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Phys. Scr. 96 (2021)065207 https://doi.org/10.1088/1402-4896/abecf8
PAPER
Discrete symmetry approach to exact bound-state solutions for a
regular hexagon Dirac billiard
Wajdi A Gaddah
Department of Physics, Faculty of Science, University of Tripoli, P.O. Box 84441, Tripoli, Libya
E-mail: wajdigaddah@yahoo.co.uk
Keywords: Hexagon quantum billiards, Planar Dirac equation, Group theoretic methods, Isospectral domains
Abstract
We consider solving the stationary Dirac equation for a spin-1/2 fermion confined in a two-
dimensional quantum billiard with a regular hexagon boundary, using symmetry transformations of
the point group C
6v
. Closed-form bound-state solutions for this problem are obtained and the non-
relativistic limit of our results are clearly discussed. Due to an adequate choice of confining boundary
conditions the upper components of the planar Dirac-spinor eigenfunctions are shown to satisfy the
corresponding hexagonal Schrödinger billiard, and the Dirac positive energy eigenvalues are proven
to reduce directly to their Schrödinger counterparts in the non-relativistic limit. An illustrative
application of our group theoretic method to the well-known square billiard problem has been
explicitly provided. The success of our approach in solving equilateral-triangle, square and regular
hexagon quantum billiards may well imply a possible applicability to other regular polygonal billiards.
A quick look on nodal domains of the Schrödinger eigenfunctions for the hexagon billiard is also
considered. Moreover, we have determined a number of distinct non-congruent polygonal billiards
that have the same eigenvalue spectrum as that of the regular hexagon.
1. Introduction
A billiard system is a dynamicalsystem of a point particle constrained to move freely in a two-dimensional closed
region with an arbitrary geometrical shapeand perfectly reflecting boundary. The study of billiard systems provides
the opportunity to investigate quantum manifestations of chaos [1–3], and enhance our understanding of the
quantum confinement effects on electrons trapped in two-dimensional semiconductor quantum dots. Thespatial
confinement of electrons insideplanar billiards of different shapes has attracted considerable attention in various
branches of physics, particularly in the studyof the graphene quantum dots [4–7]. Charge carriers in a graphene
sheet mimic relativistic particles with zero rest mass, and are accordingly governed by the Dirac equation rather
than the usual Schrödinger equation [8–10]. The amazing similarity between graphene electrons and massless
relativistic particles suggests a new area ofpractical applications to the theory of the two-dimensional Dirac
equation,especially in the investigation of complexrelativistic quantumphenomena on graphene [11].
Berry and Mondragon in an earlier work on neutrino billiards [12]proposed an infinite-mass potential to
confine a massless Dirac fermion to a closed planar domain Ωof an arbitrary shape. They assumed the rest mass
of the particle to be position-dependent, having the value of zero inside Ωand infinity outside. In this way, they
avoided the Klein-paradox problem and obtained an appropriate boundary condition leading to a vanishing
outward normal component of the probability current at the billiard boundary. In general, the vanishing
requirement of the normal current component at the boundary is a sufficient condition for the spatial
confinement of a Dirac particle, but not necessarily always sufficient for satisfying the self-adjointness property
of the Dirac billiard Hamiltonian [13]. Imposing a zero Dirichlet boundary condition on the Dirac spinor
eigenfunctions guarantees the impenetrability of the probability current across the billiard boundary but, unlike
non-relativistic quantum mechanics, does not fulfill the self-adjointness condition for the Hamiltonian operator
of the associated Dirac billiard. This condition is very important to be seriously considered because observables
RECEIVED
12 November 2020
REVISED
10 February 2021
ACCEPTED FOR PUBLICATION
9 March 2021
PUBLISHED
5 April 2021
© 2021 IOP Publishing Ltd
in quantum mechanics correspond only to self-adjoint operators in Hilbert space. In a recent work on a massive
Dirac billiard system [13], confining boundary conditions that comply with self-adjointness are introduced, and
a unique choice of them is applied to an equilateral triangular billiard problem, yielding exact analytical bound
state solutions.
The degree of difficulty in analytically solving the Dirac equation for a billiard system depends largely on the
geometrical shape of the billiard and the confining boundary condition being imposed by the self-adjointness
requirement of the Dirac Hamiltonian. All quantum billiards with regular polygon shapes, except the one with a
square-shaped boundary, can not be solved solely by the separation of variables method, which is the most
widely used technique for partial differential equations. Because this method is based on the restrictive
assumption that the billiard solution is separable, i.e., represented as a product of two functions each of which
depends only on a single independent variable. This form of a single-term solution can not satisfy the confining
boundary condition at all sides of the polygonal perimeter of the billiard. An illustrative example of this case is
the quantum equilateral triangular billiard whose zero Dirichlet boundary condition can only be satisfied by a
three-term solution [14]. This motivates us to search for another analytical method that is more suitable for
obtaining non-separable solutions to a wide class of quantum polygonal billiards. The hexagon billiard, in
particular, is of primary importance since it mimics the primitive unit cell of graphene’s honeycomb lattice
whose study has received great attention recently. In addition to its relevance to graphene quantum dots, this
problem is of interest to mathematical physics as well. We are still not aware of any analytical approach, based on
symmetry group considerations only, that yields the exact bound state solutions to the Dirac hexagon billiard in
a systematically straightforward manner.
The principle aim of the present article is to develop a rigorous and simple analytical approach to solving the
massive Dirac equation for a hexagon billiard system using mainly irreducible matrix representations of the C
6v
point group that describe the symmetry of a regular hexagon. By employing an adequately selected boundary
condition, which respects the probability current confinement and self-adjointness property of the Dirac
operator, we have managed with the help of group-theory arguments to derive the exact two-dimensional Dirac
spinors whose upper components are also proven to be closed-form solutions to the corresponding hexagonal
Schrödinger billiard. In the non-relativistic limit, the Dirac positive energy eigenvalues are shown to reduce to
their Schrödinger counterparts in an obvious and direct way. We also provide a concrete illustration of our
approach, in its simplest application, by solving the well-known square billiard problem in appendix C, giving
the same solution derived from the separation of variables method. In this article, we have also included an
illustrative description of some nodal domain patterns of the Schrödinger hexagon-billiard eigenfunctions
together with a brief discussion on the nodal domain counts. Furthermore, we have extended our investigation
to include the construction of non-congruent isospectral polygonal billiards.
The organization of this paper is as follows. In section 2, we present the problem and introduce a symmetry
group method for the solution, giving an explicit derivation of the exact spinor eigenfunctions of the Dirac
hexagon billiard system. In section 3, our results are discussed in full details. Finally, we briefly summarize our
conclusions in section 4.
2. The Dirac eigenproblem of a regular hexagon billiard
This section is devoted to solving the Dirac equation for a spin-1/2 fermion confined in a quantum billiard of a
regular hexagonal shape with impenetrable boundary, using group theoretical method.
The spatial domain Ωof a regular hexagon of side length L, depicted in figure 1, can be defined as:
W= Î - xy x L y L x
,3
2,3.1
2
⎧
⎨
⎩
⎫
⎬
⎭
() ∣∣∣ ∣∣ ∣∣ ()
On this domain, we consider finding the exact analytical bound-state solutions of the stationary Dirac equation:
Y=YHxy Exy,,, 2() () ()
subject to an appropriate confining boundary condition. Here, Eis the fermion energy and
H
is the two-
dimensional Dirac Hamiltonian operator [12,15,16]:
sb=- +
Hcmci. , 3
02
()
with cand m
0
being the speed of light and fermion rest mass, respectively. Furthermore, ∇=(∂/∂x,∂/∂y),
s
ss=,
12
(
)
, and β=σ
3
, where σ
1
,σ
2
and σ
3
are the well-known Pauli spin matrices:
2
Phys. Scr. 96 (2021)065207 W A Gaddah
ss s==
-=-
01
10 ,0i
i0,10
01
.4
12 3
()
() ( )
()
The corresponding eigenfunctions of the Dirac eigenvalue equation (2)are two-component spinors of the form:
f
c
Y= = ÎWrN
r
rrxy,,, 5
⎛
⎝
⎜⎞
⎠
⎟
() ()
() () ()
where Nis a normalization constant obtained from the normalization condition:
fc+=
W
Nrrdr1. 6
2222
∬
∣ ∣ (∣ ()∣ ∣ ()∣) ()
To confine a relativistic particle inside the hexagonal domain Ω, one can introduce a scalar potential
V
r()
into the mass term m
0
c
2
in (3)by replacing m
0
c
2
with +
m
cVr
02()
, and then taking
V
r()
to be zero inside Ω
and infinite everywhere else. This method was originally used by Berry and Mondragon [12]in an earlier work
on neutrino billiards to avoid the so-called Klein-paradox. The Berry-Mondragon infinite mass confinement
corresponds to the boundary condition
f
r()
=
qc
-
-riexp i()()
for
ζWr
, where θis the angle, measured in
the counterclockwise direction, from the positive x-axis to the outward unit vector normal to the billiard
boundary ∂Ωat
r
. This condition has two important advantages: first, it preserves the self-adjointness of the
Dirac operator
H
, and second, it confines the probability current density
s=Y YJr c r r,7() () () ()
†
inside the billiard boundary ∂Ω. However, this choice of boundary condition is not the only one that possesses
these two advantages. A family of other new boundary conditions that also guarantees the impenetrability of the
probability flux
J
r()
across ∂Ω, without spoiling the self-adjointness of the Dirac operator, is introduced
in [13].
In this paper, we follow reference [13]in choosing to work with a simple boundary condition that requires
the vanishing of only one component of the Dirac spinor Ψon ∂Ω, namely
f
r()
. Besides its desirable properties
of preserving self-adjointness and probability current confinement, this choice is adopted here in order to
recover, in terms of
f
r()
, the corresponding non-relativistic quantum billiard problem, defined by the free
Schrödinger equation (or Helmholtz equation)with the zero Dirichlet boundary condition
f
=r0()
at
ζWr
. In this way, by taking the non-relativistic limit c→∞, we obtain directly the energy spectrum of the
Schrödinger hexagon billiard. Moreover, this choice of boundary condition provides an adequate degree of
simplicity and convenience in the analytical computation of the Dirac spinors and their corresponding energy
eigenvalues.
Figure 1. Regular hexagon of side length L, and centre Oat the origin of the xy-Cartesian coordinate system.
3
Phys. Scr. 96 (2021)065207 W A Gaddah
Henceforth, we shall consider imposing on the upper component of Ψthe Dirichlet boundary condition:
ff== ζWrxy xy,0,forall, . 8() () () ()
The Dirac equation (2)can be decomposed, with the help of (5), into a pair of coupled first-order differential
equations:
fc=
++
Pr k r,9() () ()
cf=
--
Pr k r,10() () ( )
where k
±
and
P
are identified by:
=
kEmc c,11
02
() ()
=- ¶
¶¶
¶
Pxy
ii. 12
⎛
⎝
⎜⎞
⎠
⎟()
By eliminating cr(
)
from (9)and (10), we obtain a second-order partial differential equation in f:
ff
¶
¶+¶
¶=-
xy
xy k xy,,, 13
2
2
2
2
2
⎛
⎝
⎜⎞
⎠
⎟() () ()
where
=-
kE mc c.14
20
241 2
() ()
Here, kis positive or zero for |E|m
0
c
2
, and pure imaginary for |E|<m
0
c
2
. By solving equation (13)for f(x,y)
under the boundary condition (8), we can obtain χ(x,y)from f(x,y)via (9)by direct differentiation. As there are
three cases to consider for the values of k, we shall investigate separately each case in turn, beginning with the case
of positive kin which |E|>m
0
c
2
.
The symmetry group C
6v
of a regular hexagon [17,18]serves as our main tool in solving the boundary value
problem (13)for f(x,y). This is because the differential equation (13)admits invariance under the same
geometrical symmetry transformations of the regular hexagon. The defining irreducible representations of the
C
6v
group elements consist of a set of twelve 2 ×2 orthogonal matrices, given by [17]:
L= =¼
=¼
-
n
n
R
S
,if 1,2,,6,
,if 7,8,,12, 15
nn
n6
⎧
⎨
⎩()
where
pp
pp
=
---
--
nn
nn
R
cos 1
3sin 1
3
sin 1
3cos 1
3
,16
n
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
() ()
() () ()
and
pp
pp
=
--
---
nn
nn
S
cos 1
3sin 1
3
sin 1
3cos 1
3
,17
n
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
() ()
() () ()
Here, R
n
represents a counterclockwise rotation of a regular hexagon (see figure 1)about its centre by an angle of
π(n−1)/3, whereas S
n
represents a reflection in the line of symmetry that passes through the centre of the
hexagon and makes an angle of π(n−1)/6 with the positive x-axis, where n=1, 2,K,6.
The symmetry group transformations which leave the differential equation (13)invariant can be obtained
from the C
6v
group representations (15)in the form of the orthogonal coordinate transformations:
=¢=
¢
¢=Lrx
yrx
yr,18
n
⎛
⎝
⎜⎞
⎠
⎟
()
⟼()
for
μ
n
1, 2, ,12{}
. Under these transformations, the gradient operator ∇transforms as:
=
¶
¶
¶
¶
¢ =
¶
¶¢
¶
¶¢
=L
x
y
x
y
.19
n
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
⟼()
This accordingly leaves the Laplacian ∇
2
invariant in form, which in turn leads to the invariance of the
Helmholtz equation (13)under the coordinate transformations (18), giving:
4
Phys. Scr. 96 (2021)065207 W A Gaddah
f
¶
¶+¶
¶+L=
-
xy
kr0. 20
n
2
2
2
2
21
⎛
⎝
⎜⎞
⎠
⎟() ()
This is the same as the original equation (13)but with the corresponding solution
f
r()
being transformed as:
ff ff=GL = L
-
rr r r,21
nn1
()⟼() ( ) () ( ) ( )
where
G
Ln
()
is a linear operator defined to act on fin such a way that it transforms
f
r()
to
f
L-r
n1
(
)
for n=1,
2,K,12. If
f
r()
is a non-invariant function under the coordinate transformations (18), i.e.
f
fL¹
-rr
n1
()()
,
then
G
Ln
()
is said to generate from the old solution
f
r()
a new different solution
f
L-r
n1
(
)
. The old and new
solutions will share the same domain of definition Ω, because the hexagonal boundary ∂Ωdoes not change in
shape or size under the hexagon symmetry transformations (18).
Since the differential equation (13)is linear, a linear combination of all the solutions obtained by the
symmetry transformations (18)via (21)is also a solution. Therefore, we can write the C
6v
symmetry-generated
solution of equation (13)as:
å
jf=L
=
-
rrc, 22
n
nn
1
12
1
() ( ) ( )
where c
n
are arbitrary constants. Now, if we obtain one analytical solution
f
r()
for (13)that is not invariant
under (18)we can immediately construct (22). To achieve this for the case |E|>m
0
c
2
, we begin by considering a
well-known preliminary solution to equation (13)in the form of a traveling plane wave:
f== +rkxkyeexpi , 23
kri12
() [( )] ( )
·
which does not satisfy the desired boundary condition (8), where
=+kk kij
12
ˆˆ
with k
1
and k
2
being real
constants satisfying the condition:
+== -
kkk E mc c.24
1
22
2220
2422
() ()
If we then operate on (23)by
G
Ln
()
as in (21)for n=1, 2,K,12, we generate twelve linearly independent
solutions whose linear combination yields the sought solution (22)in terms of distinct plane waves:
j=+ + +
++++
+++ +
ll ww
ll ww ww
ll ww ll
++-+-+
-+ + - -
-- -- -- -
xy c c c c
cccc
ccc c
,e e e e
eeee
eee e,25
kx ky x y x y kx ky
xy xy kxky xy
xy kxky xy xy
1i2i3i4i
5i6i7i8i
9i10 i11 i12 i
12 1 2 1 2 12
12 12 12 12
12 12 12 12
()
()
() ( ) ( ) ()
() () () ()
() () ( ) ()
where
ll
ww
=- = +
=+ = -
kk kk
kk kk
32, 32,
3 2, 3 2. 26
11 2 2 2 1
11 2 2 2 1
() ()
() () ()
This generated solution of equation (13)can now be manipulated via the arbitrary constants c
n
and (k
1
,k
2
)to
make it satisfy the given boundary condition in (8).
The lower and upper oblique sides of the hexagon are described, as shown in figure 1, by the equations
=- yLx3
and =yL x3, respectively, whereas the two vertical sides are identified by
=xL32
. By imposing the Dirichlet boundary condition (8)on
j
r(
)
at the right and left lower oblique
sides of the hexagon; namely
j
- =xL x,30()
,wefind that:
=====-
=- =- =- =- =-
*
*
caccbccbccabcc bc
c acc cc cc abcc bc
,,, ,,
,,, , , 27
3142516 27 2
8 1 9 2 10 1 11 2 12 1
()
where
l=
a
Lexp i 3
1
(
)
,
w=-bLexp i 3
1
(
)
and the superscript asterisk symbol
*
denotes complex
conjugation. Substituting (27)in (25)for the corresponding values of c
n
, for n3, yields:
j=+ = +
*
rcWrcbWrN Wr N WrRe Im , 28
12 1 2
() () () { ()} { ()} ( )
where N
1
=c
1
+c
2
b,N
2
=i(c
1
−c
2
b)and
ww ll
ww ll
=++-++-+
-- - - - - -
Wxy kx ky a x y b x y
kx ky a x y b x y
, exp i exp i exp i
exp i exp i exp i . 29
12 1 2 1 2
12 1 2 1 2
( ) [ ( )] [ ( )] [ ( )]
[ ( )] [ ( )] [ ( )] ( )
Here, W(x,y)satisfies the two Dirichlet boundary conditions
- =
W
xL x,30()
. This implies that both
the real and imaginary parts of W, which constitute two independent solutions to (13), comply also with the
same boundary conditions. It is worth noting that the boundary condition
ζW =
W
r0()
is equivalent
to
j
ζW =r0()
.
5
Phys. Scr. 96 (2021)065207 W A Gaddah
To avoid having the zero solution
j
=r0()
in (28), or equivalently
=
W
r0()
, throughout the interior of
the hexagonal boundary ∂Ω, we restrict the values of k
1
and k
2
to the constraints:
¹¹¹kk k k k,0,0, 0, 3. 30
12 1 1 2
()() ()
The allowed values of k
1
and k
2
can be obtained by imposing the Dirichlet boundary condition on W(x,y)at
either one of the two vertical sides of the hexagon in figure 1, namely:
=WLy32, 0. 31() ()
Since W(x,y)is an odd function of x, the boundary condition
-=
W
Ly32, 0()
holds true whenever
=
W
Ly32, 0()
is satisfied and vice versa. The application of this boundary condition gives:
== =kL kL kL kLsin 3
2sin 3
20, cos 3
2cos 3
2,32
12 1 2
⎜⎟ ⎜⎟
⎛
⎝
⎜⎞
⎠
⎟⎛
⎝
⎞
⎠
⎛
⎝
⎜⎞
⎠
⎟⎛
⎝
⎞
⎠()
from which we obtain k
1
and k
2
as:
pp
==+kLnk
Lnn
2
3,2
32, 33
11212
() ()
where n
1
and n
2
are integer numbers. It follows from (30)and (33)that:
¹¹¹-nnnn0, 0, . 34
1212
()
Now, if we apply the Dirichlet boundary condition on W(x,y)at the right and left upper oblique sides of the
hexagon in figure 1,wefind that =
W
xL x,30()
, or equivalently
j
=xL x,30()
, only if:
== =kL kL kL kLsin 3 sin 3 0, cos 3 cos 3 . 35
12 1 2
()() ()() ()
These are related to the constraints (32)by the trigonometric double-angle formulas
qq
q
=sin 2 2 sin cos
and
qq=-cos 2 2 cos 1
2() . Hence, the restricting conditions in (35)are automatically satisfied once the
constraints (32), or equivalently (33), are fulfilled, but not vice versa.
Introducing (33)into (29)allows us to express (29)in the form of a simple determinant:
=--
--
Wr
111
ee e
ee e
,36
nn nu n v u nv
nu n u v nv
,ii i
ii i
12 11 1
22 2
() ( )
()
()
where
pp=+ =-u xyLv xyL23 3, 23 3. 37() () ()
It is evident that the constraints (34)guarantee the non-vanishing of the determinant (36)as they exclude the
cases in which any two rows of the determinant are equal. The real and imaginary parts of
W
xy,
nn,
12
(
)
are
separately two different solutions of the same equation (13), with the same eigenvalue (14), and both individually
satisfy the same prescribed Dirichlet boundary condition (8). Hence, the possible solutions of the Helmholtz
equation (13), which satisfy (8), can be identified by two different classes of eigenfunctions:
j==
=
xy Wxy j
Wxy j
,Re , , 1,
Im , , 2, 38
nn
jnn
nn
,
1,
2,
12
12
12
⎧
⎨
⎩
() {()}
{()} ()
()
where
1and
2
are normalization constants. In compact trigonometric forms,
j
nn,
1
12
() and
j
nn,
2
12
() can be written
as:
j=+
-+ -
-+
xy qnx q n n y
qn n x qn n y
qn x q n n y
, sin 3 sin 2
sin 3 sin
sin 3 sin 2 , 39
nn,
11212
12 21
112
12
() [( )(( ))
( ( )) (( ))
( ) (( ))] ( )
()
and
j=+
-+ -
++
xy qnx q n n y
qn n x qn n y
qn x q n n y
,sin3cos2
sin 3 cos
sin 3 cos 2 , 40
nn,
22212
12 21
112
12
() [( )(( ))
( ( )) (( ))
( ) (( ))] ( )
()
where q=2π/3L,=
2
11
and =
2
2
2
. Here, the values n
1
=n
2
are not allowed for
j
xy,
nn,
1
12
()
() because
they lead to the trivial zero solution. Moreover, the eigenfunctions in each of the four pairs:
jj jj jj jj
-- - + --
,,,,, ,, , 41
nn
j
nn
j
nn
j
nn
j
nn
j
nn n
j
nn
j
nnn
j
,, ,, ,, ,,
12 1 2 12 21 12 21 2 12 1 1 2
{}{}{ }{ } ()
() () () () () () () ()
6
Phys. Scr. 96 (2021)065207 W A Gaddah
are not linearly independent, because they satisfy the following relations:
jjjj
jjjj
=- =-
=- =- "=
+--
+-+ --
rrrr
rrrrj
1, 1,
1 , 1 , 1, 2. 42
nn
jj
nn
j
nn
jj
nn
j
nn
jj
nn n
j
nn
jj
nnn
j
,
1
,, ,
,
1
,, ,
12 1 2 12 21
12 21 2 12 1 1 2
() ( ) () () ( ) ()
() ( ) () () ( ) () ( )
() () () ()
() () () ()
Hence, we impose the restrictions:
>> >nn nn0, and 0, 43
21 21 ()
on
j
r
nn,
1
12
()
()
and
j
r
nn,
2
12
()
()
, respectively, in order to exclude all duplicate solutions. A quick inspection of the
analytic expressions (39)and (40)reveals the fact that
j
r
nn,
1
12
()
()
is symmetric whereas
j
r
nn,
2
12
()
()
is antisymmetric
under reflection through the origin; that is
j
j-=rr
nn nn,
1
,
1
12 12
() ()
() ()
and
j
-r
nn,
2
12
()
()
=
j
-
r
nn,
2
12
(
)
()
. This divides our
solutions into two distinct classes; one of even symmetry and the other of odd symmetry about the origin.
In figure 2, we represent the allowed values of the ordered pair of quantum numbers (n
1
,n
2
)by solid dots
and small circles in the n
1
n
2
-lattice space. Each solid dot on the off-diagonal line in figure 2corresponds to a
distinct pair of eigenfunctions jj,
nn nn,
1
,
2
12 12
{
}
() () with the same eigenvalue, whereas the small circles on the diagonal
line n
1
=n
2
correspond only to one class of eigenfunctions, namely the antisymmetric solutions
j
nn,
2
12
() .
The correctness of our solutions (39)and (40)has been confirmed by directly substituting the solutions into
the Helmholtz equation (13), and by explicitly introducing them to the Dirichlet boundary condition (8).
The constants
1
and
2in equations (39)and (40), respectively, can be determined from the normalization
condition:
j=
W
rdr 1, 44
nn
j
,
22
12
∬
∣()∣ ()
()
which, after making use of the evenness property of jxy,
nn
j
,
2
12
∣
()∣
() in xand y, can be rewritten as:
òò
j=
-
x y dydx4,1, 45
LLx
nn
j
0
32
0
3
,
2
12
∣()∣ ()
()
for j=1,2. By evaluating this integral, we find that:
== >>
Lnn
22
9
3,for 0, 46
12
34
21
()
Figure 2. Geometric representation of the allowed values of the quantum numbers (n
1
,n
2
), denoted here by solid dots and small
circles. The circles on the diagonal line n
1
=n
2
represent one set of the quantum numbers that is allowed only for jnn,
2
12
() , but not for
jnn,
1
12
() .
7
Phys. Scr. 96 (2021)065207 W A Gaddah
==
Lnn
2
9
3,for . 47
2
34
21 ()
The lower component cr
nn
j
,
12
()
() of the Dirac spinor in (5)can be obtained from
j
r
nn
j
,
12
()
()
by direct
differentiation using (9)as:
cj=+
¶
¶-¶
¶
rc
Emc y x ri, 48
nn
j
nn
j
,02,
12 12
⎛
⎝
⎜⎞
⎠
⎟
() () ( )
() ()
where |E|>m
0
c
2
and j=1,2.
Looking back at the Helmholtz equation (13)for the case where |E|m
0
c
2
, we have proven that there are no
(nontrivial)solutions f(x,y)to this equation that can satisfy the prescribed Dirichlet boundary condition (8).
The full details of this proof are provided in appendix A.
3. Results and discussion
In this section, we determine the exact energy expression for a Dirac massive particle inside a hexagon billiard
and then investigate its non-relativistic limit. Moreover, we present the corresponding Dirac probability density
together with its non-relativistic counterpart, and show analytically the vanishing of the Dirac current along the
billiard boundary. We also give a brief account of the nodal domains of the Schrödinger energy eigenfunctions,
and finally discuss the isospectrality of polygonal billiards.
To find the energy eigenvalues, we begin by substituting (33)into the energy relation (24)to obtain:
p
-=++
Emc
cL
nnnn
16
9.49
20
24
22
2
21
212 2
2
() ()
This can be solved algebraically for Eto yield two distinct real roots:
a= + + +
Emc nnnn1, 50
nn,022
1
212 2
2
12
() ()
()
where αis a dimensionless positive constant given by:
ap
=
Lm c
4
3,51
0
()
and the quantum numbers (n
1
,n
2
)are non-zero positive integers restricted by the inequality n
2
n
1
.
It follows from (50)that the energy spectrum consists of an infinite number of discrete energy levels
corresponding to bound states of positive and negative energies lying outside the energy gap [−m
0
c
2
,m
0
c
2
],
and not being bounded from below. The interpretation of the two possible values of the energy E, which only
differ in sign, is the same as in the familiar case of a relativistic free particle [19]. Namely, the positive energy
solutions correspond to a particle with energy
>
+
E
mc
nn,02
12
()
, whereas the negative energy solutions correspond
to an anti-particle with energy
<-
-
E
mc
nn,0
2
12
() .Infigure 3, we show the spectrum of the relativistic energy ratio
=
Emc
nn nn,,
0
2
12 12
() ()
as a function of αfor any arbitrary ordered pair of quantum numbers (n
1
,n
2
). The figure
illustrates the dependence of
E
nn,
12
()
on the billiard size through α, and shows the growth of the energy gap
(between the positive and negative energy levels)with the decrease of Ldue to the increase of α.
To investigate the non-relativistic energy limit, we expand the positive energy eigenvalue +
E
nn,
12
()
in powers of
α
2
as:
aa=+ ++-+++
+
E m c n nn n n nn n11
2
1
8.52
nn
,022
1
212 2
241
212 2
22
12
⎡
⎣
⎢⎤
⎦
⎥
()() ()
()
Then, subtracting the rest energy m
0
c
2
from this expansion and retaining only the term of first order in α
2
(or
alternatively, taking the limit c→∞)yields the non-relativistic limit of our energy expression:
p
=++
EmL nnnn
8
9.53
nn
,
NR 22
021
212 2
2
12
() ()
Following the same argument of Sewell [20]on the non-relativistic limit approach, we shall show that our
solutions
j
r
nn
j
,
12
()
()
for the upper component of the Dirac spinor are identical to their non-relativistic
counterparts obtained from the corresponding Schrödinger equation. The central aspect of this method is to
expand the relativistic energy eigenvalues together with the upper and lower components of the Dirac spinor in
inverse powers of c, the speed of light, and then substituting these expansions into the coupled equations (9)
and (10), and equating coefficients of the equal powers of c. In this way, the zeroth-order approximation in c
−1
yields the corresponding Schrödinger equation.
Inspection of the energy expression (50)shows that the positive relativistic energy Eis an even function of the
parameter c, and hence can be expanded in a perturbation series in powers of c
−2
, as suggested by Sewell [20],in
8
Phys. Scr. 96 (2021)065207 W A Gaddah
the following form:
=++ + +¼Emc E cEcE
11,54
020
2
1
4
2
()
() () ()
where
== -
¥
E
EEmclim
c
0NR 02
()
() , and E
(i>0)
are higher-order corrections. Likewise, the invariance of
the Helmholtz equation (13), under the transformation c→−c, indicates the evenness of its solutions
j
rc,()
with respect to c. On this basis, it follows from (48)that the lower component of the Dirac spinor crc,()
is an
odd function of c. In accordance with the non-relativistic limit approach, proposed first by Sewell [20]in a more
general context and adapted by several other authors [21–23], we expand
j
rc,()
and crc,()
perturbatively in
inverse powers of cas:
å
jy=
=
¥
rc cr,1,55
nn
n
02
() () ()
()
å
cx=
=
¥
+
rc cr,1,56
nn
n
021
() () ()
()
where ψ
(n)
and ξ
(n)
are correction functions independent of c. Then substituting equations (54),(55)and (56)
into the coupled equations (9)and (10), and after that equating coefficients of equal powers of c
−1
on both sides
of the resulting identities, we finally obtain for the zeroth power of c
−1
:
yx=
+
Prmr2, 57
000
() () ( )
() ()
xy=
-
PrE r.58
000
() () ( )
() () ()
By eliminating
x
r
0()
() from these two equations, we find that the leading term
y
r
0
()
()
in the expansion of the
upper component
j
rc,()
satisfies the non-relativistic Schrödinger equation:
yy- =
mrE r
2.59
2
0
20 0 0
() () ( )
() () ()
This, in principle, allows us to extract the stationary Schrödinger eigenfunction
y
r
0
()
()
from the knowledge of
j
rc,()
by taking the limit c→∞. Note that since
j
=rc,0()
on the billiard boundary ∂Ω, we have:
Figure 3. Spectrum of the relativistic energy ratio
=
Emc
nn nn,,
0
2
12 12
() () . The upper and lower curves represent
+
nn,
12
()
and
-
nn,
12
()
,
respectively, as a function of αfor any arbitrary ordered pair of quantum numbers (n
1
,n
2
).
9
Phys. Scr. 96 (2021)065207 W A Gaddah
jy==
¥ ζW ζW
rc rlim , 0. 60
crr
0
()∣ ()∣ ()
()
This means that both
j
rc,()
and
y
r
0
()
()
obey the same boundary condition.
It is worth mentioning that the dependence of the constructed solution (25)on cstems naturally from the
wave-number components (k
1
,k
2
)as they satisfy the c-dependent energy relation (24). But, due to the
imposition of the Dirichlet boundary condition (8)the c-dependence of k
1
and k
2
disappears completely, as is
clear from (33), and hence the solution (25)becomes independent of c. Therefore, regardless of the non-
relativistic limit under which
j
rc,()
reduces to
y
r
0
()
()
, our solutions
j
r
nn
j
,
12
()
()
for the upper component of the
Dirac spinor turn out to be the exact eigenfunctions of the stationary Schrödinger equation (59)subjected to the
zero Dirichlet boundary condition. This can be simply verified by directly substituting
j
r
nn
j
,
12
()
()
for
y
r
0
()
()
in (59). In doing so, we find that the energy expression for E
(0)
in (59)has exactly the same form as the one
obtained in the non-relativistic limit, namely
E
nn,
NR
12
.
Now, we briefly summarize our solutions for the Dirac equation (2)under the Dirichlet boundary
condition (8). The positive and negative energy states are described, with the help of (5),(39),(40)and (48),by
the Dirac spinors:
j
c
Y=
rN r
r,61
nn
jnn
nn
j
nn
j
,,,
,
,
,
12 12
12
12
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
() ()
() ()
() ()
()
()
for j=1,2, where the plus and minus signs (±)correspond to states with positive and negative energies
E
nn,
12
()
,
respectively,
N
nn,
12
()
is the normalization constant being determined in appendix B from (6)as:
a
=
+++
N
nnnn
1
211
1
,62
nn
,21
212 2
2
12
12
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
() ()
()
and
c
r
nn
j
,
,
12
(
)
()
is given by:
ch j=¶
¶-¶
¶
ryx ri, 63
nn
j
nn nn
j
,
,
,,
12 12 12
⎛
⎝
⎜⎞
⎠
⎟
() () ( )
() () ()
with
h
nn,
12
()being defined as:
h
a
=+ + +
mc nnnn
1
11
.64
nn
,021
212 2
2
12 () ()
()
To each ordered pair of quantum numbers (n
1
,n
2
)with the restriction n
2
>n
1
>0, there corresponds two
Dirac spinors
Y
nn,,1
12
()
and
Y
nn,,2
12
()
with the same energy eigenvalue
E
nn,
12
()
. However, for n
1
=n
2
there is only one
allowed class of spinors to either sign of the energy, namely
Y
nn,,2
12
()
.
The probability density
r
=Y Y
nn
jnn
jnn
j
,
,,,,,
12 12 12
() ()()
†and the corresponding probability current in (7)can be
constructed by
j
r
nn
j
,
12
()
()
, for j=1, 2, as:
rjhj=+
rN r r,65
nn
jnn nn
j
nn nn
j
,
,,2
,
2
,,
2
12 12 12 12 12
( ) [∣ ()∣ ∣ ()∣] ( )
() () () ( ) ()
and
jj=
Jrc
Err,66
nn
j
nn
nn
j
nn
j
,
,2
,
,,
12
12
12 12
() ()ˆ() ( )
()
()
() ()
D
for positive and negative energy states; denoted here by the plus and minus signs, respectively, where
D
is given
by:
=¶
¶-¶
¶yx
ij.67
ˆˆ ()D
Since -
E
nn,
12
()
=
-
+
Enn,
12
()
, we have
-
J
r
nn
j
,
,
12()
()
=
-
+
Jr
nn
j
,
,
12
(
)
()
. Hence, we denote by
J
r
nn
j
,
12
(
)
() the magnitude of the
probability current density
Jr
nn
j
,
,
12
∣
()
∣
()
for positive and negative energy levels:
jj=
Jr c
Err,68
nn
j
nn nn
j
nn
j
,
2
,,,
12
12
12 12
() ∣ () ()∣ ( )
() () ()
where =
E
E
nn nn
,,
12 12
∣
∣
()
. The fact that the probability current does not leak out through the boundary walls of the
billiard is ensured by the Dirichlet boundary condition (8). This can be easily seen by imposing on (66)or (68)
the condition that
j
=xy,0
nn
j
,
12
()
() for all points (x,y)on the billiard boundary ∂Ω.
The magnitude of the probability current
J
xy,
nn
j
,
12
()
() is illustrated in figure 4by density plots for the cases
corresponding to the first five lowest positive energy levels along with the seventh level.
10
Phys. Scr. 96 (2021)065207 W A Gaddah
Namely the cases in which n
1
,n
2
=1, 2, 3 such that n
2
>n
1
for j=1 and n
2
n
1
for j=2, using the natural
system of units (c=ÿ=m
0
=1)together with the choice of letting L=1, where m
0
is the rest mass of the
electron. In figure 5, we show density plots of the normalized probability density
r
+
xy,
nn
j
,
,
12
()
() in the xy-plane for
bound states with positive energy levels corresponding to the quantum numbers (n
1
,n
2
)with n
1
,n
2
=1, 2, 3
such that n
2
>n
1
for j=1, and n
2
n
1
for j=2, using the same units and the same choice of L=1asinfigure 4.
It is interesting to note that the graphs of all density plots of the probability current magnitude
J
xy,
nn
j
,
12
()
() and
the corresponding probability density
r
+
xy,
nn
j
,
,
12
()
() ,infigures 4and 5, exhibit six-fold rotational and six-fold
reflectional symmetry. This can be proven analytically by investigating the invariance of
J
xy,
nn
j
,
12
()
() and
r
+
xy,
nn
j
,
,
12
()
() under the coordinate transformations of rotation:
¢
¢=
x
yx
y
x
y
R,69
n
⎛
⎝
⎜⎞
⎠
⎟
() ()
⟼()
Figure 4. Density plots of the magnitude of the probability current
Jxy,
nn
j
,
12
()
()
in the xy-plane. The cases in which (n
1
,n
2
;j)=(1, 1; 2),
(1, 2; 1)and (1, 2; 2)are illustrated respectively from left to right in the upper row, and those corresponding to (n
1
,n
2
;j)=(2, 2; 2),(1,
3; 1)and (1, 3; 2)are illustrated respectively from left to right in the middle row, whereas the cases for (n
1
,n
2
;j)=(2, 3; 1),(2, 3; 2)and
(3, 3; 2)are shown respectively from left to right in the lower row. The color bar scale at the bottom indicates the range of intensity
values of
Jxy,
nn
j
,
12
()
()
, starting from the color spectrum of yellowish orange, as the minimum value of zero, to the red color as the
maximum value. Here, the natural units c=ÿ=m
0
=1 with L=1 are employed.
11
Phys. Scr. 96 (2021)065207 W A Gaddah
and reflection:
¢
¢=
x
yx
y
x
y
S.70
n
⎛
⎝
⎜⎞
⎠
⎟
() ()
⟼()
where R
n
and S
n
are given by (16)and (17), respectively, with n=1, 2,K,6. Since
j
r
nn,
1
12
()
()
and
j
r
nn,
2
12
()
()
satisfy
the identities:
jjj j
jjjj
==-
=- =-
--
-+-+
rr r r
rrr r
RS
RS
,,
1, 1, 71
nn nnn nn nnn
nn nn
nn nn nn
nn
,
11
,
1
,
11
,
1
,
21 1
,
2
,
21 1
,
2
12 12 12 12
12 12 12 12
( ) () ( ) ()
( )() () ( )() () ()
() () () ()
() () () ()
it follows that the symmetry transformations (69)and (70)map both
J
r
nn
j
,
12
(
)
() and
r
r
nn
j
,
,
12
()
()
into themselves,
exhibiting clear rotational and reflectional invariance in our results as:
Figure 5. Density plots of the probability density
r
+xy,
nn
j
,
,
12
(
)
() in the xy-plane for positive energy states. The cases in which (n
1
,
n
2
;j)=(1, 1; 2 ),(1, 2; 1)and (1, 2; 2)are illustrated respectively from left to right in the upper row, and those corresponding to (n
1
,
n
2
;j)=(2, 2; 2 ),(1, 3; 1)and (1, 3; 2)are illustrated respectively from left to right in the middle row, whereas the cases for (n
1
,
n
2
;j)=(2, 3; 1 ),(2, 3; 2)and (3, 3; 2)are shown respectively from left to right in the lower row. The color bar scale at the bottom
indicates the range of intensity values of the probability density, starting from the color spectrum of yellowish orange, as the minimum
value of zero, to the red color as the maximum value. Here, the natural units c=ÿ=m
0
=1 with L=1 are employed.
12
Phys. Scr. 96 (2021)065207 W A Gaddah
rrrr==
==
- -
--
rr r r
J rJrJ rJr
RS
RS
,,
,, 72
nn
jnnn
j
nn
jnnn
j
nn
jnnn
jnn
jnnn
j
,
,1
,
,
,
,1
,
,
,1,,
1,
12 12 12 12
12 12 12 12
( ) () ( ) ()
( ) () ( ) () ( )
() () () ()
() () () ()
for j=1, 2.
In the non-relativistic limit where c→∞, the second term in equation (65)vanishes completely for positive
energy levels, and
+
N
1
nn,
12
() . Hence, the probability density
r
+
xy,
nn
j
,
,
12
()
() reduces directly to its non-relativistic
counterpart jxy,
nn
j
,
2
12
∣
()∣
() which is also invariant under the symmetry transformations (69)and (70)as can be
seen from (71).Infigure 6, we show density plots of the probability density
r
+
xy,
nn
j
,
,
12
()
() in the non-relativistic
limit for bound states with positive energy levels that correspond to the same quantum numbers (n
1
,n
2
), and the
same order of arrangement as in figure 5. The relativistic probability density
r
+
xy,
nn
j
,
,
12
()
() and its non-relativistic
counterpart jxy,
nn
j
,
2
12
∣
()∣
() share the same rotational and reflectional symmetry as illustrated in figures 5and 6.In
other words, the Dirac probability density does not lose its symmetry in the non-relativistic limit.
Nodal domains of the Schrödinger eigenfunctions
j
r
nn
j
,
12
()
()
, i.e. the connected regions where
j
r
nn
j
,
12
()
()
has a
constant sign inside Ω, are illustrated in figure 7for the cases in which n
1
,n
2
=1, 2, 3 such that n
2
>n
1
for j=1
and n
2
n
1
for j=2, using L=1 for the side length of the billiard.
Figure 6. Density plots of the non-relativistic probability densities
jjjxy xy xy,, ,, ,
1,1
22
1,2
12
1,2
22
{
∣()∣∣()∣∣()∣}
() () ()
(upper row),
jjjxy xy xy,, ,, ,
2,2
22
1,3
12
1,3
22
{
∣()∣∣()∣∣()∣}
() () ()
(middle row), and
jjjxy xy xy,, ,, ,
2,3
12
2,3
22
3,3
22
{
∣()∣∣()∣∣()∣}
() () ()
(lower row), lined up
horizontally (in each row)from left to right respectively. The color bar scale at the bottom is the same measuring scale used in figures 4
and 5with L=1.
13
Phys. Scr. 96 (2021)065207 W A Gaddah
The white and black regions in the checkerboard pattern of the nodal domains correspond to the positive
and negative values of
j
r
nn
j
,
12
()
()
, respectively. Since all the eigenfunctions
j
r
nn
j
,
12
()
()
are odd under the mirror-
reflection of coordinates (x,y)about the y-axis, i.e.
j
-xy,
nn
j
,
12
(
)
()
=j
-
xy,
nn
j
,
12
()
() , there must exist for each non-
zero (positive or negative)value of
j
nn
j
,
12
() at (x,y)the same value in magnitude for
j
nn
j
,
12
() at (−x,y)but with an
opposite sign. This implies that all black areas of the nodal domains inside the hexagon Ωare equal in number
and similar in shape to the white ones. Because each black or white nodal domain on the left side of the y-axis has
areflected image on the right side but with an opposite color. Hence, the total number of nodal domains for any
eigenfunction
j
r
nn
j
,
12
()
()
is always an even number. This fact can be easily checked by directly counting the total
number of nodal domains in each subplot of figure 7.
Furthermore, by looking at figure 7we quickly note that all the Schrödinger eigenfunctions
j
r
nn
j
,
12
()
()
share a
common feature of being zero on the three long diagonals (circumdiameters)of the hexagon, defined as:
=Î=
=Î=
=Î=-
lxy yx x L
lxy x yL
lxy yx x L
,3,32,
,0,,
,3,32. 73
12
22
32
{( ) ∣ ∣ ∣ }
{( ) ∣ ∣ ∣ }
{( ) ∣ ∣ ∣ } ( )
Figure 7. Nodal domains of the hexagon-billiard Schrödinger eigenfunctions jjjxy xy xy,, ,, ,
1,1
2
1,2
1
1,2
2
{
() () ()}
() () () (upper row),
jjjxy xy xy,, ,, ,
2,2
2
1,3
1
1,3
2
{
() () ()}
() () () (middle row), and jjjxy xy xy,, ,, ,
2,3
1
2,3
2
3,3
2
{
() () ()}
() () () (lowerrow), lined up horizontally (in each
row)from left to right respectively. The eigenfunctions are positive (negative)in the white (black)regions.
14
Phys. Scr. 96 (2021)065207 W A Gaddah
This is because all the eigenfunctions
j
r
nn
j
,
12
()
()
are antisymmetric with respect to mirror reflection of coordinates
about the long diagonals l
1
,l
2
and l
3
, i.e. satisfying the following properties:
jj=- " =
-rrnS, 2,4,6, 74
nn
jnnn
j
,
1
,
12 12
() () ()
() ()
where -r
S
21,-r
S
41and -r
S
61stand for the point images
+-xyxy232,32 2
(
)
,(−x,y)and
---xyxy232,32 2
(
)
, respectively, which are produced by reflection of a point (x,y)across the
long diagonals l
1
,l
2
and l
3
, successively. Taking this into account and the fact that the long diagonals partition the
regular hexagon into six congruent equilateral triangles whose perimeters are continuous loci of zeros for
j
r
nn
j
,
12
()
()
,wefind that the nodal domain patterns inside any two adjacent equilateral triangles are inverted
mirror-images of each other, i.e. the checkerboard-like patterns in any two neighboring triangles have the same
form and structure but with the black and white colors being inverted as shown in figure 7. This leads to the
conclusion that the number of nodal domains of
j
r
nn
j
,
12
()
()
inside a regular hexagon is six times the number of its
corresponding counterpart inside a single equilateral triangle. An exact counting of the total number of nodal
domains for a non-separable system such as the equilateral triangular billiard is a difficult problem, in general,
and beyond the scope of this paper. For a detailed analysis on nodal counts for eigenfunctions of an equilateral
triangular billiard, we refer to the works done by Samajdar and Jain [24,25].
It is worth noting from the implications of equation (74)that all the Schrödinger eigenfunctions
j
r
nn
j
,
12
()
()
vanish on the perimeter of each polygon that has one or two of its edges lying on the long diagonals {l
1
,l
2
,l
3
}
while the rest being located on the surrounding hexagonal boundary ∂Ωas shown in figure 8. Hence, it follows
that the solutions
j
r
nn
j
,
12
()
()
we obtained for the regular hexagon quantum billiard are also exact solutions (but
unnormalized)for other polygonal billiards such as the ones shaded in red in figure 8, namely the billiards in the
shape of equilateral triangle, rhombus with acute angles of π/3, isosceles trapezoid, irregular hexagon, and
irregular heptagon.
In 1966 Mark Kac [26]posed the famous question: ‘Can one hear the shape of a drum?’, which can be
restated for more clarity as: ‘Can the frequency spectra of two or more non-congruent flat drums be identical?’.
He was referring to the problem of whether a two-dimensional Helmholtz equation with a zero Dirichlet
boundary condition, such as our eigenvalue equation (13), could have identical spectra (eigenvalues)on (at least)
two non-congruent simply-connected planar domains. The question of Kac raises the issue of whether it is
Figure 8. A sample of different polygonal billiards, shaded in red, with one or two of their edges lying on the long diagonals {l
1
,l
2
,l
3
}
while the rest being located on the edges of the surrounding hexagonal boundary ∂Ω. The upper row shows convex polygonal billiards
in the shape of equilateral triangle, rhombus with acute angles of π/3, and isosceles trapezoid whereas the lower row depicts concave
polygonal billiards in the shape of irregular hexagons (the two on the left)and irregular heptagon.
15
Phys. Scr. 96 (2021)065207 W A Gaddah
possible to construct two or more billiards that have different shapes (non-isometric domains)on a Euclidean plane
but possess the same eigenvalues. Many authors have worked on this problem [27–31], providing several examples of
isospectral planar billiards, that is, non-congruent billiards with the same eigenvalue spectrum. In this regard, we
have constructed from an equilateral triangle a set of isospectral polygonal domains by simply gluing or sewing
together two or more identical copies of equilateral triangles side-by-side with one vertex of each located at the origin
so that they form a whole or part of a regular hexagon as illustrated in figure 8.Thiscanbeverified by considering the
fact that since the eigenfunctions (39)and (40)foraregularhexagonbilliardwithadomainΩsatisfy the Helmholtz
equation (13)andgiveexactlythesamesetofeigenvalues on all of the simply-connected domains in figure 8,the
isospectrality for these domains is confirmed by showing that the eigenfunctions on such domains obey the zero
Dirichlet boundary condition, which they do as they all vanish on the long diagonals {l
1
,l
2
,l
3
}and the hexagonal
boundary ∂Ω.Infigure 8we present five distinct non-congruent simply-connected domains on which the
Helmholtz equation (13), under a zero Dirichlet boundary condition, has exactly the same eigenvalue spectrum as
that of the regular hexagon billiard of figure 1. With this, we provide clear examples for the non-existence of a one-to-
one bijective correspondence between polygonally-shaped drums and their sound frequency eigenvalues.
4. Conclusions
We have presented a rigorous group-theoretic method for deriving exact analytical solutions of the planar Dirac
equation for a massive point-like particle confined to a flat billiard with the shape of a regular hexagon. We have
shown that, by applying an adequate boundary condition of our previous work [13], the upper component of the
two-dimensional Dirac spinor isfound to satisfy the Helmholtz equation with zero Dirichlet boundary condition.
This has accordingly reduced the problem to solving the hexagonal Schrödinger billiard. The basic idea of our
approach is to generate new independent solutions from an old well-known plane-wave solution by making use of
the invariance property of the Helmholtz equation under the C
6v
point group that describes the symmetry of a
regular hexagon. The linear combination of the C
6v
symmetry-generated solutions is shown to produce all the
required stationary Schrödinger eigenfunctions in simple non-separable trigonometric forms which satisfy the
prescribed Dirichlet boundary condition.The contribution of the lower spinor-component can be easily calculated
from the knowledge of the associated upper component, obtained independently from the corresponding
Schrödinger billiard. Our solution method is conceptuallysimple and can be possibly tested on a number of regular
n-sided polygonal billiards which are invariant under the symmetry group C
nv
. Practically, it has been proven to be
very successful in producing exact analytical solutions for equilateraltriangular billiards [13],squarebilliards(see
appendix C for details), and regular hexagon billiards. Our approach demonstrates the important role of the
geometric symmetry groups of the regularly-shaped polygons in the derivation of the exact stationary states and
energy spectrum for the corresponding Schrödinger and Dirac billiards.
We have shown that there exists no solutions for the hexagon billiard inside the familiar Dirac energy gap
|E|m
0
c
2
just like in the free particle case. The solutions obtained outside the energy gap were found to have
simple but non-separable trigonometric forms. The non-relativistic limit of our results are also discussed and
the exact Schrödinger energy eigenvalues are shown to be recovered correctly from their relativistic
counterparts. Illustrative density plots of the probability current and the corresponding probability density for a
selected set of positive energy states have been provided, and their symmetry have been examined. We have also
provided a number of demonstrative density plots for the probability density in the non-relativistic limit.
Moreover, we have presented checkerboard-like patterns that describe the nodal domains for the hexagon-
billiard Schrödinger eigenfunctions, and have given a simple nodal-count relation between billiards with regular
hexagon shapes and those with equilateral triangular shapes. By investigating the symmetry properties of the
bound state solutions for a regular hexagon billiard, we manged to construct a set of isospectral quantum
billiards with various polygonal shapes. With this, we have proven that the eigenvalue spectra for a number of
polygonal billiards do not correspond uniquely (bijectively)to their bounded domain shapes.
Appendix A. A proof of the non-existence of acceptable solutions inside the energy gap
|E|m
0
c
2
In this appendix, we show that there are no acceptable solutions to the Helmholtz equation (13)that can satisfy
the Dirichlet boundary condition (8)for the case in which |E|m
0
c
2
.
We begin by considering Green’sfirst identity in a two-dimensional space:
ff f f ff+ =
W¶W
dr dln,A.1
12212
212
∬∮
[·] ·
ˆ()
16
Phys. Scr. 96 (2021)065207 W A Gaddah
where f
1
and f
2
are arbitrary differentiable scalar fields, dl is the infinitesimal arc length of the perimeter ∂Ωof
the given domain Ω, oriented counterclockwise, and n
ˆis the outward-pointing unit normal vector to ∂Ω.By
letting
f
f=r
1*()
and
f
f=r
2
()
in (A.1), with the assumption that fis a normalized solution to (13), and then
substituting the right-hand side of (13)for ∇
2
f, we obtain the integral equation:
fff=
-+
W¶W
dr Emc
cdln.A.2
22 20
24
22 *
∬∮
∣∣ ·
ˆ()
Now, if we impose the Dirichlet boundary condition (8)on ffor the case E=±m
0
c
2
, equation (A.2)reduces to:
f=
W
dr 0. A.3
22
∬
∣∣ ()
Since |∇f|
2
is a non-negative integrand, the only possibility for this integral relation to hold is when |∇f|
2
=0, or
rather ∇f=0, inside Ω. Thisimplies that fis a constant within Ω. But since fis continuous on Ωand ∂Ω,the
value of the constant is equal to the value of fon the boundary ∂Ω, which is zero. Consequently, we have f(x,
y)=0 throughout the whole region Ω. This trivial solution is not admissible as it implies the nonexistence of the
particle inside the given domain Ω.
For the case |E|<m
0
c
2
, the imposition of the Dirichlet boundary condition (8)in (A.2)results in a
contradictory inequality, namely the integral of a non-negative integrand |∇f|
2
is negative:
f<
Wdr 0. A.4
22
∬∣∣ ()
This contradiction indicates that the eigenvalue equation (13)has no solution that can satisfy the Dirichlet
boundary condition (8)in the case |E|<m
0
c
2
.
Appendix B. Determination of the normalization constant of the Dirac spinors
In this appendix, we show how to determine the normalization constant given in equation (62)for the two-
dimensional Dirac spinors.
We begin by substituting
j
r
nn
j
,
12
()
()
and (48)in the integral identity of the normalization condition (6). Then,
by using the fact that the functions
j
r
nn
j
,
12
()
()
are all normalized to unity, we obtain:
++=
Nc
Emc I11, B.1
222
022
⎡
⎣
⎢⎤
⎦
⎥
∣∣ () ()
where, for notational simplicity, we let
j=
W
Irdr,B.2
22
∬
∣()∣ ()
and denote
E
nn,
12
()
and
j
nn
j
,
12
() by Eand j, respectively. Since ∇·(j∇j)=j∇
2
j+|∇j|
2
, we can rewrite Ias:
jj jj=- +
WW
Idr dr.B.3
22 2
∬∬
·( ) ( )
With the help of the planar divergence theorem, one can show that the second term in (B.3)assumes the form:
jj jj =
W¶W
dr dln,B.4
2
∬∮
·( ) ·ˆ()
where dl is the infinitesimal arc length of the perimeter ∂Ωof the hexagonal domain Ω, oriented
counterclockwise, and n
ˆis a unit vector normal to ∂Ωand pointing outward. Due to the Dirichlet boundary
condition
j
ζW =r0()
, the integral in (B.4)vanishes, and Ireduces to:
jj=-
W
Idr.B.5
22
∬
()
By using the fact that
j
r(
)
is normalized, and satisfies the Helmholtz equation (13), i.e. ∇
2
j=−k
2
j, one can
evaluate the integral in (B.5)as:
== -
Ik Emc
c.B.6
220
24
22
()
Inserting this result for Iinto (B.1), and then solving for Nyields:
=+Nmc
E
1
21, B.7
0212
⎡
⎣
⎢⎤
⎦
⎥()
where, without loss of generality, we have assumed Nto be a real positive constant. Now, by substituting (50)for
Ein (B.7)we derive (62)as required.
17
Phys. Scr. 96 (2021)065207 W A Gaddah
Appendix C. A square billiard as an illustrative example
In this appendix, we present an illustrative and simple example of the application of our method for the
derivation of the well-known solutions of the square-shaped quantum billiard without the use of separation of
variables. We start by considering the Helmholtz equation (13)on a square-shaped domain Ωof side length L
centered at the origin:
W= Î xy x LyL
,2,2,C.1
2
{}
() ∣∣∣ ∣∣ ()
with the Dirichlet boundary conditions:
ff==Ly xL2, , 2 0. C.2()() ()
As a first step towards the solution of this Helmholtz boundary-value problem, we proceed as in section 2 by
considering the orthogonal coordinate transformations:
=¢=
¢
¢=Lrx
yrx
yr,C.3
n
⎛
⎝
⎜⎞
⎠
⎟
()
⟼()
which leave the Helmholtz equation (13)invariant, where Λ
n
denotes the irreducible orthogonal-matrix
representations of the C
4v
point group that describe the symmetry of the square billiard, namely:
L= =
=
-
n
n
R
S
, if 1,2,3,4,
, if 5,6,7,8, C.4
nn
n4
⎧
⎨
⎩()
where
pp
pp
=
---
--
nn
nn
R
cos 1
2sin 1
2
sin 1
2cos 1
2
,C.5
n
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
() ()
() () ()
and
pp
pp
=
--
---
nn
nn
S
cos 1
2sin 1
2
sin 1
2cos 1
2
.C.6
n
⎛
⎝
⎜
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
⎟
() ()
() () ()
Since a plane wave of the form:
f== +rkxkyeexpi , C.7
kri12
() [( )] ( )
·
with
+=kkk
1
22
2
2
solves the Helmholtz equation (13)exactly, we use this particular solution to generate seven
more new solutions to (13)by applying on (C.7)the C
4v
symmetry transformations (C.3). In this way we obtain,
without solving (13)directly, a finite set of linearly independent solutions whose linear combination yields the
required solution:
j=+ + +
++ + +
+---+ -
-+---+
xy c c c c
ccc c
,e e e e
eee e, C.8
kx ky ky kx kx ky ky kx
kx ky ky kx kx k y ky kx
1i2i3i4i
5i6i7i8i
12 12 12 12
12 12 12 12
()
()
() () () ()
() () () ()
which can be made to satisfy the Dirichlet boundary conditions (C.2)by appropriately choosing the values of the
arbitrary constants {c
1
,c
2
,K,c
8
}and (k
1
,k
2
).
By imposing on (C.8)the two boundary conditions j(L/2, y)=0 and j(x,L/2)=0 consecutively, we find
that:
===-
=- =- =-
+-
-
cc cc c c
cc cc c c
e, e, e,
e , e, e. C.9
kkL kkL kL
kL kL kL
31
i42
i51
i
62
i71
i82
i
12 21 2
11 2 ()
() ()
Inserting (C.9)into (C.8)yields:
j=- -
+- -
xy N kx L k y L
NkxL kyL
, sin 2 sin 2
sin 2 sin 2 , C.10
11 2
22 1
( ) [ ( )] [ ( )]
[ ( )] [ ( )] ( )
where N
1
and N
2
are mere constants given by:
=- =
+-
Nc Nc4e , 4e . C.11
kkL kkL
11
i2
22
i2
12 21 ()
() ()//
18
Phys. Scr. 96 (2021)065207 W A Gaddah
Applying now the boundary condition j(−L/2, y)=0to(C.10)results in:
-+ -=NkLkyL NkLkyLsin sin 2 sin sin 2 0. C.12
11 2 2 2 1
( ) [ ( )] ( ) [ ( )] ( )//
Since this condition is to be valid for all values of yon the interval [−L/2, L/2], we must demand that:
==NkL N kLsin 0, and sin 0, C.13
11 22
() () ( )
simultaneously. It is clear that we can not allow both N
1
and N
2
to vanish at the same time, since this would give
an unacceptable trivial solution j(x,y)=0 inside the billiard. Hence, there are three possible cases to consider,
namely: (i)N
1
=0 with
=kLsin 0
2
()
,(ii)N
2
=0 with =kLsin 0
1
() , and (iii)=kLsin 0
1
() with
=kLsin 0
2
()
.
For the first case, the eigenfunction j(x,y)in (C.10)reduces to:
j=- -xy N k x L ky L, sin 2 sin 2 , C.14
22 1
( ) [ ( )] [ ( )] ( )
with k
2
being restricted to the discrete values:
p==¼knL n, for 1, 2, 3, . C.15
2
()/
To determine the allowed values of k
1
in this case, we apply the last remaining boundary condition j(x,−L/
2)=0to(C.14). This yields:
p==¼kmL m, for 1, 2, 3, . C.16
1
()/
Applying the normalization condition (44)to the eigenfunctions j(x,y)in (C.14)over the billiard domain (C.1)
determines the constant N
2
as N
2
=2/L. Now, we can rewrite the solution (C.14)explicitly in terms of its
characteristic quantum numbers (n,m)as:
jpp=- -xy Lnx
Lmy
L
,2sin 1
2sin 1
2.C.17
nm,
⎜⎟ ⎜⎟
⎡
⎣
⎢⎛
⎝
⎞
⎠
⎤
⎦
⎥⎡
⎣
⎢⎛
⎝
⎞
⎠
⎤
⎦
⎥
() ( )
With the help of (C.15)and (C.16), we determine the exact eigenvalues of the Helmholtz equation (13)on a
square billiard as:
p
=+kLnm.C.18
nm,22
()
For the second case in which N
2
=0 and =kLsin 0
1
() , we obtain the same eigenfunctions and eigenvalues as
the ones derived in the first case. However, for the third case, we find that the solution obtained is not a new one
but a linear combination of two degenerate eigenfunctions j
n,m
(x,y)and j
m,n
(x,y), each given by (C.17).
ORCID iDs
Wajdi A Gaddah https://orcid.org/0000-0003-1750-1410
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