1. Introduction
The quatum Hall effect has had a great deal of physical and experimental importance since its discovery [1] [2] [3] [4], and one of the basic elements to understand this effect is the Landau’s levels (eigenvalues of the eigenvalue problem), which has shown being correct even if the eigenfunctions are not totally right since the eigenvalue problem is not separable in all of its variables [5]. These eigenfunctions have been also found in different form [6], but both of them result to be equivalents [7]. However, a correct non separable solution of the eigenvalue problem has already been given on references [5] [8], where it is shown that for the eigenvalue problem with the Hamiltonian
(1)
with the Landau’s gauge
(B is constant) and with the symmetric gauge
(inverse magnetic field), the Landau’s levels are gotten
(2)
the magnetic flux
is quantized
(3)
and the eigenfunctions for Landau’s gauge (ignoring the z-variable) are given by
(4)
where
represents the length of the box in the y-direction, and
is the harmonic oscillator solutions. For the symmetric gauge the eigenfunctions are
(5)
where
is a complex constant, and
is a normalized constant. In addition, for the Landau’s gauge case one has
, and for the symmetric gauge case one has
. These facts allow to have the following additionals generated functions
(6)
and
(7)
which are also eigenfunctions of the Hamiltonian
(8)
and
(9)
from these relations, it was thought that Landau’s levels were doubly degenerated. However, we will see that this result is deeper than it was first thought since it allows to have numerably degeneration for the Landau’s levels. The reason for example (6) is also an eigenfunction, as shown in (8), is the following: from the expression
, one has
(10)
and the result (8) follows.
2. Analysis of the Degeneration
Let us make first some mathematical statements that will help to understand the situation. Let
be our Hilbert space, normally the set of quadratic integrable function in some set
contained in some dimensional space
, and let
the set of linear operators acting in the space
. Thus, one has the following proposition.
Prop. 1.- Let
be linear operators such that
, and let
be the solutions of the eigenvalue problem
. If
is not proportional to
, then
is an eigenfunction of H with the same eigenvalue
(Therefore, the spectrum is numerably degenerated).
Proof: The fact
implies that
for
, where
means j-applications of the operator A (
). Therefore one has that
. Since
is not proportional to
, it represents a new function, and by induction
represents a new function for
. So, defining
and
, one has a set functions
which are eigenfunctions with the same eigenvalue
(11)
In addition to the above proposition, one has the following: If
is an Hermitian operator, that is
with the inner product
, one has the known proposition
Prop. 2.- Let
be an Hermitian operator, and let
the set of solutions of the eigenvalue problem
, where the spectrum is degenerated (this degeneration is represente by the index “j”). Then, the functions
are orthogonal with respect the index “n”, but the orthogonality is undetermined with respect the index “j”.
Proof: The relation
implies that
. Then, for
one has necessarily that
(orthogonality, independently of
and
), but if
the expression
is undetermined
Of course, given a non orthogonal set of functions
, one can construct an orthogonal set
through the Gram-Schmidt process [9]. Now, the results presented in (4), (5), (6), (7), (8) and (9) state exactly the conditions for the application of the Prop. 1 above. Therefore, the Landau’s levels are numerably degenerated in both cases with the Landau and symmetric gauges. The states associated to each Landau’s level are
(12)
and
(13)
It is easy to see, for example, from (4) and (6) that
. Therefore, the set defined by (12) is non orthogonal, and the same happens with the set (13). Of course, the general solution of the Schödinger’s equation (
) for this problem should be written for the Landau’s gauge (ignoring the z-variable) as
(14)
and for the symmetric gauge as
(15)
being
and
constant, and
is the Landau’s levels (2).
Now, from the result (11) and the expression (12) it is not difficult to see that one has the following relation
(16)
and
(17)
Thus, from the result (11), one must have that
(18)
which it is not difficult to check it directly.
Similarly, from (5) and (7), one can get
(19)
where one has defined the constants
and
as
and
, being
the binomial coefficient. In addition, one has the following action
(20)
Then, using (11), it follows that
(21)
which it is also not difficult to verify directly.
3. Conclusion
Due to previous results (6), (7), (8) and (9), obtained in [8], and the Prop. 1, we must conclude that the Landau’s levels are numerably degenerated. This degeneration may have important consequences in the quantum dynamics of the quantum Hall Effect and topological insulators.