Charged particle in a flat box with static electromagnetic field and Landau's levels
CCharged particle in a flat box with staticelectromagnetic field and Landau’s levels
Gustavo V. L´opez * , Jorge A. Lizarraga ** .Departamento de F´ısica, Universidad de Guadalajara,Blvd. Marcelino Garc´ıa Barragan y Calzada Ol´ımpica,CP 44200, Guadalajara, Jalisco, M´exico,2 de abril de 2020 Resumen
We study the quantization of the motion of a charged particle without spin inside a flat box under astatic electromagnetic field. Contrary to Landau’s solution with constant magnetic field transverse to thebox, we found a non separable variables solution for the wave function, and this fact remains when staticelectric field is added. However, the Landau’s Levels appear in all cases.
Key words:
Landau’s Levels, quantum Hall effect.
PACS: * [email protected] ** [email protected] a r X i v : . [ phy s i c s . g e n - ph ] M a r . Introduction Landau’ solution [1] of a charged particle in a flat surface with magnetic field has become of greatimportance in understanding integer hall effect [2–6], fractional Hall effect [6–9], and topological insulators[10–16]. This last elements promise to become essential for future nanotechnology devices [17–19]. Due tothis considerable application of the Landau’s levels, it is worth to re-study this problem and its variationswith an static electric field. In this paper, we show that there exists a non separable solution for this typeof quantum problems, but having the same Landau’s levels. In our cases, instead of having a flat surface, weconsider to have a flat box with lengths L x , L y , and L z such that L z (cid:28) L x , L y
2. Analytical approach for the case B = (0 , , B ) Let us consider a charged particle “q”with mass “m¨ın a flat box with a constant magnetic field orthogonalto the flat surface, B = (0 , , B ), as shown in the next figure.Figura 1: Electric charged in a flat box with magnetic fieldFor a non relativistic charged particle, the Hamiltonian of the system (units CGS) is H = ( p − q A /c ) m , (1)where p is the generalized linear momentum, A is the magnetic potential such that B = ∇ × A , and “c¨ıs thespeed of light. We can choose the Landau’s gauge to have the vector potential of the form A = ( − By, , H = ( p x + qBy/c ) m + p y m + p z m . (2)To quantize the system, we need to solve the Schr¨odinger’s equation [20] i (cid:126) ∂ Ψ ∂t = (cid:40) (ˆ p x + qBy/c ) m + ˆ p y m + ˆ p z m (cid:41) Ψ . (3)where Ψ = Ψ( x , t ) is the wave function, (cid:126) is the Plank’s constant divided by 2 π , ˆ p i are the momentumoperators such that [ x i , ˆ p j ] = i (cid:126) δ ij . Now, the argument used by Landau is that due to commutation relation2ˆ p x , ˆ H ] = 0, between the operators ˆ p x and the Hamiltonian ˆ H (implying that ˆ p x is a constant of motion), itis possible to replace this component of the momentum by (cid:126) k x , having a solution for the eigenvalue problemof separable variable type, f ( t ) f ( x ) f ( y ) f ( z ). However, we will see that this type of commutation does notimply necessarily separability of the solution. Since the Hamiltonian ˆ H does not depend explicitly on time,the proposition Ψ( x , t ) = e − iEt/ (cid:126) Φ( x ) (4)reduces the equation to an eigenvalue problem (cid:98) H Φ = E Φ . (5)Then, this equation is written as (cid:40) m (cid:18) ˆ p x + 2 qBc y ˆ p x + q B c y (cid:19) + ˆ p y m + ˆ p z m (cid:41) Φ = E Φ . (6)The variable “z¨ıs separable through the propositionΦ( x ) = φ ( x, y ) e − ik z z , k z ∈ (cid:60) , (7)resulting the following equation (cid:40) m (cid:18) ˆ p x + 2 qBc y ˆ p x + q B c y (cid:19) + ˆ p y m (cid:41) φ = E (cid:48) φ, (8)where E (cid:48) is E (cid:48) = E − (cid:126) k x m . (9)That is, the resulting partial differential equation is of the form12 m (cid:26) − (cid:126) ∂ φ∂x − i qB (cid:126) c y ∂φ∂x + q B c y φ (cid:27) − (cid:126) m ∂ φ∂y = E (cid:48) φ. (10)This equation does not admit a separable variable solution ( φ ( x, y ) = f ( x ) g ( y )) as Landau’ solution is, butwe can use Fourier transformation [21] on the variable “x”,ˆ φ ( k, y ) = F [ φ ] = 1 √ π (cid:90) (cid:60) e ikx φ ( x, y ) dx, (11)to solve this equation. Applying Fourier transformation to this equation, knowing its property F [ ∂φ/∂x ] =( − ik ) ˆ φ , we get the ordinary differential equation − (cid:126) m d ˆ φdy + m ω c ( y − y ) ˆ φ = E (cid:48) ˆ φ, (12)where ω c is the cyclotron frequency ω c = qBmc (13a)and y is the displacement parameter y = (cid:126) cqB k. (13b)This equation is just the quantum harmonic oscillator in the “y”direction displaced by a amount y . So, thesolution is ˆ φ n ( k, y ) = ψ n ( ξ ) , ξ = (cid:114) mω c (cid:126) ( y − y ) , ψ n ( ξ ) = A n e − ξ H n ( ξ ) , (14)3eing H n ( ξ ) the Hermit polynomials, and A n is a constant of normalization, An = ( mω c /π (cid:126) ) / / √ n n !. and E (cid:48) n = (cid:126) ω c ( n + 1 / . (15)Now, the solution in the real space φ n ( x, y ) is gotten by using the inverse Fourier transformation, φ n ( x, y ) = F − [ φ n ( k, y )] = 1 √ π (cid:90) (cid:60) e − ikx ψ n (cid:18)(cid:114) mω c (cid:126) ( y − (cid:126) ck/qB ) (cid:19) dk. (16)Making the change of variable σ = (cid:112) mω c / (cid:126) ( y − (cid:126) ck/qB ), and knowing that the Fourier transformation ofthe harmonic oscillator solution is another harmonic oscillator solution, we get φ n ( x, y ) = − qB √ mc (cid:126) ω c e − i qB (cid:126) c xy ψ n (cid:18) qB x √ mc (cid:126) ω c (cid:19) . (17)This is indeed the non separable solution of (8). Therefore, the normalized eigenfunctions of the eigenvalueproblem (5) are (ignoring the sign)Φ n,k z ( x , t ) = √ qB ( mc (cid:126) ω c ) / e − i ( qB (cid:126) c xy − k z z ) ψ n (cid:18) qB x √ mc (cid:126) ω c (cid:19) . (18a)and E n,k z = (cid:126) ω c ( n + 12 ) + (cid:126) k z m . (18b)These eigenvalues represent just the Landau’s levels , but its solution (18a) is totally different to that givenby Landau since it is of non separable type. Note that there is not displacement at all in the harmonicoscillation solution. Now, assuming a periodicity in the z-direction, Φ n,k z ( x , t ) = Φ n,k z ( x, y, z + L z , t ), theusual condition k z L z = 2 πn (cid:48) , n (cid:48) ∈ Z makes the eigenvalues to be written as and the general solution ofSchr¨odinger’s equation (3) can be written as E n,n (cid:48) = (cid:126) ω c ( n + 1 /
2) + (cid:126) π mL z n (cid:48) . (19)We must observed that this quantum numbers correspond to the degree of freedom in the “y (n). a nd“z(n’)”directions. The quantization conditions of the magnetic flux appears rather naturally since by askingperiodicity in the y direction Ψ( x , t ) = Ψ( x, y + L y , z, t ), this one must be satisfied for any x ∈ [0 , L x ]. So, inparticular for x = L x . Thus, it follows from the phase term that qBL x L y (cid:126) c = 2 πj, j ∈ Z , (20)where BL x L y is the magnetic flux crossing the surface with area L x L y , and (cid:126) c/q is the so called quantumflux [22]. Then, equation (18a) isΦ nn (cid:48) j ( x , t ) = √ qB ( mc (cid:126) ω c ) / e − i ( πjLxLy xy − πn (cid:48) Lz z ) ψ n (cid:18) qB x √ mc (cid:126) ω c (cid:19) . (21)The degeneration of the eigenvalues (19) comes from the degree of freedom in “x. a nd can be obtained bymaking use the following quasi-classical argument: given the energy of the harmonic oscillator E o = (cid:126) ω c ( n +1 / x max = ± (cid:112) E o /mω c ,and since the periodicity in the variable ‘y”mentioned before is valid for any “x”value, we must have thatthe maximum value of the quantum number “j”must be∆ j = qBL y π (cid:126) c x max = qBL y π (cid:126) c (cid:115) (cid:126) ( n + 1 / mω c , (22)4nd this represents the degeneration, D ( n ), we have in the system D ( n ) = (cid:20) qBL y π √ mc (cid:126) ω c √ n + 1 (cid:21) . (23)where [ ξ ] means the integer part of the number ξ . Therefore, the general solution (absorbing the sign in theconstants) isΨ( x , t ) = (cid:88) n, (cid:48) n D ( n ) (cid:88) j =0 C nn (cid:48) j (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / e − i ( πjLxLy xy − πn (cid:48) Lz z ) e − i En,n (cid:48) (cid:126) t ψ n (cid:18)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) x (cid:19) , (24)where the constants C nn (cid:48) j must satisfy that (cid:80) n,n (cid:48) ,j | C nn (cid:48) j | = 1. The Landau’s levels E n,n (cid:48) are given byexpression (19).
3. Analytical approach for the case B ⊥ E. This case is illustrated on the next figure,Figura 2: Electric charged in a flat box with magnetic and electric fieldswhere the magnetic and electric constant fields are given by B = (0 , , B ) and E = (0 , E , A = ( − By, ,
0) and φ = −E y . Then, our Hamiltonian is [23–25]ˆ H = (ˆ p − qc A ) m + qφ ( x , ) (25)and the Schr¨odinger’s equation, i (cid:126) ∂ Ψ ∂t = ˆ H Ψ , (26)is written as i (cid:126) ∂ Ψ ∂t = (cid:40) m (cid:18) ˆ p x + qBc y (cid:19) + ˆ p y m + ˆ p z m (cid:41) Ψ − q E y Ψ . (27)5sing the definition ˆ p j = − i (cid:126) ∂/∂x j and the commutation relation [ x k , ˆ p j ] = i (cid:126) δ j k , the above expression iswritten as the following partial differential equation i (cid:126) ∂ Ψ ∂t = − (cid:126) m ∂ Ψ ∂x − i qB (cid:126) mc ∂ Ψ ∂x + q B mc y Ψ − (cid:126) m ∂ Ψ ∂y − (cid:126) m ∂ Ψ ∂z − q E y Ψ . (28)Taking the Fourier transformation with respect the x-variable, ˆΨ( k, y, z, t ) = F x [Ψ( x , t )], the resultingexpression is i (cid:126) ∂ ˆΨ ∂t = (cid:20) (cid:126) k m − (cid:18) qB (cid:126) kmc + q E (cid:19) y + q B mc y (cid:21) ˆΨ − (cid:126) m ∂ ˆΨ ∂y − (cid:126) m ∂ ˆΨ ∂z . (29)By proposing a solution of the form ˆΨ( k, yz, t ) = e − iEt/ (cid:126) + ik z z Φ( k, y ) (30)and after some rearrangements, the resulting equation for Φ is − (cid:126) m d Φ dy + 12 mω c ( y − y ) Φ = E (cid:48) Φ , (31)where ω c is the cyclotron frequency (13a), and we have made the definitions y = (cid:126) cqB k + mc E qB (32)and E (cid:48) = E − (cid:126) k m − (cid:126) k z m + 12 m ( (cid:126) k + mc E B ) . (33)This equation is again the quantum harmonic oscillator on the variable “y”with a cyclotron frequency ω c and displaced by a quantity y . Therefore, the solution (14) isΦ( k, y ) = ψ n (cid:18)(cid:114) mω c (cid:126) ( y − y ) (cid:19) (34)and E (cid:48) n = (cid:126) ω c ( n + 1 / . (35)Thus, the solution in the Fourier space isˆΨ( k, y, z, t ) = e − iE n,kz t/ (cid:126) + ik z z ψ n (cid:18)(cid:114) mω c (cid:126) ( y − y ) (cid:19) (36)with the energies E n,k z given by E n,k z = (cid:126) ω c ( n + 1 /
2) + (cid:126) k z m − mc E B − c E (cid:126) B k. (37)The solution in the space-time is obtained by applying the inverse Fourier transformation,Ψ n,k z ( x , t ) = F [ ˆΨ n,k z ( k, y, z, t )] = 1 √ π (cid:90) (cid:60) e − ixk ˆΨ n,k z ( k, y, z, t ) dk, (38)which after a proper change of variable and rearrangements , we get the normalized function (ignoring thesign) Ψ n,k z ( x , t ) = √ qB ( mc (cid:126) ω c ) / e − iφ n,kz ( x ,t ) ψ n (cid:18) qB √ mc (cid:126) ω c (cid:0) x − c E tB (cid:1)(cid:19) , (39)6here the phase φ n,k z ( x , t ) has been defined as φ n,k z ( x , t ) = (cid:20) (cid:126) ω c ( n + 1 /
2) + (cid:126) k z m − mc E B (cid:21) t (cid:126) − k z z + qB (cid:126) c (cid:18) x − c E tB (cid:19) (cid:18) y − mc E qB (cid:19) . (40)asking for the periodicity with respect the variable “z”, Ψ n,k z ( x , t ) = Ψ n,k z ( z, y, z + L z , t ), it follows that k z L z = 2 πn (cid:48) where n (cid:48) is an integer number , and the above phase is now written as φ nn (cid:48) ( x , t ) = (cid:20) (cid:126) ω c ( n + 1 /
2) + (cid:126) π n (cid:48) mL z − mc E B (cid:21) t (cid:126) − πn (cid:48) L z z + qB (cid:126) c (cid:18) x − c E tB (cid:19) (cid:18) y − mc E qB (cid:19) . (41)Note from this expression that the term e − iφ ( x ,t ) contains the element e i qB (cid:126) c xy , and by assuming the periodiccondition Ψ( x , t ) = Ψ( x, y + L y , z, t ), will imply that Ψ( x , t ) will be periodic with respect the variable “y”,for any “x. a t any time “t.¨In particular, this will be true for x = L x . This bring about the quantization of themagnetic flux of the form qBL x L y (cid:126) c = 2 πj, J ∈ Z , (42)obtaining the same expression as (20), and this phase is now depending of the quantum number “j” φ nn (cid:48) j ( x , t ) = e nn (cid:48) t/ (cid:126) − πn (cid:48) L z z + 2 πjL x L y xy − πjL x L y (cid:20) mc E qB x + c E B ty (cid:21) . (43)where e nn (cid:48) is the energy associated to the system, e n,n (cid:48) = (cid:126) ω c ( n + 1 /
2) + 2 π (cid:126) mL z n (cid:48) + mc E B . (44)In this way, from these relations and the expression (39) we have a family of solutions { Ψ nn (cid:48) j ( x , t ) } n,n (cid:48) ,j ∈Z of the Schr¨odinger equation (27),Ψ nn (cid:48) j ( x , t ) = (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / e − iφ nn (cid:48) j ( x ,t ) ψ n (cid:18)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) (cid:0) x − c E tB (cid:1)(cid:19) , (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 formΨ( x , t ) = (cid:88) n,n (cid:48) D ( n ) (cid:88) j =0 (cid:101) C nn (cid:48) j Ψ nn (cid:48) j ( x , t ) . (46)7 . Analytical approach for the case B (cid:107) E. The following figure shows this case.Figura 3: Electric charged in a flat box with parallel electric and magnetic fieldsThe fields are of the form B = (0 , B,
0) and E = (0 , E , A = ( Bz, ,
0) and φ = −E y . The Shr¨odinger equation is for this case as i (cid:126) ∂ Ψ ∂t = (cid:40) (ˆ p x − qBz/c ) m + ˆ p y m + ˆ p z m − q E y (cid:41) Ψ , (47)which defines the following partial differential equation i (cid:126) ∂ Ψ ∂t = − (cid:126) m ∂ Ψ ∂x + i qB (cid:126) zmc ∂ Ψ ∂x + q B mc z Ψ − (cid:126) m ∂ Ψ ∂y − (cid:126) m ∂ Ψ ∂z − q E y Ψ . (48)Proposing a solution of the form Ψ( x , t ) = e − iEt/ (cid:126) Φ( x ), we get the following eigenvalue problem E Φ = − (cid:126) m ∂ Φ ∂x + i qB (cid:126) zmc ∂ Φ ∂x + q B mc z Φ − (cid:126) m ∂ Φ ∂y − (cid:126) m ∂ Φ ∂z − q E y Φ . (49)Applying the Fourier transformation over the x-variable, ˆΦ( k, y, z ) = F x [Φ( x )], the following equation arisesafter some rearrangements E ˆΦ = ( (cid:126) k + qBz/c ) m ˆΦ − (cid:126) m ∂ ˆ φ∂z − (cid:126) m ∂ ˆΦ ∂y − q E y ˆΦ , (50)which can be written as − (cid:126) m ∂ ˆΦ ∂z + 12 mω c ( z + z ) ˆΦ − (cid:126) m ∂ ˆΦ ∂y − q E y ˆΦ , (51a)where ω c is the cyclotron frequency (13a), and z has been defined as z = (cid:126) cqB k. (51b)8his equation admits a variable separable approach since by the proposition ˆΦ( k, y, z ) = f ( k, z ) g ( y ), thefollowing equations are bringing about − (cid:126) m d fdz + 12 mω c ( z + z ) = E (1) f (52a)and − (cid:126) m d gdy − g E yg = E (2) g, (52b)where E = E (1) + E (2) . The solutions of these equations are, of course, the quantum harmonic oscillator andthe quantum bouncer, which are given by f n ( k, z ) = A n e − ξ / H n ( ξ ) , ξ = (cid:114) mω c (cid:126) ( z + z ) , E (1) n = (cid:126) ω c ( n + 1 / . (53a)and g n (cid:48) ( y ) = Ai ( ˜ ξ − ˜ ξ n (cid:48) ) | Ai (cid:48) ( − ˜ ξ n (cid:48) ) | , ˜ ξ = y/l, E (2) n = − q E l ˜ ξ n (cid:48) , (53b)where A n = ( mω c /π (cid:126) ) / / √ n n !, l = ( (cid:126) / ( − mq E )) / , Ai ( − ˜ ξ n (cid:48) ) = 0, and Ai (cid:48) ( ξ ) is the differentiation ofthe Airy function. In this way, we haveˆΦ n,n (cid:48) ( k, y, z ) = a n (cid:48) ψ n (cid:18)(cid:114) mω c (cid:126) ( z + z ) (cid:19) Ai ( l − ( y − y n (cid:48) )) , E n,n (cid:48) = (cid:126) ω c ( n + 1 / − q E y n (cid:48) , (54)where we have defined a n (cid:48) as a n (cid:48) = 1 / | Ai (cid:48) ( − l − y n (cid:48) ) | . Now, the inverse Fourier transformation will affectonly the quantum harmonic oscillator function ψ n through the k-dependence on the parameter z , and theresulting expression is Φ n,n (cid:48) ( x ) = a n (cid:48) qB √ mc (cid:126) ω c e i qB (cid:126) c xz ψ n (cid:18) qBx √ mc (cid:126) ω c (cid:19) Ai (cid:0) l − ( y − y n (cid:48) ) (cid:1) . (55)Now, asking for the periodicity condition of the above solution with respect the z-variable, Ψ( x , t ) = Ψ( x, y, z + L z , t ), the periodicity must satisfy for any x-values, and in particular for x = L x . Thus it follows thequantization expression for the magnetic flux qBL x L z (cid:126) c = 2 πj, 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 function (55) is given by (normalized)Φ nn (cid:48) j ( x ) = a n (cid:48) (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / e i πjLxLz xz ψ n (cid:18)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) x (cid:19) Ai (cid:0) l − ( y − y n (cid:48) ) (cid:1) . (57)Then, we have obtained a family of solution of the Schr¨odinger equation (48),Ψ n,n (cid:48) ( x , t ) = e − iE n,n (cid:48) t/ (cid:126) Φ nn (cid:48) j ( x ) , (58)where the energies E n,n (cid:48) are given by the expression (54). The general solution of (48) can be written asΨ( x , t ) = (cid:88) n,n (cid:48) D ( n ) (cid:88) j =0 C ∗ n,n (cid:48) e − iE n,n (cid:48) t/ (cid:126) e i πjLxLz xz ˜ u n,n (cid:48) ( x, y ) , (59)with the condition (cid:80) n,n (cid:48) | C ∗ n,n (cid:48) | = 1, and where it has been defined the functions ˜ u n,n (cid:48) as˜ u n,n (cid:48) ( x, y ) = a n (cid:48) (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / ψ n (cid:18)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) x (cid:19) Ai (cid:0) l − ( y − y n (cid:48) ) (cid:1) . (60)9 .1. Same system but with new magnetic gauge. Let us consider the magnetic gauge given such that the vector potential is of the form A = (0 , , − Bx ),and the potential is the same φ = −E y . Passing directly to the eigenvalue problem for the Schr¨odingerequation when we select the wave function of the form Ψ( x , t ) = e − iEt/ (cid:126) Φ( x ), the resulting equation is − (cid:126) m ∂ Φ ∂x − (cid:126) m ∂ Φ ∂y − (cid:126) m ∂ Φ ∂z − i qB (cid:126) mc x ∂ Φ ∂z + q B mc x Φ − q E y Φ = E Φ . (61)Taking the Fourier transformation with respect the z-variable, ˆΦ( x, y, k ) = F z [Φ( x )], and making somerearrangements, it follows that − (cid:126) m ∂ ˆΦ ∂x + 12 m (cid:0) (cid:126) k − qBc x (cid:1) ˆΦ − (cid:126) m ∂ ˆΦ ∂y − q E y ˆΦ = E ˆΦ . (62)This equation admits a variable separable solution of the form ˆΦ( x, y, k ) = φ ( k, x ) φ ( y ), where the functions φ and φ satisfy the equations − (cid:126) m d φ dx + ( (cid:126) k − qBc x ) m φ = E (1) φ (63)and − (cid:126) m ∂ φ ∂y − q E yφ = E (2) φ , (64)where E = E (1) + E (2) . The solution of these equations are φ n ( k, x ) = ψ n ( ξ ) = A n e − ξ / H n ( ξ ) , ξ = (cid:114) mω c (cid:126) ( x − x ) , E (1) n = (cid:126) ω c ( n + 1 /
2) (65)and φ n (cid:48) ( y ) = a n (cid:48) Ai ( l − ( y − y n (cid:48) )) , l = (cid:18) (cid:126) − mq E (cid:19) / , E (2) n (cid:48) = − q E y n (cid:48) , (66)where ω c is the cyclotron frequency (13a), x is the displacement x = (cid:126) ck/qB , a n (cid:48) = 1 / | Ai (cid:48) ( l − y n (cid:48) ) | is aconstant, and A n the constant associated to the quantum harmonic oscillator solution. The inverse Fouriertransformation affect only the function φ , and we have φ n ( z, x ) = F − [ φ n ( k, x )] = − qB √ mc (cid:126) ω c e − i qB (cid:126) c xz ψ n (cid:18) qBz √ mc (cid:126) ω c (cid:19) . (67)The periodic condition on the variable “x”, Ψ( x , t ) = Ψ( 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 magneticflux ( BL x L y ), qBL x L z (cid:126) c = 2 πj, j ∈ Z . (68)Thus, we have a family of solutions { Ψ nn (cid:48) j ( x , t ) } of the Shcr¨odinger equation of the formΨ nn (cid:48) j ( x , t ) = e − iE n,n (cid:48) t/ (cid:126) Φ nn (cid:48) j ( x ) , (69)or (normalized and ignoring the sign)Ψ nn (cid:48) j ( x , t ) = a n (cid:48) (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / e − i ( E n,n (cid:48) t (cid:126) + πjLxLz xz ) ψ n (cid:18)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) z (cid:19) Ai ( l − ( y − y n (cid:48) )) . (70)10y the same arguments about the degenerationn of the systems, the general solution is just a combinationof all of these, Ψ( x , t ) = (cid:88) n,n (cid:48) A nn (cid:48) j e − i ( E n,n (cid:48) t (cid:126) + πjLxLz xz ) v nn (cid:48) j ( y, z ) , (71)where the condition (cid:80) n,n (cid:48) | A nn (cid:48) j | = 1 must be satisfied, and the function v nn (cid:48) j is given by v nn (cid:48) j ( y, z ) = a n (cid:48) (cid:115) πjL x L y (cid:18) (cid:126) mω c (cid:19) / ψ n (cid:32)(cid:114) (cid:126) mω c (cid:18) πjL x L y (cid:19) z (cid:33) Ai (cid:0) l − ( y − y n (cid:48) ) (cid:1) . (72)
5. Conclusions and comments
We have studied the quantization of a charged particle in a flat box and under constants magnetic andelectric fields for several cases and have shown that a full separation of variable solution is not admitted inthese cases (contrary to Landau’s solution in one of these cases). This situation arises since the commutationof a component of the generalized linear momentum operator with the Hamiltonian of the system does notimply necessarily that a variable separation of its associated variable must exist in the Schr¨odinger equation.However, using the Fourier transformation, we were be able to find the full solution of the problems. Asexpected, Landau’s level appears in all these cases, and a characteristic phase which help us to find thequantization of the magnetic flux in a natural way. We consider that the approach given here maybe veryuseful to understand quantum Hall effect and related phenomena.
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