Eur. J. Phys. 17 (1996) 19–24. Printed in the UK 19
Relativistic particle in a box
P Alberto†, C Fiolhais† andVMSGil‡
†Departamento de F
´
ısica da Universidade de Coimbra, P-3000 Coimbra, Portugal
‡Departamento de Qu
´
ımica da Universidade de Coimbra, P-3000 Coimbra, Portugal
Received 3 August 1995
Abstract. The problem of a relativistic spin 1/2 particle
confined to a one-dimensional box is solved in a way that
resembles closely the solution of the well known
quantum-mechanical textbook problem of a non-relativistic
particle in a box. The energy levels and probability density
are computed and compared with the non-relativistic case.
Resumo. O problema de uma part
´
ıcula de spin 1/2 confinada
por uma caixa a uma dimens
˜
ao
´
e resolvido de uma maneira
muito semelhante
`
a da resolu¸c
˜
ao do problema de uma
part
´
ıcula no-relativista numa caixa referido em muitos livros
introdut
´
orios de Mec
ˆ
anica Qunt
ˆ
aica. Os n
´
ıveis de energia e a
densidade de probabilidade s
˜
ao calculados e comparados com
os valores n
˜
ao-relativistas.
1. Introduction
Energy quantization in atoms and molecules plays an
essential role in the physical sciences.
From which theoretical arguments does the quanti-
zation of energy comes from? The axioms of quantum
theory were built to explain that phenomenon. A simple
example of application of these rules, which has great
pedagogical value, is the study of a particle in a one-
dimensional box, i.e., an infinite one-dimensional square
well. Indeed, in many introductory courses of physics
or chemistry the study of electronic wavefunctions in
atoms and molecules is made by analogy with a parti-
cle in a box, without having to solve the more involved
Schr
¨
odinger equation for systems ruled by the Coulomb
potential. In this way one is able to introduce the con-
cepts of energy quantization and orbitals for atoms and
molecules without being lost in the mathematical de-
tails of solving the Schr
¨
odinger equation for a central
potential.
We note that for obtaining quantization it is not so
much the type of differential equation which must be
solved but the boundary condition which must be obeyed
by the solution: a particle confined to a finite region
of space does have discrete energy levels. In order to
solve the time-independent Schr
¨
odinger equation for a
free non-relativistic particle of mass m inside a one-
dimensional box of length L in the z-axis,
−
¯h
2
2m
d
2
ψ
dz
2
= Eψ (1)
which is the stationary wave (0 z L)
ψ(z) = C sin(kz) (2)
one has to impose the boundary conditions
ψ(0) =
ψ(L) = 0. These same conditions give rise to the
quantization rule for the wavenumber, k = nπ/L, n =
1, 2,... . Outside the box the wavefunction vanishes
which means that the derivative of the wavefunction is
discontinuous at the well walls (z = 0 and z = L). This
is related to the fact that the potential has a infinite jump
at the well walls. It is worth stressing this point as we
move later to the relativistic case. The energy levels of
equation (1) are given by
E
n
=
p
2
2m
=
¯h
2
k
2
2m
=
¯h
2
n
2
π
2
2mL
2
(3)
where p = ¯hk = nπ/L is the quantized momentum.
In a similar fashion, we propose to solve the Dirac
equation for a particle in a box, emphasizing the role
of the boundary conditions in the solution of the wave
equation in this paper ourselves. At the same time, we
will get the solutions in a way which closely resembles
the non-relativistic approach. The method we shall
use set ourselves apart from either the one-dimensional
Dirac equation approach [3, 4] and the calculations of
Greiner [5] both using a finite square well. By using
the three-dimensional Dirac with a one-dimensional
potential well in the z-axis, we are able to include
spin in the wavefunction without increasing much the
mathematical burden. The crucial points are the use
of a Lorentz scalar potential, which avoids the Klein
paradox problem, and boundary conditions which assure
the continuity of the probability current rather than that
of wavefunction itself. In this way we are able to avoid
the problems referred in [3–5] when the depth of the
potential goes to infinity.
In the next section we solve the Dirac equation in
an one-dimensional infinite square well, leaving the
mathematical details for an appendix. In the conclusions
we present some comment on our solution and compare
it with other approaches found in the literature.
0143-0807/96/010019+06$19.50
c
1996 IOP Publishing Ltd & The European Physical Society
20
P Alberto
et al
2. Solution of the Dirac equation in an
one-dimensional infinite square well
Let us consider the time-independent relativistic
equation for the wavefunction of a free electron of mass
m moving along the z direction. This Dirac equation
can be written as
ˆ
H
ψ = (α
z
ˆp
z
c + βmc
2
)ψ = Eψ (4)
where
ˆ
H = α
z
ˆp
z
c + βmc
2
is the Dirac energy operator
(Hamiltonian), ˆp
z
=−i¯h
d
dz
is the z component of the
momentum operator and α
z
and β the matrices
α
z
=
0 σ
z
σ
z
0
β =
I 0
0 −I
. (5)
Here, σ
z
isa2×2 Pauli matrix and I is the 2 × 2 unit
matrix.
We wish now to consider the solutions of (4) for a
free particle moving in the direction of the positive z-
axis with momentum ¯hk. These can be taken from a
relativistic quantum mechanics textbook (for example,
[1]), giving the following normalized† wavefunction
ψ
k
(z) =
1
2π
r
E + mc
2
2mc
2
e
i kz
χ
σ
z
¯hkc
E+mc
2
χ
(6)
where E =
p
(¯hkc)
2
+ m
2
c
4
is the energy of the particle
and χ an arbitrary two-component normalized spinor,
i.e., χ
†
χ = 1(χ =
1
0
for spin ‘up’, χ =
0
1
for
spin ‘down’, or any linear combination of these two).
Equation (4) admits also negative energy solutions with
energy −
p
(¯hkc)
2
+ m
2
c
4
. We will come back to this
point. We notice that for non-relativistic momenta,
i.e., ¯hk mc, the lower two-component spinor in (6)
vanishes and we get a wavefunction which is a plane-
wave solution of the Schr
¨
odinger equation for a free
particle with spin 1/2
ψ
k
(z) ∝ e
i kz
χ
0
. (7)
One can show that the wavefunction (6) is an
eigenstate of the square of spin operator
P
3
j=1
ˆ
S
2
j
with
eigenvalue
3
4
¯h
2
. Actually, this wavefunction is also an
eigenstate of
ˆ
S
z
, given by
S
z
=
¯h
2
σ
z
0
0 σ
z
(8)
if χ has the spin up or spin down form, but this is an
artifact of having restricted the motion only to the z-
axis. This means that each state is be twice degenerate.
Note, however, that it is not an eigenstate of the square
of the total angular momentum operator
P
3
j=1
(
ˆ
S
j
+
ˆ
L
j
)
2
.
We are now interested in finding solutions describing
a relativistic particle in a infinitely deep one-dimensional
† In the sense that for wavefunctions with wavevectors k
and k
0
, denoted by ψ
k
(z) and ψ
k
0
(z) respectively, one has
R
∞
−∞
dz ψ
†
k
(z)ψ
k
0
(z) = δ(k −k
0
), where δ(k − k
0
) is the Dirac
delta function.
z
m(z)
M
L
E
O
m
IIIIII
Figure 1. Plot of the mass as function of position
m
(
z
)
showing the three different zones I, II, III in which the
solution of the Dirac is evaluated. Eventually we take the
limit
M
→∞.
well. By analogy with the non-relativistic case, we
could proceed by adding to the Dirac Hamiltonian in
(4) a potential V(z) such that
V(z)=
V
0
z 0
00<z<L
V
0
zL
(9)
and then letting V
0
go to infinity. As said in the
introduction, the solution for the time-independent
Schr
¨
odinger equation with this potential, with V
0
→∞,
is a standing wavefunction in the region 0 <z<L
ψ
k
n
(z) ∝ sin(k
n
z) (10)
with k
n
= (nπ )/L, n = 1, 2,..., the energy eigenvalues
being given by (3), and a null function for z 0 and
z L.
This procedure, however, leads to the so-called
‘Klein paradox’: the flux of the reflected plane wave
in the walls of the potential is larger than the flux
of the incident wave [6]. This happens because the
wavefunction starts to pick up components from the
negative energy states once |V
0
− E| >m.AsV
0
continues to increase, more and more negative energy
states contribute as these are unbounded from below.
A way out of this problem is to assume that the mass
of the particle is itself a function of z. This is equivalent
of using a Lorentz scalar potential instead of a vector
potential. The infinite well is then introduced replacing
the mass in (4) by the function m(z) defined by
m(z) =
Mz0
m 0<z<L
MzL
(11)
where M is a constant which we will let go to infinity
later. The function m(z) is represented in figure 1.
This method was used in the MIT bag model of
hadrons [2], in which an infinite spherical well (bag)
confines the otherwise free constituent quarks inside the
bag. The solutions of the Dirac equation with m(z)
given by (11) can be considered separately in regions
I, II and III, corresponding to z<0, 0 z L
and z>L, respectively. In each of these regions the
Relativistic particle in a box 21
solution of the Dirac equation is of the form of (6)
since the function m(z) is constant in each of them.
We consider that inside the well we have two plane
waves travelling in opposite directions (an incident and
a reflected wave on the walls of the well). Outside the
well, we just consider one wave travelling outwards,
anticipating the condition lim
z→±∞
ψ(z) = 0 for a
bound solution. Thus, we may write for the three
regions:
ψ
I
(z) = A e
−ik
0
z
χ
−σ
z
¯hk
0
c
E+Mc
2
χ
ψ
II
(z) = B e
ikz
χ
σ
z
¯hkc
E+mc
2
χ
+ Ce
−ikz
χ
−σ
z
¯hkc
E+mc
2
χ
ψ
III
(z) = De
ik
0
z
χ
σ
z
¯hk
0
c
E+Mc
2
χ
(12)
where k
0
=
p
E
2
/c
2
− M
2
c
2
/¯h, and A, B, C and D are
constants.
In the appendix we compute the solutions in the case
M →∞with suitable boundary conditions (the outward
flux of probability at walls is zero). The result is that
ψ
I
(z) and ψ
III
(z) vanish identically, and ψ
II
(z) is given
by
ψ
II
(z) = B e
iδ/2
2 cos
kz −
δ
2
χ
2iP sin
kz −
δ
2
σ
z
χ
(13)
where
δ = arctan
2P
P
2
− 1
P =
¯hkc
E + mc
2
. (14)
The need for a special boundary condition instead of
simply requiring that the wavefunction be continuous
at the walls is explained in the appendix. Actually,
the wavefunction turns out to be discontinuous at the
boundaries of the box, as can be seen from (13) This is
not surprising if one realizes that the Dirac equation is
a first order differential equation with a infinite jump in
the potential in this case. In the non-relativistic particle
in a box, we have a second-order differential equation,
leading to a discontinuity in the first derivative of the
wavefunction.
For non-relativistic momenta, ¯hk mc and P ∼ 0
so that
δ ∼ arctan
0
−1
= π. (15)
Setting δ = π and P = 0 in (13) we get
ψ
II
(z) = B e
iπ/2
2 cos
kz −
π
2
χ
0
!
= B e
iπ/2
2 sin(kz)χ
0
(16)
Figure 2. Graphical solution of (16) when
L
= ¯
h
/(
mc
).
The functions tan(
kL
) and −
kL
are plotted. The values
of
kL
for which the two curves intersect are the solutions
of the equation tan(
kL
) = −
kL
.
which is exactly the non-relativistic result (10) apart
from a normalization factor and the spinor χ.
In the appendix we show that the wavenumber k is
provided by the transcendental equation
tan(kL) =
2P
P
2
− 1
=−
¯hk
mc
. (17)
From the discrete set of values of k that satisfy (17)
equation we get the discrete energy eigenvalues E =
p
¯h
2
c
2
k
2
+ m
2
c
4
. We can check that for non-relativistic
momenta tan(kL) ∼ 0 and kL ∼ nπ , n = 1, 2,...,
recovering the non-relativistic result. The size L of the
box provides a criterion to decide whether a particle
inside a well is relativistic or not, at least for the
first energy eigenvalues. The distance scale is set
by the length L
0
= ¯h/(mc), which is the Compton
wavelength of the particle divided by 2π. For L ∼ L
0
or smaller the particle is relativistic and we must apply
the relation (17). This length is about 0.004
˚
A for
an electron. Figure 2 shows the graphic solution of
(17) when L = L
0
.IfLis much bigger than L
0
,
the slope of the line approaches zero, in which case
the two curves cross at kL = π, 2π, 3π, ,...,
corresponding to the non-relativistic behaviour. Figure
3 shows the values of kL/π for boxes with L = L
0
,
L = 10L
0
and L = 100L
0
, in comparison with the
non-relativistic case. One sees that a particle inside a
box with L = 100L
0
is already non-relativistic, at least
for the lower spectrum. It is clear from Figure 4 that
the relativistic levels are lower than the corresponding
non-relativistic ones.
It is interesting to note that the relativistic probability
density constructed from the wavefunction (13) has no
22
P Alberto
et al
Figure 3. Solutions of (16) for boxes of sizes
L
=
L
0
,
L
=10
L
0
and
L
= 100
L
0
, together with the non-relativistic
case (for which
kL
=
n
π).
Figure 4. Relativistic spectrum of kinetic energies for the
same box sizes of figure 3
(
E
kin
=
p
(¯
hkc
)
2
+
m
2
c
4
−
mc
2
) in comparison with the
non-relativistic spectrum (
E
kin
= (
n
¯
h
π)
2
/(2
mL
2
)). The
values shown are the logarithm of the kinetic energy in
units of
mc
2
for the lowest nine levels.
zeros. Indeed, if we compute ψ
II
(z)
†
ψ
II
(z) we obtain
ψ
II
(z)
†
ψ
II
(z) = 4|B|
2
cos
2
kz −
δ
2
+ P
2
sin
2
kz −
δ
2
= 4|B|
2
1 + (P
2
− 1) sin
2
kz −
δ
2
.
(18)
This quantity is never zero for k 6= 0 (if k = 0 the
wavefunction is zero inside the well) because 0 <
P<1. This could only happen if P → 0, i.e., in
the non-relativistic limit. Figure 5 shows the densities
corresponding to the first three levels of a particle in a
box with sizes L
0
,10L
0
and 100L
0
. The latter case is
non-relativistic and can be taken as a reference, showing
the characteristic nodes of a stationary wave in a closed
box.
3. Conclusions
We have analysed the relativistic particle in a one-
dimensional box and compared it with the non-
relativistic case. The quantized momentum values
and corresponding energies emerge as solutions of a
simple transcendental equation, in contrast with the
more involved case of the relativistic particle in the
three dimensional Coulomb problem. We have checked
that both the spectrum and the wavefunctions go to their
non-relativistic values as the size of the box grows. The
basic scale is set by the length L
0
which is related to
the Compton wavelength for the particle in the box.
The probability density shows no nodes in a relativistic
regime (L ∼ L
0
) but its minima approach zero as the
box becomes larger.
Finally, let us compare the boundary conditions we
have used with other approaches. For instance, (17)
similar in form to the one obtained by Greiner [5].
However, one cannot make a correspondence between
the two equations because of the way the potential well
is defined in [5] and the problems that the approach
used has when the depth of the well is allowed to go
to infinity. In that work the wavefunction is required
to be continuous at the walls of the well. One should
also mention a textbook treatment of the same problem
[7] in which two components of the wavefunction
were required to be continuous at the box walls, while
the other two were not. We think that there is no
reason to treat the components of the wavefunction
differently, especially when one is far from the non-
relativistic regime. However, if one would demand that
all components be continuous and also to be zero outside
the box, the wavefunction would be identically zero
everywhere. This means that the treatment of [7] is
in fact restricted to the non-relativistic regime. On the
other hand, the continuity of the wavefunction, which
was crucial for solving the Schr
¨
odinger equation (a
second order equation), does not need to be imposed for
solving the Dirac equation (a first order equation), with
a mass going to infinity outside a certain region. This
allows one to impose a boundary condition in which the
flux of probability is continuous at the box walls but the
wavefunction is not. By doing that, one is still able to
guarantee the fulfillment of the continuity equation for
the probability density.
Relativistic particle in a box 23
Figure 5. Plot of the probability density ψ
†
(
z
)ψ(
z
) normalized to unity for the first three levels of a relativistic particle
in boxes of sizes
L
=
L
0
,
L
=10
L
0
and
L
= 100
L
0
. The latter case is almost identical with the non-relativistic limit, as it
is apparent from Figs. 4 and 5. Note that the density is discontinuous at the walls in the relativistic regime.
Appendix
In this appendix we present the mathematical formalism
leading to (17).
Let us start by considering the solutions of the Dirac
equation given by (12). Since we are going to let
M →∞, we have E<Mc
2
, so that the quantity
k
0
=
1
¯h
r
E
2
c
2
− M
2
c
2
=
i
¯h
r
M
2
c
2
−
E
2
c
2
(19)
is an imaginary number, and thus the exponentials in k
0
are real decreasing exponentials, making ψ
I
(z) → 0
and
ψ
III
(z) → 0asM →∞. We have absorbed
the normalization constants into the arbitrary complex
constants A, B, C and D and set the same spinor χ for
all plane waves for simplicity.
To find the relationship between the constants B and
C and to get the energy eigenvalues we need to impose
boundary conditions at z = 0 and z = L. We might
be tempted to set the usual boundary conditions of
the non-relativistic case, i.e., set the wavefunction to
zero at those points. However, this implies that the
wavefunction inside the well is zero. One alternative
is to demand that the outward flux of probability at the
walls of the well be zero, as was done in the MIT bag
model [2]. The boundary condition is
±(−i)βα
z
ψ = ψ (20)
where the minus sign corresponds to z = 0 and the plus
sign to z = L. We can check that the flux is zero by
multiplying (20) at the left by
¯
ψ = ψ
†
β. We obtain
±(−i)
¯
ψβα
z
ψ =
¯
ψψ . (21)
Since (−i)
¯
ψβ Eα.
E
k/|
E
k|
ψ represents the probability
current density for a Dirac spinor in the direction of
the wave vector
E
k =Ep/¯h, this condition just states that
the probability current density at z = 0 and z = L is
equal to the value of
¯
ψψ at those points. Applying now
(20) to
ψ
II
(z) at z = 0 gives the condition
C = B
iP − 1
iP + 1
(22)
with
P =
¯hkc
E + mc
2
. (23)
On the other hand,
¯
ψ
II
ψ
II
at z = 0 is given by
¯
ψ
II
ψ
II
z=0
=|B+C|
2
−|B−C|
2
P
2
=0 (24)
taking (22) into account. This means that, according to
(21), there is no outward flow of probability at z = 0.
The same result is obtained if we apply the boundary
condition at z = L.
Let us look at the resulting wavefunction. Noticing
that (iP − 1)/(iP + 1) has unit modulus, we can write
iP − 1
iP + 1
= e
iδ
δ = arctan
2P
P
2
− 1
. (25)
Replacing condition (22) in the wavefunction
ψ
II
(z) in
the expressions (12) we obtain
ψ
II
(z) =B
e
ikz
+ e
−ikz
e
iδ
χ
P
e
ikz
− e
−ikz
e
iδ
σ
z
χ
=B e
iδ/2
2 cos
kz −
δ
2
χ
2iP sin
kz −
δ
2
σ
z
χ
.
(26)
