Charged particle in a flat box with static electromagnetic field and Landau’s levels

Landau’ solution [1] of a charged particle in a flat surface with magnetic field has become of great importance in understanding integer hall effect [2, 3, 4, 5, 6], fractional Hall effect [6, 7, 8, 9], and topological insulators [10, 11, 12, 13, 14, 15, 16]. This last elements promise to become essential for future nanotechnology devices [17, 18, 19]. Due to this considerable application of the Landau’s levels, it is worth to re-study this problem and its variations with an static electric field. In this paper, we show that there exists a non separable solution for this type of quantum problems, but having the same Landau’s levels. In our cases, instead of having a flat surface, we consider to have a flat box with lengths $L_{x}$, $L_{y}$, and $L_{z}$ such that $L_{z}\ll L_{x},L_{y}$

Let us consider a charged particle “q”with mass “mïn a flat box with a constant magnetic field orthogonal to the flat surface, ${\bf B}=(0,0,B)$, as shown in the next figure.

For a non relativistic charged particle, the Hamiltonian of the system (units CGS) is

$$H=\frac{({\bf p}-q{\bf A}/c)^{2}}{2m},$$

(1)

where ${\bf p}$ is the generalized linear momentum, ${\bf A}$ is the magnetic potential such that ${\bf B}=\nabla\times{\bf A}$, and “cïs the speed of light. We can choose the Landau’s gauge to have the vector potential of the form ${\bf A}=(-By,0,0)$. Therefore, the Hamiltonian has the following form

$$H=\frac{(p_{x}+qBy/c)^{2}}{2m}+\frac{p_{y}^{2}}{2m}+\frac{p_{z}^{2}}{2m}\mathchar 46\relax$$

(2)

To quantize the system, we need to solve the Schrödinger’s equation [20]

$$i\hbar\frac{\partial\Psi}{\partial t}=\left\{\frac{(\hat{}p_{x}+qBy/c)^{2}}{2m}+\frac{\hat{p}_{y}^{2}}{2m}+\frac{\hat{p}_{z}^{2}}{2m}\right\}\Psi\mathchar 46\relax$$

(3)

where $\Psi=\Psi({\bf x},t)$ is the wave function, $\hbar$ is the Plank’s constant divided by $2\pi$, $\hat{p}_{i}$ are the momentum operators such that $[x_{i},\hat{p}_{j}]=i\hbar\delta_{ij}$. Now, the argument used by Landau is that due to commutation relation $[\hat{p}_{x},\hat{H}]=0$, between the operators $\hat{p}_{x}$ and the Hamiltonian $\hat{H}$ (implying that $\hat{p}_{x}$ is a constant of motion), it is possible to replace this component of the momentum by $\hbar k_{x}$, having a solution for the eigenvalue problem of separable variable type, $f_{1}(t)f_{2}(x)f_{3}(y)f_{4}(z)$. However, we will see that this type of commutation does not imply necessarily separability of the solution. Since the Hamiltonian $\hat{H}$ does not depend explicitly on time, the proposition

$$\Psi({\bf x},t)=e^{-iEt/\hbar}\Phi({\bf x})$$

(4)

reduces the equation to an eigenvalue problem

$$\widehat{H}\Phi=E\Phi\mathchar 46\relax$$

(5)

Then, this equation is written as

$$\left\{\frac{1}{2m}\left(\hat{p}_{x}^{2}+\frac{2qB}{c}y\hat{p}_{x}+\frac{q^{2}B^{2}}{c^{2}}y^{2}\right)+\frac{\hat{p}_{y}^{2}}{2m}+\frac{\hat{p}_{z}^{2}}{2m}\right\}\Phi=E\Phi\mathchar 46\relax$$

(6)

The variable “zïs separable through the proposition

$$\Phi({\bf x})=\phi(x,y)e^{-ik_{z}z},\quad\quad k_{z}\in\Re,$$

(7)

resulting the following equation

$$\left\{\frac{1}{2m}\left(\hat{p}_{x}^{2}+\frac{2qB}{c}y\hat{p}_{x}+\frac{q^{2}B^{2}}{c^{2}}y^{2}\right)+\frac{\hat{p}_{y}^{2}}{2m}\right\}\phi=E^{\prime}\phi,$$

(8)

where $E^{\prime}$ is

$$E^{\prime}=E-\frac{\hbar^{2}k_{x}^{2}}{2m}\mathchar 46\relax$$

(9)

That is, the resulting partial differential equation is of the form

$$\frac{1}{2m}\left\{-\hbar^{2}\frac{\partial^{2}\phi}{\partial x^{2}}-i\frac{2qB\hbar}{c}y\frac{\partial\phi}{\partial x}+\frac{q^{2}B2}{c^{2}}y^{2}\phi\right\}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\phi}{\partial y^{2}}=E^{\prime}\phi\mathchar 46\relax$$

(10)

This equation does not admit a separable variable solution ($\phi(x,y)=f(x)g(y)$) as Landau’ solution is, but we can use Fourier transformation [21] on the variable “x”,

$$\hat{\phi}(k,y)={\cal F}[\phi]=\frac{1}{\sqrt{2\pi}}\int_{\Re}e^{ikx}\phi(x,y)dx,$$

(11)

to solve this equation. Applying Fourier transformation to this equation, knowing its property ${\cal F}[\partial\phi/\partial x]=(-ik)\hat{\phi}$, we get the ordinary differential equation

$$-\frac{\hbar^{2}}{2m}\frac{d^{2}\hat{\phi}}{dy^{2}}+\frac{m}{2}\omega_{c}^{2}(y-y_{0})^{2}\hat{\phi}=E^{\prime}\hat{\phi},$$

(12)

where $\omega_{c}$ is the cyclotron frequency

	$$\omega_{c}=\frac{qB}{mc}$$		(13a)
and $y_{0}$ is the displacement parameter
	$$y_{0}=\frac{\hbar c}{qB}k\mathchar 46\relax$$		(13b)

This equation is just the quantum harmonic oscillator in the “y”direction displaced by a amount $y_{0}$. So, the solution is

$$\hat{\phi}_{n}(k,y)=\psi_{n}(\xi),\quad\xi=\sqrt{\frac{m\omega_{c}}{\hbar}}(y-y_{0}),\quad\psi_{n}(\xi)=A_{n}e^{-\xi^{2}}H_{n}(\xi),$$

(14)

being $H_{n}(\xi)$ the Hermit polynomials, and $A_{n}$ is a constant of normalization,$An=(m\omega_{c}/\pi\hbar)^{1/4}/\sqrt{2^{n}n!}$. and

$$E^{\prime}_{n}=\hbar\omega_{c}(n+{1}/{2})\mathchar 46\relax$$

(15)

Now, the solution in the real space $\phi_{n}(x,y)$ is gotten by using the inverse Fourier transformation,

$$\phi_{n}(x,y)={\cal F}^{-1}[\phi_{n}(k,y)]=\frac{1}{\sqrt{2\pi}}\int_{\Re}e^{-ikx}\psi_{n}\biggl{(}\sqrt{\frac{m\omega_{c}}{\hbar}}(y-\hbar ck/qB)\biggr{)}dk\mathchar 46\relax$$

(16)

Making the change of variable $\sigma=\sqrt{m\omega_{c}/\hbar}(y-\hbar ck/qB)$, and knowing that the Fourier transformation of the harmonic oscillator solution is another harmonic oscillator solution, we get

$$\phi_{n}(x,y)=\frac{-qB}{\sqrt{mc^{2}\hbar\omega_{c}}}e^{-i\frac{qB}{\hbar c}xy}\psi_{n}\biggl{(}\frac{qB~{}x}{\sqrt{mc^{2}\hbar\omega_{c}}}\biggr{)}\mathchar 46\relax$$

(17)

This is indeed the non separable solution of (8). Therefore, the normalized eigenfunctions of the eigenvalue problem (5) are (ignoring the sign)

	$$\Phi_{n,k_{z}}({\bf x},t)=\frac{\sqrt{qB}}{\left(mc^{2}\hbar\omega_{c}\right)^{1/4}}e^{-i(\frac{qB}{\hbar c}xy-k_{z}z)}\psi_{n}\biggl{(}\frac{qB~{}x}{\sqrt{mc^{2}\hbar\omega_{c}}}\biggr{)}\mathchar 46\relax$$		(18a)
and
	$$E_{n,k_{z}}=\hbar\omega_{c}(n+\frac{1}{2})+\frac{\hbar^{2}k_{z}^{2}}{2m}\mathchar 46\relax$$		(18b)

These eigenvalues represent just the Landau’s levels , but its solution (18a) is totally different to that given by Landau since it is of non separable type. Note that there is not displacement at all in the harmonic oscillation solution. Now, assuming a periodicity in the z-direction, $\Phi_{n,k_{z}}({\bf x},t)=\Phi_{n,k_{z}}(x,y,z+L_{z},t)$, the usual condition $k_{z}L_{z}=2\pi n^{\prime},\quad n^{\prime}\in{\cal Z}$ makes the eigenvalues to be written as and the general solution of Schrödinger’s equation (3) can be written as

$$E_{n,n^{\prime}}=\hbar\omega_{c}(n+1/2)+\frac{\hbar^{2}2\pi^{2}}{mL_{z}^{2}}n^{\prime 2}\mathchar 46\relax$$

(19)

We must observed that this quantum numbers correspond to the degree of freedom in the “y (n).and “z(n’)”directions. The quantization conditions of the magnetic flux appears rather naturally since by asking periodicity in the y direction $\Psi({\bf x},t)=\Psi(x,y+L_{y},z,t)$, this one must be satisfied for any $x\in[0,L_{x}]$. So, in particular for $x=L_{x}$. Thus, it follows from the phase term that

$$\frac{qBL_{x}L_{y}}{\hbar c}=2\pi j,\quad\quad j\in{\cal Z},$$

(20)

where $BL_{x}L_{y}$ is the magnetic flux crossing the surface with area $L_{x}L_{y}$, and $\hbar c/q$ is the so called quantum flux [22]. Then, equation (18a) is

$$\Phi_{nn^{\prime}j}({\bf x},t)=\frac{\sqrt{qB}}{\left(mc^{2}\hbar\omega_{c}\right)^{1/4}}e^{-i(\frac{2\pi j}{L_{x}L_{y}}xy-\frac{2\pi n^{\prime}}{L_{z}}z)}\psi_{n}\biggl{(}\frac{qB~{}x}{\sqrt{mc^{2}\hbar\omega_{c}}}\biggr{)}\mathchar 46\relax$$

(21)

The degeneration of the eigenvalues (19) comes from the degree of freedom in “x.and can be obtained by making use the following quasi-classical argument: given the energy of the harmonic oscillator $E_{o}=\hbar\omega_{c}(n+1/2)$, we know the the maximum displacement of the particle (classically) is given by $x_{max}=\pm\sqrt{2E_{o}/m\omega_{c}^{2}}$, and since the periodicity in the variable ‘y”mentioned before is valid for any “x”value, we must have that the maximum value of the quantum number “j”must be

$$\Delta j=\frac{qBL_{y}}{\pi\hbar c}x_{max}=\frac{qBL_{y}}{\pi\hbar c}\sqrt{\frac{2\hbar(n+1/2)}{m\omega_{c}}},$$

(22)

and this represents the degeneration, $D(n)$, we have in the system

$$D(n)=\biggl{[}\frac{qBL_{y}}{\pi\sqrt{mc^{2}\hbar\omega_{c}}}\sqrt{2n+1}\biggr{]}\mathchar 46\relax$$

(23)

where $[\xi]$ means the integer part of the number $\xi$. Therefore, the general solution (absorbing the sign in the constants) is

$$\Psi({\bf x},t)=\sum_{n,^{\prime}n}\sum_{j=0}^{D(n)}C_{nn^{\prime}j}\sqrt{\frac{2\pi j}{L_{x}L_{y}}}\left(\frac{\hbar}{m\omega_{c}}\right)^{1/4}e^{-i(\frac{2\pi j}{L_{x}L_{y}}xy-\frac{2\pi n^{\prime}}{L_{z}}z)}e^{-i\frac{E_{n,n^{\prime}}}{\hbar}t}\psi_{n}\biggl{(}\sqrt{\frac{\hbar}{m\omega_{c}}}\left(\frac{2\pi j}{L_{x}L_{y}}\right)x\biggr{)},$$

(24)

where the constants $C_{nn^{\prime}j}$ must satisfy that $\sum_{n,n^{\prime},j}|C_{nn^{\prime}j}|^{2}=1$. The Landau’s levels $E_{n,n^{\prime}}$ are given by expression (19).

This case is illustrated on the next figure,

where the magnetic and electric constant fields are given by ${\bf B}=(0,0,B)$ and ${\bf E}=(0,{\cal E},0)$. We select Landau’s gauge for the magnetic field such that the vector and scalar potentials are ${\bf A}=(-By,0,0)$ and $\phi=-{\cal E}y$. Then, our Hamiltonian is [23, 24, 25]

$$\hat{H}=\frac{(\hat{\bf p}-\frac{q}{c}{\bf A})^{2}}{2m}+q\phi({\bf x},)$$

(25)

and the Schrödinger’s equation,

$$i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi,$$

(26)

is written as

$$i\hbar\frac{\partial\Psi}{\partial t}=\left\{\frac{1}{2m}\biggl{(}\hat{p}_{x}+\frac{qB}{c}y\biggr{)}^{2}+\frac{\hat{p}_{y}^{2}}{2m}+\frac{\hat{p}_{z}^{2}}{2m}\right\}\Psi-q{\cal E}y\Psi\mathchar 46\relax$$

(27)

Using the definition $\hat{p}_{j}=-i\hbar\partial/\partial x_{j}$ and the commutation relation $[x_{k},\hat{p}_{j}]=i\hbar\delta_{j}k$, the above expression is written as the following partial differential equation

$$i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}-i\frac{qB\hbar}{mc}\frac{\partial\Psi}{\partial x}+\frac{q^{2}B^{2}}{2mc^{2}}y^{2}\Psi-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial y^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial z^{2}}-q{\cal E}y\Psi\mathchar 46\relax$$

(28)

Taking the Fourier transformation with respect the x-variable, $\hat{\Psi}(k,y,z,t)={\cal F}_{x}[\Psi({\bf x},t)]$, the resulting expression is

$$i\hbar\frac{\partial\hat{\Psi}}{\partial t}=\left[\frac{\hbar^{2}k^{2}}{2m}-\biggl{(}\frac{qB\hbar k}{mc}+q{\cal E}\biggr{)}y+\frac{q^{2}B^{2}}{2mc^{2}}y^{2}\right]\hat{\Psi}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}{\hat{\Psi}}}{\partial y^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\hat{\Psi}}{\partial z^{2}}\mathchar 46\relax$$

(29)

By proposing a solution of the form

$$\hat{\Psi}(k,yz,t)=e^{-iEt/\hbar+ik_{z}z}\Phi(k,y)$$

(30)

and after some rearrangements, the resulting equation for $\Phi$ is

$$-\frac{\hbar^{2}}{2m}\frac{d^{2}\Phi}{dy^{2}}+\frac{1}{2}m\omega_{c}^{2}(y-y_{0})^{2}\Phi=E^{\prime}\Phi,$$

(31)

where $\omega_{c}$ is the cyclotron frequency (13a), and we have made the definitions

$$y_{0}=\frac{\hbar c}{qB}k+\frac{mc^{2}{\cal E}}{qB^{2}}$$

(32)

and

$$E^{\prime}=E-\frac{\hbar^{2}k^{2}}{2m}-\frac{\hbar^{2}k_{z}^{2}}{2m}+\frac{1}{2m}(\hbar k+\frac{mc{\cal E}}{B})^{2}\mathchar 46\relax$$

(33)

This equation is again the quantum harmonic oscillator on the variable “y”with a cyclotron frequency $\omega_{c}$ and displaced by a quantity $y_{0}$. Therefore, the solution (14) is

$$\Phi(k,y)=\psi_{n}\biggl{(}\sqrt{\frac{m\omega_{c}}{\hbar}}(y-y_{0})\biggr{)}$$

(34)

and

$$E^{\prime}_{n}=\hbar\omega_{c}(n+1/2)\mathchar 46\relax$$

(35)

Thus, the solution in the Fourier space is

$$\hat{\Psi}(k,y,z,t)=e^{-iE_{n,k_{z}}t/\hbar+ik_{z}z}\psi_{n}\biggl{(}\sqrt{\frac{m\omega_{c}}{\hbar}}(y-y_{0})\biggr{)}$$

(36)

with the energies $E_{n,k_{z}}$ given by

$$E_{n,k_{z}}=\hbar\omega_{c}(n+1/2)+\frac{\hbar^{2}k_{z}^{2}}{2m}-\frac{mc^{2}{\cal E}^{2}}{2B^{2}}-\frac{c{\cal E}\hbar}{B}k\mathchar 46\relax$$

(37)

The solution in the space-time is obtained by applying the inverse Fourier transformation,

$$\Psi_{n,k_{z}}({\bf x},t)={\cal F}[\hat{\Psi}_{n,k_{z}}(k,y,z,t)]=\frac{1}{\sqrt{2\pi}}\int_{\Re}e^{-ixk}\hat{\Psi}_{n,k_{z}}(k,y,z,t)dk,$$

(38)

which after a proper change of variable and rearrangements , we get the normalized function (ignoring the sign)

$$\Psi_{n,k_{z}}({\bf x},t)=\frac{\sqrt{qB}}{\left(mc^{2}\hbar\omega_{c}\right)^{1/4}}e^{-i\phi_{n,k_{z}}({\bf x},t)}\psi_{n}\biggl{(}\frac{qB}{\sqrt{mc^{2}\hbar\omega_{c}}}\bigl{(}x-\frac{c{\cal E}t}{B}\bigr{)}\biggr{)},$$

(39)

where the phase $\phi_{n,k_{z}}({\bf x},t)$ has been defined as

$$\phi_{n,k_{z}}({\bf x},t)=\left[\hbar\omega_{c}(n+1/2)+\frac{\hbar^{2}k_{z}^{2}}{2m}-\frac{mc^{2}{\cal E}^{2}}{2B^{2}}\right]\frac{t}{\hbar}-k_{z}z+\frac{qB}{\hbar c}\left(x-\frac{c{\cal E}t}{B}\right)\left(y-\frac{mc^{2}{\cal E}}{qB^{2}}\right)\mathchar 46\relax$$

(40)

asking for the periodicity with respect the variable “z”, $\Psi_{n,k_{z}}({\bf x},t)=\Psi_{n,k_{z}}(z,y,z+L_{z},t)$, it follows that $k_{z}L_{z}=2\pi n^{\prime}$ where $n^{\prime}$ is an integer number , and the above phase is now written as

$$\phi_{nn^{\prime}}({\bf x},t)=\left[\hbar\omega_{c}(n+1/2)+\frac{\hbar^{2}2\pi^{2}n^{\prime 2}}{mL_{z}^{2}}-\frac{mc^{2}{\cal E}^{2}}{2B^{2}}\right]\frac{t}{\hbar}-\frac{2\pi n^{\prime}}{L_{z}}z+\frac{qB}{\hbar c}\left(x-\frac{c{\cal E}t}{B}\right)\left(y-\frac{mc^{2}{\cal E}}{qB^{2}}\right)\mathchar 46\relax$$

(41)

Note from this expression that the term $e^{-i\phi({\bf x},t)}$ contains the element $e^{i\frac{qB}{\hbar c}xy}$, and by assuming the periodic condition $\Psi({\bf x},t)=\Psi(x,y+L_{y},z,t)$, will imply that $\Psi({\bf x},t)$ will be periodic with respect the variable “y”, for any “x.at any time “t.Ïn particular, this will be true for $x=L_{x}$. This bring about the quantization of the magnetic flux of the form

$$\frac{qBL_{x}L_{y}}{\hbar c}=2\pi j,\quad J\in{\cal Z}\ ,$$

(42)

obtaining the same expression as (20), and this phase is now depending of the quantum number “j”

$$\phi_{nn^{\prime}j}({\bf x},t)=e_{nn^{\prime}}t/\hbar-\frac{2\pi n^{\prime}}{L_{z}}z+\frac{2\pi j}{L_{x}L_{y}}xy-\frac{2\pi j}{L_{x}L_{y}}\left[\frac{mc^{2}{\cal E}}{qB^{2}}x+\frac{c{\cal E}}{B}ty\right]\mathchar 46\relax$$

(43)

where $e_{nn^{\prime}}$ is the energy associated to the system,

$$e_{n,n^{\prime}}=\hbar\omega_{c}(n+1/2)+\frac{2\pi^{2}\hbar^{2}}{mL_{z}^{2}}n^{\prime 2}+\frac{mc^{2}{\cal E}^{2}}{2B^{2}}\mathchar 46\relax$$

(44)

In this way, from these relations and the expression (39) we have a family of solutions $\{\Psi_{nn^{\prime}j}({\bf x},t)\}_{n,n^{\prime},j\in{\cal Z}}$ of the Schrödinger equation (27),

$$\Psi_{nn^{\prime}j}({\bf x},t)=\sqrt{\frac{2\pi j}{L_{x}L_{y}}}\left(\frac{\hbar}{m\omega_{c}}\right)^{1/4}e^{-i\phi_{nn^{\prime}j}({\bf x},t)}\psi_{n}\biggl{(}\sqrt{\frac{\hbar}{m\omega_{c}}}\left(\frac{2\pi j}{L_{x}L_{y}}\right)\bigl{(}x-\frac{c{\cal E}t}{B}\bigr{)}\biggr{)},$$

(45)

Now, by the same arguments we did in the previous case, the degeneration of the systems would be given by (23), and the general solution would be of the form

$$\Psi({\bf x},t)=\sum_{n,n^{\prime}}\sum_{j=0}^{D(n)}\widetilde{C}_{nn^{\prime}j}\Psi_{nn^{\prime}j}({\bf x},t)\mathchar 46\relax$$

(46)

The following figure shows this case.

The fields are of the form ${\bf B}=(0,B,0)$ and ${\bf E}=(0,{\cal E},0)$. The scalar and vector potentials are chosen as ${\bf A}=(Bz,0,0)$ and $\phi=-{\cal E}y$. The Shrödinger equation is for this case as

$$i\hbar\frac{\partial\Psi}{\partial t}=\left\{\frac{(\hat{p}_{x}-qBz/c)^{2}}{2m}+\frac{\hat{p}_{y}^{2}}{2m}+\frac{\hat{p}_{z}^{2}}{2m}-q{\cal E}y\right\}\Psi,$$

(47)

which defines the following partial differential equation

$$i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+i\frac{qB\hbar z}{mc}\frac{\partial\Psi}{\partial x}+\frac{q^{2}B^{2}}{2mc^{2}}z^{2}\Psi-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial y^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial z^{2}}-q{\cal E}y\Psi\mathchar 46\relax$$

(48)

Proposing a solution of the form $\Psi({\bf x},t)=e^{-iEt/\hbar}\Phi({\bf x})$, we get the following eigenvalue problem

$$E\Phi=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Phi}{\partial x^{2}}+i\frac{qB\hbar z}{mc}\frac{\partial\Phi}{\partial x}+\frac{q^{2}B^{2}}{2mc^{2}}z^{2}\Phi-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Phi}{\partial y^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Phi}{\partial z^{2}}-q{\cal E}y\Phi\mathchar 46\relax$$

(49)

Applying the Fourier transformation over the x-variable, $\hat{\Phi}(k,y,z)={\cal F}_{x}[\Phi({\bf x})]$, the following equation arises after some rearrangements

$$E\hat{\Phi}=\frac{(\hbar k+qBz/c)^{2}}{2m}\hat{\Phi}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\hat{\phi}}{\partial z^{2}}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\hat{\Phi}}{\partial y^{2}}-q{\cal E}y\hat{\Phi},$$

(50)

which can be written as

	$$-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\hat{\Phi}}{\partial z^{2}}+\frac{1}{2}m\omega_{c}(z+z_{0})^{2}\hat{\Phi}-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\hat{\Phi}}{\partial y^{2}}-q{\cal E}y\hat{\Phi},$$		(51a)
where $\omega_{c}$ is the cyclotron frequency (13a), and $z_{0}$ has been defined as
	$$z_{0}=\frac{\hbar c}{qB}k\mathchar 46\relax$$		(51b)

This equation admits a variable separable approach since by the proposition $\hat{\Phi}(k,y,z)=f(k,z)g(y)$, the following equations are bringing about

	$$-\frac{\hbar^{2}}{2m}\frac{d^{2}f}{dz^{2}}+\frac{1}{2}m\omega_{c}^{2}(z+z_{0})^{2}=E^{(1)}f$$		(52a)
and
	$$-\frac{\hbar^{2}}{2m}\frac{d^{2}g}{dy^{2}}-g{\cal E}yg=E^{(2)}g,$$		(52b)

where $E=E^{(1)}+E^{(2)}$. The solutions of these equations are, of course, the quantum harmonic oscillator and the quantum bouncer, which are given by

	$$f_{n}(k,z)=A_{n}e^{-\xi^{2}/2}H_{n}(\xi),\quad\quad\xi=\sqrt{\frac{m\omega_{c}}{\hbar}}(z+z_{0}),\quad\quad E_{n}^{(1)}=\hbar\omega_{c}(n+1/2)\mathchar 46\relax$$		(53a)
and
	$$g_{n^{\prime}}(y)=\frac{Ai(\tilde{\xi}-\tilde{\xi}_{n^{\prime}})}{|Ai^{\prime}(-\tilde{\xi}_{n^{\prime}})|},\quad\quad\tilde{\xi}=y/l,\quad\quad E_{n}^{(2)}=-q{\cal E}l\tilde{\xi}_{n^{\prime}},$$		(53b)

where $A_{n}=(m\omega_{c}/\pi\hbar)^{1/4}/\sqrt{2^{n}n!}$, $l=(\hbar^{2}/(-2mq{\cal E}))^{1/3}$, $Ai(-\tilde{\xi}_{n^{\prime}})=0$, and $Ai^{\prime}(\xi)$ is the differentiation of the Airy function. In this way, we have

$$\hat{\Phi}_{n,n^{\prime}}(k,y,z)=a_{n^{\prime}}\psi_{n}\biggl{(}\sqrt{\frac{m\omega_{c}}{\hbar}}(z+z_{0})\biggr{)}Ai(l^{-1}(y-y_{n^{\prime}})),\quad\quad E_{n,n^{\prime}}=\hbar\omega_{c}(n+1/2)-q{\cal E}y_{n^{\prime}},$$

(54)

where we have defined $a_{n^{\prime}}$ as $a_{n^{\prime}}=1/|Ai^{\prime}(-l^{-1}y_{n^{\prime}})|$. Now, the inverse Fourier transformation will affect only the quantum harmonic oscillator function $\psi_{n}$ through the k-dependence on the parameter $z_{0}$, and the resulting expression is

$$\Phi_{n,n^{\prime}}({\bf x})=\frac{a_{n^{\prime}}qB}{\sqrt{mc^{2}\hbar\omega_{c}}}e^{i\frac{qB}{\hbar c}xz}\psi_{n}\biggl{(}\frac{qBx}{\sqrt{mc^{2}\hbar\omega_{c}}}\biggr{)}Ai\bigl{(}l^{-1}(y-y_{n^{\prime}})\bigr{)}\mathchar 46\relax$$

(55)

Now, asking for the periodicity condition of the above solution with respect the z-variable, $\Psi({\bf x},t)=\Psi(x,y,z+L_{z},t)$, the periodicity must satisfy for any x-values, and in particular for $x=L_{x}$. Thus it follows the quantization expression for the magnetic flux

$$\frac{qBL_{x}L_{z}}{\hbar c}=2\pi j,\quad\quad j\in{\cal Z}\mathchar 46\relax$$

(56)

Using the same arguments shown above for the degeneration of the system, we have the same expression (23) for the degeneration of the system and the function (55) is given by (normalized)

$$\Phi_{nn^{\prime}j}({\bf x})=a_{n^{\prime}}\sqrt{\frac{2\pi j}{L_{x}L_{y}}}\left(\frac{\hbar}{m\omega_{c}}\right)^{1/4}e^{i\frac{2\pi j}{L_{x}L_{z}}xz}\psi_{n}\biggl{(}\sqrt{\frac{\hbar}{m\omega_{c}}}\left(\frac{2\pi j}{L_{x}L_{y}}\right)x\biggr{)}Ai\bigl{(}l^{-1}(y-y_{n^{\prime}})\bigr{)}\mathchar 46\relax$$

(57)

Then, we have obtained a family of solution of the Schrödinger equation (48),

$$\Psi_{n,n^{\prime}}({\bf x},t)=e^{-iE_{n,n^{\prime}}t/\hbar}\Phi_{nn^{\prime}j}({\bf x}),$$

(58)

where the energies $E_{n,n^{\prime}}$ are given by the expression (54). The general solution of (48) can be written as

$$\Psi({\bf x},t)=\sum_{n,n^{\prime}}\sum_{j=0}^{D(n)}C_{n,n^{\prime}}^{*}e^{-iE_{n,n^{\prime}}t/\hbar}e^{i\frac{2\pi j}{L_{x}L_{z}}xz}\tilde{u}_{n,n^{\prime}}(x,y),$$

(59)

with the condition $\sum_{n,n^{\prime}}|C_{n,n^{\prime}}^{*}|^{2}=1$, and where it has been defined the functions $\tilde{u}_{n,n^{\prime}}$ as

$$\tilde{u}_{n,n^{\prime}}(x,y)=a_{n^{\prime}}\sqrt{\frac{2\pi j}{L_{x}L_{y}}}\left(\frac{\hbar}{m\omega_{c}}\right)^{1/4}\psi_{n}\biggl{(}\sqrt{\frac{\hbar}{m\omega_{c}}}\left(\frac{2\pi j}{L_{x}L_{y}}\right)x\biggr{)}Ai\bigl{(}l^{-1}(y-y_{n^{\prime}})\bigr{)}\mathchar 46\relax$$

(60)