# Fractional Schrödinger equation

The fractional Schrödinger equation is a fundamental equation of fractional quantum mechanics. It was discovered by Nick Laskin (1999) as a result of extending the Feynman path integral, from the Brownian-like to Lévy-like quantum mechanical paths. The term fractional Schrödinger equation was coined by Nick Laskin.[1]

## Fundamentals

The fractional Schrödinger equation in the form originally obtained by Nick Laskin is:[2]

 $i\hbar \frac{\partial \psi (\mathbf{r},t)}{\partial t}=D_\alpha (-\hbar ^2\Delta )^{\alpha /2}\psi (\mathbf{r},t)+V(\mathbf{r},t)\psi (\mathbf{r},t)$

Further,

• Dα is a scale constant with physical dimension [Dα] = [energy]1 − α·[length]α[time]α, at α = 2, D2 =1/2m, where m is a particle mass,
• the operator (−ħ2Δ)α/2 is the 3-dimensional fractional quantum Riesz derivative defined by (see, Ref.[2]);
$(-\hbar ^2\Delta )^{\alpha /2}\psi (\mathbf{r},t)=\frac 1{(2\pi \hbar )^3}\int d^3pe^{i\frac{\mathbf{pr}}\hbar }|\mathbf{p}|^\alpha \varphi ( \mathbf{p},t),$

Here, the wave functions in the position and momentum spaces; $\psi(\mathbf{r},t)$ and $\varphi (\mathbf{p},t)$ are related each other by the 3-dimensional Fourier transforms:

$\psi (\mathbf{r},t)=\frac 1{(2\pi \hbar )^3}\int d^3pe^{i \mathbf{p}\cdot\mathbf{r}/\hbar}\varphi (\mathbf{p},t),\qquad \varphi (\mathbf{p},t)=\int d^3re^{-i \mathbf{p}\cdot\mathbf{r}/\hbar }\psi (\mathbf{r},t).$

The index α in the fractional Schrödinger equation is the Lévy index, 1 < α ≤ 2. Thus, the fractional Schrödinger equation includes a space derivative of fractional order α instead of the second order (α = 2) space derivative in the standard Schrödinger equation. Thus, the fractional Schrödinger equation is a fractional differential equation in accordance with modern terminology.[3] This is the main point of the term fractional Schrödinger equation or a more general term fractional quantum mechanics.[4] At α = 2 fractional Schrödinger equation becomes the well-known Schrödinger equation.

The fractional Schrödinger equation has the following operator form

 $i\hbar \frac{\partial \psi (\mathbf{r},t)}{\partial t}=\widehat{H}_\alpha \psi (\mathbf{r},t)$

where the fractional Hamilton operator $\widehat{H}_\alpha$ is given by

$\widehat{H}_\alpha =D_\alpha (-\hbar ^2\Delta )^{\alpha /2}+V(\mathbf{r},t).$

The Hamilton operator, $\widehat{H}_\alpha$ corresponds to the classical mechanics Hamiltonian function intoduced by Nick Laskin

$H_\alpha (\mathbf{p},\mathbf{r})=D_\alpha |\mathbf{p}|^\alpha +V(\mathbf{r},t),$

where p and r are the momentum and the position vectors respectively.

### Time-independent fractional Schrödinger equation

The special case when the Hamiltonian $H_\alpha$ is independent of time

$H_\alpha =D_\alpha (-\hbar ^2\Delta )^{\alpha /2}+V(\mathbf{r}),$

is of great importance for physical applications. It is easy to see that in this case there exist the special solution of the fractional Schrödinger equation

$\psi (\mathbf{r},t)=e^{-(i/\hbar )Et}\phi (\mathbf{r}),$

where $\phi (\mathbf{r})$ satisfies

$H_\alpha \phi (\mathbf{r}) = E\phi (\mathbf{r}),$

or

$D_\alpha (-\hbar ^2\Delta )^{\alpha /2}\phi (\mathbf{r})+V(\mathbf{r})\phi ( \mathbf{r})=E\phi (\mathbf{r}).$

This is the time-independent fractional Schrödinger equation.

Thus, we see that the wave function $\psi (\mathbf{r},t)$ oscillates with a definite frequency. In classical physics the frequency corresponds to the energy. Therefore, the quantum mechanical state has a definite energy E. The probability to find a particle at $\mathbf{r}$ is the absolute square of the wave function $| \psi (\mathbf{r},t) |^2 .$ Because of time-independent fractional Schrödinger equation this is equal to $| \phi (\mathbf{r})|^2$ and does not depend upon the time. That is, the probability of finding the particle at $\mathbf{r}$ is independent of the time. One can say that the system is in a stationary state. In other words, there is no variation in the probabilities as a function of time.

### Probability current density

The continuity equation for probability current and density follows from the fractional Schrödinger equation:

$\frac{\partial \rho (\mathbf{r},t)}{\partial t}+\nabla \cdot \mathbf{j}( \mathbf{r},t)=0,$

where $\rho (\mathbf{r},t)=\psi ^{\ast }(\mathbf{r},t)\psi (\mathbf{r},t)$ is the quantum mechanical probability density and the vector $\mathbf{j}(\mathbf{r},t)$ can be called by the fractional probability current density vector

$\mathbf{j}(\mathbf{r},t)=\frac{D_\alpha \hbar }i\left( \psi ^{*}(\mathbf{r} ,t)(-\hbar ^2\Delta )^{\alpha /2-1}\mathbf{\nabla }\psi (\mathbf{r},t)-\psi ( \mathbf{r},t)(-\hbar ^2\Delta )^{\alpha /2-1}\mathbf{\nabla }\psi ^{*}( \mathbf{r},t)\right) ,$

where we use the notation (see also matrix calculus): $\mathbf{\nabla =\partial /\partial r}$.

Introducing the momentum operator $\widehat{\mathbf{p}}=\frac{\hbar }{i} \frac{\partial }{\partial \mathbf{r}}$ we can write the vector $\mathbf{j}$ in the form (see, Ref.[2])

$\mathbf{j=}D_{\alpha }\left( \psi (\widehat{\mathbf{p}}^{2})^{\alpha /2-1} \widehat{\mathbf{p}}\psi ^{\ast }+\psi ^{\ast }(\widehat{\mathbf{p}}^{\ast 2})^{\alpha /2-1}\widehat{\mathbf{p}}^{\ast }\psi \right).$

This is fractional generalization of the well-known equation for probability current density vector of standard quantum mechanics (see, Ref.[7]).

#### Velocity operator

The quantum mechanical velocity operator $\widehat{\mathbf{v}}$ is defined as follows:

$\widehat{\mathbf{v}}=\frac{i}{\hbar }(H_{\alpha }\widehat{\mathbf{r}}\mathbf{ -}\widehat{\mathbf{r}}H_{\alpha }),$

Straightforward calculation results in (see, Ref.[2])

$\widehat{\mathbf{v}}=\alpha D_{\alpha }|\widehat{\mathbf{p}}^{2}|^{\alpha /2-1}\widehat{\mathbf{p}}\,.$

Hence,

$\mathbf{j=}\frac{1}{\alpha }\left( \psi \widehat{\mathbf{v}}\psi ^{\ast }+\psi ^{\ast }\widehat{\mathbf{v}}\psi \right) ,\qquad 1<\alpha \leq 2.$

To get the probability current density equal to 1 (the current when one particle passes through unit area per unit time) the wave function of a free particle has to be normalized as

$\psi (\mathbf{r},t)=\sqrt{\frac{\alpha }{2\mathrm{v}}}\exp \left[\frac{i}{\hbar }( \mathbf{p}\cdot\mathbf{r}-Et)\right],\qquad E=D_{\alpha }|\mathbf{p}|^{\alpha },\qquad 1<\alpha \leq 2,$

where $\mathrm{v}$ is the particle velocity, $\mathrm{v}=\alpha D_{\alpha }p^{\alpha -1}$.

Then we have

$\mathbf{j=}\frac{\mathbf{v}}{\mathrm{v}},\qquad \mathbf{v}=\alpha D_{\alpha }|\mathbf{p}^{2}|^{\frac{\alpha }{2}-1}\mathbf{p,}$

that is, the vector $\mathbf{j}$ is indeed the unit vector.

## Physical applications

### Fractional Bohr atom

Main article: Bohr atom

When $V(\mathbf{r})$ is the potential energy of hydrogenlike atom,

$V(\mathbf{r})=-\frac{Ze^{2}}{|\mathbf{r}|},$

where e is the electron charge and Z is the atomic number of the hydrogenlike atom, (so Ze is the nuclear charge of the atom), we come to following fractional eigenvalue problem,

$D_{\alpha }(-\hbar ^{2}\Delta )^{\alpha /2}\phi (\mathbf{r})-\frac{Ze^{2}}{| \mathbf{r|}}\phi (\mathbf{r})=E\phi (\mathbf{r}).$

This eigenvalue problem has first been solved by Nick Laskin in.[5]

Using the first Niels Bohr postulate yields

$\alpha D_{\alpha }\left( \frac{n\hbar }{a_{n}}\right) ^{\alpha }=\frac{Ze^{2}}{a_{n} },$

and it gives us the equation for the Bohr radius of the fractional hydrogenlike atom

$a_{n}=a_{0}n^{\alpha /(\alpha -1)}.$

Here a0 is the fractional Bohr radius (the radius of the lowest, n = 1, Bohr orbit) defined as,

$a_{0}=\left( \frac{\alpha D_{\alpha }\hbar ^{\alpha }}{Ze^{2}}\right) ^{1/(\alpha -1)}.$

The energy levels of the fractional hydrogenlike atom are given by

$E_{n}=(1-\alpha )E_{0}n^{-\alpha/(\alpha -1)},\qquad 1<\alpha \leq 2,$

where E0 is the binding energy of the electron in the lowest Bohr orbit that is, the energy required to put it in a state with E = 0 corresponding to n = ∞,

$E_{0}=\left( \frac{Ze^{2}}{\alpha D_{\alpha }^{1/\alpha }\hbar }\right) ^{\alpha/(\alpha -1)}.$

The energy (α − 1)E0 divided by ħc, (α − 1)E0/ħc, can be considered as fractional generalization of the Rydberg constant of standard quantum mechanics. For α = 2 and Z = 1 the formula $(\alpha -1)E_{0}/\hbar c$ is transformed into

$\mathrm{Ry}=me^{4}/2\hbar ^{3}c$,

which is the well-known expression for the Rydberg formula.

According to the second Niels Bohr postulate, the frequency of radiation $\omega$ associated with the transition, say, for example from the orbit m to the orbit n, is,

$\omega =\frac{(1-\alpha )E_{0}}{\hbar }\left[ \frac{1}{n^{\frac{\alpha }{ \alpha -1}}}-\frac{1}{m^{\frac{\alpha }{\alpha -1}}}\right]$.

The above equations are fractional generalization of the Bohr model. In the special Gaussian case, when (α = 2) those equations give us the well-known results of the Bohr model.[6]

### The infinite potential well

A particle in a one-dimensional well moves in a potential field $V({x})$, which is zero for $-a\leq x\leq a$ and which is infinite elsewhere,

$V(x)=\infty ,\qquad x<-a\qquad \qquad (\mathrm{i})$
$V(x)=0,\quad -a\leq x\leq a\quad \quad \quad \ (\mathrm{ii})$
$V(x)=\infty ,\qquad \ x>a\qquad \qquad \ (\mathrm{iii})$

It is evident a priori that the energy spectrum will be discrete. The solution of the fractional Schrödinger equation for the stationary state with well-defined energy E is described by a wave function $\psi (x)$, which can be written as

$\psi (x,t)=\left(-i\frac{Et}{\hbar }\right)\phi(x)$,

where $\phi (x)$, is now time independent. In regions (i) and (iii), the fractional Schrödinger equation can be satisfied only if we take $\phi (x)=0$. In the middle region (ii), the time-independent fractional Schrödinger equation is (see, Ref.[5]).

$D_\alpha (\hbar \nabla )^\alpha \phi (x)=E\phi (x).$

This equation defines the wave functions and the energy spectrum within region (ii), while outside of the region (ii), x<-a and x>a, the wave functions are zero. The wave function $\phi (x)$ has to be continuous everywhere, thus we impose the boundary conditions $\phi (-a)=\phi (a)=0$ for the solutions of the time-independent fractional Schrödinger equation (see, Ref.[5]). Then the solution in region (ii) can be written as

$\phi (x)=A\exp(ikx) +B\exp(-ikx).$

To satisfy the boundary conditions we have to choose

$A = -B\exp(-i2ka),$

and

$\sin(2ka) =0.$

It follows from the last equation that

$2ka = n\pi.$

Then the even ($\phi _n^{\mathrm{even}}(-x) = \phi _n^{\mathrm{even}}(x)$ under reflection $x\rightarrow -x$) solution of the time-independent fractional Schrödinger equation $\phi ^{\mathrm{even}}(x)$ in the infinite potential well is

$\phi _n^{\mathrm{even}}(x)=\frac 1{\sqrt{a}}\cos \left[ \frac{ n\pi x}{2a}\right] , \quad n = 1, 3, 5, ....$

The odd ($\phi _n^{\mathrm{odd}}(-x) = -\phi _n^{\mathrm{odd}}(x)$ under reflection $x\rightarrow -x$) solution of the time-independent fractional Schrödinger equation $\phi ^{\mathrm{even}}(x)$ in the infinite potential well is

$\phi _n^{\mathrm{odd}}(x)=\frac 1{\sqrt{a}}\sin \left[ \frac{ n\pi x}{2a}\right] , \quad n = 2, 4, 6, ....$

The solutions $\phi ^{\mathrm{even}}(x)$ and $\phi ^{\mathrm{odd}}(x)$ have the property that

$\int\limits_{-a}^{a}dx\phi _{m}^{\mathrm{even}}(x)\phi _{n}^{\mathrm{even} }(x)=\int\limits_{-a}^{a}dx\phi _{m}^{\mathrm{odd}}(x)\phi _{n}^{\mathrm{odd} }(x)=\delta _{mn},$

where $\delta _{mn}$ is the Kronecker symbol and

$\int\limits_{-a}^{a}dx\phi _{m}^{\mathrm{even}}(x)\phi _{n}^{\mathrm{odd} }(x)=0.$

The eigenvalues of the particle in an infinite potential well are (see, Ref.[5])

$E_n=D_\alpha \left( \frac{\pi \hbar }{2a}\right) ^\alpha n^\alpha ,\qquad \qquad n=1,2,3....,\qquad 1<\alpha \leq 2.$

It is obvious that in the Gaussian case (α = 2) above equations are transformed into the standard quantum mechanical equations for a particle in a box (for example, see Eq.(20.7) in [7])

The state of the lowest energy, the ground state, in the infinite potential well is represented by the $\phi _n^{\mathrm{even}}(x)$ at n=1,

$\phi _{\mathrm{ground}}(x)\equiv \phi _1^{\mathrm{even}}(x)=\frac 1{\sqrt{a} }\cos \left(\frac{\pi x}{2a}\right),$

and its energy is

$E_{\mathrm{ground}}=D_{\alpha }\left( \frac{\pi \hbar }{2a}\right) ^{\alpha }.$

### Fractional quantum oscillator

Fractional quantum oscillator introduced by Nick Laskin (see, Ref.[2]) is the fractional quantum mechanical model with the Hamiltonian operator $H_{\alpha ,\beta }$ defined as

$H_{\alpha,\beta}=D_{\alpha }(-\hbar ^{2}\Delta )^{\alpha /2}+q^{2}|\mathbf{ r}|^{\beta },\quad 1<\alpha \leq 2,\quad 1<\beta \leq 2,$,

where q is interaction constant.

The fractional Schrödinger equation for the wave function $\psi (\mathbf{r},t)$ of the fractional quantum oscillator is,

$i\hbar \frac{\partial \psi (\mathbf{r},t)}{\partial t}=D_{\alpha }(-\hbar ^{2}\Delta )^{\alpha /2}\psi (\mathbf{r},t)+q^{2}|\mathbf{r}|^{\beta }\psi ( \mathbf{r},t)$

Aiming to search for solution in form

$\psi (\mathbf{r},t)=e^{-iEt/\hbar }\phi (\mathbf{r}),$

we come to the time-independent fractional Schrödinger equation,

$D_{\alpha }(-\hbar ^{2}\Delta )^{\alpha /2}\phi (\mathbf{r},t)+q^{2}|\mathbf{ r}|^{\beta }\phi (\mathbf{r},t)=E\phi (\mathbf{r},t).$

The Hamiltonian $H_{\alpha,\beta}$ is the fractional generalization of the 3D quantum harmonic oscillator Hamiltonian of standard quantum mechanics.

#### Energy levels of the 1D fractional quantum oscillator in semiclassical approximation

The energy levels of 1D fractional quantum oscillator with the Hamiltonian function $H_{\alpha}=D_{\alpha }|p|^{\alpha }+q^{2}|x|^{\beta }$ were found in semiclassical approximation by Nick Laskin (2002) (see, Ref.[2]).

We set the total energy equal to E, so that

$E=D_{\alpha }|p|^{\alpha }+q^{2}|x|^{\beta },$

whence

$|p|=\left( \frac{1}{D_{\alpha }}(E-q^{2}|x|^{\beta })\right) ^{1/\alpha }$.

At the turning points $p=0$. Hence, the classical motion is possible in the range $|x|\leq (E/q^{2})^{1/\beta }$.

A routine use of the Bohr-Sommerfeld quantization rule yields

$2\pi \hbar (n+\frac{1}{2})=\oint pdx=4\int\limits_{0}^{x_{m}}pdx=4\int\limits_{0}^{x_{m}}D_{\alpha }^{-1/\alpha }(E-q^{2}|x|^{\beta })^{1/\alpha }dx,$

where the notation $\oint$ means the integral over one complete period of the classical motion and $x_{m}=(E/q^{2})^{1/\beta }$ is the turning point of classical motion.

To evaluate the integral in the right hand we introduce a new variable $y=x(E/q^{2})^{-1/\beta }$. Then we have

$\int\limits_0^{x_m}D_\alpha ^{-1/\alpha }(E-q^2|x|^\beta )^{1/\alpha }dx=\frac 1{D_\alpha ^{1/\alpha }q^{2/\beta }}E^{\frac 1\alpha +\frac 1\beta }\int\limits_0^1dy(1-y^\beta )^{1/\alpha }.$

The integral over dy can be expressed in terms of the Beta-function,

$\int\limits_{0}^{1}dy(1-y^{\beta })^{1/\alpha }=\frac{1}{\beta } \int\limits_{0}^{1}dzz^{\frac{1}{\beta }-1}(1-z)^{\frac{1}{\alpha }}=\frac{1 }{\beta }\Beta \left(\frac{1}{\beta },\frac{1}{\alpha }+1\right).$

Therefore

$2\pi \hbar (n+\frac 12)=\frac 4{D_\alpha ^{1/\alpha }q^{2/\beta }}E^{\frac 1\alpha +\frac 1\beta }\frac 1\beta \Beta\left(\frac 1\beta ,\frac 1\alpha +1\right).$

The above equation gives the energy levels of stationary states for the 1D fractional quantum oscillator (see, Ref.[2]),

$E_{n}=\left( \frac{\pi \hbar \beta D_{\alpha }^{1/\alpha }q^{2/\beta }}{2\Beta( \frac{1}{\beta },\frac{1}{\alpha }+1)}\right) ^{\frac{\alpha \beta }{\alpha +\beta }}\left(n+\frac{1}{2}\right)^{\frac{\alpha \beta }{\alpha +\beta }}.$

This equation is generalization of the well-known energy levels equation of the standard quantum harmonic oscillator (see, Ref.[7]) and is transformed into it at α = 2 and β = 2. It follows from this equation that at $\frac{1}{\alpha }+\frac{1}{\beta }=1$ the energy levels are equidistant. When $1<\alpha \leq 2$ and $1<\beta \leq 2$ the equidistant energy levels can be for α = 2 and β = 2 only. It means that the only standard quantum harmonic oscillator has an equidistant energy spectrum.