Fractional Schrödinger equation

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

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

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

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

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

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

or

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

This is the time-independent fractional Schrödinger equation (see, Ref.[2]).

Thus, we see that the wave function ${\displaystyle \psi (\mathbf {r} ,t)}$ 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 ${\displaystyle \mathbf {r} }$ is the absolute square of the wave function ${\displaystyle |\psi (\mathbf {r} ,t)|^{2}.}$ Because of time-independent fractional Schrödinger equation this is equal to ${\displaystyle |\phi (\mathbf {r} )|^{2}}$ and does not depend upon the time. That is, the probability of finding the particle at ${\displaystyle \mathbf {r} }$ 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.

The conservation law of fractional quantum mechanical probability has been discovered for the first time by D.A.Tayurskii and Yu.V. Lysogorski [5]

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

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

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

and

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

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

It has been found in Ref.[5] that there are quantum physical conditions when the new term ${\displaystyle {\mathit {K}}(\mathbf {r} ,t)}$ is negligible and we come to the continuity equation for quantum probability current and quantum density (see, Ref.[2]):

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

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

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

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

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

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

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

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

Hence,

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

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

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

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

Then we have

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

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

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

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

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,

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

This eigenvalue problem has first been introduced and solved by Nick Laskin in.[6]

Using the first Niels Bohr postulate yields

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

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

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

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

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

The energy levels of the fractional hydrogenlike atom are given by

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

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 = ∞,

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

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 ${\displaystyle (\alpha -1)E_{0}/\hbar c}$ is transformed into

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

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

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

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

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.[7]

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

${\displaystyle V(x)=\infty ,\qquad x<-a\qquad \qquad (\mathrm {i} )}$

${\displaystyle V(x)=0,\quad -a\leq x\leq a\quad \quad \quad \ (\mathrm {ii} )}$

${\displaystyle V(x)=\infty ,\qquad \ x>a\qquad \qquad \ (\mathrm {iii} )}$

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 ${\displaystyle \psi (x)}$, which can be written as

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

where ${\displaystyle \phi (x)}$, is now time independent. In regions (i) and (iii), the fractional Schrödinger equation can be satisfied only if we take ${\displaystyle \phi (x)=0}$. In the middle region (ii), the time-independent fractional Schrödinger equation is (see, Ref.[6]).

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

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 ${\displaystyle \phi (x)}$ has to be continuous everywhere, thus we impose the boundary conditions ${\displaystyle \phi (-a)=\phi (a)=0}$ for the solutions of the time-independent fractional Schrödinger equation (see, Ref.[6]). Then the solution in region (ii) can be written as

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

To satisfy the boundary conditions we have to choose

${\displaystyle A=-B\exp(-i2ka),}$

and

${\displaystyle \sin(2ka)=0.}$

It follows from the last equation that

${\displaystyle 2ka=n\pi .}$

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

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

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

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

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

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

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

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

The eigenvalues of the particle in an infinite potential well are (see, Ref.[6])

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

It is obvious that in the Gaussian case (α = 2) above equations are ö transformed into the standard quantum mechanical equations for a particle in a box (for example, see Eq.(20.7) in [8])

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

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

and its energy is

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

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

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

where q is interaction constant.

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

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

Aiming to search for solution in form

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

we come to the time-independent fractional Schrödinger equation,

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

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

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

We set the total energy equal to E, so that

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

whence

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

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

A routine use of the Bohr-Sommerfeld quantization rule yields

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

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

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

${\displaystyle \int \limits _{0}^{x_{m}}D_{\alpha }^{-1/\alpha }(E-q^{2}|x|^{\beta })^{1/\alpha }dx={\frac {1}{D_{\alpha }^{1/\alpha }q^{2/\beta }}}E^{{\frac {1}{\alpha }}+{\frac {1}{\beta }}}\int \limits _{0}^{1}dy(1-y^{\beta })^{1/\alpha }.}$

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

${\displaystyle \int \limits _{0}^{1}dy(1-y^{\beta })^{1/\alpha }={\frac {1}{\beta }}\int \limits _{0}^{1}dzz^{{\frac {1}{\beta }}-1}(1-z)^{\frac {1}{\alpha }}={\frac {1}{\beta }}\mathrm {B} \left({\frac {1}{\beta }},{\frac {1}{\alpha }}+1\right).}$

Therefore,

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

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

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

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 ${\displaystyle {\frac {1}{\alpha }}+{\frac {1}{\beta }}=1}$ the energy levels are equidistant. When ${\displaystyle 1<\alpha \leq 2}$ and ${\displaystyle 1<\beta \leq 2}$ 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.

The effective mass of states in solid state systems can depend on the wave vector k, i.e. formally one considers m=m(k). Polariton Bose-Einstein condensate modes are examples of states in solid state systems with mass sensitive to variations and locally in k fractional quantum mechanics is experimentally feasible .

Self-accelerating beams, such as the Airy beam, are known solutions of the conventional free Schrödinger equation (with ${\displaystyle \alpha =2}$ and without a potential term). Equivalent solutions exist in the free fractional Schrödinger equation. The time-dependent fractional Schrödinger equation in momentum space (assuming ${\displaystyle \hbar =1}$ and with one spatial coordinate) is:

${\displaystyle i{\frac {\partial \varphi (p,t)}{\partial t}}=D_{\alpha }|p|^{\alpha }\varphi (p,t)}$.

In position space, an Airy beam is typically expressed using the special Airy function, although it possesses a more transparent expression in momentum space:

${\displaystyle \psi (x,t=0)=\mathrm {Ai} (bx)\exp(ax),\qquad \varphi (p,t=0)={\frac {1}{2b{\sqrt {2\pi }}}}\exp \left({\frac {(a-ip)^{3}}{3b^{3}}}\right)}$.

Here, the exponential function ensures the square-integrability of the wave function, i.e. that the beam possesses a finite energy, in order to be a physical solution. The parameter ${\displaystyle a}$ controls the exponential cut-off on the beam’s tail, while the parameter ${\displaystyle b}$ controls the width of the peaks in position space. The Airy beam solution for the fractional Schrödinger equation in momentum space is obtained from simple integration of the above equation and initial condition:

${\displaystyle \varphi (p,t)={\frac {1}{2b{\sqrt {2\pi }}}}\exp \left(-iD_{\alpha }|p|^{\alpha }t+{\frac {(a-ip)^{3}}{3b^{3}}}\right)}$.

This solution self-accelerates at a rate proportional to ${\displaystyle t^{\frac {2}{3-\alpha }}}$.[9] When taking ${\displaystyle \alpha =2}$ for the conventional Schrödinger equation, one recovers the original Airy beam solution with a parabolic acceleration (${\displaystyle \propto t^{2}}$).