Three dimensional Non-commutative Quantum Mechanics with ...

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College of Natural Science, Gyeongsang National University, Jinju 660-701, Korea. Abstract ... gate the quantum mechanics in non-commutative space.
Three dimensional Non-commutative Quantum Mechanics with Generalized uncertainty relation Won Sang Chung∗ Department of Physics and Research Institute of Natural Science, College of Natural Science, Gyeongsang National University, Jinju 660-701, Korea

Abstract In this paper we discuss the three dimensional non-commutative quantum mechanics with generalized uncertainty relation. We discuss a harmonic oscillator problem in the three dimensional non-commutative space with generalized uncertainty relation. We use the perturbation method to solve the harmonic oscillator problem in three dimensional non-commutative quantum space with generalized uncertainty relation.



Electronic address: [email protected]

1

I.

INTRODUCTION

Since Sydner [1] and Yang [2] introduced the idea of non-commutative geometry in spacetime, there has been much effort to apply this idea to the quantization of gravity [3-7]. To understand how non-commutative coordinates may play in physics, it is helpful to investigate the quantum mechanics in non-commutative space. The simplest model is a system of particles moving in two dimensional non-commutative spaces. Recently some progress has been accomplished in applying the non-commutativity of the space to the quantum mechanics which is called a non-commutative quantum mechanics (NCQM) [8-14]. The perturbative non-commutative quantum mechanics have been extensively studied [15-24]. The perturbation approach is based on the Weyl- Moyal correspondence, according to which the usual product of functions should be replaced by the star-product. In this paper we discuss the three dimensional non-commutative quantum mechanics with generalized uncertainty relation. This paper is organized as follows: In section II we discuss the extension of the NCQM algebra so that it involves the non-commutativity between x ˆi and pˆj for a different i, j. We express the non-commutative position and momentum operators in terms of the commutative position and momentum operators. In section III we use the results of section II to discuss a harmonic oscillator problem in the three dimensional noncommutative space with generalized uncertainty relation. We use the perturbation method to solve the harmonic oscillator problem in three dimensional non-commutative quantum space with generalized uncertainty relation.

II.

NCQM ALGEBRA IN THREE DIMENSIONAL NONCOMMUTATIVE

SPACE

In the usual commutative three dimensional space, the coordinates and momenta in quantum mechanics have the following commutation relations: [xi , xj ] = 0,

[pi , pj ] = 0,

[xi , pj ] = i¯ hδij ,

(i, j = 1, 2, 3),

(1)

At very tiny scales, say string scale, the three dimensional space may not commute anymore. Let us denote the operators of coordinates and momenta in three dimensional noncommutative space as xˆi and pˆi respectively. Then, we can consider the following commutation 2

relations: [ˆ xi , xˆj ] = i²ijk θk ,

[ˆ pi , pˆj ] = i²ijk θ¯k ,

[ˆ xi , pˆj ] = i¯hδij (i, j = 1, 2, 3),

(2)

where ²ijk is Levi-Civita symbol and the summation convention is used. Here θi is called spatial noncommutative parameters of dimension of (length)2 and θ¯i is called momentum noncommutative parameters of dimension of (momentum)2 . In this paper we consider the following extension of the relation (2): [ˆ xi , xˆj ] = i²ijk θk ,

[ˆ pi , pˆj ] = i²ijk θ¯k ,

[ˆ xi , pˆj ] = i¯hef f δij + i²ijk ηk , [ˆ xi , pˆj ] = i¯ hef f δij − i²ijk ηk ,

(i < j)

(i > j), (i, j = 1, 2, 3)

(3)

where ηi ’s are related to the uncertainty relation between xˆi and pˆj for i 6= j and h ¯ ef f is a effective Planck constant to be determined later. The NCQM algebra (3) has different possible perturbation realizations [10]. Here we consider the following ansatz of the perturbation expansions of xˆi and pˆi : xˆi = ξ(xi + Aij pj ) pˆi = ξ(pi + Bij xj ),

(4)

where xi and pi satisfy the ordinary Heisenberg algebra [xi , xj ] = [pi , pj ] = 0, [xi , pj ] = i¯hδij . Inserting the realization (4) into the NCQM algebra, we get h ¯ ξ 2 (Aji − Aij ) = ²ijk θk h ¯ ξ 2 (Bji − Bij ) = −²ijk θ¯k h ¯ ef f (i is not summed) h ¯ 1 Ail Bjl = − 2 ²ijk ηk , (i < j) h ¯ξ 1 Ail Bjl = 2 ²ijk ηk , (i > j) h ¯ξ

ξ 2 (1 − Ail Bil ) =

(5)

The eq.(5) can be expressed in terms of the matrices A, B whose entries are Aij , Bij , respectively. (AB T )ii = 1 − 3

h ¯ ef f h ¯ ξ2

1 ²ijk ηk (i < j) h ¯ ξ2 1 (AB T )ij = 2 ²ijk ηk (i > j) h ¯ξ

(AB T )ij = −

(6)

Thus we have 18 unknown parameters but we have 12 equations. Therefore, in order to fix the unknown parameters, we need to impose some additional conditions. We assume that all diagonal elements of the matrix A are taken to be zero and Aij are proportional to Aji for different pairs of i, j. Then we can set 

A=

1 h ¯ ξ2

   

0

(l − 1)θ3

lθ3

0

(m − 1)θ2

nθ1



mθ2

  

(n − 1)θ1 

(7)

0

From the eq.(6), we have the following linear equations: Ã

(l − 1)θ3 B12 + mθ2 B13

!

h ¯ ef f =h ¯ ξ − h ¯ 2

(l − 1)θ3 B22 + mθ2 B23 = −η3 (l − 1)θ3 B32 + mθ2 B33 = η2 lθ3 B11 + (n − 1)θ1 B13 = −η3 Ã

lθ3 B21 + (n − 1)θ1 B23

h ¯ ef f =h ¯ ξ − h ¯

!

2

lθ3 B31 + (n − 1)θ1 B33 = −η1 (m − 1)θ2 B11 + nθ1 B12 = η2 (m − 1)θ2 B21 + nθ1 B22 = −η1 Ã

(m − 1)θ2 B31 + nθ1 B32

h ¯ ef f =h ¯ ξ2 − h ¯

!

(8)

To solve the eq.(8), we assume that all diagonal elements of the matrix B are taken to be zero. Then, from the second relation and the eq.(5), we can set n = m = l = 1/2. Solving the eq.(8), we have 2 η2 θ1 2 B21 = η1 θ2 2 B23 = − η3 θ2 B12 =

4

2 η2 θ3 2 B31 = − η1 θ3 2 B13 = η3 θ1 B32 = −

(9)

Using the eq.(8) to solve ηi ’s, we get η1 = −

θ¯1 θ2 θ3 + θ1 θ¯2 θ3 − θ1 θ2 θ¯3 4¯hξ 2 θ1

θ¯1 θ2 θ3 + θ1 θ¯2 θ3 − 3θ1 θ2 θ¯3 4¯hξ 2 θ2 θ¯1 θ2 θ3 − θ1 θ¯2 θ3 − θ1 θ2 θ¯3

η2 = − η3 =

4¯hξ 2 θ3

(10)

Inserting the eq.(10) into the eq.(8), we have the following condition: θi = θ, θ¯i = θ¯ Thus, we have η1 = η3 = − In this case we have the

θθ¯ , 4¯hξ 2

Ã

h ¯ ef f h ¯ ξ2 − h ¯

(11)

η2 =

!

=−

θθ¯ 4¯ hξ 2

θθ¯ 2¯hξ 2

(12)

(13)

Here we have the following two cases: Case I : ξ = 1 In this case one obtain Ã

h ¯ ef f

!

θθ¯ =h ¯ 1+ 2 , 2¯ h

η=−

θθ¯ 4¯ h

(14)

Case II : h ¯ ef f = h ¯ In this case one obtain ξ=

v q u u th ¯+ h ¯ 2 − 2θθ¯

2¯h

(15)

Throughout this paper we will restrict our concern to the case I. For case I, we have the following realization: 1 ²ijk θj pk 2¯hef f 1 pˆi = pi − ²ijk θ¯j xk 2¯hef f

xˆi = xi +

5

(16) (17)

³

where h ¯ ef f = h ¯ 1+

θθ¯ 2¯ h2

´

. The length squared is then given by 1 1 xˆ2i = r2 − (θ · L) + 2 [θ2 p2 − (θ · p)2 ] h ¯ 4¯h

(18)

and the momentum squared is then given by 1 1 pˆ2i = p2 − (θ¯ · L) + 2 [θ¯2 r2 − (θ · r)2 ] h ¯ 4¯h

(19)

where θ · L = θi Li ,

θ2 = θi θi ,

θ¯2 = θ¯i θ¯i ,

r 2 = xi xi ,

p2 = pi pi

Li = ²ijk xj pk

III.

HARMONIC

OSCILLATOR

IN

(20)

THREE

DIMENSIONAL

NON-

COMMUTATIVE QUANTUM SPACE

In this section we discuss the harmonic oscillator problem in three dimensional noncommutative quantum space. Now let us consider the isotropic harmonic Hamiltonian ˆ = 1 pˆi pˆi + 1 µw2 xˆi xˆi H 2µ 2

(21)

ˆ in the commutative space Using the eq.(16) and the eq.(17), we obtain the expression of H as ˆ = H0 + H 0 , H

(22)

where H0 = Ã

!

1 2 1 2 2 p + µw r 2µ i 2

(23)

2 1¯ 1 ¯2 2 ¯ · r)2 ] + µw [θ2 p2 − (θ · p)2 ] θ + µw2 θ · L + [ θ r − ( θ (24) µ 8µ¯ h2 8¯h2 The angular momentum operator in three dimensional non-commutative quantum space

1 H =− 2¯h 0

satisfies the following commutation relations: [Li , H0 ] = 0,

[Li , r2 ] = 0,

[Li , xj ] = −i¯ h²iaj xa ,

[Li , p2 ] = 0

(25)

[Li , pj ] = i¯ h²ijb pb

(26)

Using the spherical coordinate, we have ·

¸

h ¯2 1 1 1 1 H0 = − ∂r (r2 ∂r ) + 2 ∂θ (sin θ∂θ ) + 2 2 ∂φ2 + µw2 r2 2 2µ r r sin θ 2 r sin θ 6

(27)

If we set ψ = R(r)Plm (cos θ)eimφ , l = 0, 1, 2, · · · ,

− l ≤ m ≤ l,

(28)

the radial equation reads (r2 R0 )0 + (²r2 − α2 r4 − l(l + 1))R = 0,

(29)

where ²=

2µE , h ¯2

α=

µw h ¯

(30)

a

If we set r2 = x and R = e− 2 x y(x), we have ³

´

4x2 y 00 + −4αx2 + 6x y 0 + [(² − 3α) x − l(l + 1)]y = 0

(31)

The solution of the eq.(31) which is non-singular at x = 0 is then given by Ã

yN,l = x

l/2

1 F1

l 3 ² 3 + − ; l + ; αx 2 4 4α 2

!

(32)

The polynomial solution can be obtained by demanding l 3 ² + − = −N, 2 4 4α

(33)

where N = 0, 1, 2, · · ·. In this case we know Ã

l 3 ² + − 2 4 4α Ã

!

6= 0 (i = 0, 1, 2, · · · , N ) i

3 ² l + − 2 4 4α

!

=0

(34)

N +1

Thus, y becomes a polynomial of order N multiplied by xl/2 . Now we denote these polynomials by yN (x) = 1 F1 (−N ; l + 3/2; x)

(35)

The first few yN ’s are y0 = 1 y1 = 1 − y2 = 1 −

α x l + 3/2

2α α2 x+ x2 l + 3/2 (l + 3/2)(l + 5/2)

7

(36)

Thus, the wave function takes the following form à − µw r2 2¯ h

ψN lm = cN lm e

1 F1

!

3 h ¯ 2 −N ; l + ; r Plm (cos θ)eimφ , 2 µw

(37)

where cN lm is a normalization constant given by cN lm = r

1

(38)

4π(l+m)! I (2l+1)(l−m)! N

and the explicit expression of IN is given in Appendix. From the eq.(33), we have the energy eigenvalue

Ã

EN,l,m

!

l 3 = 2¯hw N + + , 2 4

N = 0, 1, 2, · · ·

(39)

The ground state energy is then given by 3 ¯ w, E0,0,0 = h 2

(40)

which is independent of l and m. The eq.(40) coincides with the one in the commutative QM. Up to a first order in θi we have 1 H =− 2¯ h

Ã

0

!

1¯ θ + µw2 θ · L µ

(41)

Since the unperturbed Hamiltonian H0 possesses spherical symmetry, we can use the perturbation method to obtain the exact result for the energy level to the first order in θ’s by choosing the x3 ( or z)-direction to be along the direction of the composite vector 1 θ¯ + µw2 θ. µ

Thus, we have 1 H =− 2¯h 0

Ã

!

1¯ θ + µw2 θ L3 µ

(42)

The expectation value of H 0 with respect to the eigenstates of H0 is then given by !

Ã

1 ¯ µw2 θ+ θ m¯h hH i = − 2¯ hµ 2¯h 0

(43)

¯ the energy eigenvalue takes the following form: Therefore, up to a first order in θ, θ, Ã

EN lm

!

Ã

!

l 3 1 ¯ µw2 = 2¯hw N + + − θ+ θ m¯ h 2 4 2¯hµ 2¯h

8

(44)

IV.

CONCLUSION

In this paper we discussed the three dimensional non-commutative quantum mechanics with generalized uncertainty relation. The generalized uncertainty relation involves the noncommutativity between xˆi and pˆj for a different i, j. We expressed the non-commutative position and momentum operators in terms of the commutative position and momentum operators. We used the results of section II to discuss a harmonic oscillator problem in the three dimensional non-commutative space with generalized uncertainty relation. We used the perturbation method to solve the harmonic oscillator problem in three dimensional noncommutative quantum space with generalized uncertainty relation. We found that the energy of the harmonic Hamiltonian in three dimensional non-commutative space with generalized uncertainty relation depends on two quantum numbers; one is principal quantum number and another a magnetic quantum number.

Acknowledgement

This Work (GNUDFF- 2014-25) was Supported by Academy-oriented Research Funds of Development Fund Foundation, Gyeongsang National University, 2014.

V.

APPENDIX

In this appendix we compute IN , which is defined as IN = where α =

µw . h ¯

Z ∞ 0

2 drr2 RN =

Z ∞ 0

2

drr2 e−αr (1 F1 (−N ; l + 3/2; r2 /α))2

(45)

Replacing r2 = x, we get 1 Z ∞ √ −αx dx xe (1 F1 (−N ; l + 3/2; x/α))2 IN = 2 0

(46)

From the fact that 1 F1 (−N ; l + 3/2; x/α) =

N X N !(−1)k 22k (2l + 1)!(l + k)! k=0

k!l!(2l + 2k + 1)!(N − k)!

(x/α)k

(47)

we have N X N 1 Z ∞ √ −αx X (N !)2 (−1)k+k 22k+2k (2l + 1)!2 0 dx xe (x/α)k+k IN = 0 2 0 0 2 0 k=0 k0 =0 k!k !l! (2l + 2k + 1)!(2l + 2k + 1)!(N − k)!(N − k )! (48) 0

9

0

Setting k + k 0 = s, we have k ≤ s because k 0 = s − k ≥ 0. Thus, we have IN =

2N X s 1X (N !)2 (−1)s 22s (2l + 1)!2 (s + 1/2)! (1/α)2s+3/2 2 2 s=0 k=0 l! k!(s − k)!(2l + 2k + 1)!(2l + 2s − 2k + 1)!(N − k)!(N − s + k)! (49)

Because (s − k)! goes to infinity when k = s + 1, s + 2, · · ·, we have IN =

2N 1X N !(2l + 1)! (−1)s 22s (s + 1/2)!(1/α)2s+3/2 J 2 s=0 s!(N − s)!l!2 (2l + 2s + 1)!

(50)

where J=

∞ X

s!(N − s)!N !(2l + 1)!(2l + 2s + 1)! k=0 k!(s − k)!(2l + 2k + 1)!(2l + 2s − 2k + 1)!(N − k)!(N − s + k)!

(51)

Using the formulas 1 22k = (−s − l − 1/2)k (−s − l)k (2s + 2l − 2k + 1)! (2s + 2l + 1)!

(52)

(2k + 1)! = 22k (3/2)k k!

(53)

J = 5 F2 (−s, −N + s, −N, −s − l − 1/2, −s − l; l + 3/2, l + 1; −1)

(54)

and

we have

Refernces

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