A study of fractional Schrödinger equation composed of Jumarie fractional derivative

Abstract

In this paper we have derived the fractional-order Schrödinger equation composed of Jumarie fractional derivative. The solution of this fractional-order Schrödinger equation is obtained in terms of Mittag–Leffler function with complex arguments, and fractional trigonometric functions. A few important properties of the fractional Schrödinger equation are then described for the case of particles in one-dimensional infinite potential well. One of the motivations for using fractional calculus in physical systems is that the space and time variables, which we often deal with, exhibit coarse-grained phenomena. This means infinitesimal quantities cannot be arbitrarily taken to zero – rather they are non-zero with a minimum spread. This type of non-zero spread arises in the microscopic to mesoscopic levels of system dynamics, which means that, if we denote x as the point in space and t as the point in time, then limit of the differentials dx (and dt) cannot be taken as zero. To take the concept of coarse graining into account, use the infinitesimal quantities as (Δx)^α (and (Δt)^α) with 0 < α < 1; called as ‘fractional differentials’. For arbitrarily small Δx and Δt (tending towards zero), these ‘fractional’ differentials are greater than Δx (and Δt), i.e. (Δx)^α > Δx and (Δt)^α > Δt. This way of defining the fractional differentials helps us to use fractional derivatives in the study of dynamic systems.

Appendices

Appendix

1.1 A.1 Fractional mass

Fractional mass m _α is defined as $m_{\alpha } =\int {\rho \mathrm {d}} x^{\alpha }$, where ρ is the fractional linear mass density in one dimension; when ρ is constant, $m_{\alpha } =\rho \int \mathrm {d} x^{\alpha }$. We have considered that the density is the same as the case α = 1.

1.2 A.2 Fractional velocity

The fractional change of displacement, i.e. d^α x = Γ(1 + α)dx per unit change in fractional time differential (dt)^α is the fractional velocity, i.e. with 0 < α < 1

$${\vec{v}_{\alpha} } =\frac{\mathrm{d}^{\alpha} x}{\mathrm{d}t^{\alpha} }=\alpha !\frac{\mathrm{d}x}{\left( {\mathrm{d}t} \right)^{\alpha} }. $$

Then we can write the following equation:

$$\frac{\mathrm{d}^{1-\alpha} }{\mathrm{d}t^{1-\alpha} }(\vec{v}_{\alpha} )=\frac{\mathrm{d}\vec{x}} {\mathrm{d}t}=\vec{v} . $$

1.3 A.3 Fractional wavelength

Fractional wavelength is demonstrated in figure A1 of a fractional wave of the order α = 0.8.

Figure A1

Fractional wavelength of a fractional wave of the order α = 0.8.

The wavelength is the distance AB. That is the distance covered by a fractional wave in a full fractional cycle. Fractional wavelength is not a fixed quantity. It changes with the evolution of fractional time like a damped oscillating wave.

1.4 A.4 Fractional time period

The time taken N _α for a wave to cover the distance AB (figure A1) is the fractional time period. We should note that this is the first-order time period. As wavelength changes, the time period also changes with the wave propagation. But we assume that λ _α = v _α N _α .

1.5 A.5 Fractional angular frequency

Figure A2 is the polar plot of the fractional wave of the order α = 0.8. In this polar plot we can easily see that the wave is returned to the same point after completing a fractional cycle, i.e. to its origin. From the polar plot we can say that the angle traversed in a full fractional cycle is 2π. Thus, the fractional angular frequency can be assigned as

$$\omega_{\alpha} =\frac{2\pi} {N_{\alpha} } . $$

In figure A2 the initial line is along the horizontal line. From this, we can see that the product of fractional angular momentum and fractional time period N _α is always 2π though both are varying. In limiting condition we have

$$\lim_{\alpha \to 1} \omega_{\alpha} =\omega \quad \text{or}\quad \lim_{\alpha \to 1} \frac{\omega_{\alpha} } {\omega} =1. $$

Figure A2

The plot showing the concept of fractional angular frequency.

1.6 A.6 Fractional wave constant (or vector in three dimensions)

From the analysis of fractional wave of eq. (12) if k _α x ^α − ω _α t ^α = 0, that is the phase part of the wave is zero, we get k _α = ω _α (t ^α/x ^α) = ω _α /v _α as x ^α/t ^α = v _α is the fractional velocity. We have k _α = (ω _α /v _α ) = 2π/ (v _α N _α ), using Appendix (A.5). Now by using Appendix (A.4) we have k _α = 2π/λ _α .

1.7 A.7 Fractional reduced Planck’s constant

In this paper we have introduced fractional Planck’s constant $\hslash _{\alpha }$ as a basic constant. For the limiting condition of α this constant is of the form of reduced Planck’s constant $\hslash$. Consider integer (i.e. classical) and fractional energies, which are $E=\hslash \omega =h\upsilon$ and $\varepsilon _{\alpha } =\hslash _{\alpha } \omega _{\alpha } =h_{\alpha } \upsilon _{\alpha }$ respectively. Now we can write for the limiting condition, i.e. with α → 1,

$$E=h\upsilon =\lim_{\alpha \to 1} (h_{\alpha} \upsilon_{\alpha}). $$

Here, υ is the integer-order frequency and υ _α is the fractional-order frequency. We write the following expression with the above description:

$$h=\lim_{\alpha \to 1} \left( {\frac{\upsilon_{\alpha} } {\upsilon} } \right)h_{\alpha}, \quad \hslash =\lim_{\alpha \to 1} \frac{\omega_{\alpha}} {\omega} \hslash_{\alpha} . $$

Using Appendix A.5, we have, $\hslash =\lim _{\alpha \to 1} \hslash _{\alpha }$. We assume that the energy is proportional to angular frequency and this assumption does not depend upon the value of α. Therefore, we have, E ∝ ω and ε _α ∝ ω _α . If we remove the proportionality, we can have a constant, such that the following condition is satisfied:

$$\frac{\varepsilon_{0.73}} {\omega_{0.73}} =\frac{\varepsilon_{0.79}} {\omega_{0.79}} =\frac{\varepsilon_{0.8}} {\omega_{0.8}} =\cdots=\frac{E}{\omega} =\hslash . $$

Therefore, we have $\hslash _{\alpha } =\hslash$ and hence h _α = h. We can conclude that the fractional Planck’s constant is nothing but Planck’s constant. The reduced fractional Planck’s constant is also the reduced Planck’s constant.

A.8 Theorem

If a function, i.e. f(x,y)is fractionally differentiable, with order α with respect to both the variables x and y, then $D_{y}^{\alpha } D_{x}^{\alpha } f(x,y)=D_{x}^{\alpha } D_{y}^{\alpha } f(x,y)$ or $f_{xy}^{2\alpha } (x,y)=f_{yx}^{2\alpha } (x,y)$ are equivalent. where 0 ≤ α ≤ 1 and D ^α is the Jumarie derivative operator.

Proof.

Consider a function ϕ(x) = f(x,y + k) − f(x,y),k > 0. The fractional mean value theorem states, for 0 < 𝜃 < 1 and 0 ≤ α ≤ 1, the following [14] equation:

$$\phi (x+h)-\phi (x)=\frac{h^{\alpha} }{\Gamma (1+\alpha )}\phi_{x}^{\alpha} (x+\theta h), $$

$$\begin{array}{@{}rcl@{}} &&\phi (x+h)-\phi (x)\,=\,\frac{h^{\alpha}}{\Gamma (1+\alpha )} [f_{x}^{\alpha} (x+\theta h,y+k)\\ &&\qquad\qquad\qquad\qquad -f_{x}^{\alpha} (x+\theta h,y)]. \end{array} $$

Let $F(y)=f_{x}^{\alpha } (x+\theta h,y)$. Then using fractional mean value theorem, we have the following expression:

$$\begin{array}{@{}rcl@{}} && \phi (x+h)-\phi (x)=\frac{h^{\alpha} }{\Gamma (1+\alpha )}[F(y+k)-F(y)]\\ &&\quad =\frac{h^{\alpha} }{\Gamma (1+\alpha )}\frac{k^{\alpha} }{\Gamma (1+\alpha )}[F_{y}^{\alpha} (y+\theta_{1} k)-F_{y}^{\alpha} (y)] \\ &&\quad =\frac{h^{\alpha} }{\Gamma (1+\alpha )}\frac{k^{\alpha}}{\Gamma (1+\alpha )}[f_{yx}^{\alpha} (x+\theta h,y+\theta_{1} k)]\\ &&\text{where } 0<\theta,\theta_{1} <1,0\le \alpha \le 1.\ \end{array} $$

On the other hand, we have

$$\phi (x+h)=f(x+h,y+k)-f(x+h,y). $$

Therefore,

$$\begin{array}{@{}rcl@{}} && \phi (x+h)-\phi (x)=f(x+h,y+k)-f(x+h,y) -f(x,y+k)+f(x,y)\\ && f_{y}^{\alpha} (x,y)={\Gamma} (1+\alpha )\lim_{k\to 0} \frac{f(x,y+k)-f(x,y)}{k^{\alpha}}\\ && \text{or}\\&& f_{xy}^{2\alpha} (x,y)={\Gamma} (1+\alpha )\lim_{h\to 0} \frac{f_{y}^{\alpha}(x,y+h)-f_{y}^{\alpha}(x,y)}{h^{\alpha} } \\ &&\qquad\quad~ =\left( {\Gamma (1+\alpha )} \right)^{2}\lim_{h\to 0} \lim_{k\to 0}\frac{f(x+h,y+k)-f(x+h,y)-f(x,y+k)+f(x,y)}{k^{\alpha} h^{\alpha}}\\ && \qquad\quad~=\left( {\Gamma (1+\alpha )} \right)^{2}\lim_{h\to 0} \lim_{k\to 0} \frac{\phi (x+h)-\phi (x)}{k^{\alpha} h^{\alpha} }\\ &&\qquad\quad~=\left( {\Gamma (1+\alpha )}\right)^{2}\lim_{h\to 0} \lim_{k\to 0}f_{yx}^{2\alpha} (x+\theta h,y+\theta_{1} k) \\ && \text{or } \\&& f_{xy}^{2\alpha} (x,y)=f_{yx}^{2\alpha} (x,y). \end{array} $$

Hence the theorem is proved. □

We have used this theorem in our derivation.