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Physica E 16 (2003) 286–296

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Con ned states in parabolic cylinder quantum dots

M. Willatzena ;∗, L.C. Lew Yan Voonb

aMads Clausen Institute, University of Southern Denmark, Grundtvigs Alle 150, DK-6400 S#nderborg, DenmarkbDepartment of Physics, Worcester Polytechnic Institute, 100 Institute Road, Worcester, MA 01609, USA

Received 7 November 2002; accepted 19 November 2002

Abstract

An in nite-barrier parabolic cylindrical quantum dot is solved using parabolic cylinder coordinates. An additional condition,beyond the conventional boundary condition, is required in order to solve the eigenvalue problem. The volume dependenceof the energy spectrum is shown to be di5erent from spheroidal and elliptical dots and to depend strongly on the shape.? 2002 Elsevier Science B.V. All rights reserved.

PACS: 03.65.Ge; 73.21.La

Keywords: Quantum dots; Parabolic cylinder coordinates

1. Introduction

Semiconductor quantum dots (QDs) have alwaysbeen of great interest [1]. Recently, there has beena lot of work in obtaining analytic or semi-analyticresults for various QD shapes. The latter is oftendone using an in nite barrier and within a one-bande5ective-mass approach; this approximation is validfor electron states in isolated QDs made from largeband-gap materials and of large radii (at least ten’s ofAA). One goal of the latter approach is to obtain a morephysical picture of the role of the shape on the elec-tronic and optical properties. As a recent example ofthe latter methodology, Cantele et al. [2] discoveredtopological surface states in spheroidal QDs. Someof the other shapes considered so far are spheres [3],cones [4], rectangles [5], discs or cylinders [6], domesor lens [7,8].

∗ Corresponding author. Tel.: +45-6550-1682; fax: +45-6550-1660.

E-mail address: willatzen@mci.sdu.dk (M. Willatzen).

This paper is motivated by recent work [2,8–10]in which an appropriate orthogonal curvilinear co-ordinate system is chosen in order to solve the QDproblem exactly, and also in renewed interest innanowires [11–14] and quantum rods [15,16]. Thus,van den Broek and Peeters [9] used elliptic coordi-nates in two dimensions for an elliptical dot, Canteleet al. [2] used spheroidal coordinates for a spheroidaldot, and we used parabolic rotational coordinates fora lens-shaped dot [8]. In the above, the ellipse andspheroid have a one-coordinate surface, and the lenshas a three-coordinate surface. Here we consider aQD with a four-coordinate surface. The example wehave chosen is that of a parabolic cylinder shapedQD in parabolic cylinder coordinates (PCC). We willshow that the mathematics has some similarity to theproblem studied in parabolic rotational coordinates[8]; however, the four-coordinate surface in PCCleads to an additional level of complexity. Also, thepresence of a singular curve in the coordinate systemmakes this problem di5erent from the standard pre-scription of just applying the boundary conditions at

1386-9477/03/$ - see front matter ? 2002 Elsevier Science B.V. All rights reserved.PII: doi:10.1016/S1386-9477(02)00927-X

M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296 287

the boundary for the eigenvalues. Nevertheless, wewere able to solve the problem exactly.We use this parabolic cylinder shaped QD as a

model system to study the energy dependence onshape and volume [2,8,9,17–21]. In particular, Henset al. [21] recently demonstrated experimentally thatthe energy spectrum of a Lattened cube is very di5er-ent in terms of energy and degeneracy from that of asphere and of a cube (all of the same volume), whilethe latter two have similar energy spectra.In the next section, we show how SchrModinger’s

equation is separated using parabolic cylinder coor-dinates and present a series solution of the resultingordinary di5erential equations. We then present nu-merical results for various symmetrical and asymmet-rical QDs.

2. Theory

2.1. Schr:odinger’s equation in parabolic cylindercoordinates

In this section we show how to calculate the eigen-functions and eigenvalues of an electron con ned ina closed region of space de ned by two confocalparabolic cylinders (|�|=�0 and �=�0) and two planesurfaces z = z1 and z = z2 (Fig. 1). The relation be-tween the parabolic cylinder coordinates and cartesiancoordinates is as follows:

x = 12

(�2 − �2

);

y = ��;

z = z;

06 �¡∞;−∞¡�¡+∞;−∞¡z¡+∞: (1)

We have chosen the range of � to include negativevalues in order to cover the y¡ 0 region. The expres-sions for the various geometrical parameters given inFig. 1 are as follows:

R= xmax − xmin = 12

(�20 + �20

); (2)

H = 12(ymax − ymin) = �0�0; (3)

V = 23L�0�0(�

20 + �20) =

43LRH; (4)

where L is the axial length of the QD (along z-axis),R is the radius (along x-axis) H is the height (along

-30 -20 -10 0 10 20 30

-200

-100

0

100

200

300

400

500

RH

=20

=10

x

y

Fig. 1. Geometry of quantum dot (cross-sectional view in z-plane).

y-axis), and V is the QD volume. The volume isobtained by integrating a volume element in PCC:

V =∫ z2

z1

∫ �0

0

∫ �0

−�0(�2 + �2) dz d� d�; (5)

and use is made of L= z2 − z1.The bound states of an electron in the QD are

obtained by solving SchrModinger’s equation

− ˝22m∗∇2 (r) + V (r) (r) = E (r); (6)

with m∗ the electron e5ective mass inside the QD, Ethe energy, the wave function, and subject to thehard-wall boundary condition

(�; �; z)|�=�0 ;�=�0 ;z=z1 ;z=z2 = 0: (7)

SchrModinger’s equation is known to be separable inparabolic cylinder coordinates [22]. Writing

(�; �; z) =M (�)N (�)Z(z); (8)

one gets the following separated ordinary di5erentialequations:

d2Md�2

− (�2 + �3�2)M = 0; (9)

d2Nd�2

+ (�2 − �3�2)N = 0; (10)

d2Zdz2

+ (k2 + �3)Z = 0; (11)

where k2 = 2m∗E=˝2.

288 M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296

Series solutions to the ordinary di5erential Eqs. (9)–(11) can be found using the Frobenius method [23].Since there are no singular points in Eqs. (9)–(11), twoindependent and well-behaved solutions exist for eachseparated equation. Consider the di5erential equationin M rst. Expanding M about � = 0 one writes:

M (�) =∞∑n=0

an�n+k : (12)

Inserting Eq. (12) in Eq. (9) and equating terms pro-portional to �k−2 and �k−1 gives two equations:

k(k − 1)a0 = 0; (13)

k(k + 1)a1 = 0: (14)

Thus, k = 0 or 1 if we choose a0 = 1 and a1 = 0.The two choices for k will give us two independentseries solutions as we will prove next. Applying theidentity theorem for power series to terms �k+n wheren= 0; 1; 2; : : : give the recursion formula:

a2 =�22; a3 = 0;

an+4 =�2an+2 + �3an

(n+ 4)(n+ 3);

where n= 0; 1; 2; 3; : : : ; (15)

if k = 0. This solution for M (�) ≡ M1(�) is clearlyeven in �.Similarly, when k = 1, the following recursion for-

mula is obtained:

b0 = 1; b1 = 0; b2 =�26; b3 = 0;

bn+4 =�2bn+2 + �3bn

(n+ 5)(n+ 4);

where n= 0; 1; 2; 3; : : : ; (16)

and the associated solution for M (�) ≡ M2(�) is anodd function of �. The general solution to Eq. (9) isaccordingly

M (�) =M (�2; �3; �) = AM1(�) + BM2(�); (17)

where A and B are arbitrary constants, and

M1(�) =M1(�2; �3; �) =∞∑n=0

a2n�2n; (18)

M2(�) =M2(�2; �3; �) =∞∑n=0

b2n�2n+1: (19)

Consider next the di5erential equation in N (Eq.(10)). This di5erential equation is analogous to thedi5erential equation inM (Eq. (9)) except that �2 mustbe replaced by −�2. We obtain

N (�) = N (�2; �3; �) = CN1(�) + DN2(�); (20)

where C and D are arbitrary constants, and

N1(�) = N1(�2; �3; �) =∞∑n=0

c2n�2n; (21)

N2(�) = N2(�2; �3; �) =∞∑n=0

d2n�2n+1: (22)

The coePcients cn satisfy the recurrence relations

c0 = 1; c0 = 0; c2 =−�22

; c3 = 0;

cn+4 =−�2cn+2 + �3cn(n+ 4)(n+ 3)

;

where n= 0; 1; 2; 3; : : : : (23)

For dn we obtain

d0 = 1; d1 = 0; d2 =−�26

; d3 = 0;

dn+4 =−�2dn+2 + �3dn

(n+ 5)(n+ 4);

where n= 0; 1; 2; 3; : : : : (24)

The remaining di5erential equation in Z can besolved immediately so as to give the general solution

Z(z) = E sin([√

k2 + �3]z)

+F cos([√

k2 + �3]z); (25)

where E and F are arbitrary constants.The hard-wall boundary condition (Eq. (7)) now

becomes:

M (�2; �3; �0) = 0; (26)

N (�2; �3; �0) = N (�2; �3;−�0) = 0; (27)

Z(k; �3; z1) = Z(k; �3; z2) = 0; (28)

where (�0; �0; z1; z2) de nes the QD boundary. It turnsout that Eqs. (26)–(28) together with normalizationare not suPcient to solve the eigenvalue problem. Thisis due to the vanishing of the Jacobian for the trans-formation to PCC. It was shown by Lebedev [24] that

M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296 289

one also requires that ∇̃ is nite everywhere insidethe QD region. Using this result and

(∇̃ )2 =1

�2 + �2

[(9 9�

)2+(9 9�

)2]

+(9 9z

)2; (29)

leads to[(9 9�

)2+(9 9�

)2]∣∣∣∣∣�=�=0

= 0: (30)

Eigenstates and associated energies E (or k2) willbe found by imposing the four conditions given byEqs. (26)–(28) and (30).Combining the general solution (Eq. (20)) with

Eq. (27) and making use of the fact that N1 and N2

are even and odd, respectively, allows us to write

N (�2; �3; �0) + N (�2; �3;−�0) = 2CN1(�0) = 0: (31)

This condition can be ful lled either if

C = 0; (32)

or

N1(�0) = 0: (33)

However, if C=0, then D �= 0 for to be non-trivial(di5erent from zero). Thus, N2(�0)=0 so as to satisfyEq. (27). Instead, if N1(�0) = 0, we obtain N (�0) =DN2(�0) = 0 leaving two possibilities open: D= 0 orN2(�0) = 0.The condition in Eq. (30) can be restated as

B2C2 + A2D2 = 0; (34)

by use of

N (0) = C;

M (0) = A;(9M9�

)2∣∣∣∣∣�=0

= B2;

(9N9�

)2∣∣∣∣∣�=0

= D2: (35)

The latter set of relations follow immediately fromEqs. (17) and (20) and the series expansions ofM1(�),M2(�), N1(�), and N2(�).

Next, let us consider the three possible cases (c)C �= 0, D = 0; (b) D �= 0, C = 0; and (c) C �= 0,D �= 0; separately.

2.2. Case (a)

In case (a): C �= 0 and D=0. Eq. (33) then impliesN1(�0)=0. The condition that ∇̃ is nite at �=�=0now gives B = 0 (refer to Eq. (34)). If B = 0 thenA �= 0 and M1(�0) = 0 so as to satisfy Eq. (26). Inconclusion, case (a) demands

N1(�0) = 0;

M1(�0) = 0: (36)

2.3. Case (b)

In case (b): D �= 0 and C = 0. Thus N2(�0) = 0(refer to the discussion following Eqs. (32) and (33).The condition that ∇̃ is nite at �= �=0 now givesA = 0 (refer to Eq. (34)). If A = 0 then B �= 0 andM2(�0) = 0 so as to satisfy Eq. (26). In conclusion,case (b) demands

N2(�0) = 0;

M2(�0) = 0: (37)

2.4. Case (c)

It follows immediately from the discussion follow-ing Eqs. (32) and (33) that when C �= 0, D �= 0, thetwo conditions

N1(�0) = 0;

N2(�0) = 0; (38)

must be imposed. If M1(�0) = 0 by accidence thenM2(�0)=0 so as to satisfy Eq. (26) and Eq. (34) (Eq.(34) forces B= 0 if A= 0, C �= 0, and D �= 0, whichis not possible for a non-trivial solution). Similarly, ifM2(�0)= 0 by accidence then M1(�0)= 0. These twospecial cases therefore require four conditions to beful lled: M1(�0) =M2(�0) = N1(�0) = N2(�0) = 0. Ifinstead bothM1(�0) �= 0 andM2(�0) �= 0, it is alwayspossible to nd non-zero (complex) coePcients A andB such that Eqs. (26) and (30) are satis ed simultane-ously. In the latter case, the two conditions given by

290 M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296

Eq. (38) are the only conditions that must be satis edfor an eigenstate to be found.

2.5. Relation between k and �3

Consider next the Z equation (Eq. (25)) and theassociated boundary conditions (Eq. (28)). Two pos-sible cases can occur if we choose the origin of thez-axis such that z2=−z1=L=2: (i) E=0 or (ii) F=0.If E=0, then the following relation between k and �3must hold:

k2 = (2p+ 1)2("L

)2− �3;

where p= 0; 1; 2; 3; : : : : (39)

If, instead, F = 0, we nd

k2 = (2p)2("L

)2− �3;

where p= 1; 2; 3; : : : : (40)

The procedure for determining eigenstates is as fol-lows. For each case (a), (b), and (c), solution sets(�2; �3) are found by solving Eqs. (36), (37), and (38),respectively. The values found for �3 can nally beinserted in Eqs. (39) and (40) to get k.

3. Results and discussions

3.1. Shapes studied

Di5erent structures were modeled, including sym-metrical and asymmetrical QDs. The symmetrical QDis obtained when �0=�0 while asymmetrical dots cor-respond to any QD with parameters �0 �= �0. In Fig. 2,three QD cross-sections are shown. The three QDs areof the same volume (6:533×103 AA2×L) with parame-ters (i) �20=70 AA; �20=70 AA; (ii) �20=130 AA; �20=29 AA;and (iii) (i) �20 = 500 AA; �20 = 0:77 AA; correspondingto the upper, middle, and lower gures, respectively.For the symmetrical quantum dot, the height andradius are the same.

3.2. Energies

In searching for energies, we started with �2 andscanned in �3 for zeroes of M1 and N1, M2 and N2,and N1 and N2 corresponding to case (a), (b), and (c)

(refer to the Section on Theory), respectively. Thistechnique is illustrated in Fig. 3 for the symmetricalQD. In Fig. 3a, (�2; �3) values are plotted where M1

and N1 have zeroes corresponding to case (a). Linecodings for zeroes of M1 and N1 are squares and dia-monds, respectively. The simultaneous zeroes are theintersection points in a (�2; �3) plot; the solutions areeven about y=0. In a similar way, Fig. 3b shows the(�2; �3) values where M2 and N2 are zero correspond-ing to case (b). Again, the simultaneous zeroes are theintersection points in a (�2; �3) plot; the solutions areodd about y = 0. In Fig. 3c, zeroes of N1 and N2 areplotted as a function of (�2; �3) corresponding to case(c). Evidently, no simultaneous zeroes are found (aclose inspection of the curves will show that the N1

and N2 curves do not intersect). Thus, there will beno eigenstates of the type considered in case (c). Thisis to be expected on the basis of symmetry since thesolutions must have de nite parity about y = 0.Some actual �2 and �3 values obtained for a sym-

metrical QD are given in Table 1. Note that energyeigenvalues E=˝2k2=2m∗ are related to �3 accordingto Eqs. (39) and (40). A given bound state is distigu-ished by three labels p, �2, and k. Conventionally, in-stead of k, one uses the counting index of k and labelsit by n, the principal quantum number. We will fol-low this convention here. The index �2 does not havea direct analog and it will, in general, be non-integral.Hence, we also use the counting index l of �2 to labelthe states. We also adopt the convention that degener-ate states with the same p index will be assigned suchthat the state with a negative �2 value has the smallerl value.As mentioned earlier, it follows from Eqs. (9) and

(10) that

M1(�2; �3; $) = N1(−�2; �3; $) (41)

M2(�2; �3; $) = N2(−�2; �3; $); (42)

for any value $. Thus,

M1(�2; �3; $)N1(�2; �3; $)

=N1(−�2; �3; $)M1(−�2; �3; $); (43)

M2(�2; �3; $)N2(�2; �3; $)

=N2(−�2; �3; $)M2(−�2; �3; $): (44)

The latter relation applied to the case of a sym-metric QD ($ = �0 = �0) shows that if (�2; �3) is a

M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296 291

-60 -40 -20 0 20 40 60

-100

0

100

200

300

y [Angstrom]

x [A

ngst

rom

]

-60 -40 -20 0 20 40 60

-100

0

100

200

300

y [Angstrom]

x [A

ngst

rom

]

-60 -40 -20 0 20 40 60

-100

0

100

200

300

y [Angstrom]

x [A

ngst

rom

]

Fig. 2. Plot of three quantum dots having the same volume (cross-sectional view in z-plane). The upper gure shows the QD withparameters: �20 = �20 = 70 AA. The middle gure is for �20 = 130 AA and the lower gure is for �20 = 500 AA.

Table 1Energies (in meV) for a symmetrical QD. (Using m∗ = 0:067m0 and p = 1; 2; 3; 4; : : :). �20 = 70 AA, �20 = 70 AA

�2 �3 E − ˝2=2m∗(p"=L)2 n l

0 −0:00327 186 1 10 −0:00631 359 2 1

−0:126 −0:0103 586 3 10:126 −0:0103 586 3 2

−0:253 −0:0152 864 4 10:253 −0:0152 864 4 20 −0:0214 1217 5 10 −0:0285 1621 6 1

−0:2043 −0:0365 2075 7 10:2043 −0:0365 2075 7 2

simultaneous zero point for M1 and N1 (or M2

and N2) then (−�2; �3) is also a simultaneous zeropoint for M1 and N1 (or M2 and N2). Indeed, thisis in agreement with results shown in Figs. 3a andb, and Table 1, e.g., when �20 = �20 = 70 AA, thereare two simultaneous intersection points (zeroesof M1 and N1) at (�2; �3) = (−0:126;−0:0103)and (�2; �3) = (0:126;−0:0103). Applied to an

asymmetric QD, it shows that the (�0; �0) dot has thesame energy spectrum as the (�0; �0), as expectedfrom the mirror image shapes.

3.3. Wave functions and symmetry

Examples of wave functions for both symmetricaland asymmetrical QDs are given in Fig. 4. The three

292 M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296

-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4

-0.14

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02 α

3 [A

ngst

rom

-2]

α2 [Angstrom-1]

crossings M1,N1

-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4

-0.14

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02

α3

[Ang

stro

m-2

]

α2 [Angstrom-1]

crossings M2,N2

-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4

-0.14

-0.12

-0.1

-0.08

-0.06

-0.04

-0.02

α3

[Ang

stro

m-2

]

α2 [Angstrom-1]

crossings N1,N2

Fig. 3. Plot of the zeroes of M1 and N1 (upper gure left), M2 and N2 (upper gure right), and N1 and N2 (lower gure) for a symmetric QD.

columns are for the symmetrical QD with �20 = �20 =70 AA (left), the asymmetrical QD with �20 = 130 AAhaving the same volume as the symmetrical QD (mid-dle), and the asymmetrical QD with �20 = 170 AA hav-ing the same volume as the symmetrical QD (right).The rows are the lowest four states, with the groundstate at the top. Thus, in the upper left corner, wehave the contour plot of the ground state of the sym-metric quantum dot �20 = �20 = 70 AA. This state corre-sponds to case (a) where B = D = 0 is imposed (andM1(�0) = N1(�0) = 0), i.e.,

(x; y; z) = (�; �; z) =M1(�)N1(�)Z(z)

=M1(�)N1(−�)Z(z)

= (�;−�; z) = (x;−y; z); (45)

and (if �2 = 0),

(x; y; z) = (�; �; z) =M1(�)N1(�)Z(z)

=N1(�)M1(�)Z(z)

= (�; �; z) = (−x; y; z); (46)

where use has been made of Eq. (41) in obtainingthe third equality. Thus, the ground state is symmetricwith respect to a mirror reLection in the x=0 and y=0planes. A plot of the rst excited state of the same QDis also shown. This state is the lowest energetic state(for a xed value of p) corresponding to case (b) forwhich A= C = 0, i.e.,

(x; y; z) = (�; �; z) =M2(�)N2(�)Z(z)

=−M2(�)N2(−�)Z(z)

=− (�;−�; z) =− (x;−y; z): (47)

In addition, �2 = 0 for the rst excited state, and thus

(x; y; z) = (−x; y; z); (48)

following steps analogous to those used in deriving Eq.(46). Therefore, this state is symmetric with respect toa mirror reLection in the x=0 plane and antisymmetricwith respect to a mirror reLection in the y = 0 plane.Note that for a state for which �2 �= 0 (states 3 and 4 inFig. 4), the state is found not to be even with respect

M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296 293

Fig. 4. Contour plots of wave functions (all for p=1) in the xy-plane. The three columns are for the symmetrical QD with �20 = �20 =70 AA(left), the asymmetrical QD with �20 = 130 AA having the same volume as the symmetrical QD (middle), and the asymmetrical QD with�20 = 170 AA having the same volume as the symmetrical QD (right). The rows are the lowest four states, with the ground state at the top.

to the x = 0 plane. However, this is a consequenceof the double degeneracy of these states (both ±�2allowed for the same �3); one can then form linearcombinations that have de nite parities.In the middle column and top row of Fig. 4, the

ground state of the asymmetrical QD with �20 =130 AA(having the same volume as the symmetrical QD with�20 = 70 AA) is shown. Mirror reLection in the y =0 plane is still a symmetry because the ground statecorresponds to case (a) by use of Eq. (41). However,

since �2 �= 0 for this state, mirror reLection in thex=0 plane is not a symmetry. In the middle column,second row of Fig. 4, the rst excited state of theasymmetrical QD with �20 = 130 AA is plotted. As forthe rst excited state of the symmetrical QD, this statebelongs to case (b) and thus obeys the antisymmetrywith respect to a mirror reLection in the y = 0 plane.Again, as �2 = 0:2818 �= 0 for this state, symmetrywith respect to a mirror reLection in the x = 0 planedoes not exist.

294 M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296

0 20 40 60 80 100 120 140 160 180 2000

100

200

300

400

500

600

700

800

900

1000

1100

µ02(Angstrom)

Ene

rgy

(meV

)

Fig. 5. Plot of the lowest four states for a series of quantum dots having the same volume. Parameters for the symmetrical QD are:�20 = �20 = 70 AA.

3.4. Shape dependence of energies

We next consider the energy dependence on QDasymmetry—as speci ed by �0—for a given volumeequal to the volume of a symmetrical QD with para-meters �20 = �20 = 70 AA (Fig. 5). One can observe therich structure of the energy levels as a function ofthe shape in the xy-plane (there is additional depen-dence on L which we do not consider here to sim-plify the discussion). The basic observation is that theground-state energy curve has a double minimum, thatfor the rst excited state has a single minimum, andthe next state has a double degeneracy that splits forthe asymmetric structure.For the ground state, it is evident that a local en-

ergy maximum at �20 = 70 AA exists between two localenergy minima: �A and �B where �B ¿�A (the re-gion: �A6 �06 �B is region I in the following). Atlarger asymmetries, i.e., if �0 ¡�A or �0 ¿�B, theground-state energy increases and no additional ex-trema are found (the region �0 ¡�A or �0 ¿�B isregion II in the following). The reason is that the QDwith parameter �20 =70 AA is more Lat along the x-axisthan the asymmetrical QDs with �0 parameters in

region I. Flatness along the x-axis can be measuredby the radius R= 1

2(�20 + �20) (refer to Eq. (2)). How-

ever, at larger asymmetries (�0 values in region II),the quantum dot becomes more Lat along the y-axisas compared to the x-axis (note that Latness alongthe y-axis is represented by the value of the heightH = �0�0 and H decreases with increasing asymme-try). Hence, in region II, the ground-state energy in-creases with increasing asymmetry as Fig. 5 reveals. Asimilar behavior is observed in the case of lens-shapedquantum dots in parabolic rotational coordinates [8].For spheroidal QDs, the situation is di5erent, how-ever, as shown by Cantele et al. [2]. In this case, thesymmetrical QD is spherical such that any asymmet-rical (spheroidal) QD is more Lat than the symmetri-cal QD and the ground-state energy has an absoluteminimum for the symmetrical QD. Our result is alsodi5erent from that obtained for a QD with an ellipticalcross-section [9,10] for the same reason, despite thefact that an elliptical QD is as close as one can get tothe parabolic cylinder QD. This reveals the individu-ality of each QD shape.The absence, for the excited states shown in

Fig. 5, of the complications outlined above for the

M. Willatzen, L.C. Lew Yan Voon / Physica E 16 (2003) 286–296 295

ground state can be readily explained by looking at thewave function in the xy-plane (Fig. 4). For example, itcan be seen that the rst excited state has a node alongy = 0. Thus, when the asymmetry is introduced intothe symmetric structure, there is a larger impact on thecon nement energy in the x direction. The dimensionin that direction is the height H (Fig. 1) which getssmaller with increasing asymmetry; hence the energyincreases. The slightly di5erent behavior of the nextstate is due to the repulsion between originally de-generate states. While the upper state has a similarbehavior to the second excited state (of the symmetricstructure), the lower state displays an energy maxi-mum and even crosses the second excited state. Bothfeatures are evident when one compares the wavefunctions given in Fig. 4. Thus, the third state for�20 = 130 AA is seen to have the nodal surface roughlyperpendicular to the second state for �20 =70 AA, hencethe opposite behavior in the shape dependence of theenergy. The crossing of the second and third statesnear �20=160 AA is also reLected in the wave functionsgiven in Fig. 4.The strong shape dependence of the energies and

wave functions will have signi cant consequences on,for example, the optical properties of the QDs sincethe transition energies and polarization selection ruleswill be di5erent. For example, if one is interested inthe fundamental intersubband transition, the transitionenergy is seen to change in a non-monotonic fash-ion as the shape is changed. If we also assume thatthe length of the QD is small enough that the secondstate represented in Figs. 4 and 5 is, indeed, the sec-ond excited state, then one can see that the polariza-tion selection rule will change at the cross-over near�20 = 160 AA. Based upon the contour plots in Fig. 4,it is anticipated that the dominant linear polarizationwill rotate by roughly 900 at the cross-over. This be-havior is absent in intersubband transitions in quantumwells [25] and demonstrates the versality of QDs inoptoelectronic applications. Detailed numerical calcu-lations of the optical properties are being performedand will be published in the future.

4. Conclusions

The in nite-barrier parabolic cylinder quantum dotsin parabolic cylinder coordinates have been solved

exactly. Energies and wave functions are given forsymmetrical and asymmetrical quantum dots. It wasfound that the ground state energy and nature of thesecond state are very sensitive to the shape of thequantum dot at constant volume. Signi cant changesin the optical properties are also anticipated.

Acknowledgements

This work was supported by an NSF CAREERaward (NSF Grant No 9984059).

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