The New Type of Extended Uncertainty Principle and Some Applications in Deformed Quantum Mechanics

Abstract

In this paper we present a new type of extended uncertainty principle (EUP) of the form [X, P] = i(1 − q|X|) and show that it has the non-zero minimal momentum. For this EUP we discuss the classical mechanics in the curved space, deformed calculus, deformed quantum mechanics and Bohr-Sommerfeld quantization.

Appendices

Appendix A

When x > 0, the deformed exponential function can also be written as

$$ e_q (1: x) = \sum\limits_{n=0}^{\infty} \frac{\left( \frac{1}{q} \right)_n}{n!} (q x)^n ={}_1 F \left( \frac{1}{q} ; ; qx \right) $$

(123)

When x < 0, the deformed exponential function can also be written as

$$ e_q (1: x) = \sum\limits_{n=0}^{\infty} \frac{\left( -\frac{1}{q} \right)_n}{n!} (q x)^n ={}_1 F \left( -\frac{1}{q} ; ; qx \right) $$

(124)

If we introduce the q-number as

$$ \{ N \}_q = \frac{N}{ 1 + q (N-1)} $$

(125)

we have

$$ e_q (1: x) = \left\{\begin{array}{lll} {\sum}_{N=0}^{\infty} \frac{x^N}{ \{ N\}_q! }~~~~ (x\ge 0)\\ {\sum}_{N=0}^{\infty} \frac{x^N}{ \{ N\}_{-q}! }~~~~ (x< 0) \end{array}\right. $$

(126)

where

$$ \{ 0\}_q! =1 $$

(127)

and

$$ \{ N\}_q! = \{ N\}_q \{ N-1\}_q {\cdots} \{ 2\}_q \{ 1\}_q, ~~~ (N \ge 1) $$

(128)

The q-Gamma function is defined by

$$ {\Gamma}_q (z) = {\int}_0^{1/q } \left( \frac{1-q}{1 - q t}\right) e_q (-1: t) t^{z-1} d_q t, $$

(129)

Limiting q → 0, the (129) reduces to an ordinary Gamma function. For z = N, (N = 0, 1, 2, ⋯), we have

$$ {\Gamma}_q (N+1 ) = \{ N\}_q! $$

(130)

The q-Gamma function satisfies the following recurrence relation:

$$ {\Gamma}_q(z+1) = \{ z \}_q {\Gamma}_q(z) $$

(131)

This is the q-analogue of basic functional relation for the ordinary Gamma function. The q-Gamma function for arbitrary number z can also be written as follows:

$$ {\Gamma}_q(z) = \lim\limits_{n \rightarrow \infty} \frac{ \{n\}_q! \{n\}_q^z }{ \{z\}_q \{ z+1 \}_q \{ z+2 \}_q{\cdots} \{ z+n \}_q } $$

(132)

Indeed, replacing z with z + 1, we have

$$ {\Gamma}_q(z+1) = \{ z \}_q {\Gamma}_q(z) $$

(133)

A glance at the (132) shows

$$ {\Gamma}_q(- n ) = \pm \infty, ~~~ for~ n \in Z_+ $$

(134)

and

$$ {\Gamma}_q(z ) =0 ~~~ for ~~ z = -\frac{1}{q}+1, -\frac{1}{q}, -\frac{1}{q}-1 , -\frac{1}{q}-2 , {\cdots} $$

(135)

From the definition (132), we have the third definition of q-Gamma function as follows:

$$ {\Gamma}_q(z) = \frac{\{1\}_q^z}{ \{z\}_q} \prod\limits_{k=1}^{\infty} \left( 1 + \frac{1}{k} \right)^{z} \left( 1 + \frac{z}{k} \right)^{-1} \left( 1 - \frac{q}{1-qk} \right)^{-z} \left( 1 - \frac{qz}{1-qk} \right) $$

(136)

Using the (136), we find the following relations for the product of two q-Gamma functions:

$$ \{ z \}_q ! \{ -z \}_q ! = \frac{ \pi z }{ \sin \pi z } \prod\limits_{k=1}^{\infty} \left[ 1 - \left( \frac{q}{1-qk} z \right)^2 \right] $$

(137)

Inserting z = 1/2 into the (137), we have

$$ {\Gamma}_q \left( \frac{1}{2} \right)= \sqrt{\frac{\pi}{q}} \frac{ {\Gamma} \left( \frac{1}{2} + \frac{1}{q} \right)}{ {\Gamma} \left( 1 + \frac{1}{q} \right)} $$

(138)

The q-deformed digamma function is defined as

$$ F(z) = D_z \ln_q \{z\}_q! $$

(139)

Using the property of the q-log function, we have the following result:

$$ F(z) = \frac{ (1 +q z ) {\prod}_{k=1}^{\infty}\{k \}_q^q }{ {\prod}_{k=1}^{\infty}\{ z + k \}_q^q } \lim\limits_{n \rightarrow \infty} \left[ \{n \}_q^{qz} \ln \{ n \}_q - \sum\limits_{k=1}^n \left( \frac{1}{z+k} - \frac{ q}{ 1 + q (z + k ) } \right)\right] $$

(140)

Inserting z = 0 into the (140) yields

$$ F(0) = - \gamma_q $$

(141)

where the q-analogue of Euler constant is then given by

$$ \gamma_q = \lim\limits_{n \rightarrow \infty} \left[ \sum\limits_{k=1}^{\infty} \left( \frac{1}{k } - \frac{ q}{ 1 + qk} \right)- \ln \{ n \}_q \right] $$

(142)

Appendix B: Nikiforov - Uvarov Method

In this technique, we solve a second-order differential equation in the form of a hypergeometric-type or a general Schrodinger-type equation [1, 2],

$$ \left[ \frac{d^2}{ds^2} + \frac{\alpha_1-\alpha_2 s }{s(1 - \alpha_3 s) } \frac{d}{ds} + \frac{ -\lambda_1 s^2 + \lambda_2 s -\lambda_3 }{[s(1 - \alpha_3 s)]^2} \right] \psi_n (s)=0 $$

(143)

For calculation of eigenvalues and corresponding wave functions of the energy by using this approach, we try to obtain the following parameters

$$ \begin{array}{@{}rcl@{}} \alpha_4 = \frac{1}{2} (1 - \alpha_1), ~\alpha_5 = \frac{1}{2} (\alpha_2 - 2 \alpha_3), ~ \alpha_6 = \alpha_5^2 + \lambda_1, ~ \alpha_7 = 2 \alpha_4 \alpha_5 -\lambda_2, ~\alpha_8 =\alpha_4^2 + \lambda_3 \\\alpha_9 =\alpha_3 \alpha_7 + \alpha_3^2 \alpha_8 + \alpha_6, ~ \alpha_{10} =\alpha_1 + 2 \alpha_4 + 2 \sqrt{ \alpha_8}, ~ \alpha_{11} = \alpha_2 - 2 \alpha_5 +2 (\sqrt{ \alpha_9} + \alpha_3 \sqrt{ \alpha_8}), \\\alpha_{12} = \alpha_4 + \sqrt{\alpha_8} , ~ \alpha_{13} = \alpha_5 - (\sqrt{\alpha_9}+\alpha_3 \sqrt{\alpha_8}) \end{array} $$

(144)

The energy relation using the set of parameters mentioned above forms an energy equation as,

$$ \alpha_2 n - (2n+1) \alpha_5 + (2n+1) (\sqrt{\alpha_9}+\alpha_3 \sqrt{\alpha_8}) + n (n-1) \alpha_3 + \alpha_7 + 2 \alpha_3 \alpha_8 + 2 \sqrt{ \alpha_8 \alpha_9} =0 $$

(145)

Also, the final wave function relation in NU method will be of the following form,

$$ \psi_n(s) = s^{\alpha_{12}} (1 - \alpha_3 s)^{- \alpha_{13} /\alpha_3} P_n^{\left( \alpha_{10} -1, \alpha_{11}/\alpha_3 - \alpha_{10} -1 \right)} (1 - 2 \alpha_3 s) $$

(146)

If α₃ = 0 , the relation (147) will be reduced to

$$ \psi_n(s) = s^{\alpha_{12}} e^{\alpha_{13}s} L_n^{\alpha_{10} -1} (\alpha_{11} s) $$

(147)