Isomorphism of Analytical Spectrum between Noncommutative Harmonic Oscillator and Landau Problem

In symmetric gauge, the vector potential can be written as

Then by substituting (2.2) into (2.1),

Note that we define the z-axis such that $qB>0$. For the case of the electron where $q<0$, it means that the magnetic field is along the negative z-axis. Alternatively, $qB>0$ is also true for the proton where $q>0$ when the magnetic field is along the positive z-axis. Hence we will now adopt $qB=eB$ instead, where $eB$ is also greater than $0$. After a few algebraic steps,

where

is the z-component of the orbital angular momentum operator. Let

which represents the cyclotron frequency in SI units.

However in some works in literature such as [21, 24], the Hamiltonian can also be written as

which leads to

As a result,

which still represents the cyclotron frequency but in Gaussian units as the Lorentz force differs by a factor of $1/c$ where $c$ is the speed of light. Both are valid representations of the Hamiltonian for the system. In this work, the Hamiltonian which considers the SI units of the cyclotron frequency will be considered.

Returning to our initial problem of calculating the eigenstates and eigenvalues of (2.4), by applying a few change of variables and orthogonality relation, the solution to the time-independent Schródinger equation is

where $n_{r}$ and $m_{l}$ are radial and angular momentum quantum numbers respectively [25] and is in agreement with [26]. As for the eigenvalues, they are given as

The first term mimics the energy eigenvalues of the standard two-dimensional harmonic oscillator (with the frequency $\frac{\omega_{c}}{2}$ instead). The second term comes from the coupled momentum-gauge term that is manifested in the Hamiltonian after expanding the kinetic momentum operator. This term represents the interaction between the canonical momenta and the magnetic field, and what distinguishes this problem from a standard planar isotropic oscillator [26].

Now, the interesting question that could be pondered is, what if we choose to use $qB=-eB$ instead? At first, this may look trivial but a careful calculation reveals that the resulting energy levels are actually different from the case that we obtain previously. In fact, it can be shown that

while the eigenfunction maintains the same form as in (2.1). Notice that the sign of the second term is different than before which implies that, the mere assumption that we make on the product, $qB$ can actually change the pattern of degeneracy. We refer to [26] for a detailed derivation of the eigenstates and eigenenergies obtained when $qB=-eB$.

As for the Landau gauge, it can be written in two different ways;

From here thereon, we are only going to focus our attention on the first Landau gauge, as the discussion that follows will also be applied to the second Landau gauge in a similar manner to the first Landau gauge but with a minute difference as shall be pointed out later. Then by substituting (2.15) into (2.1)

Just like in the symmetric gauge, we will define the z-axis such that $qB>0$ and hence, $qB=eB$. After a few algebraic steps,

The derivation of the solution of the Schrödinger equation involving the Hamiltonian in (2.18) is very well known as it has the mathematical structure of a shifted harmonic oscillator [27] and thus, will not be repeated here. It is

where $H_{n}$ is the Hermite polynomial. This method of solution has introduced two quantum numbers; $k_{x}$, a real number, and $n_{y}$, a non-negative integer. This latter quantum number counts the number of nodes of the probability density. On the other hand, the eigenvalues are

Note the lack of dependence on the quantum number $k_{x}$. Consequently, the energies are infinitely degenerate for every non-negative integer $n_{y}$. The energy eigenvalues of the second Landau gauge will also take the form of (2.20). In regard to the energy eigenstates, we need to change the sign of the shifted term of the position of the harmonic oscillator to have $-\frac{\hbar k_{y}}{eB}$ with quantum numbers, $k_{y}$ and $n_{x}$. In contrast to our result, in [26], an assumption of $qB=-eB$ is analysed instead. The only difference of our results with those in [26] is, we need to change the sign of the spatial shift of the eigenfunction of the oscillator in both Landau gauges while having the same energy.

From Section 2, we have determined the energy spectra and wavefunctions of Landau levels. The question now is, what are the conditions to be obeyed in order for Landau levels and noncommutative oscillator to be equivalent? We can naively compare the Hamiltonian of the two systems, (3.10) and (2.4) then try to infer from there. Thus we have,

Realize that, the set of equations that we have from the comparison of the coefficients of the Hamiltonians needs to be true simultaneously. In other words, (3.13) altogether has to be consistent but, that is not the case here. For the two systems to be isomorphic, we need to invoke some mathematical scheme by redefining $M$ and $\Omega$ in other terms instead of $m$ and $\omega$. This is allowable as $M$ and $\Omega$ are respectively the parameters that characterize the effective mass and frequency of the oscillator in a commutative space. With reference to [17] which is generalized for any noncommutative quantum mechanical problem in central field, we are going to let the effective mass and frequency to be

and

which are $\theta$-dependent. Due to (3.14) and (3.15), the resulting Hamiltonian of the noncommutative oscillator is the following,

Realize that by imposing the oscillator to have the new effective mass and frequency in terms of $\theta$ alone, the Hamiltonian (3.16) lack of parameters permits an opportunity of introducing a factor to make the two systems consistent. By multiplying each of the coefficients of the Hamiltonian by a factor $\zeta$ to be determined, then

Now, by comparing the newly-defined Hamiltonian of the noncommutative oscillator and the Landau levels, hence

Unlike the comparison from (3.13), a close inspection reveals that (3.18) is altogether consistent and in accordance with [17]. As a result, together with factor $\zeta$, the effective mass and frequency of the oscillator will then be equal to those of Landau levels as shown below

where $\zeta=\frac{e^{2}B^{2}}{8\mu}$ is a controlling factor. By calculation, we can evaluate that when Landau levels have $\mu=m_{e}=9.109\times 10^{-31}$ and $\omega_{c}=\frac{eB}{\mu}=2.110\times 10^{12}$, then the oscillator would have $M=4.620\times 10^{-37}$ and $\Omega=2.081\times 10^{18}$ due to (3.14) and (3.15). Since we have established the relationship above, we can always examine a different mass and frequency of either of the systems and be able to find the same parameters of the other system studied for them to be isomorphic.

Let us contemplate the effect of a constant $B$ and $\theta$ on their respective systems. In the quantum Hall effect experiments, the magnetic fields are typically about $B=12T$  [17, 30, 31]. Using the second and third equations in (3.18),

in $m^{2}$ or it can also be written as $\theta=0.2195\times 10^{-11}$ in $cm^{2}$ which is in conformance with [17]. For this value of $\theta$, one cannot distinguish between noncommutative quantum mechanics (in this case, it would be the noncommutative isotropic harmonic oscillator) and the usual Landau problem.

Since we have determined the value of $\theta$, we should also compute the value of the controlling factor $\zeta$, that is

which implies that just like $\theta$, there is a dependence on the charge of the particle and the magnetic field which acts on it in the Landau problem with an additional dependence on the mass of the particle.

Hence, since their Hamiltonians are now similar, analysing Landau levels is tantamount to analysing the noncommutative oscillator. Let us revisit the energy eigenvalues of Landau levels in symmetric gauge for $qB=eB>0$. They are

For the noncommutative oscillator, we can express (3.23) in terms of $\theta$ and $\zeta$ as follows

Therefore, we conclude that the noncommutative oscillator and Landau levels are isomorphic provided that the pair of quantum numbers $n_{r}$ and $m_{l}$ of both systems have the same values. Besides, the Hamiltonian of the noncommutative oscillator has to be parametrized by the controlling factor $\zeta$ so that it is consistent with the Hamiltonian of Landau levels in symmetric gauge. In regards to the wavefunctions of Landau levels, we recall that

For the noncommutative oscillator, again by expressing (3.1) in terms of $\theta$ and $\zeta$, one obtains

With the above results, we can find the analytical spectra of the noncommutative oscillator in the context of Landau problem in a symmetric gauge. However, the results obtained are by no means representing the general spectra of the solutions of the noncommutative oscillator but rather specific to the case when $M$ and $\Omega$ are $\theta-$dependent designated in (3.1) and (3.1). We refer to [32, 33, 34, 35] as reading materials that deliver more elaborate research on the spectra of the noncommutative oscillator using path integral formulation.

The same question as raised before can still be asked here. What if we impose the following condition, $qB=-eB$? We will encounter a situation where the comparison cannot be made as

which is clearly false as the left-hand side is negative while the right-hand side is positive. This is the result when comparing the third term of their Hamiltonians. One might argue that why do we not account for this with a new $\zeta$. The problem is, the $\zeta-$ term is also present in the first and second terms of the Hamiltonian of the noncommutative oscillator. If we change them, it will generate a false statement like in (3.27). Hence, we can say that this is also a condition for isomorphism. We want to emphasize again that the results of the energy spectra and wavefunctions of the noncommutative oscillator that we obtain here are applied only when $M$ and $\Omega$ are $\theta-$dependent which is the limitation of this formalism.

As for the two Landau gauges, we cannot directly compare the coefficients of their Hamiltonians as their variables or operators are clearly different from one another. Nonetheless, an assumption can be made for each to account for this. For the first Landau gauge, assume that the position operator, $\hat{x}$ to be the zero operator, $\hat{0}$. The same supposition goes to the second Landau gauge as well but with $\hat{y}$. Hence, the correlation can then be made as follows,

Notice that the set of equations in (3.28) is reminiscent to the one in symmetric gauge and therefore, we will not repeat the discussion here. However we will show how their Hamiltonians take the form as portrayed below,

After solving the time-independent Schrödinger equation, the energy eigenvalues of the noncommutative oscillator in the context of the first Landau gauge are

and the energy eigenstate is

The important thing to note here is, the system is more restrictive compared to the case of symmetric gauge since we have to lose a spatial degree of freedom. As a result, the system is nondegenerate for every quantum number, $n_{y}$. The real number $k_{x}$ will be a constant that we will denote as $k_{0}$, since the system has lost the degree of freedom associated with position $x$. This is equivalent to the statement that the position has localized to a fixed value. Consequently, it implies that it is impossible to determine $k_{0}$ as it is correlated with the momentum $p_{x}$ of the particle due to the uncertainty principle. We will treat $k_{0}$ as a parameter instead of a quantum number and its effect on the system will be analysed in Section 4.2. If we use the second Landau gauge, the energy eigenvalues resemble the ones obtained before and, the energy eigenfunction will have a negative shift in position. Otherwise, as before, it will not work if we assume $qB<0$.

By using Mathematica, the 3D plot of one of the ground-state wavefunctions and its probability density of the isomorphic system when both quantum numbers are zero are shown as follows,

In Figure 1, since $n_{r}$ and $m_{l}$ are both zero, the imaginary part of the wavefunction naturally vanishes and so are other related terms in (3.1) leaving the wavefunction to have a structure of a $e^{-r^{2}}$ function. The same argument goes to Figure 2 as well except that it follows the structure of a $e^{-2r^{2}}$ which explains its slighly different feature. Yet, the characteristic shape for both follows the two-dimensional Gaussian function. Due to its distribution, it is most likely to find the particle at the origin. Generally, the wavefunction bears no physical interpretation while the probability density function does as its own name implies, by giving us the likelihood of finding an electron (or some other system) at some given point in space based on the Born rule. Hence, in the upcoming discussion, the wavefunction (in this paper we will focus on its real part) will only serve as a visualisation while the probability distribution is subjected to further analysis. The ground-state energy is $1.112\times 10^{-22}J$ or $6.943\times 10^{-4}eV$.

For this subsection, we study the effect of the quantum number $n_{y}$ and a parameter $k_{0}$ on the wavefunctions and probability distributions of the isomorphic system in the context of the first Landau gauge. Even though the functions are two-dimensional in nature as they only depend on the position $y$, we decide to let our $k_{0}$ to be the variable of our functions to study their collective effects (with $n_{y}$) on the same set of axes. This will not create any issue on the interpretation of the plot as the effect of $k_{0}$ is trivial as shall be seen later. Recall that $k_{0}$ is also unpredictable and hence, we will not provide a table to show some exact value of energies.

The contour plot of the aforementioned functions are provided in Figure 9 and 10 below,

From Figures 9 and 10, we can see that the parameter $k_{0}$ only plays a role in shifting the harmonic oscillator to the right of the origin as $k_{0}$ increases. From the mathematical form of the wavefunction, we can think of the system as a shifted harmonic oscillator where the behaviour of such function is well understood. By right, the quantum number $n_{y}$ reflects the number of nodes and antinodes of the wavefunction and only the number of nodes for the probability density curve. However, we can clearly see that only two nodes are present even when $n_{y}$ is greater than 1 as shown in 10. This is due to the factor used in the wavefunction which makes the nodes look invisible. Besides, the amplitude of the two highest peaks designated by the brightest strips on the plot also increases as $n_{y}$ increases which implies that the amplitude of the other seemingly invisible peaks in between should decrease and thus, making it even harder to observe. This behaviour is also displayed in higher-order states.

Let us consider the following random values of the magnetic field; $10T$, $15T$ and $20T$. From Figures 11 and 12, we only plot the ground-state wavefunctions and their respective probability densities i.e, $\Psi_{0,0}$ and $|\Psi_{0,0}|^{2}$ as the effect can naturally be extended to higher-order states. We observe that the radius at which the distribution begins to flatten out decreases as $B$ increases or $\theta$ decreases. This is due to the radial symmetry of the wavefunction. To compensate for the reduction in radius, the amplitude has to increase to maintain the norm to be one.

This time we also assign the same values of the magnetic field ($10T$, $15T$ and $15T$). We only plot the ground-state wavefunctions and probability distributions just like before. Figures 13 and 14 below show that the deviation of the harmonic oscillator due to parameter $k_{0}$ from the origin to be lesser as the magnetic field increases or noncommutativity decreases. This is expected as these two parameters are present in the shifted position term in (3.2). The same behaviour is also present in higher-order states.