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

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

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López, G. and Lizarraga, J. (2022) Single Charged Particle Motion in a Flat Surface with Static Electromagnetic Field and Quantum Hall Effect. *Journal of Modern Physics*, **13**, 1324-1330. doi: 10.4236/jmp.2022.1311081.

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],

$\frac{m{\omega}_{c}}{\hslash}A=2\pi l,\text{\hspace{1em}}l\in \mathcal{Z},\text{\hspace{1em}}{\omega}_{c}=\frac{qB}{mc},$ (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\text{\hspace{0.17em}}\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

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

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

$i\hslash \frac{\partial \Psi}{\partial t}=\stackrel{^}{H}\Psi ,$ (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 ${\Psi}^{\mathrm{*}}$. Multiplying this one by $\Psi $, (4) by ${\Psi}^{\mathrm{*}}$ 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

$\rho =\Psi \cdot {\Psi}^{\mathrm{*}}$ (6)

and

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

${\varphi}_{n}=-\left[\hslash {\omega}_{c}\left(n+\frac{1}{2}\right)-\frac{m{c}^{2}\mathcal{E}}{2{B}^{2}}\right]\frac{t}{\hslash}-\frac{m{\omega}_{c}}{\hslash}\left(x-\frac{c\mathcal{E}t}{B}\right)\left(y-\frac{m{c}^{2}\mathcal{E}}{q{B}^{2}}\right)\mathrm{.}$ (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 ${f}_{n}^{j}={\left({\stackrel{^}{p}}_{x}\right)}^{j}{f}_{n}^{0}\mathrm{,}j\in Z$. 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

${J}_{n}=\left[\frac{cE}{B}\stackrel{^}{i}-{\omega}_{c}\left(x-\frac{c\mathcal{E}t}{B}\right)\stackrel{^}{j}\right]{\left|{f}_{n}^{0}\right|}^{2}\mathrm{.}$ (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

$\langle {f}_{0}^{0}|{\rho}_{H}|{f}_{0}^{0}\rangle =\frac{h}{{q}^{2}}l,\text{\hspace{1em}}l\in \mathcal{Z}.$ (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)

$=-\frac{\mathcal{E}}{q{\omega}_{c}}{\displaystyle {\int}_{0}^{{L}_{x}}}{\displaystyle {\int}_{0}^{{L}_{y}}}\frac{\text{d}x\text{d}y}{x-\frac{cEt}{B}}=-\frac{\mathcal{E}{L}_{y}}{q{\omega}_{c}}\mathrm{ln}\left(1-\frac{{L}_{x}B}{c\mathcal{E}t}\right)\approx 0$ (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 $B{L}_{x}/c\mathcal{E}=BA/c{V}_{g}~4.5\times {10}^{-8}\mathrm{sec}$. 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

${f}_{n}^{0}\left({L}_{x},y+k{l}_{y},t\right)={f}_{n}^{0}\left({L}_{x},y,t\right),\text{\hspace{1em}}k\in \mathcal{Z},$ (23)

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

$\frac{m{\omega}_{c}}{\hslash}{L}_{x}k{l}_{y}=2\pi l,\text{\hspace{1em}}l\in \mathcal{Z}$ (24)

which brings about the relation

$\frac{m{\omega}_{c}}{\hslash}a=2\pi \frac{l}{k},\text{\hspace{1em}}\text{\hspace{0.05em}}\text{with}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{\hspace{0.05em}}a={L}_{x}{l}_{y}.$ (25)

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

$B=\alpha \frac{l}{k},\text{\hspace{1em}}\text{\hspace{0.05em}}\text{with}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\alpha =\frac{hc}{qa}$ (26)

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

$\langle {f}_{0}^{0}|{\rho}_{H}|{f}_{0}^{0}\rangle =\frac{h}{{q}^{2}}\frac{l}{k},\text{\hspace{1em}}k,l\in \mathcal{Z},$ (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 ${B}_{0}\approx 5\text{\hspace{0.17em}}\text{T}$ for $l=k=1$ (corresponding to an area $a\approx 8.27\times {10}^{-4}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\mu {\text{m}}^{2}$ ), and the other FQHE are matched quite well for $l=3$ and $k=1$, that is $B\approx 15\text{\hspace{0.17em}}\text{T}$. Another example is shown on the reference [8] page 886, one sees that ${B}_{0}\approx 9.8\text{\hspace{0.17em}}\text{T}$ for $l=k=1$ (corresponding to an area $a\approx 4.22\times {10}^{-4}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\mu {\text{m}}^{2}$ ), 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 ${B}_{0}\approx 4.2\text{\hspace{0.17em}}\text{T}$ for $l=k=1$ (corresponding to an area $a=9.85\times {10}^{-4}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\mu {\text{m}}^{2}$ ), 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 ${B}_{0}\approx 5.3\text{\hspace{0.17em}}\text{T}$ for $l=k=1$ (corresponding to an area $a=7.8\times {10}^{-4}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\mu {\text{m}}^{2}$ ), and for the filling factor $\nu =3/4$ one gets $B=4{B}_{0}/3=7.06\text{\hspace{0.17em}}\text{T}$, 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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