The Symmetric Linear Potential [ ]

The Symmetric Linear Potential
The goal is to find analytical solutions to the TISE eigensystem for a potential of
the form
V ( x)
b x x ,
(1)
where b is a constant specific to the potential in question. The first thing to notice is that
the potential is symmetric, which in this case means we can classify solutions in terms of
parity. For x 0 , the TISE is
2 2
d ψ( x)
bx ( x) E ( x)
2
2m
dx
ψ ψ (2)
We change variables with the definitions 1
x as
2
a
mb
1/3
W
2
ma
2
E
( s ) (3)
W
and obtain a differential equation whose solutions are, in general, a linear combination of
3
3
the Airy functions Ai ( 2 ) and Bi ( 2 ) .
ψ''( ) 2 ψ ( )
ψ ( ) C Ai( 2 ) C Bi( 2 )
(4)
3 3
1 2
However, Bi(ξ) diverges as . This violates the requirement that ψ 0 as
, which does not make physical sense. We therefore restrict the solutions to the
form
ψ (5)
3
n
( ) Cn Ai( 2 n)
where C n are normalization factors. Again, this is for s 0 .
At this point we consider the parity of our solutions. Even and odd eigenstates for
s 0 can be extended to give the solutions for s 0 by
even
even
odd
odd
ψ ( s) ψ ( s)
ψ ( s) ψ ( s)
(6)
1 Another option is to take a
2mb
2 1/3
and
2 2
W 2ma
the Airy equation, but includes a factor of 2 -1/3 in the energy (9).
. This leads to a more common form of

With these properties, it then stands to reason that we need the boundary conditions
ψ
odd
3
k
(0) Ai( 2
k
) 0
dψ
ds
even
k
3
(0) Ai '( 2
k
) 0
(7)
for our (non-normalized) solutions. Normalization is done numerically as
1/2
1/3 2
C 2 Ai( ) d
n 3
(8)
2
n n
n
The two equations in (7) then give the eigenvalues for odd and even parity states,
respectively. Finally, we come to the conclusion that the bound state energies are given
by
E
2 2
1/3
b
n
nW
n
m
(9)
Fig 1: An example of the symmetric linear potential theoretical eigenfunctions (n = 20). Note that since
ψ is an odd state, ψ 20 (0) 0 . Observe also the decaying behavior of the wavefunction around the
20
classical turning points s
20
.