Analytical Approach for the Case B = (0, 0, ) where p is the generalized linear momentum, A is the magnetic potential such that B = ?×A, and "c" is the speed of light. We can choose the Landau's gauge to have the vector potential of the form A = (?By, 0, 0). Therefore, the Hamiltonian has the following form To quantize the system, we need to solve the Schrödinger's equation [20] i ?? ?t = (?p x + qBy/c) 2 2m + p2
H = (p x + qBy/c) 2 2m + p 2 y 2m + p 2 z 2m .(2)y 2m + p2 z 2m ?.(3)
where ? = ?(x, t) is the wave function, is the Plank's constant divided by 2?, pi are the momentum operators such that [x i , pj ] = i ? ij . Now, the argument used by Landau is that due to commutation relation [p x , ?] = 0, between the operators px and the Hamiltonian ? (implying that px is a constant of motion), it is possible to replace this component of the momentum by 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 ? does not depend explicitly on time, the proposition ?(x, t) = e ?iEt/ ?(x) (4) reduces the equation to an eigenvalue problem
H? = E?.(5)
Then, this equation is written as
1 2m p2 x + 2qB c y px + q 2 B 2 c 2 y 2 + p2 y 2m + p2 z 2m ? = E?.(6)
The variable "z" is separable through the proposition
?(x) = ?(x, y)e ?ikzz , k z ? ,(7)
resulting the following equation
1 2m p2 x + 2qB c y px + q 2 B 2 c 2 y 2 + p2 y 2m ? = E ?, (8)
where E is
E = E ? 2 k 2 x 2m .(9)
That is, the resulting partial differential equation is of the form
1 2m ? 2 ? 2 ? ?x 2 ? i 2qB c y ?? ?x + q 2 B2 c 2 y 2 ? ? 2 2m ? 2 ? ?y 2 = E ?. (10)
This equation does not admit a separable variable solution (?(x, y) = f (x)g(y)) as Landau' solution is, but we can use Fourier transformation [21] on the variable "x",
?(k, y) = F[?] = 1 ? 2? e ikx ?(x, y)dx,(11)
to solve this equation. Applying Fourier transformation to this equation, knowing its property F[??/?x] = (?ik) ?, we get the ordinary differential equation
? 2 2m d 2 ? dy 2 + m 2 ? 2 c (y ? y 0 ) 2 ? = E ?, (12)
where ? c is the cyclotron frequency
? c = qB mc (13a)
and y 0 is the displacement parameter
y 0 = c qB k. (13b)
This equation is just the quantum harmonic oscillator in the "y" direction displaced by a amount y 0 . So, the solution is
?n (k, y) = n (?), ? = m? c (y ? y 0 ), ? n (?) = A n e ?? 2 H n (?),(14)
being H n (?) the Hermit polynomials, and A n is a constant of normalization, An= (m? c /? ) 1/4 / ? 2 n n!. and
E n = ? c (n + 1/2). (15)
Now, the solution in the real space ? n (x, y) is gotten by using the inverse Fourier transformation,
? n (x, y) = F ?1 [? n (k, y)] = 1 ? 2? e ?ikx n m? c (y ? ck/qB) dk. (16)
Making the change of variable ? = m? c / (y ? ck/qB), and knowing that the Fourier transformation of the harmonic oscillator solution is another harmonic oscillator solution, we get
? n (x, y) = ?qB ? mc 2 ? c e ?i qB c xy n qB x ? mc 2 ? c .(17)
This is indeed the non separable solution of (8). Therefore, the normalized eigenfunctions of the eigenvalue problem ( 5) are (ignoring the sign)
? n,kz (x, t) = ? qB (mc 2 ? c ) 1/4 e ?i( qB c xy?kzz) n qB x ? mc 2 ? c . (18a)
and
E n,kz = ? c (n + 1 2 ) + 2 k 2 z 2m . (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, ? n,kz (x, t) = ? n,kz (x, y, z + L z , t), the usual condition k z L z = 2?n , n ? Z makes the eigenvalues to be written as and the general solution of Schrödinger's equation ( 3) can be written as
E n,n = ? c (n + 1/2) + 2 2? 2 mL 2 z n 2 . (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 ?(x, t) = ?(x, y + L y , z, t), this one must be satisfied for any x ? [0, L x ]. So, in particular for x = L x . Thus, it follows from the phase term that
qBL x L y c = 2?j, j ? Z,(20)
where BL x L y is the magnetic flux crossing the surface with area L x L y , and c/q is the so called quantum flux [22]. Then, equation (18a) is 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 = ? c (n + 1/2), we know the the maximum displacement of the particle (classically) is given by x max = ± 2E o /m? 2 c , 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 ?j = qBL y ? c
? nn j (x, t) = ? qB (mc 2 ? c ) 1/4 e ?i( 2?j Lx Ly xy? 2?n Lz z) n qB x ? mc 2 ? c .(21
x max = qBL y ? c
2 (n + 1/2) m? c ,(22)
and this represents the degeneration, D(n), we have in the system
D(n) = qBL y ? ? mc 2 ? c ? 2n + 1 . (23)
where [?] means the integer part of the number ?. Therefore, the general solution (absorbing the sign in the constants) is
?(x, t) = n, n D(n) j=0 C nn j 2?j L x L y m? c 1/4 e ?i( 2?j Lx Ly xy? 2?n Lz z) e ?i E n,n t n m? c 2?j L x L y x ,(24)
where the constants C nn j must satisfy that n,n ,j |C nn j | 2 = 1. The Landau's levels E n,n are given by expression (19).

This case is illustrated on the next figure , where the magnetic and electric constant fields are given by B = (0, 0, B) and E = (0, E, 0). We select Landau's gauge for the magnetic field such that the vector and scalar potentials are A = (?By, 0, 0) and ? = ?Ey. Then, our Hamiltonian is [23][24][25] 
? = (p ? q c A) 2 2m + q?(x, )(25)
and the Schrödinger's equation,
i ?? ?t = ??,(26)
is written as
i ?? ?t = 1 2m px + qB c y 2 + p2 y 2m + p2 z 2m ? ? qEy?. (27)
III.

Analytical Approach for the Case B E Using the definition pj = ?i ?/?x j and the commutation relation [x k , pj ] = i ? j k, the above expression is written as the following partial differential equation
i ?? ?t = ? 2 2m ? 2 ? ?x 2 ? i qB mc ?? ?x + q 2 B 2 2mc 2 y 2 ? ? 2 2m ? 2 ? ?y 2 ? 2 2m ? 2 ? ?z 2 ? qEy?.(28)
Taking the Fourier transformation with respect the x-variable, ?(k, y, z, t) = F x [?(x, t)], the resulting expression is
i ? ? ?t = 2 k 2 2m ? qB k mc + qE y + q 2 B 2 2mc 2 y 2 ? ? 2 2m ? 2 ? ?y 2 ? 2 2m ? 2 ? ?z 2 . (29)
By proposing a solution of the form
?(k, yz, t) = e ?iEt/ +ikzz ?(k, y)(30)
and after some rearrangements, the resulting equation for ? is
? 2 2m d 2 ? dy 2 + 1 2 m? 2 c (y ? y 0 ) 2 ? = E ?, (31)
where ? c is the cyclotron frequency (13a), and we have made the definitions
y 0 = c qB k + mc 2 E qB 2(32) and (33)
This equation is again the quantum harmonic oscillator on the variable "y" with a cyclotron frequency ? c and displaced by a quantity y 0 . Therefore, the solution ( 14) is
? (34)
and
E n = ? c (n + 1/2). (35)
Thus, the solution in the Fourier space is
?(k, y, z, t) = e ?iE n,kz t/ +ikzz ? n m? c (y ? y 0 ) (36)
with the energies E n,kz given by
E n,kz = ? c (n + 1/2) + 2 k 2 z 2m ? mc 2 E 2 2B 2 ? cE B k. (37) ? n,kz (x, t) = F[ ?n,kz (k, y, z, t)] = 1 ? 2? e ?ixk ?n,kz (k, y, z, t)dk,(38)
which after a proper change of variable and rearrangement, we get the normalized function (ignoring the sign)
? n,kz (x, t) = ? qB (mc 2 ? c ) 1/4 e ?i? n,kz (x,t) ? n qB ? mc 2 ? c x ? cEt B ,(39)
The solution in the space-time is obtained by applying the inverse Fourier transformation, where the phase ? n,kz (x, t) has been defined as
E = E ? 2 k 2 2m ? 2 k 2 z 2m + 1 2m ( k + mcE B )2? n,kz (x, t) = ? c (n + 1/2) + 2 k 2 z 2m ? mc 2 E 2 2B 2 t ? k z z + qB c x ? cEt B y ? mc 2 E qB 2 . (40)
asking for the periodicity with respect the variable "z", ? n,kz (x, t) = ? n,kz (z, y, z + L z , t), it follows that k z L z = 2?n where n is an integer number , and the above phase is now written as
? nn (x, t) = ? c (n + 1/2) + 2 2? 2 n 2 mL 2 z ? mc 2 E 2 2B 2 t ? 2?n L z z + qB c x ? cEt B y ? mc 2 E qB 2 . (41)
Note from this expression that the term e ?i?(x,t) contains the element e i qB c xy , and by assuming the periodic condition ?(x, t) = ?(x, y + L y , z, t), will imply that ?(x, t) will be periodic with respect the variable "y", for any "x" at any time "t." In particular, this will be true for x = L x . This bring about the quantization of the magnetic flux of the form
qBL x L y c = 2?j, J ? Z ,(42)
obtaining the same expression as (20), and this phase is now depending of the quantum number "j"
? nn j (x, t) = e nn t/ ? 2?n L z z + 2?j L x L y xy ? 2?j L x L y mc 2 E qB 2 x + cE B ty . (43)
where e nn is the energy associated to the system,
e n,n = ? c (n + 1/2) + 2? 2 2 mL 2 z n 2 + mc 2 E 2 2B 2 . (44)
In this way, from these relations and the expression (39) we have a family of solutions {? nn j (x, t)} n,n ,j?Z of the Schrödinger equation ( 27),
? nn j (x, t) = 2?j L x L y m? c 1/4
e ?i? nn j (x,t)
? n m? c 2?j L x L y x ? cEt B ,(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
?(x, t) = n,n D(n) j=0 C nn j ? nn j (x, t). (46)
The following figure shows this case.

IV.

Analytical approach for the case B E The fields are of the form B = (0, B, 0) and E = (0, E, 0). The scalar and vector potentials are chosen as A = (Bz, 0, 0) and ? = ?Ey. The Shrödinger equation is for this case as
i ?? ?t = (p x ? qBz/c) 2 2m + p2 y 2m + p2 z 2m ? qEy ?,(47)
which defines the following partial differential equation
i ?? ?t = ? 2 2m ? 2 ? ?x 2 + i qB z mc ?? ?x + q 2 B 2 2mc 2 z 2 ? ? 2 2m ? 2 ? ?y 2 ? 2 2m ? 2 ? ?z 2 ? qEy?. (48)
Proposing a solution of the form ?(x, t) = e ?iEt/ ?(x), we get the following eigenvalue problem
E? = ? 2 2m ? 2 ? ?x 2 + i qB z mc ?? ?x + q 2 B 2 2mc 2 z 2 ? ? 2 2m ? 2 ? ?y 2 ? 2 2m ? 2 ? ?z 2 ? qEy?. (49)
Applying the Fourier transformation over the x-variable, ?(k, y, z) = F x [?(x)], the following equation arises after some rearrangements
E ? = ( k + qBz/c) 2 2m ? ? 2 2m ? 2 ? ?z 2 ? 2 2m ? 2 ? ?y 2 ? qEy ?, (50)
which can be written as
? 2 2m ? 2 ? ?z 2 + 1 2 m? c (z + z 0 ) 2 ? ? 2 2m ? 2 ? ?y 2 ? qEy ?,(51a)
where ? c is the cyclotron frequency (13a), and z 0 has been defined as
z 0 = c qB k. (51b)
This equation admits a variable separable approach since by the proposition ?(k, y, z) = f (k, z)g(y), the following equations are bringing about
? 2 2m d 2 f dz 2 + 1 2 m? 2 c (z + z 0 ) 2 = E (1) f(52a)
and
? 2 2m d 2 g dy 2 ? gEyg = 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 ?? 2 /2 H n (?), ? = m? c (z + z 0 ), E(1)n = ? c (n + 1/2). (53a)
and
g n (y) = Ai( ? ? ?n ) |Ai (? ?n )| , ? = y/l, E (2) n = ?qEl ?n ,(53b)
where
A n = (m? c /? ) 1/4 / ? 2 n n!, l = ( 2 /(?2mqE)) 1/3
, Ai(? ?n ) = 0, and Ai (?) is the differentiation of the Airy function. In this way, we have where we have defined a n as a n = 1/|Ai (?l ?1 y n )|. Now, the inverse Fourier transformation will affect only the quantum harmonic oscillator function n through the k-dependence on the parameter z 0 , and the resulting expression is
?n,n (k, y, z) = a n n m? c (z + z 0 ) Ai(l ?1 (y ? y n )), E n,n = ? c (n + 1/2) ? qEy n ,(54) Global? n,n (x) = a n qB ? mc 2 ? c e i qB c xz n qBx ? mc 2 ? c Ai l ?1 (y ? y n ) .(55)
Now, asking for the periodicity condition of the above solution with respect the z-variable, ?(x, t) = ?(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
qBL x L z c = 2?j, j ? Z.(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 (55) is given by (normalized)
? nn j (x) = a n 2?j L x L y m? c 1/4 e i 2?j Lx Lz xz n m? c 2?j L x L y x Ai l ?1 (y ? y n ) .(57)
Then, we have obtained a family of solution of the Schrödinger equation ( 48),
? n,n (x, t) = e ?iE n,n t/ ? nn j (x),(58)
where the energies E n,n are given by the expression (54). The general solution of (48) can be written as 
?(x, t) = n,n D(n) j=0 C * n,n e ?iE n,n t/ e i 2?j Lx Lz xz ?n,n (x, y),(59)
Let us consider the magnetic gauge given such that the vector potential is of the form A = (0, 0, ?Bx), and the potential is the same ? = ?Ey. Passing directly to the eigenvalue problem for the Schrödinger equation when we select the wave function of the form ?(x, t) = e ?iEt/ ?(x), the resulting equation is
? 2 2m ? 2 ? ?x 2 ? 2 2m ? 2 ? ?y 2 ? 2 2m ? 2 ? ?z 2 ? i qB mc x ?? ?z + q 2 B 2 2mc 2 x 2 ? ? qEy? = E?.(61)
Taking the Fourier transformation with respect the z-variable, ?(x, y, k) = F z [?(x)], and making some rearrangements, it follows that
? 2 2m ? 2 ? ?x 2 + 1 2m k ? qB c x 2 ? ? 2 2m ? 2 ? ?y 2 ? qEy ? = E ?. (62)
This equation admits a variable separable solution of the form ?(x, y, k) = ? 1 (k, x)? 2 (y), where the functions ? 1 and ? 2 satisfy the equations
? 2 2m d 2 ? 1 dx 2 + ( k ? qB c x) 2 2m ? 1 = E (1) ? 1(63)
and where E = E (1) + E (2) . The solution of these equations are
? 2 2m ? 2 ? 2 ?y 2 ? qEy? 2 = E (2) ? 2 ,(64? 1n (k, x) = n (?) = A n e ?? 2 /2 H n (?), ? = m? c (x ? x 0 ), E (1) n = ? c (n + 1/2)(65)
and
? 2n (y) = a n Ai(l ?1 (y ? y n )), l = 2 ?2mqE 1/3 , E(2)n = ?qEy n ,(66)
where ? c is the cyclotron frequency (13a), x 0 is the displacement x 0 = ck/qB, a n = 1/|Ai (l ?1 y n )| is a constant, and A n the constant associated to the quantum harmonic oscillator solution. The inverse Fourier transformation affect only the function ? 1 , and we have
? 1n (z, x) = F ?1 [? 1n (k, x)] = ?qB ? mc 2 ? c e ?i qB c xz n qBz ? mc 2 ? c .(67)
The periodic condition on the variable "x", ?(x, t) = ?(x + L x , y, z, t), for any value of the other variables, implies that this will happen in particular for the value of z = L z . So, we get the quantization of the magnetic flux (BL x L y ),
qBL x L z c = 2?j, j ? Z.(68)
Thus, we have a family of solutions {? nn j (x, t)} of the Shcrödinger equation of the form
? nn j (x, t) = e ?iE n,n t/ ? nn j (x),(69)
or (normalized and ignoring the sign)
? nn j (x, t) = a n 2?j L x L y m? c 1/4 e ?i(E n,n t + 2?j Lx Lz xz) n m? c 2?j L x L y z Ai(l ?1 (y ? y n )). (70)
By the same arguments about the degenerationn of the systems, the general solution is just a combination of all of these, ?(x, t) = n,n A nn j e ?i(E n,n t + 2?j Lx Lz xz) v nn j (y, z),

where the condition n,n |A nn j | 2 = 1 must be satisfied, and the function v nn j is given by
v nn j (y, z) = a n 2?j L x L y m? c 1/4 n m? c 2?j L x L y z Ai l ?1 (y ? y n ) . (72)
We have studied the quantization of a charged particle in a flat box and under constants magnetic and electric fields for several cases and have shown that a full separation of variable solution is not admitted in these cases (contrary to Landau's solution in one of these cases). This situation arises since the commutation of a component of the generalized linear momentum operator with the Hamiltonian of the system does not imply necessarily that a variable separation of its associated variable must exist in the Schrödinger equation. However, using the Fourier transformation, we were be able to find the full solution of the problems. As expected, Landau's level appears in all these cases, and a characteristic phase which help us to find the quantization of the magnetic flux in a natural way. We consider that the approach given here maybe very useful to understand quantum Hall effect and related phenomena.

V. 


# Conclusions and Comments
1![Figure 1: Electric charged in a at box with magnetic field](image-2.png "Figure 1 :")
![Charged Particle in a Flat Box with Static Electromagnetic field and Landau's Levels](image-3.png "")
![Charged Particle in a Flat Box with Static Electromagnetic field and Landau's Levels](image-4.png "")
2![Figure 2: Electric charged in a flat box with magnetic and electric fields](image-5.png "Figure 2 :")
![. ?(k, y) = n m? c (y ? y 0 ) Charged Particle in a Flat Box with Static Electromagnetic field and Landau's Levels](image-6.png "")
3![Figure 3: Electric charged in a flat box with parallel electric and magnetic fields](image-7.png "Figure 3 :")
![Charged Particle in a Flat Box with Static Electromagnetic field and Landau's Levels](image-8.png "")



		
		
			

				


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