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Advanced Studies in Theoretical Physics
Vol. 11, 2017, no. 2, 37 - 75
HIKARI Ltd, www.m-hikari.com
https://doi.org/10.12988/astp.2017.61033
The α-Deformed Calculus and
Some Physical Applications
Jae Yoon Kim, Hyung Ok Yoon, Eun Ji Jang, Won Sang Chung∗
Department of Physics and Research Institute of Natural Science
College of Natural Science
Gyeongsang National University, Jinju 660-701, Korea
∗
Corresponding author
Hassan Hassanabadi
Physics Department, Shahrood University of Technology, Shahrood, Iran
c 2016 Jae Yoon Kim et al. This article is distributed under the Creative
Copyright Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
In this paper we propose a new deformed calculus called a α-deformed
calculus. We use this new derivative to formulate a new calculus theory.
We also discuss the α-deformed exponential, α-deformed logarithm and
α-deformed trigonometry. We discuss the α-deformed Laplace transformation. As applications we discuss the α-deformed chemical reaction
dynamics, α-deformed Euler equation, α-deformed mechanics and αdeformed quantum mechanics.
1
Introduction
Fractional calculus [1, 2, 3] has attracted much interest among of many
researchers either theoretically or in different fields of applications [4, 5]. The
theory of q-deformed fractional calculus was initiated in early of fifties of last
century [6, 7, 8, 9, 10]. Then, this theory has started to be developed in the
last decade or so [11]-[16].
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Jae Yoon Kim et al.
Recently, a new kind of fractional derivative called a conformable fractional
derivative was proposed by Khalil, Al Horani, Yousef and Sababheh [17]. This
derivative depends just on the basic limit definition of the derivative. Namely.
for a function f : (0, ∞) → R the (conformable) fractional derivative of order
0 < α ≤ 1 of f at x > 0 was defined by
f (x + x1−α ) − f (x)
→0
Dxα f (x) = lim
(1)
and the fractional derivative at 0 is defined as (Dxα f )(0) = limx→0 (Dxα f )(x).
Some application of this type of fractional derivative is given in some literatures
[18-30].
From the definition of the conformable fractional derivative, we can easily
obtain the α-deformed exponential function eα (x) obeying Dxα eα (x) = eα (x).
The α-deformed exponential function can be expressed in terms of the ordinary
1 α
exponential function as follows: eα (x) = e α x . For this deformed exponential,
we have eα (x)eα (y) 6= eα (x + y), instead, we have eα (x)eα (y) = eα (x ⊕ y),
where the α-deformed addition is defined as x ⊕ y = (xα + y α )1/α . Similarly,
the α-deformed subtraction is defined as x y = (xα − y α )1/α for x > y. Using
the α-deformed deformed subtraction, we can define a new kind of deformed
derivative called a α-deformed derivative as
F (y) − F (x)
y→x 1 (y x)α
α
Dxα F (x) = lim
(2)
where we assume y > x. This definition seems somewhat unnatural because
the ordinary subtraction is used in numerator while the α-deformed subtraction
is used in denominator. This fact impels us to define a new kind of deformed
derivative called a α-deformed derivative in a natural way.
In this paper we propose a new deformed calculus called a α-deformed calculus where the conformable fractional derivative given in the eq.(3) is replaced
with
F (y) F (x)
(3)
Dxα F (x) = lim
y→x
yx
We use this new derivative called a α-deformed derivative to formulate a new
calculus theory. This paper is organized as follows: In section II we discuss
the α-deformed addition and α-deformed subtraction, α-deformed number,
α-deformed derivative and α-deformed integral. In section III we find the
α-deformed exponential, α-deformed logarithm and α-deformed trigonometry
and investigate their properties. In section IV we discuss the α-deformed
Laplace transformation and investigate its properties. In section V we deal
with some application such as the α-deformed chemical reaction dynamics,
α-deformed Euler equation, α-deformed mechanics and α-deformed quantum
mechanics.
The α-deformed calculus and some physical applications
2
39
α-deformed calculus
In this section we discuss the algebraic structure related to the α-deformed
calculus.
2.1
α-deformed addition and subtraction
Now let us modify the definition of the α-deformed addition and α-deformed
subtraction so that they may work well for negative numbers.
Definition 2.1 The α-deformed addition and α-deformed subtraction are defined as follows:
x ⊕ y = ||x|α−1 x + |y|α−1 y|1/α−1 (|x|α−1 x + |y|α−1 y)
x y = ||x|α−1 x − |y|α−1 y|1/α−1 (|x|α−1 x − |y|α−1 y)
(4)
where we assume that α > 0.
When α = 1 is taken, the α-deformed addition ( or α-deformed subtraction)
reduces to an ordinary addition ( or ordinary subtraction ). We can easily
check that the α-deformed addition is commutative and associative, so we can
obtain the following formula:
N
M
k=0
1/α−1
N
N
X
X
α−1
xk = |xk | xk |xk |α−1 xk
k=1
(5)
k=1
where
N
M
ak = a0 ⊕ a1 ⊕ a2 ⊕ · · · ⊕ aN
(6)
k=0
When xk ≥ 0 for k = 1, 2, · · · , N , the relation (5) reduces to
N
M
k=0
xk =
N
X
!1/α
xαk
(7)
k=1
For the α-deformed addition, the identity is still 0; indeed we have
x⊕0=0⊕x=x
(8)
For the α-deformed addition, the inverse of x denoted by x is defined by
x ⊕ (x) = (x) ⊕ x = 0
(9)
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Jae Yoon Kim et al.
For x > 0, y > 0, the eq.(4) can be written as
x ⊕ y = (xα + y α )1/α
(
(xα − y α )1/α (x > y)
xy =
−(y α − xα )1/α (x < y)
(10)
For x > 0, y < 0, the eq.(4) can be written as
(
(xα − (−y)α )1/α (x > −y)
x⊕y =
−((−y)α − xα )1/α (x < −y)
x y = (xα + (−y)α )1/α
(11)
For x < 0, y > 0, the eq.(4) can be written as
(
(y α − (−x)α )1/α (−x < y)
x⊕y =
−((−x)α − y α )1/α (−x > y)
x y = −(y α + (−x)α )1/α
(12)
For x < 0, y < 0, the eq.(4) can be written as
x ⊕ y = −((−x)α + (−y)α )1/α
(
((−y)α − (−x)α )1/α (x > y)
xy =
−((−x)α − (−y)α )1/α (x < y)
(13)
The above relations can be unified into the following ones:
(
(|x|α−1 x + |y|α−1 y)1/α if x + y > 0
x⊕y =
−(−|x|α−1 x − |y|α−1 y)1/α if x + y < 0
(
(|x|α−1 x − |y|α−1 y)1/α if x − y > 0
xy =
−(−|x|α−1 x + |y|α−1 y)1/α if x − y < 0
(14)
(15)
From the above relations we can easily check x y = x ⊕ (−y), which implies
x = −x.
Proposition 2.1 For the α-deformed addition and α-deformed subtraction,
we have the following property:
1. Distributivity
(kx) ⊕ (ky) = k(x ⊕ y),
(kx) (ky) = k(x y),
k∈R
(16)
2. Expansion
(A ⊕ B)(C ⊕ D) = AC ⊕ BC ⊕ AD ⊕ BD
(17)
The α-deformed calculus and some physical applications
41
Proof. The proof is as follows:
(A ⊕ B)(C ⊕ D) = (A ⊕ B)C ⊕ (A ⊕ B)D
= (AC ⊕ BC) ⊕ (AD ⊕ BD)
= AC ⊕ BC ⊕ AD ⊕ BD
(18)
which completes the proof of the eq.(17). The proof of the eq.(16) is similar.
From the eq.(17) we have the following identities:
(x ⊕ y)(x y) = x2 y 2
(19)
(x ⊕ y)2 = x2 ⊕ 21/α xy ⊕ y 2
2
2
1/α
(x y) = x 2
2.2
xy ⊕ y
2
(20)
(21)
(x ⊕ y)(x2 xy ⊕ y 2 ) = x3 ⊕ y 3
(22)
(x y)(x2 ⊕ xy ⊕ y 2 ) = x3 y 3
(23)
α-deformed number ( shortly α-number)
From the associativity of ⊕, we have the following formula :
1/α
|1 ⊕ 1 ⊕ 1{z⊕ · · · ⊕ 1} = (n)α = n
(24)
n times
Here we call (n)α a α-number of n, where nα reduces to n when α goes to 1.
For real number x, we can define the α-number (x)α as follows:
(
x1/α (x > 0)
1/α−1
(x)α = |x|
x=
(25)
−(−x)1/α (x < 0)
Here we have the following:
0α = 0,
1α = 1,
(−1)α = −1
(26)
For the α-number, the following holds:
m
n
=
α
(m)α
(n)α
(27)
Proposition 2.2 For two α-numbers (x)α and (y)α , the following holds:
(x)α ⊕ (y)α = (x + y)α
(28)
(x)α (y)α = (x − y)α
(29)
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Jae Yoon Kim et al.
Proof. For x > 0, y > 0, the eq.(28) can be written as
(x)α ⊕ (y)α = x1/α ⊕ y 1/α
= (x1/α )α + (y 1/α )α
= (x + y)α
1/α
(30)
For x > 0, y < 0, x + y > 0, the eq.(28) can be written as
(x)α ⊕ (y)α = x1/α ⊕ (−(−y)1/α )
= x1/α (−y)1/α
= (x1/α )α − ((−y)1/α )α
= (x + y)α
1/α
(31)
For x > 0, y < 0, x + y < 0, the eq.(28) can be written as
(x)α ⊕ (y)α = x1/α ⊕ (−(−y)1/α )
= x1/α (−y)1/α
= − −(x1/α )α + ((−y)1/α )α
= (x + y)α
1/α
(32)
For x < 0, y > 0, x + y > 0, the eq.(28) can be written as
(x)α ⊕ (y)α = −(−x)1/α ⊕ y 1/α
= y 1/α (−x)1/α
= (y 1/α )α − ((−x)1/α )α
= (x + y)α
1/α
(33)
For x < 0, y > 0, x + y < 0, the eq.(28) can be written as
(x)α ⊕ (y)α = −(−x)1/α ⊕ y 1/α
= y 1/α (−x)1/α
= − −(y 1/α )α + ((−x)1/α )α
= (x + y)α
1/α
(34)
For x < 0, y < 0, the eq.(28) can be written as
(x)α ⊕ (y)α = −(−x)1/α ⊕ (−(−y)1/α )
= −((−x)1/α ⊕ (−y)1/α )
= − ((−x)1/α )α + ((−y)1/α )α
= (x + y)α
1/α
(35)
The α-deformed calculus and some physical applications
43
which completes the proof. For α-deformed addition, we have the identity (0)α = 0 obeying
(x)α ⊕ (0)α = (x)α
(36)
The inverse of (x)α is given by (−x)α because
(x)α ⊕ (−x)α = (0)α
(37)
Proposition 2.3 For n α-numbers (x1 )α , · · · , (xn )α , the following holds:
!
n
n
M
X
(xk )α =
xk
(38)
k=1
k=1
α
Proof. It is simple. 2.3
α-deformed derivative
Now let us define the α-deformed derivative with a help of the α-deformed
addition and α-deformed subtraction.
Definition 2.2 The α-deformed derivative ( or α-derivative ) is defined as
follows:
F (x ⊕ h) F (x)
(39)
Dxα F (x) = lim
h→0
h
or
F (y) F (x)
Dxα F (x) = lim
(40)
y→x
yx
Proposition 2.4 If F (x) is increasing in the neighborhood of x, we have
1/α
Dxα F (x) = |x|1−α |F (x)|α−1 F 0 (x)
(41)
If F (x) is decreasing in the neighborhood of x, we have
1/α
Dxα F (x) = − −|x|1−α |F (x)|α−1 F 0 (x)
(42)
Proof. Let us assume y > x in the definition (40). First let us consider the
case that F (x) is increasing near x. In this case we know F 0 (x) > 0. For
F (y) > F (x) > 0, y > x > 0, we have
1/α
[F (y)]α − [F (x)]α
α
lim
Dx F (x) =
y→x
y α − xα
1/α
α[F (y)]α−1 F 0 (y)
=
lim
y→x
αy α−1
1/α
= x1−α [F (x)]α−1 F 0 (x)
(43)
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Jae Yoon Kim et al.
For F (x) < F (y) < 0, y > x > 0, we have
Dxα F (x)
1/α
[−F (x)]α − [−F (y)]α
=
lim
y→x
y α − xα
1/α
α[−F (y)]α−1 F 0 (y)
=
lim
y→x
αy α−1
1/α
= x1−α [−F (x)]α−1 F 0 (x)
(44)
For F (y) > F (x) > 0, x < y < 0, we have
Dxα F (x)
1/α
[F (y)]α − [F (x)]α
=
lim
y→x (−x)α − (−y)α
1/α
α[F (y)]α−1 F 0 (y)
=
lim
y→x
α(−y)α−1
1/α
= (−x)1−α [F (x)]α−1 F 0 (x)
(45)
For F (x) < F (y) < 0, x < y < 0, we have
Dxα F (x)
1/α
[−F (x)]α − [−F (y)]α
=
lim
y→x
(−x)α − (−y)α
1/α
α[−F (y)]α−1 F 0 (y)
=
lim
y→x
α(−y)α−1
1/α
= (−x)1−α [−F (x)]α−1 F 0 (x)
(46)
which completes the proof of the eq.(41).
Now let us consider the case that F (x) is decreasing near x. In this case
we know F 0 (x) < 0. For F (x) > F (y) > 0, y > x > 0, we have
F (x) F (y)
Dxα F (x) = − lim
y→x
yx
1/α
[F (x)]α − [F (y)]α
= − lim
y→x
y α − xα
1/α
−α[F (y)]α−1 F 0 (y)
= − lim
y→x
αy α−1
1/α
= − −x1−α [F (x)]α−1 F 0 (x)
(47)
The α-deformed calculus and some physical applications
45
For F (x) > F (y) > 0, x < y < 0, we have
F (x) F (y)
y→x (−x) (−y)
1/α
[F (x)]α − [F (y)]α
= − lim
y→x (−x)α − (−y)α
1/α
−α[F (y)]α−1 F 0 (y)
= − lim
y→x
α(−y)α−1
1/α
= − −(−x)1−α [F (x)]α−1 F 0 (x)
Dxα F (x) = − lim
(48)
For F (y) < F (x) < 0, y > x > 0, we have
(−F (y)) (−F (x))
Dxα F (x) = − lim
y→x
yx
1/α
[−F (y)]α − [−F (x)]α
= − lim
y→x
y α − xα
1/α
−α[−F (y)]α−1 F 0 (y)
= − lim
y→x
αy α−1
1/α
= − −x1−α [−F (x)]α−1 F 0 (x)
(49)
For F (y) < F (x) < 0, x < y < 0, we have
Dxα F (x) = − lim
y→x
(−F (y)) (−F (x))
(−x) (−y)
1/α
[−F (y)]α − [−F (x)]α
= − lim
y→x
(−x)α − (−y)α
1/α
−α[−F (y)]α−1 F 0 (y)
= − lim
y→x
α(−y)α−1
1/α
= − −(−x)1−α [−F (x)]α−1 F 0 (x)
(50)
which completes the proof. The α-derivative is not linear because Dxα [F (x) + G(x)] 6= Dxα F (x) +
Dxα G(x). Instead, it is α-linear because
Dxα (F (x) ⊕ G(x)) = Dxα F (x) ⊕ Dxα G(x)
(51)
For any real number a ∈ R, we have
Dxα (aF (x)) = aDxα F (x)
(52)
46
Jae Yoon Kim et al.
Proposition 2.5 For the α -derivative, the following Leibniz rule holds:
Dxα [F (x)G(x)] = (Dxα F (x))G(x) ⊕ F (x)(Dxα G(x))
(53)
Proof. Let us assume that y > x > 0 and F (y) > F (x) > 0, G(y) > G(x) > 0.
Then, form the definition of α-derivative, we get
1/α
[F (y)]α [G(y)]α − [F (x)]α [G(x)]α
α
Dx [F (x)G(x)] =
lim
y→x
y α − xα
1/α
= x1−α [F (x)]α−1 F 0 (x)[G(x)]α + x1−α [F (x)]α [G(x)]α−1 G0 (x)
= ([Dxα F (x)]α [G(x)]α + [F (x)]α [Dxα G(x)]α )1/α
= (Dxα F (x))G(x) ⊕ F (x)(Dxα G(x))
(54)
For another cases, one can prove the Leibniz rule in a similar way. Thus, we
completed the proof. From the Leibniz rule, we have the following commutation relation:
Dxα x xDxα = 1
(55)
Dxα x = 1 ⊕ xDxα
(56)
or
Proposition 2.6 For the α -derivative, the following holds:
Dxα xn = nα xn−1 ,
n = 0, 1, 2, · · ·
(57)
Proof. Let us assume that the eq.(57) holds for n − 1. Then, for n we have
Dxα xn =
=
=
=
(1 ⊕ xDxα )xn−1
xn−1 ⊕ x(n − 1)α xn−2
(1 ⊕ (n − 1)α )xn−1
nα xn−1
(58)
which completes the proof. We can also prove this proposition from the definition of the fractional derivative. For x > 0, xn is increasing, so we have
Dxα xn = (|x|1−α |xn |α−1 nxn−1 )1/α
= nα xn−1
(59)
For x < 0, we should be careful in applying the definition of fractional derivative because x2m+1 is increasing but x2m is decreasing. For x2m+1 , we get
Dxα x2m+1 = (|x|1−α |x2m+1 |α−1 (2m + 1)x2m )1/α
= (2m + 1)α |x|2m
= (2m + 1)α x2m
(60)
The α-deformed calculus and some physical applications
47
For x2m , we get
Dxα x2m = −(−|x|1−α |x2m |α−1 (2m)x2m−1 )1/α
= −(2m)α |x|2m−1
= (2m)α x2m−1
(61)
which completes the proof. Proposition 2.7 For the α-derivative, the following holds:
Dxα (xn G(x)) = (Dxα xn )G(x) ⊕ xn Dxα G(x)
(62)
Proof. Let us assume that the eq.(62) holds for n. Then, for n + 1 we have
Dxα (xn+1 G(x)) =
=
=
=
=
(1 ⊕ xDxα )xn G(x)
xn G(x) ⊕ x(Dxα xn )G(x) ⊕ xn+1 Dxα G(x)
(1 ⊕ nα )xn G(x) ⊕ xn+1 Dxα G(x)
(n + 1)α xn G(x) ⊕ xn+1 Dxα G(x)
(Dxα xn )G(x) ⊕ xn Dxα G(x)
(63)
which completes the proof. Proposition 2.8 For the α-derivative, the following Leibnitz rule holds:
Dxα [F (x)G(x)] = (Dxα F (x))G(x) ⊕ F (x)(Dxα G(x))
1
Dxα G(x)
]=
G(x)
G(x)2
(65)
F (x)
(Dxα F (x))G(x) F (x)Dxα G(x)
]=
G(x)
G(x)2
(66)
Dxα [
Dxα [
(64)
Proof. It is simple. Proposition 2.9 For the α-derivative, the following chain rule holds:
Dxα [F (x)]n = (n)α [F (x)]n−1 Dxα F (x)
Proof. It is easy. (67)
48
2.4
Jae Yoon Kim et al.
α-deformed Integral
In this subsection we find the α-deformed integral as an inverse operation of the
α-deformed derivative and investigate some properties of fractional integral.
Definition 2.3 The α-deformed integral from 0 to x ( x > 0 ) is defined as
follows: When F (x) > 0, the α-deformed integral from 0 to x ( x > 0 ) is
given by
1/α
Z x
α−1
α
α
(68)
αdx|x| |F (x)|
I0|x F (x) =
0
When F (x) < 0, the fractional integral from 0 to x ( x > 0 ) is given by
α
I0|x
F (x)
Z
=− −
x
α−1
αdx|x|
α
1/α
|F (x)|
(69)
0
Proposition 2.10 For the α-deformed integral the following holds:
α
Dxα I0|x
F (x) = F (x)
(70)
where x > 0
Proof. For F > 0, F 0 > 0, we have
α
α
α
Dxα I0|x
F (x) = [|x|1−α |I0|x
F (x)|α−1 (I0|x
F (x))0 ]1/α
= (F (x)α )1/α = F (x)
(71)
α
For F > 0, F 0 < 0, we know (I0|x
F (x))0 > 0, so we have
α
α
α
Dxα I0|x
F (x) = [|x|1−α |I0|x
F (x)|α−1 (I0|x
F (x))0 ]1/α
= (F (x)α )1/α = F (x)
(72)
α
For F < 0, F 0 > 0, we know (I0|x
F (x))0 < 0, so we have
α
α
α
F (x) = −[−|x|1−α |I0|x
F (x)|α−1 (I0|x
F (x))0 ]1/α
Dxα I0|x
= −[(−F (x))α ]1/α = F (x)
(73)
α
For F < 0, F 0 < 0, we know (I0|x
F (x))0 < 0, so we have
α
α
α
Dxα I0|x
F (x) = −[−|x|1−α |I0|x
F (x)|α−1 (I0|x
F (x))0 ]1/α
= −[(−F (x))α ]1/α = F (x)
which completes the proof. (74)
49
The α-deformed calculus and some physical applications
Proposition 2.11 For the α-deformed integral the following holds:
α
I0|x
Dxα F (x) = F (x) F (0)
(75)
The above relation is written as follows:
1. For F > 0, F 0 > 0, we have
α
I0|x
Dxα F (x) = [(F (x))α − (F (0))α ]1/α
(76)
2. For F > 0, F 0 < 0, we have
α
I0|x
Dxα F (x) = − [(F (0))α − (F (x))α ]1/α
(77)
3. For F < 0, F 0 > 0, we have
α
I0|x
Dxα F (x) = [(−F (0))α − (−F (x))α ]1/α
(78)
4. For F < 0, F 0 < 0, we have
α
I0|x
Dxα F (x) = − [(−F (x))α − (−F (0))α ]1/α
(79)
Proof.
1. For F > 0, F 0 > 0, we have
α
I0|x
Dxα F (x)
x
Z
α−1
αdx|x|
=
|Dxα F (x)|α
1/α
0
x
Z
α−1
αdx|F (x)|
=
1/α
F (x)
0
0
= [(F (x))α − (F (0))α ]1/α
(80)
2. For F > 0, F 0 < 0, we know Dxα F (x) < 0, so we have
1/α
Z x
α−1
α
α
α
α
I0|x Dx F (x) = −
αdx|x| |Dx F (x)|
0
Z
= − −
x
α−1
αdx|F (x)|
1/α
F (x)
0
0
= − [(F (0))α − (F (x))α ]1/α
(81)
3. For F < 0, F 0 > 0, we know Dxα F (x) > 0, so we have
Z x
1/α
α
α
α−1
α
α
I0|x Dx F (x) =
αdx|x| |Dx F (x)|
0
Z
=
−
x
α−1
αdx(−F (x))
1/α
(−F (x))
0
0
= [(−F (0))α − (−F (x))α ]1/α
(82)
50
Jae Yoon Kim et al.
4. For F < 0, F 0 < 0, we know Dxα F (x) < 0, so we have
α
I0|x
Dxα F (x)
Z
x
α−1
= −
αdx|x|
|Dxα F (x)|α
1/α
0
Z
= − −
x
α−1
αdx|F (x)|
1/α
F (x)
0
0
= − [(−F (x))α − (−F (0))α ]1/α
(83)
which completes the proof. Proposition 2.12 For the α-deformed integral the following holds:
α
α
I0|x
(aF (x)) = |a|[I0|x
F (x)]
(84)
where x > 0 and a ∈ R.
Proof. It is easy. Proposition 2.13 The α-deformed integral is α-linear; For two functions
F (x), G(x), we have
α
α
α
I0|x
(F (x) ⊕ G(x)) = [I0|x
F (x)] ⊕ [I0|x
G(x)]
(85)
where x > 0.
Proof. It is easy. Proposition 2.14 When a < b < c and F (x) ≥ 0 or F (x) ≤ 0 in [a, c], for
the α-deformed integral the following holds:
α
α
α
Ia|c
F (x) = [Ia|b
F (x)] ⊕ [Ib|c
F (x)]
(86)
Proof. For F (x) ≥ 0 in [a, c] we have
1/α
α
α
F (x)]α
[Ia|b
F (x)]α + [Ib|c
Z b
1/α
Z c
α−1
α
α−1
α
=
αdx|x| |F (x)| +
αdx|x| |F (x)|
α
α
[Ia|b
F (x)] ⊕ [Ib|c
F (x)] =
a
Z
=
b
c
αdx|x|α−1 |F (x)|α
1/α
a
α
α
= [Ia|b
F (x)] ⊕ [Ib|c
F (x)]
(87)
51
The α-deformed calculus and some physical applications
For F (x) ≤ 0 in [a, c] we have
1/α
α
α
α
α
F (x)]α
F (x)]α + [−Ib|c
F (x)] = − [−Ia|b
F (x)] ⊕ [Ib|c
[Ia|b
1/α
Z b
Z c
α−1
α
α−1
α
αdx|x| |F (x)|
αdx|x| |F (x)| −
= − −
b
a
Z c
1/α
α−1
α
= − −
αdx|x| |F (x)|
a
α
α
= [Ia|b
F (x)] ⊕ [Ib|c
F (x)]
(88)
which completes the proof. Proposition 2.15 When a > 0, for the α-deformed integral the following
holds:
α
α
I−a|a
F (x) = I0|a
[F (x) ⊕ F (−x)]
(89)
If F (−x) ⊕ F (x) = 0 in [−a, a], we have
α
I−a|a
F (x) = 0
(90)
In this case F (x) is called a α-odd function. If F (−x) = −F (x) in [−a, a], we
have
α
α
I−a|a
F (x) = (2α )I0|a
F (x)
(91)
In this case F (x) is called a α-even function.
Proof. It is easy. Proposition 2.16 When x > 0 and n = 0, 1, 2, · · · , for the α-deformed integral the following holds:
xn+1
α
I0|x
xn =
(92)
(n + 1)α
α
I−x|0
x2n =
x2n+1
(2n + 1)α
α
I−x|0
x2n+1 = −
x2n+2
(2n + 2)α
(93)
(94)
Proof. It is easy. 3
α-deformed exponential, α-deformed logarithm and α-deformed trigonometry
In this section we find the α-deformed exponential, α-deformed logarithm and
α-deformed trigonometry and investigate their properties.
52
3.1
Jae Yoon Kim et al.
α-deformed exponential
Definition 3.1 The α-exponential function is defined from the following relation:
Dxα eα (x) = eα (x)
(95)
Proposition 3.1 The α-Taylor expansion is given by
F (x) =
∞
M
an x n
(96)
k=0
where
an =
1
(Dxα )n F (x)|x=0
nα !
(97)
Proof. It is simple. Proposition 3.2 The α-exponential has the following expression:
∞
M
xn
eα (x) =
(n)α !
n=0
(98)
Proof. Let us consider the case that x > 0. If we set
eα (x) =
∞
M
an x n
(99)
n=0
From Dxα eα (x) = eα (x), we get
(n)α an = an−1
(100)
which completes the proof. Proposition 3.3 The α-exponential can also be written as
1
eα (x) = e α |x|
or
α−1 x
(101)
(
(x > 0)
exp α1 xα
eα (x) =
1
α
exp − α (−x)
(x < 0)
(102)
Proof. Let us consider the case that x > 0. From the eq.(99), we get
α
[eα (x)] =
∞ n α
X
x
n=0
nα !
=
∞
X
xαn
n=0
n!
= ex
α
(103)
53
The α-deformed calculus and some physical applications
Let us consider the case that x < 0. From the eq.(99), we get
∞
M
xn
eα (x) =
n !
n=0 α
=
∞
M
(−1)n (−x)n
nα !
n=0
(−x)2 (−x)3
⊕ ···
= 1 (−x) ⊕
2α !
3α !
(104)
Thus, we have
α
[eα (x)]
α α
(−x)2
(−x)3
= 1 − (−x) +
−
− +···
2α !
3α !
(−x)2α (−x)3α
= 1 − (−x)α +
−
− +···
2!
3!
α
= e−(−x)
α
(105)
which completes the proof. Proposition 3.4 The inverse of a α- exponential called a α-logarithm is given
by
(
(α ln x)1/α
(x > 1)
lnα (x) =
(106)
1/α
− (−α ln x)
(0 < x < 1)
Indeed we know
eα (lnα (x)) = lnα (eα (x)) = x
(107)
Proof. It is simple. The α-exponential function obeys the following relations:
eα (x)eα (y) = eα (x ⊕ y)
eα (x)/eα (y) = eα (x y)
(108)
The α-logarithm obeys the following relations:
lnα (xy) = lnα x ⊕ lnα y
x
= lnα x lnα y
lnα
y
(109)
Proposition 3.5 For the α-exponential, w have
Dxα eα (ax) = aeα (ax)
Proof. It can be easily proved by using ∂x
1
|x|α−1 x
α
(110)
= |x|α−1 . 54
3.2
Jae Yoon Kim et al.
α-deformed trigonometric function
Definition 3.2 The α-deformed sine and cosine function is defined as follows:
cα (x) = | cos(|x|α−1 x)|1/α−1 cos(|x|α−1 x), sα (x) = | sin(|x|α−1 x)|1/α−1 sin(|x|α−1 x)
(111)
The eq.(111) can be written as follows:
(
[cos(|x|α−1 x)]1/α (if
cα (x) =
−[− cos(|x|α−1 x)]1/α
(
[sin(|x|α−1 x)]1/α (if
sα (x) =
−[− sin(|x|α−1 x)]1/α
cos(|x|α−1 x) > 0)
(if cos(|x|α−1 x) < 0)
(112)
sin(|x|α−1 x) > 0)
(if sin(|x|α−1 x) < 0)
(113)
Proposition 3.6 For the α-deformed sine and cosine function, the following
holds:
|sα (x)|2α + |cα (x)|2α = 1
(114)
From the eq.(114) we have the following :
s (0) =
α π 1/α
=
sα
2
h
i
sα (π)1/α =
" #
1/α
3π
sα
=
2
i
h
1/α
=
sα (2π)
0,
cα (0) = 1
π 1/α
1,
cα
=0
2
h
i
0,
cα (π)1/α = −1
" #
1/α
3π
−1,
cα
=0
2
i
h
1/α
=1
0,
cα (2π)
(115)
Proof. It is easy. The α-deformed trigonometric functions obey the following addition formulas:
sα (x ⊕ y) = sα (x)cα (y) ⊕ cα (x)sα (y)
sα (x y) = sα (x)cα (y) cα (x)sα (y)
cα (x ⊕ y) = cα (x)cα (y) sα (x)sα (y)
cα (x y) = cα (x)cα (y) ⊕ sα (x)sα (y)
(116)
Proposition 3.7 For the α-deformed trigonometric functions, the following
holds;
(117)
Dxα cα (x) = −sα (x)
Dxα sα (x) = cα (x)
(118)
55
The α-deformed calculus and some physical applications
Proof. It is easy. From the definition of the α-deformed exponential we have the following
relations:
1
eα (ix) = e α |ix|
α−1 ix
= ei
|x|α−1 x
α
1
eα (−iα x) = e α |−ix|
,
α−1 (−ix)
= e−i
|x|α−1 x
α
(119)
The eq.(119) can be written as
[eα (ix)]α = ei|x|
α−1 x
[eα (−ix)]α = e−i|x|
= cos |x|α−1 x + i sin |x|α−1 x
α−1 x
= cos |x|α−1 x − i sin |x|α−1 x
(120)
(121)
Proposition 3.8 The relation between the α-deformed exponential and αdeformed trigonometric functions are given by
eα (ix) = cα (x) ⊕ isα (x),
eα (−ix) = cα (x) isα (x)
(122)
where
(a ⊕ bi)α = (a)α + (b)α i,
(a, b ∈ R)
(123)
The α-series expansion of α-deformed trigonometric functions are given by
cα (x) =
∞
M
(−1)n x2n
n=0
(2n)α !
,
sα (x) =
∞
M
(−1)n x2n+1
n=0
(2n + 1)α !
(124)
Proof. It is easy. 3.3
α-deformed polar coordinate
Now let us find the α-deformed polar coordinate. Let us denote the α-deformed
radial coordinate and the α-deformed angular coordinate by rα and θα , respectively. Then, we have
x = rα cα (θα ),
y = rα sα (θα )
where
2α
rα = (|x|
2α 1/2α
+ |y| )
,
θ=
t−1
α
(125)
y
x
(126)
and tα (x) = sα (x)/cα (x).
We remark that the α-length rα is different from a ordinary length r unless
α = 1, instead, we have the following relation
rα = r(| sin θ|2α + | cos θ|2α )1/2α
(127)
56
Jae Yoon Kim et al.
so the α-length varies with polar angle. For 0 < θα <
between the α-deformed polar angle and polar angle is
π 1/α
,
2
θα = tan−1 (tan θ)α
1/α
the relation
(128)
Using Cartesian coordinates, an α-deformed infinitesimal area element can be
calculated as dAα = dxα dyα . The substitution rule for multiple α-integrals
states that, when using the α-deformed polar coordinate, the α-deformed Jacobian determinant of the coordinate conversion formula has to be considered:
Drα x Drα y = rα
Jα = (129)
Dθα x Dθα y α
where α-determinant is defined as
a b c d = ad bc
(130)
α
Thus, the α-deformed infinitesimal area element in the α-deformed polar coordinate is given by
dAα = rα drα dθα
(131)
Let us consider the equation of the α-deformed circle:
|x|2α + |y|2α = Rα2α ,
Rα > 0,
(132)
The α-deformed area of the α-deformed circle with α-deformed radius Rα is
then given by
Aα = π 1/α Rα2
(133)
which reduces to πR12 in the limit α → 1.
3.4
α-deformed complex number
Now let us discuss the α-deformed complex numbers. We can define the αdeformed complex number z and its conjugate as follows;
z = x ⊕ iy = rα eα (iθα ) = rα cα (θα ) ⊕ irα sα (θα ),
(134)
z ∗ = x iy = rα eα (−iθα ) = rα cα (θα ) irα sα (θα ),
(135)
The norm |z|α of the α-deformed complex number z is given by
|z|2α = zz ∗ = rα2 = x2 ⊕ y 2
which is a α-length in two dimension.
(136)
The α-deformed calculus and some physical applications
57
Let us introduce the α-deformed complex derivatives
Dzα =
1
(Dα iDyα ),
(2)α x
Dzα∗ =
1
(Dα ⊕ iDyα )
2α x
(137)
Then, we have
Dzα z = 1, Dzα z ∗ = 0,
Dz∗ z n = 0,
3.5
Dzα∗ z = 0, Dzα∗ z ∗ = 1
Dz z n = (n)α z n−1
(138)
(139)
α-deformed rotation
Now let us the α-deformed rotation in xy plane which makes rα invariant. For
two n × n square matrix A, B, let us define the α-deformed product of A and
B by
n
M
(AB)αij =
Aik Bkj
(140)
k=1
Then, the α-deformed length squared is described by
where
rα2 = (X T X)α
(141)
x
X=
y
(142)
and X T means the transpose of X. The α-deformed length invariance implies
(rα0 )2 = r2α
(143)
(x0 )2 ⊕ (y 0 )2 = x2 ⊕ y 2
(144)
or
The α-deformed rotation is given by
0 x
cα (θα ) sα (θα )
x
=
,
y0
−sα (θα ) cα (θα )
y
(145)
where the α-deformed rotation matrix forms the SOα (2) group.
4
α-deformed Laplace transformation
In this section we introduce α-deformed Laplace transformation and investigate
its properties.
58
Jae Yoon Kim et al.
Definition 4.1 α-deformed Laplace transform is defined by
α
Lα (F (x)) = I0|∞
eα (−sx)F (x),
(s > 0)
(146)
where s is assumed to be sufficiently large.
Limiting α → 1, the eq.(146) reduces to an ordinary Laplace transform.
Proposition 4.1 For the α-deformed Laplace transform, the following holds:
Lα (aF (x)) = |a|Lα (F (x))
(147)
Lα (F (x) ⊕ G(x)) = Lα (F (x)) ⊕ Lα (G(x))
(148)
Proof. It is easy. Proposition 4.2 For F (x) = xN , (N = 0, 1, 2, · · · ) , we have the following
result:
Nα !
(149)
Lα (xN ) = n+1
s
Proof. It is easy. Proposition 4.3 For sufficiently large s , the following holds:
Lα (eα (ax)) =
1
sa
(150)
Proof. From the definition we have
Z ∞
1/α
α−1
α
αdxx |eα (sx)eα (ax)|
Lα (eα (ax)) =
0
Z
∞
=
αdxx
α−1
α
αdxx
α−1 −(sa)α xα
1/α
[eα (−(s a)x)]
0
Z
∞
=
1/α
e
0
=
1
sa
(151)
which completes the proof. Proposition 4.4 When a > 0, for the α-deformed trigonometric functions,
we have
1/α
s
sα
Lα (cα (ax)) = 2
=
(152)
s a2
s2α + a2α
1/α
a
aα
Lα (sα (ax)) = 2
=
(153)
s a2
s2α + a2α
The α-deformed calculus and some physical applications
59
Proof. We have
eα (iα x) ⊕ eα (−iα x)
Lα (cα (ax)) = Lα
2α
1
1
1
⊕
=
2α s iα a s ⊕ iα a
1
s
=
2α s2 a2
(154)
The proof of the eq.(153) is similar. 5
5.1
Applications
α-deformed chemical reaction equation
In chemical kinetics, relaxation methods are used for the measurement of very
fast reaction rates. The ordinary relaxation function I(t) obeying the relaxation equation
d
1
I(t) = − I(t),
(155)
dt
τ0
which gives
t
I(t) = I(0) exp −
(156)
τ0
The above decay function was used by Becquerel in the course of study on
time evolution of luminescence.
With a help of the α-deformed calculus, the relaxation equation is turned
into
1
(157)
Dtα I(t) = − I(t)
τ0
which gives
α 1 t
t
= exp −
,
(158)
I(t) = I(0)eα
τ0
α τ0
5.2
α-deformed version of the Euler-Lagrange equation
To derive the α-deformed version of the Euler-Lagrange equation let us introduce the following α-deformed functional
α
f (y(x), Dxα y(x))
J α [y] = I0|x
0
(159)
We will find the condition that J α [y] has a local minimum. To do so, we
consider the new α-deformed functional depending on the parameter α
J α [] = I0|x
f (Y (x, ), Dxα Y (x, )),
0
(160)
60
Jae Yoon Kim et al.
where
Y (x, ) = y(x) ⊕ η(x)
(161)
Dxα Y (x, ) = Dx y(x) ⊕ Dxα η(x)
(162)
η(0) = η(x0 ) = 0
(163)
and
Then, the condition for an extreme value is that
[Dα J α []]=0 = 0,
(164)
From the chain rule of the α-deformed derivative, we have
α
α
α
α
Dα J[] = I0|x
[DYα f Dα Y ⊕ DD
α f D (Dx Y )]
0
xY
α
α
α
= I0|x
[DYα f η ⊕ DD
α f Dx η] = 0
0
xY
(165)
Using the integration by parts, we get
α
I0|x0 [(DYα f Dxα (DD
α ))η(x)] = 0,
xY
(166)
which gives the α-deformed Euler-Lagrange equation
α
Dyα f Dxα (DD
α ) = 0
xy
5.3
(167)
α-deformed version of Newton’s equation
Now let us apply the α-deformed calculus of variations to the α-deformed
version of the classical mechanics which we call the α-deformed mechanics. In
this case we consider the α-deformed action
α
S = I0|t
L(x(t), Dtα x(t)),
(168)
where L is the α-deformed Lagrangian. The minimum condition of the αdeformed action gives
∂L
α
α
(169)
Dx L = Dt
∂Dtα x
In the similar way as the ordinary mechanics, we introduce the following Lagrangian
1
L = m(Dtα x)2 V (x),
(170)
2
where m is a mass and V (x) is a potential energy. From the eq.(169) and the
eq.(170), we have the α-deformed equation of motion
m(Dtα )2 x = −Dxα V (x)
(171)
The α-deformed calculus and some physical applications
61
For free motion, we have V (x) = 0; so we know that mDtα x is a constant of
motion. Generally, we can define the α-deformed canonical momentum p as
p=
∂L
∂Dtα x
(172)
Thus, for the Lagrangian of the type (170), we have
p=
∂L
= mDtα x,
α
∂Dt x
(173)
which we will call the α-deformed linear momentum. The α-deformed equation
of motion then reads
F = Dtα p = −Dxα V (x)
(174)
The α-deformed hamiltonian H is obtained from the Legendre transformation
as follows:
H(p, x) = pDtα x L
(175)
For the Lagrangian of the type (170), we have
H(p, x) =
p2
⊕ V (x)
2m
(176)
From the definition of the α-deformed linear momentum, we can define the
α-deformed velocity v as
v(t) = Dtα x(t)
(177)
and the α-deformed acceleration a as
a(t) = Dtα v(t) = (Dtα )2 x(t),
(178)
Thus, the α-deformed version of the Newton’s law reads
F = ma = mDtα v = m(Dtα )2 x(t)
5.3.1
(179)
α-deformed harmonic oscillator
Now let us introduce the α-deformed harmonic oscillator problem whose Lagrangian is given by
1
1
(180)
L = m(Dtα x)2 − mw02 x2
2
2
The α-deformed equation of motion then reads
m(Dtα )2 x = −mw02 x
(181)
with the following initial condition
x(0) = A, Dtα x(0) = 0
(182)
62
Jae Yoon Kim et al.
Solving the eq.(181), we get
x(t) = Acα (w0 t) ,
(183)
If we denote by T the α-deformed period defined by x(t ⊕ T ) = x(t), we have
w0 =
5.4
(2π)1/α
T
(184)
α-deformed quantum mechanics
The α-deformed quantum mechanics starts with the following representation
x̂ ⇔ x,
p̂ ⇔
~
Ĥ ⇔ − Dtα
i
~ α
D ,
i x
(185)
where Ĥ is a α-deformed Hamiltonian operator. In the α-deformed quantum
mechanics, the commutator of the α-position operator x̂ and α-momentum
operator p̂ is replaced with the following α-commutator
[p̂, x̂] =
~
i
(186)
where
[A, B] = AB BA
The eq.(185) gives the following fractional Schrödinger equation
~2
α
α 2
i~Dt ψ = −
(D ) ⊕ Vα (x) ψ
2m x
(187)
(188)
where we assume that the α-Hamiltonian is defined by the α-sum of the kinetic
energy and potential energy. From the eq.(188) we can obtain the following
relation
Dtα ρα (x, t) ⊕ Dxα jα (x, t) = 0,
(189)
where the α-deformed probability density ρα (x, t) is given by
ρα (x, t) = ψ ∗ ψ
(190)
and the α-deformed probability flux jα (x, t) is given by
jα (x, t) =
If we set ψ(x, t) = eα
equation as follows;
i
Et
~
~
(ψ ∗ Dxα ψ ψDxα ψ ∗ )
2mi
(191)
u(x), we have the time-independent Schrödinger
~2
α 2
−
(D ) ⊕ Vα (x) u = Eu
2m x
(192)
63
The α-deformed calculus and some physical applications
In the Hilbert space related to one-dimensional α-deformed quantum mechanics, the α-deformed inner product is given by
∞
hf |giα = I−∞
g ∗ (x)f (x)
(193)
The α-deformed expectation value of a physical operator O with respect to
the state u(x) is defined by
∞
u∗ (x, t)Ou(x, t)
hOiα = hu|Ouiα = I−∞
(194)
and O is a Hermitian operator if it obeys
hOu|uiα = hu|Ouiα
(195)
We can easily check that both α-deformed position operator and α-deformed
momentum operator is Hermitian.
5.4.1
Infinite potential well
The infinite potential well describes a particle free to move in a small space
surrounded by impenetrable barriers. The potential energy in this model is
given as
(
0 (0 < x < L)
V (x) =
(196)
∞ (x < 0, x > L)
The wave function u(x) can be found by solving the generalized conformable
fractional time-independent Schrodinger equation for the system:
~2
(Dxα )2 u = Eu
−
2m
(197)
Solving the eq.(197), we get
r
u(x) = Acα
2mE
x
~2
!
r
⊕ Bsα
From u(0) = 0, we have A = 0, so we have
r
u(x) = Bsα
From u(L) = 0 we get
r
2mE
L = (nπ)1/α ,
2
~
2mE
x
~2
2mE
x
~2
!
(198)
!
n = 1, 2, 3, · · ·
(199)
(200)
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Jae Yoon Kim et al.
Thus, the energy levels are
~2 (nπ)2/α
, n = 1, 2, 3, · · ·
2mL2
and the normalized wave functions are
r
21/α
(nπ)1/α
sα
x
un =
L
L
Here, the probability density is given by
2
21/α (nπ)1/α Pn =
x sα
L L
(E)n =
(201)
(202)
(203)
Fig.1 and Fig.2 shows the plot of P1 and P2 for α = 1 (red), α = 0.9 (green),
α = 0.8 (blue), α = 1.1 (yellow) and α = 1.2 (brown) when L = 1. We know
the following:
1. When 0 < α < 1 the peak position of the probability moves left and the
values of peaks increases as the value of α decreases.
2. When α > 1 the peak position of the probability moves right and the
values of peaks decreases as the value of α increases.
Now let us compute the expectation values. The expectation values of position
is
1
(204)
hx̂in = 1/α L
2
and
1/α
1
1
2
−
hx̂ in =
L2
(205)
3 2n2 π 2
The expectation value of momentum is
~2 (nπ)2/α
L2
If we define the α-deformed uncertainty in x and p as
p
p
∆α x = hx̂2 i hx̂i2 , ∆α p = hp̂2 i hp̂i2
hp̂in = 0,
hp̂2 in =
we have the following uncertainty relation:
2 2
1/2α
nπ
1
∆α x∆α p = ~
−
12
2
(206)
(207)
(208)
Fig.3 shows the plot of ∆α x∆α p/~ versus n for α = 1 (red), α = 0.9 (green),
α = 0.8 (blue), α = 1.1 (yellow) and α = 1.2 (brown) when ~ = 1. This
product increases with increasing n, having a minimum value for n = 1. The
value of this product for n=1 is about equal to 0.568~ for α = 1, 0.493~ for
α = 0.8, 0.533~ for α = 0.9, 0.598~ for α = 1.1 and 0.624~ for α = 1.2.
Thus, we know that the value of ∆α x∆α p for the ground state increases with
increasing α. Especially, for α ≈ 0.8164, the value of ∆α x∆α p is nearly the
same as 21 ~.
The α-deformed calculus and some physical applications
5.4.2
65
Harmonic oscillator
The Schrödinger equation for the α-deformed harmonic oscillator is then given
by
1
~2
α 2
2 2
(209)
(D ) ⊕ mw x u = Eu
−
2m x
2
or
h
mw i
α 2
2
(Dx ) x ⊕ u=0
(210)
~
where
2mE
=
(211)
~2
If we set
u(x) = eα (−ax2 )v(x) (a > 0)
(212)
we get
Dxα u(x) = eα (−ax2 ) [−(2)α axv(x) ⊕ Dxα v(x)]
(213)
and
(Dxα )2 u(x) = eα (−ax2 ) (Dxα )2 v(x) (2)2α axDxα v(x) ⊕ ((2)2α a2 x2 (2)α a)v
(214)
Using the eq.(213) and eq.(214) and demanding that the x2 terms vanish, we
have
mw
(215)
a=
(2)α ~
Then, the Schrödinger equation for v(x) is
α 2
(Dx ) (2)2α axDxα ⊕ ( (2)α a) v = 0
(216)
Introducing the variable y through
p
y = (2)α ax =
r
mw
x
~
(217)
we get
~
α 2
α
(Dy ) (2)α yDy ⊕
1 v =0
mw
(218)
If we set
~
1 = (2n)α ,
mw
we have the following energy level
1/α−1
En = 2
n = 0, 1, 2, · · ·
1/α
1
~w n +
,
2
n = 0, 1, 2, . . .
(219)
(220)
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Jae Yoon Kim et al.
The first few energies are
1
~w
2
(3)α
=
~w
2
(5)α
~w
=
2
(7)α
=
~w
2
E0 =
E1
E2
E3
(221)
which shows that the energy is not equidistant. Indeed, we have
3
1
1
1
1/α−1
En+1 − En = 2
~w(n)α 1 F0 − ; ; −
− 1 F0 − ; ; −
(222)
α
2n
α
2n
If we consider the α-difference of two successive energies, we have
En+1 En = 21/α−1 ~w
(223)
which shows that energies are α-equidistant.
For the choice (219), the eq.(218) implies the α-deformed Hermite equation:
α 2
(Dy ) (2)α yDyα ⊕ (2n)α v = 0
(224)
where v(y) = Hnα (y) is a α-deformed Hermite polynomial. The α-deformed
Hermite polynomial is generated from the following relation:
2
g(x, t) = eα (−t )eα ((2)α xt) =
∞
M
H α (y)
n
n=0
(nα )!
tn
(225)
From the generating function we have the following two recurrence relations:
α
Dyα Hnα = (2n)α Hn−1
(y)
(226)
α
α
Hn+1
(2)α xHnα ⊕ (2n)α Hn−1
=0
(227)
One can easily check that above two relations give the α-deformed Hermite
equation. From the expansion of the generating function we have
1/α−1  n
n

b c
b c
2
2
X
(−1)m (n)α !

 X (−1)m (n)α !
Hnα (y) = ((2)α x)n−2m ((2)α x)n−2m 

m=0 (m)α !(n − 2m)α !
(m)
!(n
−
2m)
!
α
α
m=0
(228)
The α-deformed calculus and some physical applications
67
where bxc is a floor function. The first few α-deformed Hermite equations are
H0α (y)
H1α (y)
H2α (y)
H3α (y)
H4α (y)
=
=
=
=
=
1
|2y|1/α−1 2y
|4y 2 − 2|1/α−1 (4x2 − 2)
|8y 3 − 12y|1/α−1 (8y 3 − 12y)
|16y 4 − 48y 2 + 12|1/α−1 (16y 4 − 48y 2 + 12)
(229)
Fig.4,5,6,7 show the plots of H1α (y), H2α (y), H3α (y), H4α (y), respectively, for α =
1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1 (yellow) and α = 1.2 (brown).
Thus, we have the following eigenfunction corresponding to the energy En :
1/4 1
mwx2
mw
α mw
un (x) =
x
,
n = 0, 1, 2, . . .
e
−
H
α
n
(2)nα (n)α ! (π)α ~
(2)α ~
~
(230)
where we used
p
α
I−∞|∞
eα (−x2 ) = (π)α
(231)
Thus, the probabilities are given by
Pn = |un |2
(232)
Fig.8, 9, 10 shows P0 , P1 , P2 , respectively, for α = 1 (red), α = 0.9 (green),
α = 0.8 (blue), α = 1.1 (yellow) and α = 1.2 (brown).
Now let us compute the expectation values for the position and momentum.
For the momentum operator, we have
1/α
1
1/α
2 1/α
hpin = 0, hp in = m~w n +
(233)
2
For the position operator, we have
hxin = 0,
~
hx in =
mw
2
1/α
1
n+
2
Therefore we have the following uncertainty relation
1/α
1
(∆x)n (∆p)n = ~ n +
2
(234)
(235)
For ground state, we have
1/α
1
(∆x)0 (∆p)0 = ~
2
Thus, for 0 < α < 1, we have ∆x∆p <
~
2
(236)
while for α > 1, we have ∆x∆p > ~2 .
68
6
Jae Yoon Kim et al.
Conclusion
In this paper we proposed a new deformed calculus called a α-deformed calculus
where the α-deformed derivative is replaced with
Dxα F (x) = lim
y→x
F (y) F (x)
yx
(237)
We used this new derivative to formulate a new calculus theory. Starting
with the definition of the α-deformed addition and α-deformed subtraction,
we formulate the theory of α-deformed number, α-deformed derivative and
α-deformed integral. We found many properties related to the α-deformed
derivative and α-deformed integral. We also proposed the α-deformed exponential, α-deformed logarithm and α-deformed trigonometry and investigate
their properties. We also found the relation between the α-deformed trigonometric function and α-deformed exponential function. We also discussed the
α-deformed polar coordinate. We discussed the α-deformed Laplace transformation and investigated its properties. As applications we discussed the αdeformed chemical reaction dynamics, α-deformed Euler equation, α-deformed
mechanics and α-deformed quantum mechanics. In the α-deformed mechanics, the energy was shown to be expressed in terms of the α-deformed addition of the kinetic energy and potential energy. In the α-deformed quantum
mechanics, we discussed two examples; Infinite potential well and harmonic
oscillator problem. For the infinite potential well we found that for 0 < α < 1
the peak position of the probability moves left and the values of peaks increases as the value of α decreases while for α > 1 the peak position of the
probability moves right and the values of peaks decreases as the value of α
increases. We also found that the value of ∆α x∆α p for the ground state increases with increasing α. Especially, for α ≈ 0.8164, the value of ∆α x∆α p is
nearly the same as 12 ~. For the harmonic oscillator, we obtained the energy
1/α
level En = 21/α−1 ~w n + 21
,
n = 0, 1, 2, . . . and the wave equation by
using the α-deformed Hermite equation. From computation of the expectation
values for the position and momentum, we found that for 0 < α < 1, we have
∆x∆p < ~2 while for α > 1, we have ∆x∆p > ~2 .
Acknowledgements. This work was supported by the National Research Foundation of Korea Grant funded by the Korean Government (NRF2015R1D1A1A01057792) and by the Gyeongsang National University Fund for
Professors on Sabbatical Leave, 2016.
The α-deformed calculus and some physical applications
69
Figure 1: Plot of P1 for α = 1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1
(yellow) and α = 1.2 (brown) when L = 1.
Figure 2: Plot of P2 for α = 1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1
(yellow) and α = 1.2 (brown) when L = 1.
Figure 3: Plot of ∆α x∆α p/~ versus n for α = 1 (red), α = 0.9 (green), α = 0.8
(blue), α = 1.1 (yellow) and α = 1.2 (brown) when ~ = 1 .
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Jae Yoon Kim et al.
Figure 4: Plot of H1α (y) for α = 1 (red), α = 0.9 (green), α = 0.8 (blue),
α = 1.1 (yellow) and α = 1.2 (brown).
Figure 5: Plot of H2α (y) for α = 1 (red), α = 0.9 (green), α = 0.8 (blue),
α = 1.1 (yellow) and α = 1.2 (brown).
Figure 6: Plot of H3α (y) for α = 1 (red), α = 0.9 (green), α = 0.8 (blue),
α = 1.1 (yellow) and α = 1.2 (brown).
The α-deformed calculus and some physical applications
71
Figure 7: Plot of H4α (y) for α = 1 (red), α = 0.9 (green), α = 0.8 (blue),
α = 1.1 (yellow) and α = 1.2 (brown).
Figure 8: Plot of P0 for α = 1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1
(yellow) and α = 1.2 (brown).
Figure 9: Plot of P1 for α = 1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1
(yellow) and α = 1.2 (brown).
72
Jae Yoon Kim et al.
Figure 10: Plot of P2 for α = 1 (red), α = 0.9 (green), α = 0.8 (blue), α = 1.1
(yellow) and α = 1.2 (brown).
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Received: October 30, 2016; Published: January 6, 2017