1

Bound states of x2
potential

Khitish Biswal
September 30, 2018

1
We are going to study the x2 potential. Consider a particle of mass m in a potential

∞ x≤0
V (x) = −a
 x>0
x2
Here a is a constant with the dimension J m2 . From dimensional analysis we see that we cannot construct
the dimensions of energy using the parameters a, ~ and m. But we choose to ignore this fact and try to obtain
the bound states by solving the Schrodinger’s equation:

~2 d2 ψ a
− 2
− 2 ψ = Eψ (1)
2m dx x
By rearranging the terms to make the equation look nice we get

d2 ψ α
2
+ 2 ψ = k 2 ψ, (2)
dx x
2m
where α = ~2 a and k = − 2mE
~2 . The boundary condition

• ψ → 0 for x → 0 is necessary for ψ to be continuous at x = 0

• ψ → 0 for x → ∞ to make the wavefunction normalizable.

d2 ψβ α
+ 2 ψβ = k 2 ψβ (3)
dx2 x
d2 ψβ α
=⇒ 2
+ 2 ψβ = β 2 k 2 ψβ (4)
dx x
So if there exists any bound state, then there is a bound state for every negative energy. In other words ,
the potential has no ground state. One inference that can be drawn from this discussion is that there are no
negative energy states. In the following discussion we will prove that this is indeed true for α < 41 . For α < 0
this is trivial since the potential becomes repulsive.

1
Claim: There does not exist any bound state for 0 < α < 4

Proof : We will prove this result by factoring the Hamiltonian.

let α = ν(1 − ν)

−~2 d2 a
H= 2
− 2
2m dx x
−~2 d
  
ν d ν
= + − (5)
2m dx x dx x

1
 ν∗
   
d ν d
Lemma : Hermitian conjugate of + is − +
dx x dx x
Proof:
  Z ∞  
d ν dg ν
hf | + |gi = f∗ + g
dx x dx x
Z0 Z
dg ν
= f∗ + f∗ g
dx x
Z ∞ ∗ Z ∞  ∗ ∗
∗ ∞ df ν
= f g|0 − gdx + f gdx
0 dx 0 x
−d ν ∗
 
= hf | + |gi (6)
dx x

E = hψ|H|ψi
−~2
  
d ν d ν
= hψ| + − |ψi
2m dx x dx x
~2 ν∗
     
d  d ν
= − ψ
 − ψ (7)
2m dx x dx x
if ν is real eqn(7) becomes
Z ∞ 2
~2  dψ ν 
E=  − ψ dx > 0 (8)
2m 0  dx x 
which tells us that there does not exist any negative energy states for real ν. Using the definition of ν we get

ν(1 − ν) = α
ν2 − ν + α = 0
√
1 ± 1 − 4α
=⇒ ν =
r2
1 1
= ± −α (9)
2 4
=⇒ there does not exist any negative energy states for α < 14 .
Now we would like to solve the Schrödinger’s equation for α > 14 . We start by writing a series solution for
the wave-function.
∞
X
ψ(x) = xs aj xj , a0 6= 0 (10)
j=0

d2 s X α X X
2
(x aj xj ) + 2 xs aj xj = k 2 aj xj
dx x
X X X
(s + j)(s + j − 1) aj xj−2 + α aj xj−2 = k 2 aj xj
X
xj−2 aj [(s + j)(s + j − 1) + α] = k 2 aj xj (11)
By equating like powers
for j = 0
x−2 (a0 [(s)(s − 1) + α]) = 0, a0 6= 0
r
1 1
s=ν= ± −α (12)
2 4
Near origin i.e. x → 0 the solution looks like
≈ a0 xs
1
√1
≈ x 2 ± 4 −α
r
√ ±ιg ln x 1
= xe , g := α−
4
(13)

2
The normalizable general solution is given by
√
ψk (x) = A xKig (kx) (14)

In the neighborhood of 0 the sinusoidal dependence on g ln x causes the wavefunction to oscillate rapidly. As we
interpret the number of zero crossings as number of lower energy states, we see that there are infinite number
of lower energy states for this potential.

Regularization
We apply a cut-off at a distance of  from 0. and redefine the potential as

∞ x≤
V (x) = −a
 x>
x2
By using this cut-off we get a new parameter . Using dimensional analysis we see that we are now able to
create the dimensions for energy using m, ~, a and .
a
f (α) (15)
2
where α = 2ma
~2 is a dimensionless quantity and f is some function.
By introducing  we broke the scale invariance of the system. We would now go on to prove that this system
has a ground state and also discrete bound states.

Our condition of wavefunction now becomes

ψk () = 0
=⇒ Kig (k) = 0 (16)

This gives us a condition in which leads to the quantization of the energy as we will see below.
for k << 1 we get a rather simple expression for Kig .
h z  i
Kig (z) ≈ A sin g ln − arg Γ(1 + ig) , z << 1 (17)
2
From eqn (16) we get that

 
k
g ln − arg Γ(1 + ig) + nπ = 0
2
2
or kn = e[arg Γ(1+ig)−nπ]/g (18)

2~2
or En = 2 e[arg Γ(1+ig)−nπ]/g (19)
 m