

14 −

Ejβ2

√V0

2 β (β |x| + c)2

if j = 0, 2, 4, . . .

[θ(x)− θ(−x)]

1

β |x| + cJ

14 −

Ejβ2

√V0

2 β (β |x| + c)2

if j = 1, 3, 5, . . .

.Note that the definite parity of the solutions is required in the sameway as in [13]. The correspondingenergy eigenvalues cannot be given in closed form, but must be obtained numerically from Eqs. (18)and (30) for odd and even parity states, respectively.

4. Application

We will now come up with an example motivated by the studies of δ-doped semiconductorstructures [13]. More precisely, we will use our results to compute explicit energy eigenvalues(subband bottom energies) and their corresponding bound states for a specific set of parameters β, cand V0, taken from the above reference:

β =2

15 πc =

1

(5 π)45

23

35

V0 =

2

15 π

2

. (35)

We are now ready to determine our energies Ej, j = 0, 1, 2, . . . . To this end, we first solve thecondition (30) that applies to the even-parity solutions, that is, to functions Ψj with even index j.The first three solutions E0, E2 and E4 are numerically found to be

E0 = −28.970281 E2 = −22.829174 E4 = −18.960034, (36)

where all energies are given in reduced units (Rydberg), and numbers have been rounded to six digitsafter the decimal point. The corresponding probability densities |Ψ0|

2, |Ψ2|2, and |Ψ4|

2 are displayedin Fig. 2. Next, we compute the energies E1, E3 and E5 by solving the condition (18):

E1 = −25.245851 E3 = −20.716053 E5 = −17.358223. (37)

332 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

Fig. 2. Probability densities |Ψ0|2, |Ψ2|

2 , and |Ψ4|2 .

Fig. 3. Probability densities |Ψ1|2, |Ψ3|

2 , and |Ψ5|2 .

As before, energies are given in Rydberg, numbers have been rounded to six digits after the decimalpoint. The associated probability densities |Ψ1|

2, |Ψ3|2, and |Ψ5|

2 are shown in Fig. 3. In order tocompare our subband bottom energies to their counterparts obtained in the conventional, constant-mass situation [13], we collect both sets of energies in the following table:

Subband Constant mass energy (Ry) [13] Effective mass energy (Ry)0 −19.2 −28.9702811 −7.44 −25.2458512 −3.37 −22.8291743 −1.18 −20.7160534 −0.36 −18.9600345 −0.054 −17.358223

Note that Ry stands for the unit Rydberg. We observe that in comparison with [13], our subbandbottom energies (36), (37) lie much lower. More precisely, in the constant-mass scenario the energiesraise according to a power-law function with increasing subband order, while for position-dependentmass we have an almost constant gap between neighboring subband bottom energies. This behavior,clearly due to the presence of a position-dependent mass, is visualized in Fig. 4, showing theThomas–Fermi potential, the energies obtained by our method, and their counterparts constructedin [13]. Next, let us comment on the solutions that are associated with the above shown subbandbottom energies that are shown in Figs. 2, 3 and 5, together with the Thomas–Fermi potential(2). Unfortunately we are unable to display the corresponding solutions of the constant-mass case,as those were obtained using a numerical method [13]. A qualitative comparison of the solutionsyields that in both the constant- and the position-dependent mass case parity and number of zeroscorrespond to each other, while in our position-dependent-mass context the solutions decrease morerapidly than in [13].

5. Concluding remarks

In this note we have presented an exactly-solvable model involving position-dependent mass anda Thomas–Fermi potential in one dimension. After constructing the general solution of the governingSchrödinger equation,we requestedparity to obtain the bound state solutions and their correspondingenergy eigenvalues. The problem discussed here contains the general characteristics of the modeldiscussed in [13] for constant mass. Despite this fact, we cannot obtain solutions for the constant-mass model [13] as special cases of our findings, because exact solvability is only generated if the

A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334 333

Fig. 4. The Thomas–Fermi potential (2) together with subband bottom energies obtained in the present note (solid lines), andin [13] for the constant-mass case (dashed lines).

Fig. 5. Probability densities |Ψj|2, j = 0, . . . , 5 and the Thomas–Fermi potential (2) for the settings (35).

position-dependentmass is present. Let us point out that in the latter reference parity of the solutionsis enforced by imposing conditions on the semiclassical wavefunction, while in the present work weuse different types of boundary conditions. As far as a generalization of the present work to higherdimensions is concerned, the technically most challenging problem is to maintain exact solvabilityof the model. Therefore, in order to obtain results in closed form, it would be most straightforwardto impose a particular symmetry that leads to a separable Schrödinger equation, the components ofwhich can be dealt with separately. As an example, one could consider spherical symmetry in threedimensions, which leads to a known angular equation and a radial equation that is to be solved.Since the radial equation contains the usual centrifugal term, it must be checkedwhether closed-formsolutions are admissible. If this approach fails, one must resort to numerical methods.

Acknowledgments

CPG gratefully acknowledges hospitality during his Ph.D. studies in Science and Technology atthe Escuela Superior de Física y Matemáticas, Instituto Politécnico Nacional. This work was partiallysupported by EDI-IPN, COFAA-IPN and SNI-Mexico.

References

[1] W. Kohn, Solid State Phys. 5 (1957) 257.[2] R. Resta, in: P.N. Butcher, N.H. March, M.P. Tosi (Eds.), Crystalline Semiconducting Materials and Devices, Plenum Press,

1986, p. 217.[3] M. Altarelli, W.Y. Hsu, Phys. Rev. Lett. 43 (1979) 1346.[4] G. Bastard, Phys. Rev. B 24 (1981) 4714.[5] V. Mlinar, M. Tadic, B. Partoens, F.M. Peeters, Phys. Rev. B 71 (2005) 205305.[6] C. Priester, G. Allan, M. Lannoo, Phys. Rev. B 28 (1983) 7194.[7] G.T. Einevoll, Phys. Rev. B 42 (1990) 3497.

334 A. Schulze-Halberg et al. / Annals of Physics 333 (2013) 323–334

[8] E.F. Schubert, A. Fischer, K. Ploog, IEEE Trans. Electron Devices ED-33 (1986) 625.[9] E.F. Schubert, C.W. Tu, R.F. Kopf, J.M. Kuo, L.M. Lunardi, Appl. Phys. Lett. 54 (1989) 2592.

[10] M.H. Degani, Phys. Rev. B 44 (1991) 5580.[11] L.M.R. Scolfaro, D. Beliaev, P. Enderlein, J.R. Leite, Phys. Rev. B 50 (1994) 8699.[12] L.M. Gaggero-Sager, R. Perez-Alvarez, J. Appl. Phys. 78 (1995) 4566.[13] L. Ioriatti, Phys. Rev. B 41 (1990) 8340.[14] L.M. Gaggero-Sager, M.E. Mora-Ramos, D.A. Contreras-Solorio, Phys. Rev. B 57 (1998) 6286.[15] O. Oubram, L.M. Gaggero-Sager, Prog. Electromagn. Res. Lett. 2 (2008) 81.[16] G.H. Herling, M.L. Rustgi, J. Appl. Phys. 69 (1991) 2328.[17] X.H. Qi, X.J. Kong, J.J. Liu, Phys. Rev. B 58 (1998) 10578.[18] Yu-Xian Li, Jian-Jun Liu, Xiao-Jun Kong, J. Appl. Phys. 88 (2000) 2588.[19] G.N. Watson, A Treatise on the Theory of Bessel Functions, Cambridge University Press, Cambridge, England, 1995.[20] I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series and Products, Academic Press, New York, 2007.[21] S. Rajashabala, K. Navaneethakrishnan, Braz. J. Phys. 37 (2007) http://dx.doi.org/10.1590/S0103-97332007000700011.

Article ♯ 3b.[22] S. Rajashabala, K. Navaneethakrishnan, Modern Phys. Lett. B 20 (2006) 1529.[23] C. Pacheco-Garcia, J. Garcia-Ravelo, J. Morales, J.J. Pena, Int. J. Quantum. Chem. 110 (2010) 2880.