Modified Dirac delta function and modified dirac delta potential in the quantum mechanics

Abstract

Some properties of q-delta distribution is discussed. The q-deformed Gauss distribution corresponding to the q-delta function is constructed. The q-delta potential problem in the quantum mechanics is discussed.

Appendix A

In this appendix we will show

$$\begin{aligned} \lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } P_q(x:x_0, \sigma ) F(x) \mathrm{d}x = F(x_0) + \frac{q}{2} x_0^2 F''(x_0) \end{aligned}$$

(56)

Now let us consider the Taylor expansion of the function F(x) around $x=x_0$ as

$$\begin{aligned} F(x) =\sum _{n=0}^{\infty } F_n (x-x_0)^n \end{aligned}$$

(57)

Then we have

$$\begin{aligned}&\lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{\sqrt{ 2\pi \sigma ^2 }} e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} F(x) =\lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{\sqrt{ 2\pi \sigma ^2 }} e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} \sum _{n=0}^{\infty } F_n x^n\nonumber \\&\quad =\lim _{\sigma \rightarrow 0} \frac{1}{\sqrt{ 2\pi \sigma ^2 }} \sum _{n=0}^{\infty } F_{2n} 2^{ n+1/2} \Gamma (n + 1/2) \sigma ^{2n+1}\nonumber \\&\quad = F_0 = F(x_0) \end{aligned}$$

(58)

and

$$\begin{aligned}&\lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{\sqrt{ 2\pi \sigma ^2 }} \left( x \partial _x e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} \right) F(x) =- \lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{\sigma ^2 \sqrt{ 2\pi \sigma ^2 }} x (x - x_0) e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} \sum _{n=0}^{\infty } F_n x^n\nonumber \\&\quad =- \lim _{\sigma \rightarrow 0} \frac{1}{\sigma ^2 \sqrt{ 2\pi \sigma ^2 }} \left( x_0 \sum _{n=0}^{\infty } F_{2n+1} 2^{n+3/2} \Gamma (n +3/2) \sigma ^{ 2n +3} + \sum _{n=0}^{\infty } F_{2n} 2^{n+3/2} \Gamma (n +3/2) \sigma ^{ 2n +3} \right) \nonumber \\&\quad = - x_0 F_1 - F_0 = -x_0 F'(x_0) - F(x_0) \end{aligned}$$

(59)

and

$$\begin{aligned}&\lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{\sqrt{ 2\pi \sigma ^2 }} \left( x^2 \partial _x^2 e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} \right) F(x) = \lim _{\sigma \rightarrow 0} \int _{-\infty }^{\infty } \mathrm{d}x \frac{1}{ \sqrt{ 2\pi \sigma ^2 }} \left( - \frac{x^2}{\sigma ^2} + \frac{x^2 (x - x_0)^2}{\sigma ^4} \right) e^{ -\frac{(x-x_0)^2}{2 \sigma ^2}} \sum _{n=0}^{\infty } F_n x^n\nonumber \\&\quad = 2F_0 + 2 F_2 x_0^2 + 4 x_0 F_1= 2 F(x_0) + x_0^2 F''(x_0) + 4 x_0 F'(x_0), \end{aligned}$$

(60)

which completes proof.