A Quantum Physics Jeopardy Problem of a One-Bound-State Double Dirac Delta Potential with a Centrifugal, Angular-Momentum-Like Tail

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1 15 April 2018
A Quantum Physics Jeopardy Problem of a One-Bound-State
Double Dirac Delta Potential with a Centrifugal, Angular-
Momentum-Like Tail
Spiros Konstantogiannis
spiroskonstantogiannis@gmail.com
15 April 2018
Abstract
Starting from a simple, one-parameter, symmetric wave function, we
derive an attractive double Dirac delta potential with a centrifugal,
angular-momentum-like tale, which has only one bound state, the energy
of which can be set to zero.
Keywords: rational wave functions, one-bound-state potentials, double Dirac delta
potential, centrifugal tail, zero-energy states, quantum physics jeopardy

15 April 2018
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Contents
A Quantum Physics Jeopardy Problem of a One-Bound-State Double Dirac Delta
Potential with a Centrifugal, Angular-Momentum-Like Tail .............................. 1
Contents ............................................................................................................ 2
1. Introduction................................................................................................... 3
2. The wave function......................................................................................... 3
3. The potential.................................................................................................. 8
4. Appendix......................................................................................................13
5. References....................................................................................................18

15 April 2018
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1. Introduction
In [1], Van Heuvelen and Maloney describe a new, reverse type of physics problems
in which the answer is given and the question is asked. Such problems, which require
from the solver to follow a working-backward approach, are called physics Jeopardy
problems (the name was taken from a game show called Jeopardy).
In quantum physics Jeopardy problems, wave functions describing energy eigenstates
are given and the respective potentials are asked [2].
Herein, we use a simple, one-parameter, symmetric wave function to derive an
attractive, equal-coupling, double Dirac delta potential, which vanishes in the interior
region (between the delta functions) and has a centrifugal, angular-momentum-like
tail in the exterior region (outside the delta functions). The potential admits only one
bound state, the energy of which can be chosen zero.
For a detailed study of the double Dirac delta potential, the reader may refer to [3] and
to references therein, while in [4] the reader can find a detailed analysis of the Dirac
delta potentials as pedagogical and physical models in quantum physics.
2. The wave function
We consider the wave function
( )
( )
1 1
;
s
A
x s
x x x x
y
=- + +
%
% % % %
(1)
where
3 2
s>
[5],
1
0
x
>
%
,
A
the normalization constant, and
0
x x x
=
%
, with
x
being
the position and
0
0
x
>
a length scale.
Since
1
0
x
¹
%
, the expression
1 1
x x x x
- + +
% % % %
is strictly positive. Then, the wave
function (1) has no singularities, it is everywhere continuous and finite.
Also, we have
( )
( )
( ) ( )
( )
}
( )
1 1 1 1
1 1
;
x x
s s s
A A A
x s
x x x x x x x x
x x x x
y
- =
- = = =
- - + - + + + -
- + + - -
%
% % % % % % % %
% % % %
Thus
(
)
(
)
; ;
x s x s
y y
- =
% %

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4
That is,
(
)
;
x s
y
%
is symmetric (i.e. of even parity).
In the region
1
x x
>
% %
, we have
i. If
1
x x
< -
% %
, then
1
0
x x
+ <
% %
and since
1
0
x
>
%
,
1 1
0
x x x x
- < + <
% % % %
.
Thus
(
)
1 1
x x x x
- = - -
% % % %
and
(
)
1 1
x x x x
+ = - +
% % % %
.
Then (1) becomes
( ) ( ) ( )
( )
( ) ( )
1 1
;2 2
s s s
s
A A A
x s
x x
x x x x
y
= = =
- -
- - - +
%
% %
% % % %
That is
( ) ( )
;2
s
s
A
x s
x
y
=-
%
%
(2)
ii. If
1
x x
>
% %
, then
1
0
x x
- >
% %
and since
1
0
x
>
%
,
1 1
0
x x x x
+ > - >
% % % %
.
Thus
1 1
x x x x
- = -
% % % %
and
1 1
x x x x
+ = +
% % % %
.
Then (1) becomes
( ) ( ) ( )
1 1
;2
2
s s
s s
A A A
x s
x
x x x x x
y
= = =
- + +
%
%
% % % % %
That is
( )
;
2
s s
A
x s
x
y
=
%
%
(3)
Combining (2) and (3), we obtain
( )
;2
s
s
A
x s
x
y
=
%
%
(4)
for
1
x x
>
% %
.
In the region
1
x x
£
% %
, we have
1 1
x x x
- £ £
% % %
, thus
1
0
x x
- £
% %
and
1
0
x x
+ ³
% %
.
Then
(
)
1 1
x x x x
- = - -
% % % %
and
1 1
x x x x
+ = +
% % % %
.
Thus (1) becomes

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( ) ( )
( )
( )
1
1 1
;2
s s
A A
x s
x
x x x x
y
= =
- - + +
%
%
% % % %
That is
( ) ( )
1
;2
s
A
x s
x
y
=
%
%
(5)
The wave function is constant in the region
1
x x
£
% %
.
Combining (4) and (5), we write the wave function as
( ) ( )
1
1
1
,
2
;
,
2
s
s
s
A
x x
x
x s A
x x
x
y
ì
£
ï
ï
=í
ï
>
ï
î
% %
%
%
% %
%
(6)
Since it is continuous,
(
)
;
x s
y
%
is Riemann integrable. Also,
(
)
;
x s
y
%
is square
integrable, as it decays as
1
s
x
%
, with
3 2
s>
.
The normalization constant
A
can be easily calculated by applying the normalization
condition, i.e.
( )
2
1
dx x
y
¥
-¥
=
ò
Using that
0
x x x
=
%
, we obtain
0
dx x dx
=
%
, and since
0
0
x
>
, the previous integral is
written as
( )
2
0
1
x dx x
y
¥
-¥
=
ò
% %
or
( )
2
0
0
2 1
x dx x
y
¥
=
ò
% %
(7)
since
(
)
x
y
%
is symmetric.
Using (6), the integral in the left-hand side of (7) is written as

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6
( ) ( ) ( )
1
1
1
2 2 2 2
2
1
2 2
2 2 2 2 1
0 0 1 1
1 1
2 2 2 1
2 2
x
s s
s s s s x
x
A A A A
dx x dx dx x
x s x
x x
y
¥
¥ ¥
-
æ ö
= + = + - =
ç ÷
ç ÷
-
è ø
ò ò ò
%
%
%
% % % % %
% %
% %
}
( )
2 2 2 2 2
2 1 0
1
22 2 1 2 2 1 2 2 1 2 2 1
1 1 1 1
1
1 1 1 1
1
2 2 1 2 2 1 2 2 1 2
2
s
ss s s s s s s s
A A A A A
xs x x s x s x
x
- >
- - - -
æ ö
= + = + = + =
ç ÷
- - -
è ø
%% % % %
%
( ) ( )( )
2 2
2 1
2 2 1
11
2
2 1 2 2 1 2
s
s s
s A s A
s x s x
-
-
= =
--
%%
That is
( ) ( )( )
2
2
2 1
01
2 1 2
s
s A
dx x s x
y
¥
-
=-
ò
% % %
Substituting into (7) yields
( )( )
( )( ) ( )( )
2
2 1 2
2
0 1 1
2 1 0 0 1
1
2 2 1 2 2 1 2
12 4
2 1 2
s s
s
x s A s x s x
Ax s x x s
s x
-
-
- -
= Þ = =
-
% %
%
%
Thus
( )
1
0 1
2
2 1
2
s
x
s
A
x x s
-
=%
%
Then, up to a constant phase, the normalization constant is
( )
1
0 1
2
2 1
2
s
x
s
A
x x s
-
=%
%
(8)
By means of (8), (6) becomes
( )
1
0 1
1
1
0 1
1 2 1,
2
;1 2 1 ,
2
s
sx x
x x s
x s xs
x x
x x s x
y
ì-£
ï
ï
=íæ ö
-
ï
>
ç ÷
ïç ÷
è ø
î
% %
%
%%
% %
% %
(9)
Using (9), we calculate the wave function derivative, with respect to
x
%
.
In the region
1
x x
<
% %
, the wave function is constant, thus its derivative vanishes.
In the region
1
x x
< -
% %
, the wave function is, from (9),

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7
( )
1
0 1
1 2 1
;2
s
x
s
x s
x x s x
y
-
æ ö
=
ç ÷
-
è ø
%
%
% %
Thus
( ) ( )( )
( )
( )
( )
( )
1
2
1 1
11 1
0 1 0 1 0 1 1
1 2 1 2 1
1 2 1 1 1 1
;2 2 2
s
s
ss s
s s s s s
x x
s
x s x
x x s x x s x x x x
x x
y
+
+ +
- - - -
-æ ö
¢
= = = =
ç ÷
-
è ø
- -
% %
% %
% % % % %
% %
( )
1
1
3
0 1
2 1
1
2
s
s s x
x x x
+
-
æ ö
=
ç ÷
-
è ø
%
% %
In the region
1
x x
>
% %
, the wave function is, from (9),
( )
1
0 1
1 2 1
;2
s
x
s
x s
x x s x
y
-
æ ö
=
ç ÷
è ø
%
%
% %
Then
( ) ( )
1
1
11 3
0 1 0 1
2 1
1 2 1 1
;2 2
s
s
s
s s xs s
x s x
x x s x x x x
y
+
+
-
- -
æ ö
¢= = -
ç ÷
è ø
%
% %
% % % %
Thus, the wave function derivative is
( ) ( )
( )
1
1
1
1
3
0 1
1
1
1
3
0 1
0,
2 1
1
; ,
2
2 1
1,
2
s
s
x x
s s x
x s x x
x x x
s s x
x x
x x x
y
+
+
ì
ï<
ï
ï-
ïæ ö
¢
= < -
íç ÷
-
è ø
ï
ï-æ ö
ï
- >
ç ÷
ïè ø
î
% %
%
% % %
% %
%
% %
% %
(10)
(
)
;
x s
y
¢
%
is odd, as expected, since
(
)
;
x s
y
%
is even.
At
1
x
±
%
,
(
)
;
x s
y
¢
%
is discontinuous, since from (10) we obtain
( )
( )
13
0 1
2 1
1
;2
s s
x s x x
y
-
-
¢- =
%%
,
(
)
1
; 0
x s
y
+
¢
- =
%
and
(
)
1
; 0
x s
y
-
¢
=
%
,
( )
( )
13
0 1
2 1
1
;2
s s
x s x x
y
+
-
¢= -
%%

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8
Thus, the wave function derivative has at
1
x
-
%
and
1
x
%
finite discontinuities, which are
equal, since
( ) ( )
( )
1 1 3
0 1
2 1
1
; ; 2
s s
x s x s x x
y y
+ -
-
¢ ¢
- - - = -
% % %
(11)
( ) ( )
( )
1 1 3
0 1
2 1
1
; ; 2
s s
x s x s x x
y y
+ -
-
¢ ¢
- = -
% % %
To summarize, the wave function
(
)
;
x s
y
%
is continuous, while its first derivative has
equal discontinuities at
1
x
±
%
.
3. The potential
Since
(
)
;
x s
y
%
has no zeros
1
, it can be the ground-state wave function of a potential
consisting of a term having, at most, finite discontinuities, and of a sum of two Dirac
delta functions with one of them acting at
1
x
-
%
and the other at
1
x
%
(see appendix).
1. By zeros, we mean real zeros.
The first term of the potential, which has, at most, finite discontinuities, does not
induce discontinuities in the wave function derivative [6, 7]. We’ll refer to this term
as the regular part of the potential.
A finite discontinuity in the wave function derivative is induced by a Dirac delta
potential [7]. Particularly, a delta potential
(
)
x
ld
induces in the wave function
derivative a finite discontinuity at zero, which is given by [7]
( ) ( )
(
)
2
2 0
0 0 m
ly
y y
+ -
¢ ¢
- =
h
(12)
where the prime denotes differentiation with respect to the position
x
.
Since
0
x x x
=
%
, we have
0
1
d d
dx x dx
=
%
and (12) is written as

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9
( ) ( )
(
)
(
)
2
0
2 0
10 0 m
x
ly
y y
+ -
¢ ¢
- =
h
(13)
where now the prime denotes differentiation with respect to
x
%
.
Then, from (13), the finite discontinuity in
(
)
;
x s
y
¢
%
at
1
x
-
%
is induced by a delta
potential
(
)
1 1
x x
l d
+
and is given by
( ) ( )
( )
(
)
1 1
1 1 2
0
2 ;
1; ;
m x s
x s x s
x
ly
y y
+ -
-
¢ ¢
- - - =
%
% %
h
Using (11) and that
( )
1
0 1
1 2 1
;2
s
x s
x x s
y
-
- =
%
%
,
as given by (9), the above discontinuity condition is written as
( )
1
0 1
3 2
0 0 1
1 2 1
222 1
1 1
2
s
m
x x s
s s
x x x
l
-
æ ö
-
ç ÷
- =
ç ÷
è ø
%
%h
Solving the last equation for the coupling
1
l
gives, after a little algebra,
2
1
0 1
2
s
mx x
l
= -
h
%
(14)
The delta potential acting at
1
x
-
%
is thus attractive.
Let us do a dimensional check on (14).
Since
[
]
PL
=
h
,
[
]
0
x L
=
, and
[
]
1
1
x
=
%
(dimensionless), we have
[ ]
( )
22
1
PL P
L EL
ML M
l
= = =
Then, since
( )
1
dx x
d
¥
-¥
=
ò, we have
(
)
1
L x
d
=
é ù
ë û
, thus
(
)
1
x L
d
-
=é ù
ë û
, and then
(
)
1
x E
ld
=
é ù
ë û
, and we are ok.

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10
The wave function is symmetric, thus
(
)
(
)
1 1
; ;
x s x s
y y
- =
% %
, and also, the
discontinuities in the wave function derivative at
1
x
±
%
are equal. Then, the
discontinuity in the wave function derivative at
1
x
%
is induced by a delta potential
acting at
1
x
%
, which has the same coupling as the delta potential acting at
1
x
-
%
, i.e. it is
induced by the delta potential
(
)
1 1
x x
l d
-
.
The sum of the two delta potentials is then
( ) ( ) ( )
( )
2
1 1
0 1
;2
delta
s
V x s x x x x
mx x
d d
= - + + -
h
%
(15)
Using that
(
)
(
)
ax x a
d d
=
, we have, since
0
0
x
>
,
( ) ( )
0
0
x
x x x
x
d d d
æ ö
= =
ç ÷
è ø
%
That is
( )
(
)
0
x
x
x
d
d
=
%
Then
( )
(
)
1
1
0
x x
x x x
d
d
±
± =
% %
where
1 1 0
x x x
=
%
.
Then (15) is written as
( ) ( ) ( )
( )
2
1 1
2
0 1
;2
delta
s
V x s x x x x
mx x
d d
= - + + -
h
% % % % %
%
(16)
In order to find the regular part of the potential, we’ll calculate the second derivative
of the wave function.
Using (10), we have

15 April 2018
11
( ) ( ) ( )
( ) ( )
1
2
1
1
5
0 1
2
1
1
5
0 1
0,
1 2 1
; ,
2
1 2 1 ,
2
s
s
x x
s s s x
x s x x
x x x
s s s x
x x
x x x
y
+
+
ì
ï<
ï
ï+ -
ïæ ö
¢¢
= < -
íç ÷
-
è ø
ï
ï+ - æ ö
ï>
ç ÷
ïè ø
î
% %
%
% % %
% %
%
% %
% %
or
( ) ( ) ( )
1
2
1
1
5
0 1
0,
;1 2 1 ,
2
s
x x
x s s s s x
x x
x x x
y
+
ì <
ï
ï
¢¢ =æ ö
í + -
>
ç ÷
ïç ÷
ïè ø
î
% %
%%
% %
% %
(17)
Since the wave function describes the ground state of the potential we are looking for,
it satisfies the energy eigenvalue equation (in position space)
2
, i.e.
( ) ( )
( )
( )
0
2
2
; ; ; 0
m
x s E V x s x s
y y
¢¢
+ - =
h
with
0
E
being the ground-state energy and
(
)
;
V x s
the potential.
2. The well-known time-independent Schrödinger equation.
Solving the energy eigenvalue equation for the potential yields
( )
(
)
( )
2
0
;
;2 ;
x s
V x s E
m x s
y
y
¢¢
= +
h
Using that
0
x x x
=
%
, we obtain
2 2
2 2 2
0
1
d d
dx x dx
=
%
,
and thus, in terms of
x
%
, the potential is written as
( )
(
)
( )
2
0
2
0
;
;2 ;
x s
V x s E
mx x s
y
y
¢¢
= +
%
h
%%
Substituting (9) and (17) into the previous equation, we obtain, after a little algebra,
that the regular part of the potential is

15 April 2018
12
( ) ( )
0 1
2
0 1
2 2
0
,
;1,
2
reg
E x x
V x s s s
E x x
mx x
ì <
ï
=+
í
+ >
ï
î
% %
%h
% %
%
(18)
From (18), we see that the regular part of the potential has, at
1
x
±
%
, finite
discontinuities, which are
( ) ( )
( )
( )
( )
2 2
1 1 0 0
2
2 2
20 1
0 1
1 1
; ; 2
2
reg reg
s s s s
V x s V x s E E
mx x
mx x
+ -
æ ö
+ +
- - - = - + = -ç ÷
ç ÷
-
è ø
h h
% %
%
%
( ) ( )
(
)
(
)
2 2
1 1 0 0
2 2 2 2
0 1 0 1
1 1
; ; 2 2
reg reg
s s s s
V x s V x s E E
mx x mx x
+ -
+ +
- = + - =
h h
% %
% %
The two discontinuities are thus opposite.
The total potential is the sum of the regular part and the delta functions, i.e.
(
)
(
)
(
)
; ; ;
reg delta
V x s V x s V x s
= +
% % %
(19)
Since
(
)
(
)
0
; ;
reg
V s V s E
±¥ = ±¥ =
, the ground-state energy
0
E
is the highest bound
energy of the potential (for a detailed explanation, see, for instance, [8]).
Thus, the potential (19) has only one bound state, which is also the ground state, of
energy
0
E
, which is described by the wave function (9).
Choosing the infinity as reference point and setting
(
)
; 0
V s
±¥ =
, we obtain
0
0
E
=
(20)
i.e. the ground-state energy becomes zero.
The regular part of the potential, given by (18), is then written as
( ) ( )
1
2
1
2 2
0
0,
;1,
2
reg
x x
V x s s s
x x
mx x
ì <
ï
=+
í
>
ï
î
% %
%h
% %
%
(21)
We’ve thus ended up at an attractive double Dirac delta potential (of equal couplings)
with an angular-momentum-like tail
(
)
2 2 2
0
1 2
s s mx x
+
%
h
, which, for every value of
3 2
s>
, has only one, zero-energy, bound state.

15 April 2018
13
4. Appendix
For a potential consisting of a part having, at most, finite discontinuities, and
of a finite sum of Dirac delta functions, if there exists a nodeless wave function
(
)
0
x
y
describing a bound energy eigenstate, then
(
)
0
x
y
is the ground-state
wave function.
Proof
Let
(
)
ˆ
a x
be the position-space operator
( ) ( ) ( )
( )
0
0 0
1
ˆ ˆ
x
a x p x i
p x
y
y
æ ö
¢
ç ÷
= +
ç ÷
è ø
h (22)
where
(
)
ˆ
p x i d dx
= -
h
is the momentum operator in position space and
0
p
is a
(positive) momentum scale.
The operator
(
)
ˆ
a x
is dimensionless.
The part of the potential which has, at most, finite discontinuities does not induce
discontinuities in the wave function derivative [6, 7], while each delta function
induces, in the wave function derivative, a finite discontinuity, at the point where the
delta function acts [7].
Since
(
)
0
x
y
is nodeless, the function
( ) ( )
0 0
x x
y y
¢ has no singularities, but it has
finite discontinuities at the points where the delta functions act.
Since the number of delta functions in the potential is finite, the function
( ) ( )
0 0
x x
y y
¢ has a finite number of finite discontinuities.
Also, since
(
)
0
x
y
is a one-dimensional bound energy eigenfunction, it is real, up to a
constant phase [7].
Thus,
( ) ( )
0 0
x x
y y
¢ is a real function, and then the Hermitian conjugate of the
operator (22) is given by
( ) ( ) ( )
( )
0
†
0 0
1
ˆ ˆ
x
a x p x i
p x
y
y
æ ö
¢
ç ÷
= -
ç ÷
è ø
h (23)
Using (22) and (23), we have

15 April 2018
14
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
0 0
†
2
0 0 0
1
ˆ ˆ ˆ ˆ
x x
a x a x p x i p x i
p x x
y y
y y
æ öæ ö
¢ ¢
ç ÷ç ÷
= - + =
ç ÷ç ÷
è øè ø
h h
( ) ( ) ( )
( ) ( )
( )
2
0 0
2
2
0 0 0
1ˆ ˆ ,x x
p x i p x i
p x x
y y
y y
æ ö
é ù æ ö
¢ ¢
ç ÷
ç ÷
ê ú
= + - =
ç ÷
ç ÷
ê ú
ç ÷
ë û è ø
è ø
h h
( ) ( ) ( )
( ) ( )
( )
2
0 0
2 2
2
0 0 0
1ˆ ˆ ,x x
p x i p x
p x x
y y
y y
æ ö
é ù æ ö
¢ ¢
ç ÷
ç ÷
ê ú
= + +
ç ÷
ç ÷
ê ú
ç ÷
ë û è ø
è ø
h h
That is
( ) ( ) ( ) ( ) ( )
( ) ( )
( )
2
0 0
†2 2
2
0 0 0
1
ˆ ˆ ˆ ˆ ,x x
a x a x p x i p x
p x x
y y
y y
æ ö
é ù æ ö
¢ ¢
ç ÷
ç ÷
ê ú
= + +
ç ÷
ç ÷
ê ú
ç ÷
ë û è ø
è ø
h h (24)
If
(
)
f x
is an arbitrary function, then
(
)
(
)
(
)
ˆ,
p x f x i f x
¢
= -é ù
ë û h
, provided that the
derivative
(
)
f x
¢
exists. This is easily shown by applying the previous commutator to
an arbitrary wave function.
Using the previous commutator, we have
( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
2
0 0 0 0
0 0 0 0
ˆ,x x x x
p x i i
x x x x
y y y y
y y y y
¢
æ ö
é ù æ ö æ ö
¢ ¢ ¢¢ ¢
ç ÷
ç ÷ ç ÷
ê ú = - = - -
ç ÷
ç ÷ ç ÷
ê ú
ç ÷
ë û è ø è ø
è ø
h h
That is
( ) ( )
( ) ( )
( ) ( )
( )
2
0 0 0
0 0 0
ˆ,x x x
p x i
x x x
y y y
y y y
æ ö
é ù æ ö
¢ ¢¢ ¢
ç ÷
ç ÷
ê ú = - -
ç ÷
ç ÷
ê ú
ç ÷
ë û è ø
è ø
h (25)
Note
At the points where
( ) ( )
0 0
x x
y y
¢
has finite discontinuities, i.e. at the points
where the delta functions act, the derivative
( ) ( )
(
)
0 0
x x
y y
¢
¢
is a delta function.
Similarly,
( )
0
x
y
¢¢
is a delta function at the points where the delta functions act.
Therefore, the function in the right-hand side of (25) contains delta functions.
By means of (25), (24) becomes

15 April 2018
15
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( )
2 2
20 0 0
†2 2
2
0 0 0 0
1
ˆ ˆ ˆ x x x
a x a x p x i
p x x x
y y y
y y y
æ ö
æ ö
æ ö æ ö
¢¢ ¢ ¢
ç ÷
ç ÷
ç ÷ ç ÷
= - - + =
ç ÷
ç ÷
ç ÷ ç ÷
ç ÷
ç ÷
è ø è ø
è ø
è ø
h h
( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
( )
2 2
0 0 0 0
2 2 2 2 2
2 2
0 0 0 0 0 0
1 1
ˆ ˆ
x x x x
p x p x
p x x x p x
y y y y
y y y y
æ ö
æ ö
æ ö æ ö æ ö
¢¢ ¢ ¢ ¢¢
ç ÷
ç ÷
ç ÷ ç ÷ ç ÷
= + - + = +
ç ÷
ç ÷
ç ÷ ç ÷ ç ÷
ç ÷
ç ÷
è ø è ø è ø
è ø
è ø
h h h
That is
( ) ( ) ( ) ( )
( )
0
†2 2
2
0 0
1
ˆ ˆ ˆ
x
a x a x p x
p x
y
y
æ ö
¢¢
ç ÷
= +
ç ÷
è ø
h (26)
Since
(
)
0
x
y
is an energy eigenfunction, it satisfies the energy eigenvalue equation
(in position space), i.e.
( ) ( )
( )
( )
0 0 0
2
2
0
m
x E V x x
y y
¢¢
+ - =
h
where
0
E
is the energy of the state described by
(
)
0
x
y
and
(
)
V x
is the potential.
The previous equation gives
( )
( ) ( )
( )
0
2
0
0
2
x
m V x E
x
y
y
¢¢ = -
h
Substituting the previous expression into (26) yields
( ) ( ) ( ) ( )
( )
( )
(
)
( )
2
†2
0 0
2 2
0 0
ˆ
1 2
ˆ ˆ ˆ 22
p x
m
a x a x p x m V x E V x E
p p m
æ ö
= + - = + -
ç ÷
è ø
Using that
(
)
( )
2
ˆ
2
p x
V x
m
+
is the Hamiltonian
(
)
ˆ
H x
(in position space), we end up to
( ) ( ) ( )
(
)
†
0
2
0
2ˆ
ˆ ˆ m
a x a x H x E
p
= -
(27)
The operator
(
)
(
)
†
ˆ ˆ
a x a x
is Hermitian, since
( ) ( )
(
)
( ) ( )
(
)
( ) ( )
† †
† † † †
ˆ ˆ ˆ ˆ ˆ ˆ
a x a x a x a x a x a x
= =
Also, using (27), we have

15 April 2018
16
( ) ( ) ( ) ( )
( )
( ) ( ) ( )
†0 0
2 2
0 0
2 2
ˆ ˆ ˆ ˆ ˆ
ˆ ˆ , , ,
m m
a x a x H x H x E H x H x E H x
p p
é ù
é ù é ù
= - = - =
ê ú
ë û ë û
ë û
( ) ( ) ( )
0
2
00 0
2ˆ ˆ ˆ
, , 0
mH x H x E H x
p
æ ö
ç ÷
é ù é ù
= - =
ë û ë û
ç ÷
è ø
1442443 14243
That is, the operator
(
)
(
)
†
ˆ ˆ
a x a x
commutes with the Hamiltonian.
Next, we’ll prove that
(
)
(
)
†
ˆ ˆ
a x a x
has non-negative eigenvalues.
Proof
Let
(
)
x
y
be an eigenfunction of
(
)
(
)
†
ˆ ˆ
a x a x
with eigenvalue
l
.
Since
(
)
(
)
†
ˆ ˆ
a x a x
commutes with the Hamiltonian
(
)
ˆ
H x
,
(
)
(
)
†
ˆ ˆ
a x a x
and
(
)
ˆ
H x
have a common set of eigenfunctions, and thus
(
)
x
y
is an energy eigenfunction.
(
)
x
y
is continuous, but its derivative
(
)
x
y
¢
has a finite number of finite
discontinuities, at the points where the delta functions of the potential act.
Let us now consider the function
( ) ( ) ( ) ( )
( ) ( )
( )
}
( ) ( ) ( )
( )
ˆ
0 0
0 0 0 0
1
ˆ ˆ
d
p x i dx
x x x
i
a x x p x i x x
p x p x
y y y
y y y
y y
=-
æ ö æ ö
¢ ¢
ç ÷ ç ÷
¢
= + = - -
ç ÷ ç ÷
è ø è ø
h
h
h
That is
( ) ( ) ( ) ( ) ( )
( )
0
0 0
ˆ
x x
i
a x x x
p x
y y
y y y
æ ö
¢
ç ÷
¢
= - -
ç ÷
è ø
h (28)
The function
(
)
(
)
ˆ
a x x
y
has no singularities, since
(
)
0
x
y
is nodeless, but it has a
finite number of finite discontinuities.
The wave functions
(
)
0
x
y
and
(
)
x
y
as well as their derivatives
( )
0
x
y
¢
and
(
)
x
y
¢
are everywhere finite, so that the respective probability densities and currents are
everywhere finite.
Then, the Riemann integral
( ) ( )
2
ˆ
dx a x x
y
¥
-¥
ò
exists
3
.
3. If a function has a finite number of discontinuities and it is everywhere finite (i.e.
if it is bounded), then it is Riemann integrable (see, for instance, [9]).

15 April 2018
17
Using that
2
*
z z z
=
, with the asterisk denoting the complex conjugate, we have
( ) ( ) ( ) ( )
( )
( ) ( )
( )
2 *
ˆ ˆ ˆ
dx a x x dx a x x a x x
y y y
¥ ¥
-¥ -¥
=
ò ò
(29)
Also, by definition
4
( ) ( ) ( ) ( )
( )
( )
( )
( )
()
( ) ( )
( )
( )
( )
( )
}
†
†
ˆ ˆ
*
†
*† †
ˆ ˆ ˆ ˆ
a x a x
dx x a x a x x dx a x x a x x
y y y y
=
¥ ¥
-¥ -¥
= =
ò ò
( ) ( )
( )
( ) ( )
( )
*
ˆ ˆ
dx a x x a x x
y y
¥
-¥
=
ò
That is
( ) ( ) ( ) ( )
( )
( ) ( )
( )
( ) ( )
( )
*
*†
ˆ ˆ ˆ ˆ
dx x a x a x x dx a x x a x x
y y y y
¥ ¥
-¥ -¥
=
ò ò
(30)
4. In position space, the Hermitian conjugate operator
(
)
†
ˆ
O x
of a linear operator
(
)
ˆ
O x
is defined by the relation [7, 10]
( ) ( ) ( ) ( ) ( )
( )
( )
*
*†
2 1 2 1
ˆ ˆ
dx x O x x dx O x x x
j j j j
¥ ¥
-¥ -¥
=
ò ò ,
where
(
)
(
)
1 2
,
x x
j j
are two arbitrary wave functions (in position space).
Comparing (29) and (30) yields
( ) ( ) ( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
( )
2*†*†
ˆ ˆ ˆ ˆ ˆ
dx a x x dx x a x a x x dx x a x a x x
y y y y y
¥ ¥ ¥
-¥ -¥ -¥
= =
ò ò ò
But
(
)
(
)
(
)
(
)
†
ˆ ˆ
a x a x x x
y ly
=
, as
(
)
x
y
is an eigenfunction of
(
)
(
)
†
ˆ ˆ
a x a x
with
eigenvalue
l
.
Thus
( ) ( ) ( ) ( ) ( )
2 2
*
ˆ
dx a x x dx x x dx x
y y ly l y
¥ ¥ ¥
-¥ -¥ -¥
= =
ò ò ò
That is

15 April 2018
18
( ) ( ) ( )
2 2
ˆ
dx a x x dx x
y l y
¥ ¥
-¥ -¥
=
ò ò
(31)
In (31), both integrands are non-negative, thus both integrals are also non-negative.
Moreover, since
(
)
x
y
is an eigenfunction, it is linearly independent, and thus it
cannot be identically zero. Thus, the integral in the right-hand side of (31) is strictly
positive and, since the integral in the left-hand side is non-negative,
l
must be non-
negative too.
Therefore, the eigenvalues of
(
)
(
)
†
ˆ ˆ
a x a x
are non-negative.
Next, using (27) and that
2
0
2 0
m p
>
, we derive that the eigenvalues of
(
)
0
ˆ
H x E
-
are also non-negative, and thus the eigenvalues of the Hamiltonian, i.e. the energies,
are greater than or equal to
0
E
.
Then, since the energy
0
E
exists, it is the ground-state energy, and the respective
wave function
(
)
0
x
y
is the ground-state wave function.
Notes
i. We point out the significance of the existence of the Riemann integral
( ) ( )
2
ˆ
dx a x x
y
¥
-¥
ò in the proof.
If the wave function
(
)
0
x
y
had nodes, the function
(
)
(
)
ˆ
a x x
y
, as given by (28),
would have singularities and the previous Riemann integral would be ill-defined.
Also, if the discontinuities in the wave function derivative
( )
0
x
y
¢
would be
infinite (in number or in magnitude), the previous Riemann integral would also be
ill-defined.
ii. From (28), we see that
(
)
(
)
0
ˆ
0
a x x
y
=
, i.e. the operator
(
)
ˆ
a x
kills the
ground-state wave function.
5. References
[1] A. Van Heuvelen and D. P. Maloney, Am. J. Phys. 67, 252–256 (1999).
[2] L. P. Gilbert, M. Belloni, M. A. Doncheski, R. W. Robinett, Am. J. Phys. 74, 1035
(2006).

15 April 2018
19
[3] Zafar Ahmed, Sachin Kumar, Mayank Sharma, Vibhu Sharma, Eur. J. Phys. 37
(2016) 045406.
[4] M. Belloni, R.W. Robinett, Phys. Rep. 540 (2014) 25–122.
[5] Zafar Ahmed, Eur. J. Phys. 37 (2016) 045404.
[6] David Branson, Am. J. Phys. 47 (11), Nov. 1979,
http://aapt.scitation.org/doi/10.1119/1.11688.
[7] David J. Griffiths, Introduction to Quantum Mechanics (Prentice Hall, Inc., 1995).
[8] Spiros Konstantogiannis, Finite Volcano Potentials Admitting a Rational
Eigenfunction,
https://www.researchgate.net/publication/323943144_Finite_Volcano_Potentials_Ad
mitting_a_Rational_Eigenfunction.
[9] John K. Hunter, An introduction to Real Analysis (2014),
https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/intro_analysis.pdf.
[10] Stephen Gasiorowicz, Quantum Physics (Wiley, 1974).