An Introduction to Complex Potentials in Quantum Mechanics

Abstract

Considering a particle moving in a general, three-dimensional complex potential, we show the following:
- The anti-Hermitian term of the Hamiltonian, i.e. the imaginary part of the potential, destroys the time-independence of the state-vector norm, which becomes time-dependent, in general.
- The total probability, which is equal to the square of the state-vector norm, is also time-dependent, and thus it is not conserved, in general.
- The continuity equation, which, in position space, expresses the time derivative of the probability to find the particle in an arbitrary closed region, i.e. in a region with a closed surface as its boundary, contains an extra term, which is proportional to the imaginary part of the potential.
- The integral of the extra term in the whole space is equal to the time derivative of the total probability. Thus, the extra term expresses the time derivative of the local probability creation or annihilation which, as we show, behaves exponentially (exponential growth or decay).
- In the last two sections, we examine the parity-time reversal transformation (PT-transformation) and we show that the energies of a PT-symmetric potential, i.e. of a complex potential that remains unchanged (invariant) under the action of a PT-transformation, are real if the respective energy eigenstates are also eigenstates of the parity-time reversal operator.

Full text

1 04 May 2018
An Introduction to Complex Potentials in Quantum Mechanics
Spiros Konstantogiannis
spiroskonstantogiannis@gmail.com
4 May 2018

2 04 May 2018
Copyright © 2018 by Spiros Konstantogiannis. All rights reserved.

3 04 May 2018
Contents
An Introduction to Complex Potentials in Quantum Mechanics ......................... 1
Contents ............................................................................................................ 3
1. Introduction................................................................................................... 4
2. The state-vector norm.................................................................................... 5
3. The total probability......................................................................................11
4. The continuity equation in a complex potential .............................................12
5. Time evolution of the energy eigenstates in a time-independent complex
potential............................................................................................................18
6. Complex potential with constant imaginary part............................................22
7. PT-transformation.........................................................................................23
8. PT-symmetric potentials and real energies ....................................................26
References........................................................................................................33

4 04 May 2018
1. Introduction
We consider a particle moving in the three-dimensional complex potential
(
)
(
)
(
)
, Re , Im ,
V r t V r t i V r t
= +
r r r
,
where
(
)
Re ,
V r t
r
and
(
)
Im ,
V r t
r
are real, scalar functions of space and time.
The previous form is general. If the potential is not in this form, it can be brought
into this form after performing algebraic operations.
The Hamiltonian of the particle is
( ) ( )
2
ˆ
ˆ ˆ
ˆ
Re , Im ,
2
p
H V r t i V r t
m
= + +
r
r r
,
where
ˆ
r
r
and
ˆ
p
r
are, respectively, the position and momentum operators.
Since the functions
(
)
Re ,
V r t
r
and
(
)
Im ,
V r t
r
are real, the respective operators
(
)
ˆ
Re ,
V r t
r
and
(
)
ˆ
Im ,
V r t
r
are Hermitian.
Thus, the term
( )
2
ˆ
ˆ
Re ,
2
p
V r t
m
+
r
r
is Hermitian, while the term
(
)
ˆ
Im ,
i V r t
r
is anti-
Hermitian. Thus, the Hamiltonian is no longer Hermitian.
We assume that, at time
t
, the state of the particle is described by the vector
(
)
t
y
.
The state vector
(
)
t
y
satisfies the Schrödinger equation, i.e.
( ) ( )
ˆ
d
i t H t
dt
y y
=h
We’ll show the following:
· The anti-Hermitian term of the Hamiltonian, i.e. the imaginary part of the
potential, destroys the time-independence of the state-vector norm, which
becomes time-dependent, in general.
· The total probability, which is equal to the square of the state-vector norm, is
also time-dependent, and thus it is not conserved, in general.
· The continuity equation, which, in position space, expresses the time
derivative of the probability to find the particle in an arbitrary closed region,

5 04 May 2018
i.e. in a region with a closed surface as its boundary, contains an extra term,
which is proportional to the imaginary part of the potential.
· The integral of the extra term in the whole space is equal to the time derivative
of the total probability. Thus, the extra term expresses the time derivative of
the local probability creation or annihilation which, as we’ll see, behaves
exponentially (exponential growth or decay).
· In the last two sections, we’ll examine the parity-time reversal transformation
(PT-transformation) and we’ll show that the energies of a PT-symmetric
potential, i.e. of a complex potential that remains unchanged (invariant) under
the action of a PT-transformation, are real if the respective energy eigenstates
are also eigenstates of the parity-time reversal operator.
2. The state-vector norm
The norm of the state vector
(
)
t
y
is, by definition,
( ) ( ) ( )
( )
1
2
,t t t
y y y
º (1)
We use the general notation
(
)
.,.
for inner products, since we have to handle non-
Hermitian operators.
The Dirac notation is mainly made for Hermitian operators.
From (1) we obtain
( ) ( ) ( )
(
)
2
,
t t t
y y y
=
Thus
( ) ( ) ( ) ( ) ( )
2
, ,
d d d
t t t t t
dt dt dt
y y y y y
æ ö æ ö
= +
ç ÷ ç ÷
è ø è ø
(2)
The state vector
(
)
t
y
satisfies the Schrödinger equation, i.e.
( ) ( )
ˆ
d
i t H t
dt
y y
=h
Thus

6 04 May 2018
( ) ( )
ˆ
d i
t H t
dt
y y
-
=
h
(3)
By means of (3), (2) becomes
( ) ( ) ( ) ( ) ( )
2
ˆ ˆ
, ,
d i i
t H t t t H t
dt
y y y y y
- -
æ ö æ ö
= +
ç ÷ ç ÷
è ø è ø
h h
Since the inner product is linear in the right argument and antilinear in the left
argument,
( ) ( ) ( ) ( )
( )
ˆ ˆ
, ,
i i
t H t t H t
y y y y
- -
æ ö =
ç ÷
è ø
h h
and
( ) ( ) ( ) ( )
( )
( ) ( )
( )
*
ˆ ˆ ˆ
, , ,
i i i
H t t H t t H t t
y y y y y y
- -
æ ö æ ö
= =
ç ÷ ç ÷
è ø è ø
h h h
Then, the derivative of the state-vector norm squared is written as
( ) ( ) ( )
(
)
( ) ( )
(
)
2
ˆ ˆ
, ,
d i i
t H t t t H t
dt
y y y y y
= -
h h
But
( ) ( )
(
)
( ) ( )
(
)
†
ˆ ˆ
, ,
H t t t H t
y y y y
=,
where
†
ˆ
H
is the Hermitian conjugate of the Hamiltonian.
Thus
( ) ( ) ( )
(
)
( ) ( )
(
)
2†
ˆ ˆ
, ,
d i i
t t H t t H t
dt
y y y y y
= - =
h h
( ) ( ) ( )
(
)
( )
(
)
( )
(
)
† †
ˆ ˆ ˆ ˆ
, ,
i i
t H t H t t H H t
y y y y y
= - - = - -
h h
That is
( ) ( )
(
)
( )
(
)
2†
ˆ ˆ
,
d i
t t H H t
dt
y y y
= - -
h
(4)
If the Hamiltonian is Hermitian, i.e. if the potential is real, the term
†
ˆ ˆ
H H
-
vanishes,
and (4) becomes

7 04 May 2018
( ) ( ) ( )
2
0 2 0
d d
t t t
dt dt
y y y
= Þ =
Assuming that the state vector
(
)
t
y
is non-trivial, it will be non-zero, thus its norm,
(
)
t
y
, will be non-zero too – remember that in a Hilbert space, only the zero vector
has zero norm – and then the last equation gives
( )
0
dt
dt
y
=
,
i.e. the state-vector norm is time-independent, i.e.
(
)
(
)
(
)
(
)
0 0
t t
y y y y
=
Thus, if the state vector is normalized at an initial time
0
0
t
=
, the time-independence
of the state-vector norm ensures that the normalization is conserved.
Besides, this is the reason for normalizing the initial wave function
(
)
(
)
,0
r r
y y
º
r r
.
We remind that
(
)
(
)
,0 0
r r
y y
=
r r
.
If the imaginary part of the potential is non-zero, the Hamiltonian is no longer
Hermitian, since then
( ) ( )
2
ˆ
ˆ ˆ
ˆ
Re , Im ,
2
p
H V r t i V r t
m
= + +
r
r r
and
( ) ( )
2
†
ˆ
ˆ ˆ
ˆ
Re , Im ,
2
p
H V r t i V r t
m
= + -
r
r r
Thus
(
)
†
ˆ
ˆ ˆ
2 Im ,
H H i V r t
- =
r
(5)
Observe the similarity to the relation *
2 Im
z z i z
- =
.
By means of (5), (4) is written as

8 04 May 2018
( ) ( )
(
)
( )
(
)
( )
(
)
( )
(
)
2
2
ˆ ˆ
,2 Im , ,Im ,
d i
t t i V r t t t V r t t
dt
y y y y y
= - =
r r
h h
That is
( ) ( )
(
)
( )
(
)
2
2ˆ
,Im ,
d
t t V r t t
dt
y y y
=
r
h
Now, we write the inner product
( )
(
)
( )
(
)
ˆ
,Im ,
t V r t t
y y
r
in the Dirac notation.
We note that the position operator is Hermitian, thus the operator
(
)
ˆ
Im ,
V r t
r
is also
Hermitian, since the function
(
)
Im ,
V r t
r
is real.
Then, the last equation is written as
( ) ( )
(
)
( )
2
2ˆ
Im ,
d
t t V r t t
dt
y y y
=
r
h
(6)
We write the expectation value of
(
)
ˆ
Im ,
V r t
r
at time
0
t
³
as
( )
( )
(
)
( )
( ) ( )
ˆ
Im ,
ˆ
Im ,
t
t V r t t
V r t t t
y y
y y
=
r
r (7)
If
(
)
0
ˆ
Im ,
V r t V
=
r
(constant), (7) becomes
(
)
(
)
( ) ( )
(
)
(
)
( ) ( )
0
0 0 0
t
t V t t t
V V V
t t t t
y y y y
y y y y
= = =
,
as it should.
If the norm of the state vector
(
)
t
y
was constant, we could normalize the vector
(
)
t
y
at an initial time, i.e. we could set
(
)
0 1
y
=
, and then
(
)
(
)
0 1
t
y y
= =
Thus
( ) ( ) ( )
2
1
t t t
y y y
= =
,
and (7) is reduced to the known relation
(
)
( )
(
)
( )
ˆ ˆ
Im , Im ,
t
V r t t V r t t
y y
=
r r

9 04 May 2018
Besides, (7) is written as
( )
( )
(
)
( )
( ) ( )
( )
(
)
( )
( )
2
ˆ ˆ
Im , Im ,
ˆ
Im ,
t
t V r t t t V r t t
V r t t t t
y y y y
y y y
º =
r r
r
Thus
( )
(
)
( )
(
)
( )
2
ˆ ˆ
Im , Im ,
t
t V r t t V r t t
y y y
=
r r
(8)
By means of (8), (6) becomes
( )
(
)
( ) ( ) ( )
(
)
( )
2 2 2
2 2
ˆ ˆ
Im , 2 Im ,
t t
d d
t V r t t t t V r t t
dt dt
y y y y y
= Þ =
r r
h h
Assuming that the state vector is non-trivial, its norm will be non-zero, thus the last
equation becomes
( )
( )
( ) ( )
( )
( )
1 1
ˆ ˆ
Im , Im ,
t t
dt
ddt
t V r t t V r t
dt t
y
y y y
= Þ =
r r
h h
Then
( )
( )
(
)
ˆ
Im ,
ln
t
V r t
dt
dt
y
=
r
h
(9)
This is the expression of the time rate of change of the state-vector norm in a complex
potential.
Integrating (9) from an initial time
0
0
t
=
up to a time
0
t
>
yields
( )
( )
(
)
( )
( )
(
)
0
0 0 0
ˆ ˆ
Im , Im ,
ln ln
t t t
t
t t
V r t V r t
d
dt t dt t dt
dt
y y
¢ ¢
¢ ¢
¢ ¢ ¢ ¢ ¢
= Þ = Þ
¢
ò ò ò
r r
h h
( )
( )
( )
( )
(
)
0
ˆ
Im ,
ln ln 0
tt
V r t
t dt
y y
¢
¢
¢
Þ - = Þ
ò
r
h
( )
( )
(
)
( )
( )
(
)
0 0
ˆ ˆ
Im , Im ,
ln exp
0 0
t t
t t
V r t V r t
t t
dt dt
y y
y y
¢ ¢
æ ö
¢ ¢
æ ö
ç ÷
ç ÷ ¢ ¢
Þ = Þ =
ç ÷
ç ÷
ç ÷
è ø
è ø
ò ò
r r
h h
Thus

10 04 May 2018
( ) ( )
(
)
0
ˆ
Im ,
0 exp
tt
V r t
t dt
y y
¢
æ ö
¢
ç ÷
¢
=
ç ÷
ç ÷
è ø
ò
r
h (10)
This is the time evolution of the state-vector norm in a complex potential.
If the potential is real, i.e. if
(
)
ˆ
Im , 0
V r t
=
r
, the expectation value
(
)
ˆ
Im ,
t
V r t
¢
¢
r
vanishes, and the definite integral
(
)
0
ˆ
Im ,
t
t
V r t
dt
¢
¢
¢
ò
r
h
vanishes too, thus the
exponential on the right-hand side of (10) becomes 1, and (10) gives
(
)
(
)
0
t
y y
=
,
i.e. the state-vector norm is time-independent.
If
(
)
0
ˆ
Im ,
V r t V
=
r
, i.e. if the imaginary part of the potential is constant, then (7) gives
(
)
0
ˆ
Im ,
t
V r t V
¢
¢
=
r
, and (10) becomes
( ) ( ) ( )
0 0
0
0 exp 0 exp
t
V V t
t dt
y y y
æ ö
æ ö
¢
= =
ç ÷
ç ÷
è ø
è ø
ò
h h
That is
( ) ( )
0
0 exp
V t
t
y y
æ ö
=
ç ÷
è ø
h
We see that if
0
0
V
>
, the state-vector norm grows exponentially to infinity, while if
0
0
V
<
, the state-vector norm decays exponentially to zero.
Observe that the exponential
(
)
0
ˆ
Im ,
t
t
V r t
dt
¢
¢
¢
ò
r
h
is dimensionless, as it should.
We note that for the previous analysis to be valid, the state-vector norm must be finite
at any time, i.e.
(
)
t
y
< ¥
for every
0
t
³
.
Therefore, the analysis concerns only bound states.

11 04 May 2018
3. The total probability
In position space, the total probability, let us denote it by
total
P
, is the probability to
find the particle in the whole space, i.e.
( ) ( )
3 *
, ,
total
P d r r t r t
y y
¥
-¥
=
ò
r r r
By means of the relation
(
)
(
)
,
r t r t
y y
=
r r
, which is the definition relation of the
wave function in position space, the total probability is written as
( ) ( )
*
3
total
P d r r t r t
y y
¥
-¥
=
ò
r r r
Using the inner product property
( ) ( )
*
r t t r
y y
=
r r
, the total probability becomes
( ) ( ) ( ) ( )
3 3
total
P d r t r r t t d r r r t
y y y y
¥ ¥
-¥ -¥
æ ö
= = ç ÷
è ø
ò ò
r r r r r r
The position operator is Hermitian, with continuous spectrum, thus the position
eigenstates form an orthogonal basis in the state space (Hilbert space), and then they
satisfy the completeness relation
3
1
d r r r
¥
-¥
=
ò
r r r
Substituting into the expression of the total probability yields
( ) ( ) ( )
2
total
P t t t
y y y
= = ,
that is
( )
2
total
P t
y
= (11)
If the potential is real, the state-vector norm is time-independent, thus, as seen from
(11), the total probability is conserved, and we can set it equal to 1
(
)
1
total
P
=
.
This is what normalization does: it sets the total probability equal to 1.
By means of (10), (11) is written as

12 04 May 2018
( )
(
)
2
0
ˆ
Im ,
0 exp 2
tt
total
V r t
P dt
y
¢
æ ö
¢
ç ÷
¢
=
ç ÷
ç ÷
è ø
ò
r
h (12)
where the expectation value
(
)
ˆ
Im ,
t
V r t
¢
¢
r
is given by (7).
Observe that the expression of the total probability, as given by (12), is independent
of the representation, since both the norm of a vector and the expectation value of
an operator are representation-free.
If
(
)
0
ˆ
Im ,
V r t V
=
r
(constant), then
(
)
0
ˆ
Im ,
t
V r t V
¢
¢
=
r
, and (12) becomes
( ) ( )
20
2
0 exp
total
V t
P t
y
æ ö
=
ç ÷
è ø
h
(13)
4. The continuity equation in a complex potential
The square of the absolute value of the position-space wave function
(
)
,
r t
y
r
is the
position probability density.
Then, the probability that the particle is found in a closed region
U
, i.e. in a region
with a closed surface as its boundary, is
( ) ( ) ( )
2
3 3 *
, , ,
U
U U
P d r r t d r r t r t
y y y
= =
ò ò
r r r r r
The time derivative of the previous probability is then
( ) ( )
(
)
( ) ( )
(
)
*
3 * 3 *
, ,
, , , ,
U
U U
r t r t
dP d d r r t r t d r r t r t
dt dt t t
y y
y y y y
æ ö
¶ ¶
= = +
ç ÷
¶ ¶
è ø
ò ò
r r
r r r r r r
That is
*
3 *
U
U
dP d r
dt t t
y y
y y
æ ö
¶ ¶
= +
ç ÷
¶ ¶
è ø
ò
r
(14)
The integrand on the right-hand side of (14) is a real function, since the first term is
the complex conjugate of the second, and thus
* *
* *
2Re 2Re
t t t t
y y y y
y y y y
æ ö
¶ ¶ ¶ ¶
æ ö
+ = =
ç ÷
ç ÷
¶ ¶ ¶ ¶
è ø
è ø

13 04 May 2018
This is expected, since the probability, and its time derivative, as measurable
quantities, must be given by real functions.
The wave function
(
)
,
r t
y
r
satisfies the Schrödinger equation, i.e.
(
)
( ) ( )
,ˆ
, ,
r t
i H r t r t
t
yy
¶=
¶
r
r r
h
(15)
where
(
)
ˆ
,
H r t
r
is the Hamiltonian in position representation, i.e.
( ) ( )
22
ˆ
, ,
2
H r t V r t
m
= - Ñ +
h
r r
(16)
In position representation,
ˆ
r r
=
r r
and
ˆ
p i
= - Ñ
r
r
h
, thus the Hamiltonian
( )
2
ˆ
ˆ
ˆ
,
2
p
H V r t
m
= +
r
r
is written as
( )
(
)
( ) ( )
222
ˆ
, , ,
2 2
i
H r t V r t V r t
m m
- Ñ
= + = - Ñ +
r
hh
r r r
By means of (16), (15) becomes
22 2
2 2
i i
i V V
t m t m
y y
y y y y
¶ ¶
= - Ñ + Þ = Ñ -
¶ ¶
h h
h
h
Thus
* *
2 2 * * *
2 2
i i i i
V V
t m m
y
y y y y
¶
æ ö æ ö
= Ñ - = - Ñ +
ç ÷ ç ÷
¶
è ø è ø
h h
h h
Time is a real variable, thus
*
*
t t
y y
¶ ¶
æ ö =
ç ÷
¶ ¶
è ø
Then, we obtain
*
2 * * *
2
i i V
t m
y
y y y yy
¶= - Ñ +
¶
h
h
(17)
Taking complex conjugates on both sides of the previous equation yields

14 04 May 2018
*
*
* 2 *
2
i i V
t m
y
y y y yy
æ ö
¶= Ñ -
ç ÷
¶
è ø
h
h
But
*
**
t t
y y
y y
æ ö
¶ ¶
=
ç ÷
¶ ¶
è ø
Combining the last two equations, we obtain
* * 2 *
2
i i V
t m
y
y y y yy
¶
= Ñ -
¶
h
h
(18)
Adding (17) and (18) yields
( ) ( )
*
* * 2 2 * * *
2
i i V V
t t m
y y
y y y y y y yy
¶ ¶
+ = Ñ - Ñ - -
¶ ¶
h
h
(19)
The term
( )
* 2 2 *
2
i
m
y y y y
Ñ - Ñ
h
comes from the kinetic term of the Hamiltonian, i.e.
from the momentum of the particle, while the term
( )
* *
iV V
yy
- -
h
involves only the
imaginary part of the potential.
Thus, the real part of the potential does not appear in the continuity equation.
Now, we want to write the term
* 2 2 *
y y y y
Ñ - Ñ as the divergence of another term.
Using the property
(
)
(
)
(
)
A A A
f f f
Ñ× = Ñ × + Ñ×
r r r
r r r
,
where
f
is a scalar function and
A
r
is a vector function, we obtain
(
)
(
)
(
)
( )
( )
*
* * * 2
y y
y y y y y y
Ñ × Ñ
Ñ× Ñ = Ñ × Ñ + Ñ
r r
r r r r
1442443
and
(
)
(
)
(
)
* * 2 *
y y y y y y
Ñ× Ñ = Ñ × Ñ + Ñ
r r r r
Subtracting the last equation from the next to last yields

15 04 May 2018
(
)
(
)
( )
* *
* * * 2 2 *
y y y y
y y y y y y y y
Ñ× Ñ - Ñ
Ñ× Ñ -Ñ × Ñ = Ñ - Ñ
r r r
r r r r
14444244443
That is
(
)
* 2 2 * * *
y y y y y y y y
Ñ - Ñ = Ñ× Ñ - Ñ
r r r
(20)
Also, we have
*
2 Im
V V i V
- =
(21)
By means of (20) and (21), (19) is written as
( )
** * * *
2 Im
2
i i i V
t t m
y y
y y y y y y yy
¶ ¶
+ = Ñ× Ñ - Ñ - =
¶ ¶
r r r
h
h
( )
2
* *
2Im
2
iV
m
y y y y y
= Ñ × Ñ - Ñ +
r r r
h
h
That is
( )
*
2
* * *
2Im
2
iV
t t m
y y
y y y y y y y
¶ ¶
+ = Ñ× Ñ - Ñ +
¶ ¶
r r r
h
h
(22)
By means of (22), (14) becomes
( )
2
3 * *
2Im
2
U
U
dP i
d r V
dt m
y y y y y
æ ö
= Ñ × Ñ - Ñ + Þ
ç ÷
è ø
ò
r r r
h
r
h
( )
2 2
3 3 * *
2
Im 0
2
U U
d i
d r d r V
dt m
y y y y y y
æ ö
Þ - Ñ × Ñ - Ñ + = Þ
ç ÷
è ø
ò ò
r r r
h
r r
h
( )
22
3 3 * * 2
Im 0
2
U U
i
d r d r V
t m
yy y y y y
¶æ ö
Þ - Ñ× Ñ - Ñ + = Þ
ç ÷
¶è ø
ò ò
r r r
hr r
h
( )
22
3 * *
2
Im 0
2
U
i
d r V
t m
yy y y y y
æ ö
¶
Þ ç - Ñ× Ñ - Ñ - ÷ =
ç ÷
¶
è ø
òr r r
hr
h
Since the region
U
is arbitrary, the integrand on the left-hand side of the last equation
must vanish, thus
( )
22
* *
2
Im 0
2
iV
t m
yy y y y y
¶
- Ñ× Ñ - Ñ - = Þ
¶
r r r
h
h
( )
2
2
* *
2Im
2V
t mi
y
y y y y y
¶æ ö
Þ + Ñ× Ñ - Ñ =
ç ÷
¶è ø
r r r
h
h
The last equation is written as

16 04 May 2018
2
Im
j V
t
r
r
¶
+Ñ× =
¶
r
r
h
(23)
where
( ) ( )
2
, ,
r t r t
r y
=
r r
is the probability density, and
( ) ( ) ( ) ( ) ( )
( )
* *
, , , , ,
2
j r t r t r t r t r t
mi
y y y y
= Ñ - Ñ
r r
r
h
r r r r r
(24)
is the probability current.
The equation (23) is the continuity equation in complex potential, in three
dimensions.
Observe that the quantity
(
)
(
)
(
)
(
)
* *
, , , ,
r t r t r t r t
y y y y
Ñ - Ñ
r r
r r r r
is imaginary, since
(
)
(
)
*
* * * *
y y y y y y y y
Ñ - Ñ = - Ñ - Ñ
r r r r
Thus, the quantity on the right-hand side of (24) is real, as the product of two
imaginary quantities.
This is expected, since the probability current, as measurable quantity, must be given
by a real function.
If the potential is real, (23) becomes
0
j
t
r
¶
+Ñ× =
¶
r
r
(25)
The equation (25) is the continuity equation in real potential, in three
dimensions.
If the potential is complex, the continuity equation contains the extra term
2 Im
V
r
h
.
Integrating the extra term in the whole space yields
( ) ( ) ( ) ( ) ( )
2
3 3 3 *
2 2 2
Im , Im , , Im , ,
d r V d r r t V r t d r r t V r t r t
r y y y
¥ ¥ ¥
-¥ -¥ -¥
æ ö = =
ç ÷
è ø
ò ò ò
r r r r r r r r
h h h
But, by definition,
(
)
(
)
,
r t r t
y y
=
r r
and, similarly
1
,

17 04 May 2018
( ) ( )
(
)
( )
ˆ
Im , , Im ,
V r t r t r V r t t
y y
=
r r r r
1. A geometric proof – explanation of similar relations is given in reference [5].
Thus
( )
( )
( )
*
3 3
2 2 ˆ
Im Im ,
d r V d r r t r V r t t
r y y
¥ ¥
-¥ -¥
æ ö
= =
ç ÷
è ø
ò ò
r r r r r
h h
( )
( )
( ) ( )
( )
( )
3 3
2 2
ˆ ˆ
Im , Im ,
d r t r r V r t t t d r r r V r t t
y y y y
¥ ¥
-¥ -¥
æ ö
= = ç ÷
è ø
ò ò
r r r r r r r r
h h
As explained in the previous section, the position eigenstates satisfy the completeness
relation
3
1
d r r r
¥
-¥
=
ò
r r r
, thus
( )
( )
( )
3
2 2 ˆ
Im Im ,
d r V t V r t t
r y y
¥
-¥
æ ö =
ç ÷
è ø
ò
r r
h h
Comparing the last equation with (6) yields
( )
23
2Im
d
t d r V
dt
y r
¥
-¥
æ ö
=
ç ÷
è ø
ò
r
h
,
and by means of (11) we end up to
3
2Im
total
dP
d r V
dt
r
¥
-¥
æ ö
=
ç ÷
è ø
ò
r
h
(26)
Therefore, the integral of the extra term in the whole space is equal to the time
derivative of the total probability.
We showed that, in the presence of complex potential, the continuity equation, in
three dimensions, is given by (23).
The term
t
r
¶
¶
is the (partial) time derivative of the probability density.
The term
j
Ñ×
r
r
expresses the local probability transport – the local probability flow –
from a point of space to another.
The (extra) term
2 Im
V
r
h
expresses the time derivative of the local probability
creation or annihilation due to the presence of complex potential. Observe that this
term is proportional to the product of the probability density and the imaginary part of

18 04 May 2018
the potential, at any point of space. This results in the local probability creation or
annihilation being exponential.
To better understand this, let us assume that
0
j
Ñ× =
r
r
, i.e. we do not have probability
flow from one point of space to another. Then, the continuity equation (23) is written
as
0 0
2 1 2 ln 2 ln 2
Im Im Im Im
t t
V V V dt dt V
t t t t
r r r r
rr
¶ ¶ ¶ ¶
¢ ¢
= Þ = Þ = Þ = Þ
¢
¶ ¶ ¶ ¶
ò ò
h h h h
(
)
( ) ( ) ( )
0 0
,2 2
ln Im , ,0 exp Im
,0
t t
r t
dt V r t r dt V
r
rr r
r
æ ö
¢ ¢
Þ = Þ =
ç ÷
è ø
ò ò
r
r r
rh h
If, additionally, we assume that the imaginary part of the potential is time-
independent, we end up to
( ) ( )
(
)
2Im
, ,0 exp
V r t
r t r
r r
æ ö
=
ç ÷
è ø
r
r r
h
(27)
The equation (27) gives the time evolution of the probability density at any point of
space, if the divergence of the probability current vanishes and the imaginary part of
the potential is time-independent.
We see that at the points where the imaginary part of the potential is negative, the
probability density decays exponentially to zero (assuming that the initial probability
density is non-zero), while at the points where the imaginary part of the potential is
positive, the probability density grows exponentially to infinity (assuming that the
initial probability density is non-zero).
On the surfaces where the imaginary part of the potential vanishes, i.e. on the surfaces
(
)
Im , , 0
V x y z
=
, the probability density is time-independent, i.e. it is stationary.
Near these surfaces, the imaginary part of the potential will have alternating sign, i.e.
at some points it will be positive and at other points it will be negative. Thus, we
observe a non-linear behaviour, with neighbouring regions where the probability
density exponentially tends to infinity or decays to zero.
5. Time evolution of the energy eigenstates in a time-independent
complex potential
In the case of a time-independent complex potential, the Hamiltonian of the particle is

19 04 May 2018
( ) ( )
2
ˆ
ˆ ˆ
ˆRe Im
2
p
H V r i V r
m
= + +
r
r r
The state vector
(
)
t
y
of the particle at time
0
t
³
satisfies the Schrödinger equation,
i.e.
( ) ( )
ˆ
=
d
i t H t
dt
y y
h
Assuming that at time
0
0
t
=
, the state of the particle is an energy eigenstate, then
(
)
(
)
ˆ
0 0
H E
y y
=
,
where
E
is the energy at time
0
0
t
=
.
Since the Hamiltonian is not Hermitian, its eigenvalues, i.e. the energies, will be, in
general, complex, i.e. they will have imaginary part.
Assuming that
(
)
(
)
(
)
0
t T t
y y
=
(28)
then
( ) ( ) ( )
( )
}
( ) ( ) ( ) ( ) ( )
ˆ is time-independent
ˆ ˆ ˆ
0 0 0
H
H t H T t T t H T t E E t
y y y y y
= = = =
That is
(
)
(
)
ˆ
H t E t
y y
=
,
thus the state
(
)
t
y
remains eigenstate with the same energy
E
.
Substituting (28) into the Schrödinger equation yields
( ) ( )
( )
( ) ( )
( )
( ) ( ) ( )
ˆ ˆ
0 = 0 0 0 0
d
i T t H T t i T TH TE
dt
y y y y y
¢
Þ = = Þ
h h
( ) ( )
( )
}
( )
}
0, since is
As an eigenstate, 0 an eigenstate, and thus
must be non-zero it is non-zero
0 0 0
T t
T
i T ET i T ET i E
T
y
y
y
¹
¢
¢ ¢
Þ - = Þ - = Þ = Þ
h h h
( )
( )
( ) ( ) ( )
ln ln exp exp
T iE iE iEt iEt
T t T t c T t C
T
¢
æ ö
¢
Þ = - Þ = - Þ = - + Þ = -
ç ÷
è ø
h h h h
The constant
(
)
exp
C
can be incorporated into the state
(
)
0
y
.

20 04 May 2018
Thus, we end up to
( )
exp
iEt
T t
æ ö
= -
ç ÷
è ø
h
Then, (28) is written as
( ) ( )
exp 0
iEt
t
y y
æ ö
= -
ç ÷
è ø
h
(29)
with the energy
E
being, in general, complex, i.e.
r im
E E iE
= + (30)
By means of (30), (29) becomes
( ) ( )
exp exp 0
im r
E t iE t
t
y y
æ ö æ ö
= -
ç ÷
ç ÷ è øè ø
h h
(31)
Using (31), the state-vector norm is
( ) ( )
exp 0
im
E t
t
y y
æ ö
=ç ÷
è ø
h
(32)
As an eigenstate,
(
)
t
y
must be non-zero, thus its norm must be non-zero too, i.e.
(
)
0
t
y
¹
, thus
(
)
0 0
y
¹
too.
Then, from (32) we see that
· If
0
im
E
>
, the norm grows exponentially to infinity.
· If
0,
im
E
<
the norm decays exponentially to zero.
· If
0
im
E
=
, then
(
)
(
)
0
t
y y
=
, i.e. the norm it time-independent.
As shown, if the potential is real, the norm of an arbitrary state is time-independent,
thus the norm of an arbitrary energy eigenstate is also time-independent.
Now, we see that the norm of an energy eigenstate can be time-independent even
if the potential is complex, provided that the energy of the eigenstate is real.
Let us see when this happens.
We showed that the norm of an arbitrary state is given by (10).

21 04 May 2018
If the state is an energy eigenstate and the potential is time-independent, then its norm
is given by (32). In this case, (10) also holds, as a more general relation.
Combining (10) and (32), we obtain that, in an energy eigenstate, when the potential
is time-independent,
( )
0
ˆ
Im
t
im
t
E t dt V r
¢
¢
=
ò
r
Differentiating both members of the previous equation with respect to time yields
( )
0
ˆ
Im
t
im
t
d
E dt V r
dt
¢
¢
=
ò
r
The integral
( )
0
ˆ
Im
t
t
dt V r
¢
¢
ò
r
is equal to the antiderivative of
(
)
ˆ
Im
t
V r
¢
r
at
t
minus
its antiderivative at 0.
Thus
( ) ( )
0
ˆ ˆ
Im Im
t
t t
d
dt V r V r
dt
¢
¢=
ò
r r
Then, we end up to
(
)
ˆ
Im
im
t
E V r
=
r
(33)
where the expectation value
(
)
ˆ
Im
t
V r
r
is given by (7).
If
(
)
0
ˆ
Im
V r V
=
r
, i.e. if the imaginary part of the potential is constant, then (7)
gives
(
)
0
ˆ
Im
t
V r V
=
r
,
and (33) is written as
0
im
E V
=
In (33),
im
E
is time-independent, since it is the imaginary part of an eigenvalue of a
time-independent Hamiltonian, thus the expectation value
(
)
ˆ
Im
t
V r
r
is also time-
independent, and we can write

22 04 May 2018
(
)
(
)
0
ˆ ˆ
Im Im
t
V r V r
=
r r
(34)
where
(
)
0
ˆ
Im
V r
r
is the expectation value at
0
0
t
=
(initial time).
The expectation value
(
)
ˆ
Im
t
V r
r
is time-independent when the particle is in an
energy eigenstate and the potential is time-independent.
Therefore, if in an energy eigenstate the expectation value of the imaginary part
of the time-independent complex potential vanishes, the energy of the eigenstate
is real.
Combining (11) and (32), we obtain that the total probability in an energy eigenstate,
when the potential is time-independent, is
( )
2
2
exp 0
im
total
E t
P
y
æ ö
=ç ÷
è ø
h
(35)
From (35), we see that
· If
0
im
E
>
, the total probability grows exponentially to infinity.
· If
0
im
E
<
, the total probability decays exponentially to zero.
· If
0
im
E
=
, the total probability is constant. As shown, this happens when the
expectation value of the imaginary part of the time-independent potential
vanishes, in the particular eigenstate.
6. Complex potential with constant imaginary part
We’ll examine the case where the imaginary part of the potential is constant, i.e.
(
)
(
)
0
ˆ ˆ
, ,
real
V r t V r t iV
= +
r r
Then, the Hamiltonian is written as
( )
2
0 0
ˆˆ
ˆ ˆ
,
2
real herm
p
H V r t iV H iV
m
= + + = +
r
r
,
where
( )
2
ˆ
ˆ
ˆ
,
2
herm real
p
H V r t
m
= +
r
r
is the Hermitian part of the Hamiltonian.
If
y
is an eigenstate of
ˆ
herm
H
, with eigenvalue (energy)
E
, then

23 04 May 2018
(
)
( )
0 0 0 0
ˆ ˆ ˆ
herm herm
H H iV H iV E iV E iV
y y y y y y y
= + = + = + = +
That is
(
)
0
ˆ
H E iV
y y
= +
Thus,
y
is an eigenstate of
ˆ
H
, with eigenvalue (energy)
0
E iV
+
.
Since
ˆ
herm
H
is Hermitian, its eigenvalues,
E
, are real.
Thus, the eigenvalues of
ˆ
H
have as real part the eigenvalues of
ˆ
herm
H
and constant
imaginary part equal to
0
iV
.
Therefore, to calculate the eigenstates and the energies of the potential
(
)
(
)
0
ˆ ˆ
, ,
real
V r t V r t iV
= +
r r
, it suffices to calculate the eigenstates and the energies of the
potential
(
)
ˆ
,
real
V r t
r
.
The eigenstates of the potential
(
)
ˆ
,
V r t
r
are the eigenstates of the potential
(
)
ˆ
,
real
V r t
r
,
with energies the energies of
(
)
ˆ
,
real
V r t
r
plus
0
iV
.
7. PT-transformation
Parity (space reflection) and time reversal are two important discrete transformations
2
,
which are represented by the operators
ˆ
P
and
ˆ
T
, respectively.
2. A transformation is discrete (i.e. non-continuous) if it cannot be composed of
infinitesimal transformations. The discrete transformations have no generators.
The operators
ˆ
P
and
ˆ
T
are such that
ˆ
P
:
ˆ ˆ
r r
® -
r r
and
ˆ
T
:
t t
® -
Both operators change the sign of the momentum operator (to understand why, you
may think classically).

24 04 May 2018
Then, the parity operator leaves the products
ˆˆ
rp
rr
and
ˆ ˆ
pr
rr
unchanged, while the time-
reversal operator changes the sign of both
ˆˆ
rp
rr
and
ˆ ˆ
pr
rr
, since it does not change the
sign of
ˆ
r
r
(time reversal is independent of space reflection).
Thus, the parity operator leaves the commutator
ˆ ˆ
,
r p
é ù
ë û
r r
unchanged, and since
[
]
ˆ ˆ
,
x p i
=
h
, the parity operator leaves the imaginary unit unchanged, i.e.
ˆ
P
:
i i
®
,
and then it leaves unchanged an arbitrary complex number.
Thus, if
1 2
a b
y y
+
is an arbitrary linear combination of two states,
(
)
(
)
(
)
1 2 1 2
ˆ ˆ ˆ
P a b a P b P
y y y y
+ = + ,
which means that
ˆ
P
is linear.
On the other hand, the time-reversal operator changes the sign of
ˆ ˆ
,
r p
é ù
ë û
r r
, and thus,
since
[
]
ˆ ˆ
,
x
x p i
=
h
, it must also change the sign of the imaginary unit, i.e.
ˆ
T
:
i i
® -
,
and then it transforms an arbitrary complex number to its complex conjugate.
Thus, if
1 2
a b
y y
+
is an arbitrary linear combination of two states,
(
)
(
)
(
)
* *
1 2 1 2
ˆ ˆ ˆ
T a b a T b T
y y y y
+ = + ,
which means that
ˆ
T
is antilinear.
The parity transformation is independent of the time-reversal transformation, thus
applying first the parity operator and then the time-reversal operator is the same as
applying first the time-reversal operator and then the parity operator, i.e.
ˆ ˆ ˆ ˆ
PT TP
=
or
ˆ ˆ
, 0
P T
é ù
=
ë û
Since
2
ˆ
P
:
(
)
ˆ ˆ ˆ ˆ
r r r r
® - ® - - =
r r r r
and

25 04 May 2018
2
ˆ
T
:
(
)
t t t t
® - ® - - =
,
we have
2 2
ˆ ˆ
1
P T
= =
,
and thus
(
)
{
{
22 2
ˆ ˆ ˆ ˆ 1
ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
1
PT TP
PT PTPT PTTP PT P P
=
= = = = =
That is
(
)
2
ˆ ˆ
1
PT
=
Also, since
ˆ
P
is linear and
ˆ
T
is antilinear,
ˆ ˆ
PT
is antilinear, i.e. the parity-time
reversal operator is antilinear.
Indeed, if
1 2
a b
y y
+
is an arbitrary linear combination of two states,
(
)
{
(
)
{
* * * *
1 2 1 2 1 2
ˆ ˆ
is antilinear is linear
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
T P
PT a b P a T b T a PT b PT
y y y y y y
+ = + = = +
That is
* *
1 2
ˆ ˆ ˆ ˆ ˆ ˆ
PT a PT b PT
y y
= +
Next, we’ll show that the eigenvalues of
ˆ ˆ
PT
are pure phases, i.e. an arbitrary
eigenvalue
l
of
ˆ ˆ
PT
is written as
(
)
exp
i
j
.
If
l
is an eigenstate of
ˆ ˆ
PT
, with eigenvalue
l
, then
ˆ ˆ
PT
l l l
=
Multiplying both members of the previous eigenvalue equation on the left by
ˆ ˆ
PT
yields
ˆ ˆˆ ˆ ˆ ˆ
PTPT PT
l l l
=
But
(
)
2
ˆ ˆˆ ˆ ˆ ˆ
1
PTPT PT
= =
and, since
ˆ ˆ
PT
is antilinear,
* *
ˆ ˆ ˆ ˆ
PT PT
l l l l l l l
= =
Thus

26 04 May 2018
2
*
l l l l l l
= =
That is
2
l l l
=
As an eigenstate,
l
is linearly independent, thus it is non-zero, and then the last
equation gives
( )
2
1 1 exp
i
l l l j
= Þ = Þ =
,
i.e. the eigenvalues of
ˆ ˆ
PT
are pure phases.
Notes
i. In the same way, if
ˆ
O
is an antilinear operator with 2
ˆ
1
O
=
, its eigenvalues are
pure phases.
ii. Also, if
ˆ
O
is a linear operator with 2
ˆ
1
O
=
, its eigenvalues are
1
±
.
As a consequence, the eigenvalues of the parity operator
ˆ
P
are
1
±
, with the
eigenvalue 1 corresponding to even-parity functions and the eigenvalue
1
-
corresponding to odd-parity functions.
8. PT-symmetric potentials and real energies
We consider the time-independent complex potential
(
)
(
)
(
)
Re Im
V r V r i V r
= +
r r r
(36)
where
(
)
Re
V r
r
and
(
)
Im
V r
r
are real, scalar functions of the position
r
r
.
As noted in the beginning of the first section, the above expression is general.
Applying a time-reversal transformation to the previous potential yields the potential
(
)
(
)
(
)
*
Re Im
V r V r i V r
= -
r r r
Applying a parity transformation to the previous potential yields the potential
(
)
(
)
(
)
*
Re Im
V r V r i V r
- = - - -
r r r
(37)

27 04 May 2018
Thus, the application of a PT-transformation to the initial potential
(
)
V r
r
yields the
potential
(
)
*
V r
-
r
.
If the initial potential remains unchanged (i.e. invariant) under the application of a PT-
transformation, the potential is called PT-symmetric, i.e. the potential
(
)
V r
r
is PT-
symmetric if
(
)
(
)
*
V r V r
- =
r r
(38)
Substituting (36) and (37) into (38) yields
(
)
(
)
Re Re
V r V r
- =
r r
(39)
and
(
)
(
)
Im Im
V r V r
- = -
r r
(40)
That is, the real part of a PT-symmetric potential is of even parity, while its
imaginary part is of odd parity.
Next, we consider an energy eigenfunction
(
)
r
y
r
of a PT-symmetric potential
(
)
V r
r
,
with energy
E
.
Then, the wave function
(
)
r
y
r
is a non-trivial (i.e. linearly independent) solution to
the time-independent Schrödinger equation, for the potential
(
)
V r
r
, with energy
E
.
We’ll show that if the energy eigenfunction
(
)
r
y
r
is also an eigenfunction of the
parity-time reversal operator
ˆ ˆ
PT
, then the energy
E
is real.
Notes
i. If a transformation is represented by a linear operator
ˆ
O
that commutes with the
Hamiltonian
ˆ
H
, then
ˆ
O
and
ˆ
H
have common eigenstates.
The parity-time reversal operator
ˆ ˆ
PT
commutes with the Hamiltonian of a PT-
symmetric potential (see note ii).
Thus, PT-symmetric potential means PT-symmetric Hamiltonian.
However,
ˆ ˆ
PT
and
ˆ
H
do not necessarily have common eigenstates, because the
operator
ˆ ˆ
PT
is not linear, it is antilinear.

28 04 May 2018
That is, the energy eigenstates are not necessarily eigenstates of the parity-time
reversal operator too.
Thus, the assumption that the energy eigenfunction
(
)
r
y
r
is also an eigenfunction
of the parity-time reversal operator is necessary, and crucial.
ii. If
(
)
,
V r t
r
is an (in general, time-dependent) PT-symmetric potential, then
ˆ ˆ ˆ
, 0
PT H
é ù
=
ë û
, i.e. the parity-time reversal operator commutes with the
Hamiltonian.
Proof
Let
(
)
,
r t
j
r
be an arbitrary wave function in position space.
In position representation,
ˆ
r r
=
r r
and
ˆ
p i
= - Ñ
r
r
h
, and the Hamiltonian
( )
2
ˆ
ˆ
ˆ
,
2
p
H V r t
m
= +
r
r
takes the form
( )
22
ˆ
,
2
H V r t
m
= - Ñ +
r
h
r
Then, the action of the operator
ˆ ˆ ˆ
PTH
on the arbitrary position-space wave
function
(
)
,
r t
j
r
yields
( ) ( ) ( ) ( )
22
ˆ ˆ ˆ ˆ ˆ
, , , ,
2
PTH r t PT r t V r t r t
m
j j j
æ ö
= - Ñ + =
ç ÷
è ø
r
hr r r r
( ) ( ) ( )
22
ˆ ˆ ˆ ˆ
, , ,
2
PT r t PTV r t r t
m
j j
= - Ñ +
r
h
r r r
That is
( ) ( ) ( ) ( )
22
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
, , , ,
2
PTH r t PT r t PTV r t r t
m
j j j
= - Ñ +
r
h
r r r r
(41)
The time-reversal operator leaves the space coordinates unchanged, thus
(
)
(
)
2 2
ˆ ˆ ˆ ˆ
, ,
PT r t P T r t
j j
Ñ = Ñ
r r
r r
(42)
On the other hand, the parity operator changes the sign of the space coordinates,
thus
( ) ( ) ( ) ( )
2
2 2 2
ˆ ˆ ˆ ˆ ˆ ˆ
, 1 , ,
P T r t PT r t PT r t
j j j
Ñ = - Ñ = Ñ
r r r
r r r
That is
(
)
(
)
2 2
ˆ ˆ ˆ ˆ
, ,
P T r t PT r t
j j
Ñ = Ñ
r r
r r
(43)
Comparing (42) and (43) yields
(
)
(
)
2 2
ˆ ˆ ˆ ˆ
, ,
PT r t PT r t
j j
Ñ = Ñ
r r
r r
(44)

29 04 May 2018
From (44), since
(
)
,
r t
j
r
is arbitrary, we derive that the parity-time reversal
operator commutes with the Laplacian, and thus it commutes with the momentum
squared operator in position representation.
Then, since the commutators are representation-free, the parity-time reversal
operator commutes with the momentum squared operator in every representation,
i.e.
2
ˆ
ˆ ˆ
, 0
PT p
é ù
=
ë û
r
(45)
Besides, since the potential is PT-symmetric,
(
)
(
)
ˆ ˆ
, ,
PTV r t V r t
=
r r
(46)
We remind that the action of the parity-time reversal operator on the time-
dependent potential
(
)
,
V r t
r
yields
(
)
(
)
*
ˆ ˆ
, ,
PTV r t V r t
= - -
r r
By means of (44) and (46), (41) becomes
( ) ( ) ( ) ( )
22
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
, , , ,
2
PTH r t PT r t V r t PT r t
m
j j j
= - Ñ + =
r
h
r r r r
( ) ( ) ( )
22
ˆ
ˆ ˆ ˆ ˆ ˆ
, , ,
2
H
V r t PT r t HPT r t
m
j j
æ ö
ç ÷
= - Ñ + =
ç ÷
ç ÷
è ø
r
h
r r r
144424443
That is
(
)
(
)
ˆ ˆ ˆ ˆ ˆ ˆ
, ,
PTH r t HPT r t
j j
=
r r
Since
(
)
,
r t
j
r
is arbitrary, we obtain
ˆ ˆ ˆ ˆ ˆ ˆ
PTH HPT
=
i.e.
ˆ ˆ ˆ
, 0
PT H
é ù
=
ë û
Therefore, in position representation, the parity-time reversal operator commutes
with the Hamiltonian, when the potential is PT-symmetric.
Since the commutators are representation-free, the previous commutator vanishes
in every representation.
We showed that the eigenvalues of
ˆ ˆ
PT
are pure phases, thus
(
)
(
)
(
)
ˆ ˆ exp
PT r i r
y j y
=
r r
(47)

30 04 May 2018
In general,
(
)
r
y
r
is a complex function, thus it can be written as
(
)
(
)
(
)
Re Im
r r i r
y y y
= +
r r r
Then, the action of the parity-time reversal operator on
(
)
r
y
r
yields
(
)
(
)
(
)
(
)
*
ˆ ˆ Re Im
PT r r r i r
y y y y
= - = - - -
r r r r
Using (47), we obtain
(
)
(
)
(
)
(
)
(
)
(
)
Re Im exp Re Imr i r i r i r
y y j y y
- - - = + =
r r r r
(
)
(
)
(
)
(
)
cos sin Re Imi r i r
j j y y
= + + =
r r
(
)
(
)
(
)
(
)
(
)
cos Re sin Im cos Im sin Re
r r i r r
j y j y j y j y
= - + +
r r r r
That is
(
)
(
)
(
)
(
)
(
)
(
)
(
)
Re Im cos Re sin Im cos Im sin Re
r i r r r i r r
y y j y j y j y j y
- - - = - + +
r r r r r r
The last equation gives
(
)
(
)
(
)
Re cos Re sin Im
r r r
y j y j y
- = -
r r r
and
(
)
(
)
(
)
(
)
Im cos Im sin Re
r r r
y j y j y
- = - +
r r r
Using the last two equations, we have
( )
(
)
( )
(
)
( ) ( )
(
)
2 2 2
Re Im cos Re sin Imr r r r
y y j y j y
- + - = - +
r r r r
( ) ( )
( )
(
)
( ) ( )
( )
22
cos Im sin Re cos Re sin Imr r r r
j y j y j y j y
+ - + = - +
r r r r
( ) ( )
(
)
( )
(
)
( )
(
)
2 2 2
2 2
cos Im sin Re cos Re sin Imr r r r
j y j y j y j y
+ + = + +
r r r r
( )
( )
( )
( )
( )
( )
( )
( )
( )
2 2 2 2
2 2 2 2
1
cos Im sin Re sin cos Re Imr r r r
j y j y j j y y
æ ö
+ + = + + =
ç ÷
ç ÷
è ø
r r r r
1442443
( )
(
)
( )
(
)
2 2
Re Im
r r
y y
= +
r r
That is
( )
(
)
( )
(
)
( )
(
)
( )
(
)
2 2 2 2
Re Im Re Im
r r r r
y y y y
- + - = +
r r r r
or

31 04 May 2018
( ) ( )
2 2
r r
y y
- =
r r
i.e. the probability density, in position space, is of even parity.
The expectation value of the imaginary part of the potential, at time
0
0
t
=
, is, from
(7),
( )
( )
(
)
( )
( ) ( )
0
ˆ
0 Im 0
ˆ
Im 0 0
V r
V r
y y
y y
=
r
r (48)
with
( ) ( ) ( )
2
0 0 0 0
y y y
= ¹
, since the state
(
)
0
y
is an energy eigenstate.
Using the completeness relation of the position eigenstates, we write the inner product
( )
(
)
( )
ˆ
0 Im 0
V r
y y
r
as
( )
( )
( ) ( )
( )
( )
3
1
ˆ ˆ
0 Im 0 0 Im 0V r d r r r V r
y y y y
¥
-¥
æ ö
ç ÷
= =
ç ÷
ç ÷
ç ÷
è ø
ò
r r r r r
14243
( )
( )
( ) ( )
( )
( )
*
3 3
ˆ ˆ
0 Im 0 0 Im 0
d r r r V r d r r r V r
y y y y
¥ ¥
-¥ -¥
= =
ò ò
r r r r r r r r
That is
( )
( )
( ) ( )
( )
( )
*
3
ˆ ˆ
0 Im 0 0 Im 0
V r d r r r V r
y y y y
¥
-¥
=
ò
r r r r r
Using that
(
)
(
)
(
)
0 ,0
r r r
y y y
= º
r r r
and
3
(
)
( ) ( ) ( )
ˆ
Im 0 Im
r V r V r r
y y
=
r r r r
,
3. A geometric proof – explanation of similar relations is given in reference [5].
we obtain
( )
( )
( ) ( ) ( ) ( ) ( ) ( )
2
3 * 3
Function
ˆ
0 Im 0 Im Im
V r d r r V r r d r r V r
y y y y y
¥ ¥
-¥ -¥
= =
ò ò
r r r r r r r r
14243

32 04 May 2018
That is
( )
( )
( ) ( ) ( )
2
3
ˆ
0 Im 0 Im
V r d r r V r
y y y
¥
-¥
=
ò
r r r r
The integrand
( ) ( )
2
Im
r V r
y
r r
is of odd parity, because
( )
2
r
y
r
is of even parity and
(
)
Im
V r
r
is of odd parity (see (40)), and the integration interval is symmetric
4
.
Therefore, the integral on the right-hand side of the last equation vanishes
5
, and then
(48) gives
(
)
0
ˆ
Im 0
V r
=
r
Then, from (34) we obtain that
(
)
ˆ
Im 0
t
V r
=
r
, and then from (33) we obtain that the
imaginary part of the energy
E
vanishes, and thus the energy
E
is real, and this
completes the proof.
If all eigenstates of a PT-symmetric Hamiltonian – i.e. of a PT-symmetric potential –
are also eigenstates of the operator
ˆ ˆ
PT
, the PT-symmetry is called unbroken.
Therefore, if the PT-symmetry of a Hamiltonian is unbroken, all energies are
real.
4. We assume that the values of the position are real, i.e. the time-independent
Schrödinger equation is solved in
3
¡
.
5. We remind that if
(
)
f x
is of odd parity,
( )
0
a
a
dxf x
-
=
ò,
0
a
>
.
In three dimensions, the integral vanishes if at least one of the three variables
, ,
x y z
is integrated on a symmetric interval.
Indeed, assuming, without loss of generality, that
x
is integrated on the symmetric
interval
[
]
,
a a
-
, then
( ) ( ) ( )
3
, , , ,
a
a
d rf r dxdydzf x y z dydz dxf x y z
-
= =
ò ò ò ò
r r
,
where, in the last equality, we changed the order of integration.

33 04 May 2018
Since
(
)
f r
r
is of odd parity, it is of odd parity in each of the three variables, and
thus
(
)
(
)
, , , ,
f x y z f x y z
- = -
, and then the integral
( )
, ,
a
a
dxf x y z
-
ò vanishes,
and thus the integral
(
)
3
d rf r
ò
r r
vanishes too.
References
[1] David J. Griffiths, Introduction to Quantum Mechanics (Prentice Hall, Inc., 1995).
[2] Martin Plenio, Quantum Mechanics (Imperial College, 2002).
[3] Sheldon Axler, Linear Algebra Done Right (Second Edition, Springer, 1997).
[4] Carl M. Bender, Contemporary Physics 46:4, 277-292 (2007).
[5] Spiros Konstantogiannis, A Geometric Presentation of the Position and
Momentum Representations in Quantum Mechanics (Preprint, ResearchGate, 2017),
https://www.researchgate.net/publication/322631571_A_Geometric_Presentation_of_
the_Position_and_Momentum_Representations_in_Quantum_Mechanics.