Single Charged Particle Motion in a Flat Surface with Static Electromagnetic Field and Quantum Hall Effect

Gustavo V. López, Jorge A. Lizarraga

Departamento de Física, Universidad de Guadalajara, Guadalajara, México.

DOI: 10.4236/jmp.2022.1311081                

Abstract

Taking into account the non separable solution for the quantum problem of the motion of a charged particle in a flat surface of lengths L_x and L_y with transversal static magnetic field B and longitudinal static electric field E, the quantum current, the transverse (Hall) and longitudinal resistivities are calculated for the state n = 0 and j = 0. We found that the transverse resistivity is proportional to an integer number, due to the quantization of the magnetic flux, and longitudinal resistivity can be zero for times t >> L_xB/cE. In addition, using a modified periodicity of the solution, a modified quantization of the magnetic flux is found which allows to have IQHE and FQHE of any filling factor of the form v = k/l, with k, l ∈ Z.

Keywords

Landau’s Gauge, Quantum Hall Effect, Degeneration

1. Introduction

There are a lot of literature dealing with the phenomenon of Quantum Hall Effect [1] - [8], and most of them use the Landau’s solution of the eigenvalue problem associated to the charged particle motion in a flat surface with static transversal magnetic field to the surface. This brings about the known Landau’s levels for the energies and a separable variable solution for the eigenfunctions [9]. However, it has been shown that a non separable of variables solution exists for this problem with the same Landau’s levels [10] [11], and these levels are numerable degenerated [12], determining the operators which causes this degeneration. In addition, the quantization of the magnetic flux appears naturally [10],

$\frac{m{\omega }_c}{\hslash }A=2\pi l,l\in \mathcal{Z},{\omega }_c=\frac{qB}{mc},$ (1)

where m is the mass of the charge q, c is the speed of light, ${\omega }_c$ is the so called cyclotron frequency, B is the magnitude of the static magnetic field, $A=L_x{L}_y$ is the area of the sample, and $2\pi \hslash =h$ is the Planck’s constant. As we mentioned before, Landau’s separable solution is normally used to try to explain the so called Integer Quantum and Fractional Quantum Hall Effects (IQHE and FQHE) [4] [5] [6] [7], which were first discovered experimentally [1] [2] [3]. The IQHE is normally explained as a single particle phenomenon; meanwhile, the FQHE is explained as a many particle event [4] [5] [6]. Experimentally, both of them occur in highly impure samples, where these impurities have the effect of extending the range of magnetic field intensity where the resistivity is quantized [2] [3] [7]. The main characteristic of the IQHE or FQHE is the resistivity (or voltage) which appears on the transverse motion of the charges, so called Hall’s resistivity ${\rho }_H$. This Hall’s resistivity acquires a constant value on certain regions of the magnetic field, and within these regions, the longitudinal resistivity is zero. The values of these constant ${\rho }_H$ turn out to be inverse to an integer number (IQHE) or proportional to an integer number (FQHE) multiplied by the constant $h/q^2$, called von Klitzing constant [2] [3] ( $h/q^2\approx 25812.80745\Omega$ ). In this paper, we calculate the quantum current and the expected value of the transverse and longitudinal resistivities for a single charged particle motion on a flat surface using the non separable solution in the lowest Landau level ( $n=0$ ) and using the first wave function ( $j=0$ ).

2. Quantum Current

The Hamiltonian associated to the motion of a charge particle q with mass m on a flat surface of lengths $L_x$ and $L_y$ with transverse magnetic field $B=\left(\mathrm{0,0,}B\right)$ and longitudinal electric field $E=\left(\mathrm{0,}E\mathrm{,0}\right)$ is given by

$\hat{H}=\frac{1}{2m}{\left(p-\frac{q}{c}A\right)}^2+qV\mathrm{,}$ (2)

where A is the vector potential, $B=\nabla \times A$, and V is the scalar potential, $E=-\nabla V$. The Schrödinger’s equation,

$i\hslash \frac{\partial \Psi }{\partial t}=\hat{H}\Psi ,$ (3)

can be written, using the operator $p=-i\hslash \nabla$, as

$i\hslash \frac{\partial \Psi }{\partial t}=\frac{1}{2m}\left[-{\hslash }^2{\nabla }^2+i\frac{\hslash q}{c}\left(\nabla \cdot A+A\cdot \nabla \right)+\frac{q^2{A}^2}{c^2}\right]\Psi +qV\Psi .$ (4)

Taking the usual complex conjugated to this expression, a similar equation is gotten for the function ${\Psi }^{\mathrm{<mo>\; *\; </mo>}}$. Multiplying this one by $\Psi$, (4) by ${\Psi }^{\mathrm{<mo>\; *\; </mo>}}$ and subtracting both, the following continuity equation is obtained

$\frac{\partial \rho }{\partial t}+\nabla \cdot J=0,$ (5)

where $\rho$ and J are defined as

$\rho =\Psi \cdot {\Psi }^{\mathrm{<mo>\; *\; </mo>}}$ (6)

and

$J=\frac{i\hslash }{2m}\left(\Psi \nabla {\Psi }^{\mathrm{<mo>\; *\; </mo>}}-{\Psi }^{\mathrm{<mo>\; *\; </mo>}}\nabla \Psi \right)-\frac{q}{mc}\rho A\mathrm{.}$ (7)

Since $\Psi$ is a scalar complex function, it can be written as $\Psi =\left|\Psi \right|{\text{e}}^{i\theta }$, where $\left|\Psi \right|$ and $\theta$ are real functions, and $\theta$ is the argument of the function. Then, the current is given by

$J=\left(\frac{\hslash }{m}\nabla \theta -\frac{q}{mc}A\right){\left|\Psi \right|}^2\mathrm{.}$ (8)

For the general solution of (3), the function $\theta$ can be very complicated expression of all variables. However, for a particular state solution of the system, say

${\psi }_n\left(x,t\right)={\text{e}}^{i{\varphi }_n\left(x,t\right)}{f}_n\left(x\right),$ (9)

the argument is just $\theta ={\varphi }_n\left(x\mathrm{,}t\right)$, and the current associated to this state of the system is given by

$J_n=\left(\frac{\hslash }{m}\nabla {\varphi }_n-\frac{q}{mc}A\right){\left|f_n\right|}^2\mathrm{.}$ (10)

3. Single Charged Particle Current

The non separable solution of (3) using the Landau’s gauge $A=B\left(-y\mathrm{,0,0}\right)$ and the longitudinal constant electric field $E=\left(\mathrm{0,}E\mathrm{,0}\right)$ was given as

$f_n^0=\frac{1}{\sqrt{2^{n}n!L_y}}{\left(\frac{m{\omega }_c}{\pi \hslash }\right)}^{1/4}{\text{e}}^{i{\varphi }_n}{\text{e}}^{-\frac{m{\omega }_c}{2\hslash }{\left(x-c\mathcal{E}t/B\right)}^2}{H}_n\left(\sqrt{\frac{m{\omega }_c}{\hslash }}\left(x-c\mathcal{E}t/B\right)\right),$ (11a)

where $\mathcal{E}=qE$, ${\omega }_c$ is the cyclotron frequency (1), and ${\varphi }_n$ is given by

${\varphi }_n=-\left[\hslash {\omega }_c\left(n+\frac{1}{2}\right)-\frac{m{c}^2\mathcal{E}}{2B^2}\right]\frac{t}{\hslash }-\frac{m{\omega }_c}{\hslash }\left(x-\frac{c\mathcal{E}t}{B}\right)\left(y-\frac{m{c}^2\mathcal{E}}{q{B}^2}\right)\mathrm{.}$ (11b)

These functions are degenerated in the sense that for each Landau’s level ( $\hslash {\omega }_c\left(n+1/2\right)$ ), one has a numerable solutions $f_n^j={\left({\hat{p}}_x\right)}^j{f}_n^0\mathrm{,}j\in Z$. Thus, the expressions (11a) define the state of the system. Using this function ${\varphi }_n$ in (10) and for the index of degeneration $j=0$, we have

$J_n=\left[\frac{cE}{B}\hat{i}-{\omega }_c\left(x-\frac{c\mathcal{E}t}{B}\right)\hat{j}\right]{\left|f_n^0\right|}^2\mathrm{.}$ (12)

In particular, for the ground state of Landau’s energy, it follows that the components of the current are

$J_0^x=\frac{c\mathcal{E}}{B}{\left|f_0^0\right|}^2\mathrm{,}$ (13)

and

$J_0^y=-{\omega }_c\left(x-\frac{c\mathcal{E}t}{B}\right){\left|f_0^0\right|}^2\mathrm{.}$ (14)

The electric conductivity along the x-axis is called Hall’s conductivity and is given by

${\sigma }_H=\frac{q}{\mathcal{E}}{J}_0^x=\frac{qc}{B}{\left|f_0^0\right|}^2\mathrm{.}$ (15)

Thus, the Hall’s resistivity is ${\rho }_H=1/{\sigma }_H$, and the expected value of the resistivity in the state $f_0^0$ is

$\langle f_0^0|{\rho }_H|f_0^0\rangle ={\displaystyle {\int }_0^{L_x}}{\displaystyle {\int }_0^{L_y}}\frac{{\left|f_0^0\right|}^2}{{\sigma }_H}\text{d}x\text{d}y=\frac{BA}{qc}.$ (16)

Now, multiplying and dividing this quantity by $m{\omega }_c/\hslash$ and making some rearrangements, one gets

$\langle f_0^0|{\rho }_H|f_0^0\rangle =\frac{\hslash }{q^2}\left(\frac{m{\omega }_c}{\hslash }A\right),$ (17)

and taking into consideration the magnetic field flux quantization (1), it follows that

$\langle f_0^0|{\rho }_H|f_0^0\rangle =\frac{h}{q^2}l,l\in \mathcal{Z}.$ (18)

The expected value in the state $f_0^0$ of the longitudinal resistivity ${\rho }_y$ is

$\langle f_0^0|{\rho }_y|f_0^0\rangle ={\displaystyle {\int }_0^{L_x}}{\displaystyle {\int }_0^{L_y}}\frac{{\left|f_0^0\right|}^2\text{d}x\text{d}y}{{\sigma }_y}=\frac{\mathcal{E}}{q}{\displaystyle {\int }_0^{L_x}}{\displaystyle {\int }_0^{L_y}}\frac{{\left|f_0^0\right|}^2\text{d}x\text{d}y}{J_0^y}$ (19)

$=-\frac{\mathcal{E}}{q{\omega }_c}{\displaystyle {\int }_0^{L_x}}{\displaystyle {\int }_0^{L_y}}\frac{\text{d}x\text{d}y}{x-\frac{cEt}{B}}=-\frac{\mathcal{E}{L}_y}{q{\omega }_c}\ln \left(1-\frac{L_{x}B}{c\mathcal{E}t}\right)\approx 0$ (20)

since one has normally in the experiments that $L_{x}B/c\mathcal{E}t\ll 1$, that is, the time in the experiments are such that

$t\gg \frac{L_{x}B}{c\mathcal{E}}.$ (21)

For example, on the reference [2] and with respect the voltage gate $V_g$, one has that $B{L}_x/c\mathcal{E}=BA/c{V}_g~4.5\times {10}^{-8}\sec$. So, the condition (21) is well satisfied in this experiment.

Note that the expression (18) implies a filling factor $\nu =1/l$, which correspond to the IQHE phenomenon for $l=1$ and to the FQHE phenomenon for $l>1$. However, this result is valid for an analysis of a single charged particle, and both QHE phenomena appear due to the quantization of the magnetic flux (1). In addition, one must note that this analysis is still valid for any $n>0$ and $j=0$.

4. Full IQHE and FQHE

The quantization of the magnetic flux (1) arises from the periodicity of the solutions of the Hamiltonian [10], which can be expressed using (11a) for $\mathcal{E}=0$ as

$f_n^0\left(L_x\mathrm{,}y+L_y\mathrm{,}t\right)=f_n^0\left(L_x\mathrm{,}y\mathrm{,}t\right)\mathrm{.}$ (22)

However (and also for $\mathcal{E}=0$ ), let us assume that $L_y=N{l}_y$ where $l_y\ll L_y$ and $N\in {\mathcal{Z}}^{+}$, that is, the total area $L_x{L}_y$ is covered with slices of area $L_x{l}_y$, with horizontal length $L_x$ and width $l_y$. Let us impose the periodicity condition of the form

$f_n^0\left(L_x,y+k{l}_y,t\right)=f_n^0\left(L_x,y,t\right),k\in \mathcal{Z},$ (23)

such that with the phase (11b), one gets

$\frac{m{\omega }_c}{\hslash }{L}_{x}k{l}_y=2\pi l,l\in \mathcal{Z}$ (24)

which brings about the relation

$\frac{m{\omega }_c}{\hslash }a=2\pi \frac{l}{k},\text{with}a=L_x{l}_y.$ (25)

Using (1) and making some rearrangements, the magnetic field can be given by

$B=\alpha \frac{l}{k},\text{with}\alpha =\frac{hc}{qa}$ (26)

and using (25) in (17), the expected value of the Hall resistivity would be

$\langle f_0^0|{\rho }_H|f_0^0\rangle =\frac{h}{q^2}\frac{l}{k},k,l\in \mathcal{Z},$ (27)

implying now a filling factor of $\nu =k/l$, which represents the full IQHE (for $l=1$ ) and FQHE (for $l>1$ ). To determine the magnetic values B where these phenomena occur, one looks for the value $B_0$ where the first IQHE ( $l=k=1$ ) appears, which intersect the normal linear dependence behavior straight line, and this defines $\alpha =B_0$. Then, one uses the resulting expression

$B=B_0\frac{l}{k}$ (28)

to find the other quantized magnetic fields which correspond to IQHE or FQHE. For example, on the experimental data shown on the reference [3], one sees that $B_0\approx 5\text{T}$ for $l=k=1$ (corresponding to an area $a\approx 8.27\times {10}^{-4}\mu {\text{m}}^2$ ), and the other FQHE are matched quite well for $l=3$ and $k=1$, that is $B\approx 15\text{T}$. Another example is shown on the reference [8] page 886, one sees that $B_0\approx 9.8\text{T}$ for $l=k=1$ (corresponding to an area $a\approx 4.22\times {10}^{-4}\mu {\text{m}}^2$ ), and the other IQHE and FQHE magnetic fields are matched quite well for $l>1$ and $k>1$. In addition, on reference [13] page 207, one sees that $B_0\approx 4.2\text{T}$ for $l=k=1$ (corresponding to an area $a=9.85\times {10}^{-4}\mu {\text{m}}^2$ ), and the other IQHE and FQHE magnetic fields are matched quite well for $l>1$ and $k>1$. Finally, on reference [14] page 156801-2, one sees that $B_0\approx 5.3\text{T}$ for $l=k=1$ (corresponding to an area $a=7.8\times {10}^{-4}\mu {\text{m}}^2$ ), and for the filling factor $\nu =3/4$ one gets $B=4B_0/3=7.06\text{T}$, which is approximately the experimental value reported.

5. Conclusion

Using the known non-separable solution for the quantum motion of a charged particle in a flat surface with static fields, in the state $n=0$ and $j=0$, the Hall and the longitudinal resistivities were calculated. For the quantization of the magnetic flux, which can appear from the simple periodicity on the y-direction, the results bring about the IQHE and FQHE phenomena since from the expression (18) it appears a filling factor of $1/l$ for a single charged particle due to the quantization of the magnetic flux. If $l=1$, one gets the IQHE phenomenon, and if $l>1$, one gets the FQHE phenomenon. However, it is not possible to say anything about filling factors of the form $\nu =k/l$. For a more extended quantization of the magnetic flux (25), which appears of the extended periodicity (23), one gets also IQHE and FQHE but with a filling factor of $\nu =k/l$.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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