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       Different quantum spectra for the same classical system
                                           Vladimir Cuesta †
                 Instituto de Ciencias Nucleares, Universidad Nacional Aut´ noma de M´ xico, 70-543, Ciudad de
                                                                          o          e
                 M´ xico, M´ xico
                   e        e

                 Abstract. I show that the usual one dimensional isotropic harmonic oscillator admits two
                 quantum spectra (one positive and another negative), I use the ket-bra formalism to find my results,
                 I write some eigenfunctions with negative spectra. I find that for the two dimensional isotropic
                 harmonic oscillator there are four different quantum spectra.



                                              INTRODUCTION

     Classical mechanics is one of the more important branches in modern physics, we can
study that subject using different approaches, using Newton’s equations, using Lagrange or
Hamiltonian formulations (see [1] for instance).
     I will study the one and two dimensional isotropic harmonic oscillator at classical and
quantum level, for the first case the phase space coordinates are (xµ ) = (x, px ) with hamiltonian
                  2                                                                        2
H = 2m p2 + mω x2 , for the second case the hamiltonian is H = 2m p2 + p2 + mω (x2 + y 2 ),
       1
          x     2
                                                                       1
                                                                            x    y       2
where the coordinates (xµ ) = (x, y, px , py ) label the phase space.
     Quantum mechanics is another branch of modern physics with great importance for the
understanding of nature, it was formulated in essentially two ways, the first one is in terms of
a Hilbert space of square-integrable functions, partial differential equations, wave functions and
so on (see [2] for instance). The second point of view is the formulation in terms of an abstract
Hilbert space of vectors (kets) and its duals (bras), abstract operators and so on (see [3] and [4] for
a profound review of the theory), with this in my mind I present my results following the second
approach. I will prove that the one and the two dimensional isotropic harmonic oscillators have
not an unique spectrum.
        QUANTUM 1D ISOTROPIC HARMONIC OSCILLATOR (ALGEBRAIC METHOD)
                                                                2
                                               ˆ      1
      I will be interested in the hamiltonian H = 2m p2 + mω x2 . Now, I define the operator
                                                        ˆx        ˆ
                                                              2
            mω         ı
ax = − 2¯ x − mω px and the commutation relations between the number operator Nx =
ˆ                ˆ        ˆ                                                                ˆ
              h
                ˆx       ˆ ˆ       ˆ      ˆ ˆx
a† ax , ax and a† are [Nx , ax ] = ax , [Nx , a† ] = −ˆ† , [ˆx , a† ] = −1, if I suppose that the
ˆx ˆ ˆ                                                 ax   a ˆx
                                            ˆ
number operator has eigenkets |nx , then Nx |nx = nx |nx , after a straightforward calculation
ax |nx =
ˆ             |nx ||nx + 1 ,      a† |nx =
                                  ˆx    |nx − 1||nx − 1 , where nx = . . . , −2, −1, 0., ax is the
                                                                                         ˆ
                    ˆx                                                  ˆ       ˆ
ascent operator and a† is the descent operator, the hamiltonian will be H = Nx − 2 hω, and
                                                                                      1
                                                                                         ¯
                                                                                     1
the spectrum of the one dimensional isotropic harmonic oscillator is Enx = nx − 2 hω, in this
                                                                                        ¯
case the vacuum ket is |0 and is defined as a|0 = 0, using the ascent creation operator I find
                                                ˆ
                             √                           √
a |0 = | − 1 , a | − 1 = 2| − 2 , a | − 2 = 3| − 3 , a† | − 3 = 2| − 4 , . . . and
ˆ †
                   ˆ †
                                              ˆ †
                                                                     ˆ
                                   |mx |
in general | − |mx | = √  1
                               a
                               ˆ †
                                         |0 , and the one dimensional isotropic oscillator has two
                               |mx |!
different quantum spectra.
  †
      vladimir.cuesta@nucleares.unam.mx
2

             QUANTUM 1D ISOTROPIC HARMONIC OSCILLATOR (EIGENSYSTEMS)

     I have shown an algebraic method to find a negative spectrum for the one dimensional
isotropic harmonic oscillator. I write some states with negative spectrum,
                      Eigenf unction                                                             Eigenvalue
                                         mω 2
                      ψ0 (x) = λ0 exp     2¯
                                           h
                                              x                                                  E0 = − ¯2
                                                                                                        hω

                      ψ−1 (x) =   λ−1 x exp mω x2
                                                2¯
                                                 h
                                                                                                 E−1 = − 3¯ ω
                                                                                                          h
                                                                                                          2
                      ψ−2 (x) =                   h
                                                  ¯
                                  λ−2 2x2 + mω exp mω x2      2¯h
                                                                                                 E−2 = − 5¯ ω
                                                                                                          h
                                                                                                          2
                      ψ−3 (x) =   λ−3 x 2x + mω exp mω x2
                                             2      3¯
                                                     h
                                                                 2¯
                                                                  h
                                                                                                 E−3 = − 7¯ ω
                                                                                                          h
                                                                                                          2
                                                 12¯ x2     3¯ 2
                      ψ−4 (x) =            4                 h
                                  λ−4 4x + mω + m2 ω2 exp mω x2
                                                    h
                                                                          2¯h
                                                                                                 E−4 = − 9¯ ω
                                                                                                          h
                                                                                                          2
                                                      h 2    15¯ 2
                      ψ−5 (x) =   λ−5 x 4x4 + 20¯ x + m2h 2 exp mω x2
                                                     mω          ω          2¯h
                                                                                                 E−5 = − 11¯ ω
                                                                                                           h
                                                                                                           2
                                                 60¯ x4    90¯ 2 x2    15h3
                      ψ−6 (x) =            6                 h
                                  λ−6 8x + mω + m2 ω2 + m3 ω3 exp mω x2
                                                    h
                                                                                   2¯
                                                                                    h
                                                                                                 E−6 = − 13¯ ω
                                                                                                           h
                                                                                                           2
                                                      h 4        h2 2         3
                      ψ−7 (x) =   λ−7 x 8x6 + 84¯ x + 210¯ωx + 105h3 exp mω x2
                                                     mω       m2 2       m3 ω         2¯
                                                                                       h
                                                                                                 E−7 = − 15¯ ω
                                                                                                           h
                                                                                                           2
                                                   224¯ x6
                                                       h     840¯ 2 x4
                                                                  h       840h3 x2  105h4 mω 2
                      ψ−8 (x) =   λ−8 16x + mω + m2 ω2 + m3 ω3 + m4 ω4 e 2¯ x
                                             8                                             h     E−8 = − 17¯ ω
                                                                                                           h
                                                                                                           2
where λ0 , . . . , λ−8 are constants.
    In this case, all the states are normalized using a weight function. I mean, the inner
                                      ∞                ∗
product will be: ψ−|m| |ψ−|n| = −∞ exp − 2mω x2 ψ−|m| (x)ψ−|n| (x)dx for m + n odd and
                                               h
                                               ¯
                        ∞                    ∗
 ψ−|m| |ψ−|n| =         −∞   x exp − 2mω x2 ψ−|m| (x)ψ−|n| (x)dx for m + n even or m = n.
                                      h
                                      ¯

                                 2D ISOTROPIC HARMONIC OSCILLATOR

     PART I.- In this case the hamiltonian for the two dimensional isotropic harmonic oscillator
is Hˆ = 1 p2 + p2 + mω2 (ˆ2 + y 2 ), the previous hamiltonian can be divided in a part
               ˆx ˆy             x     ˆ
          2m                  2
that correspond with the x direction and another in the y direction, both hamiltonians are
independent, for the x direction I can choose ax = mω x + mω px and for the y direction
                                                ˆ        2¯
                                                          h
                                                            ˆ     ı
                                                                     ˆ
ay = mω y + mω py .
ˆ          2¯
            h
                 ˆ   ı
                        ˆ
                              √                                √
      I find ax |nx , ny = nx |nx − 1, ny , a† |nx , ny = nx + 1|nx + 1, ny , ay |nx , ny =
               ˆ                               ˆx                                    ˆ
√                      †            √
  ny |nx , ny − 1 , ay |nx , ny = ny + 1|nx , ny + 1 , where nx = 0, 1, 2, . . ., ny = 0, 1, 2, . . .
                     ˆ
and N  ˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , in such a way that the hamiltonian
                                    ˆ
                          ˆ       ˆ     ˆ
can be expressed as H = Nx + Ny + 1 hω, and the spectrum for this case is Enx ,ny =
                                                ¯
(nx + ny + 1) hω, where nx = 0, 1, 2, . . ., ny = 0, 1, 2, . . ., the vacuum state is obtained as
                   ¯
                                                                             mx     my
ax ay |0, 0 = 0 and a general state is given by |mx , my = √ 1
ˆ ˆ                                                                       a†
                                                                          ˆ      a†
                                                                                 ˆ     |0, 0 .
                                                                         (mx !)(my !)
                                                        mω         ı
       PART II.- If I choose, ax
                              ˆ              =          2¯
                                                         h
                                                             x+
                                                             ˆ       p
                                                                     ˆ
                                                                  mω x
                                                                          and for the y direction ay =
                                                                                                   ˆ
                                                    √                                 √
−     mω
      2¯
       h
             y−
             ˆ        ı
                        p
                        ˆ
                     mω y
                            , I find ax |nx , ny =
                                    ˆ                   nx |nx − 1, ny , a† |nx , ny = nx + 1|nx + 1, ny ,
                                                                         ˆx
ay |nx , ny =
ˆ                  |ny ||nx , ny + 1 , a† |nx , ny = |ny − 1||nx , ny − 1 , where nx = 0, 1, 2, . . .,
                                          ˆy
ny = . . . , −2, −1, 0 and N   ˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny .
                                                           ˆ
                                                       ˆ           ˆ    ˆ ¯
     In this case the hamiltonian will be H = Nx + Ny hω, and the spectrum for
this case is Enx ,ny = (nx + ny ) hω, where nx = 0, 1, 2, . . ., ny = . . . , −2, −1, 0,
                                             ¯
the vacuum state is obtained as ax ay |0, 0 = 0 and a general state is |mx , −|my | =
                                            ˆ ˆ
                    mx        |my |
√ 1             a†
                ˆ         a†
                          ˆ         |0, 0 .
    (mx !)(|my |!)
3
                                                    mω         ı
     PART III.- I can choose, ax = −
                              ˆ                     2¯
                                                     h
                                                         x−
                                                         ˆ       p
                                                                 ˆ
                                                              mω x
                                                                       and for the y direction ay =
                                                                                               ˆ
  mω
  2¯
   h
        y+
        ˆ       ı
                  p
                  ˆ
               mω y
                   , in such a way that, ax |nx , ny =
                                                  ˆ                    |nx ||nx + 1, ny , a† |nx , ny =
                                                                                              ˆx
                                           √                                  √
  |nx − 1||nx − 1, ny , ay |nx , ny = ny |nx , ny − 1 , a† |nx , ny = ny + 1|nx , ny + 1 , where
                          ˆ                                   ˆy
nx = . . . , −2, −1, 0, ny = 0, 1, 2, . . . and Nˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , the
                                                                              ˆ
                                     ˆ         ˆ      ˆ ¯
hamiltonian can be expressed as H = Nx + Ny hω and the spectrum for this case is Enx ,ny =
(nx + ny ) hω, where nx = . . . , −2, −1, 0, ny = 0, 1, 2, . . ., the vacuum state is obtained as
             ¯
                                                                                         |mx |     my
ax ay |0, 0 = 0 and a general state is given by |−|mx |, my = √ 1
ˆ ˆ                                                                                   a†
                                                                                      ˆ         a†
                                                                                                ˆ     |0, 0 .
                                                                       (|mx |!)(my !)
    PART IV.- The hamiltonian can be divided in a part corresponding with the x direction and
another in the y direction, both hamiltonians are independent, for the x direction I can choose
ax = − mω x − mω px and for the y direction ay = − mω y − mω py .
ˆ         2¯
           h
               ˆ     ı
                       ˆ                         ˆ        2¯
                                                           h
                                                               ˆ     ı
                                                                       ˆ
     I find ax |nx , ny =
           ˆ                    |nx ||nx + 1, ny , a† |nx , ny =
                                                   ˆx                |nx − 1||nx − 1, ny , ay |nx , ny =
                                                                                           ˆ
  |ny ||nx , ny + 1 ,   a† |nx , ny
                        ˆy            =      |ny − 1||nx , ny − 1 , where nx = . . . , −2, −1, 0, ny =
                        ˆ                                ˆ
. . . , −2, −1, 0 and Nx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , in such a way that
                                          ˆ        ˆ       ˆ
the hamiltonian can be written as H = Nx + Ny − 1 hω, and the spectrum for this case is
                                                                  ¯
Enx ,ny = (nx + ny − 1) hω, where nx = . . . , −2, −1, 0, ny = . . . , −2, −1, 0, the vacuum state
                             ¯
is obtained as ax ay |0, 0 = 0, with the final expression for a general state | − |mx |, −|my | =
                  ˆ ˆ
                      |mx |    |my |
√        1
                  a†
                  ˆ         a†
                            ˆ        |0, 0 .
  (|mx |!)(|my |!)

                                 CONCLUSIONS AND PERSPECTIVES

     I have deduced that two different classical systems have more than one quantum spectra.
I have verified all these mathematical results and are correct. My personal point of view is
that there are more systems with this characteristic and in fact, using similar arguments for
the case of n-dimensional isotropic oscillators the number of different spectra is undeniable
not unique. The reader can study in a similar way the Kepler problem, a charged particle in
a constant gravitational field, problems in quantum field theory and so on. The hamiltonians are
quadratic and people can think that the spectra must be positive, I show the opposite, you can
study quadratic systems in the mathematical theory of differential equations and probably you
could find new spectra.

                                             REFERENCES

     [1] V. I. Arnold, Mathematical methods of classical mechanics, Moscow, Springer-Verlag,
New York, (1980).
     [2] W. Heisenberg, The Physical Principles of the Quantum Theory, University of Chicago
Press, Chicago, (1930).
     [3] P. A. M. Dirac, The Principles of Quantum Mechanics, Oxford University Press, Oxford,
(1981).
     [4] J. J. Sakurai, Modern Quantum Mechanics, Late, University of California, Los Angeles,
Addison-Wesley Publishing Company, Inc. (1994).

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Proceedings Different Quantum Spectra For The Same Classical System

  • 1. 1 Different quantum spectra for the same classical system Vladimir Cuesta † Instituto de Ciencias Nucleares, Universidad Nacional Aut´ noma de M´ xico, 70-543, Ciudad de o e M´ xico, M´ xico e e Abstract. I show that the usual one dimensional isotropic harmonic oscillator admits two quantum spectra (one positive and another negative), I use the ket-bra formalism to find my results, I write some eigenfunctions with negative spectra. I find that for the two dimensional isotropic harmonic oscillator there are four different quantum spectra. INTRODUCTION Classical mechanics is one of the more important branches in modern physics, we can study that subject using different approaches, using Newton’s equations, using Lagrange or Hamiltonian formulations (see [1] for instance). I will study the one and two dimensional isotropic harmonic oscillator at classical and quantum level, for the first case the phase space coordinates are (xµ ) = (x, px ) with hamiltonian 2 2 H = 2m p2 + mω x2 , for the second case the hamiltonian is H = 2m p2 + p2 + mω (x2 + y 2 ), 1 x 2 1 x y 2 where the coordinates (xµ ) = (x, y, px , py ) label the phase space. Quantum mechanics is another branch of modern physics with great importance for the understanding of nature, it was formulated in essentially two ways, the first one is in terms of a Hilbert space of square-integrable functions, partial differential equations, wave functions and so on (see [2] for instance). The second point of view is the formulation in terms of an abstract Hilbert space of vectors (kets) and its duals (bras), abstract operators and so on (see [3] and [4] for a profound review of the theory), with this in my mind I present my results following the second approach. I will prove that the one and the two dimensional isotropic harmonic oscillators have not an unique spectrum. QUANTUM 1D ISOTROPIC HARMONIC OSCILLATOR (ALGEBRAIC METHOD) 2 ˆ 1 I will be interested in the hamiltonian H = 2m p2 + mω x2 . Now, I define the operator ˆx ˆ 2 mω ı ax = − 2¯ x − mω px and the commutation relations between the number operator Nx = ˆ ˆ ˆ ˆ h ˆx ˆ ˆ ˆ ˆ ˆx a† ax , ax and a† are [Nx , ax ] = ax , [Nx , a† ] = −ˆ† , [ˆx , a† ] = −1, if I suppose that the ˆx ˆ ˆ ax a ˆx ˆ number operator has eigenkets |nx , then Nx |nx = nx |nx , after a straightforward calculation ax |nx = ˆ |nx ||nx + 1 , a† |nx = ˆx |nx − 1||nx − 1 , where nx = . . . , −2, −1, 0., ax is the ˆ ˆx ˆ ˆ ascent operator and a† is the descent operator, the hamiltonian will be H = Nx − 2 hω, and 1 ¯ 1 the spectrum of the one dimensional isotropic harmonic oscillator is Enx = nx − 2 hω, in this ¯ case the vacuum ket is |0 and is defined as a|0 = 0, using the ascent creation operator I find ˆ √ √ a |0 = | − 1 , a | − 1 = 2| − 2 , a | − 2 = 3| − 3 , a† | − 3 = 2| − 4 , . . . and ˆ † ˆ † ˆ † ˆ |mx | in general | − |mx | = √ 1 a ˆ † |0 , and the one dimensional isotropic oscillator has two |mx |! different quantum spectra. † vladimir.cuesta@nucleares.unam.mx
  • 2. 2 QUANTUM 1D ISOTROPIC HARMONIC OSCILLATOR (EIGENSYSTEMS) I have shown an algebraic method to find a negative spectrum for the one dimensional isotropic harmonic oscillator. I write some states with negative spectrum, Eigenf unction Eigenvalue mω 2 ψ0 (x) = λ0 exp 2¯ h x E0 = − ¯2 hω ψ−1 (x) = λ−1 x exp mω x2 2¯ h E−1 = − 3¯ ω h 2 ψ−2 (x) = h ¯ λ−2 2x2 + mω exp mω x2 2¯h E−2 = − 5¯ ω h 2 ψ−3 (x) = λ−3 x 2x + mω exp mω x2 2 3¯ h 2¯ h E−3 = − 7¯ ω h 2 12¯ x2 3¯ 2 ψ−4 (x) = 4 h λ−4 4x + mω + m2 ω2 exp mω x2 h 2¯h E−4 = − 9¯ ω h 2 h 2 15¯ 2 ψ−5 (x) = λ−5 x 4x4 + 20¯ x + m2h 2 exp mω x2 mω ω 2¯h E−5 = − 11¯ ω h 2 60¯ x4 90¯ 2 x2 15h3 ψ−6 (x) = 6 h λ−6 8x + mω + m2 ω2 + m3 ω3 exp mω x2 h 2¯ h E−6 = − 13¯ ω h 2 h 4 h2 2 3 ψ−7 (x) = λ−7 x 8x6 + 84¯ x + 210¯ωx + 105h3 exp mω x2 mω m2 2 m3 ω 2¯ h E−7 = − 15¯ ω h 2 224¯ x6 h 840¯ 2 x4 h 840h3 x2 105h4 mω 2 ψ−8 (x) = λ−8 16x + mω + m2 ω2 + m3 ω3 + m4 ω4 e 2¯ x 8 h E−8 = − 17¯ ω h 2 where λ0 , . . . , λ−8 are constants. In this case, all the states are normalized using a weight function. I mean, the inner ∞ ∗ product will be: ψ−|m| |ψ−|n| = −∞ exp − 2mω x2 ψ−|m| (x)ψ−|n| (x)dx for m + n odd and h ¯ ∞ ∗ ψ−|m| |ψ−|n| = −∞ x exp − 2mω x2 ψ−|m| (x)ψ−|n| (x)dx for m + n even or m = n. h ¯ 2D ISOTROPIC HARMONIC OSCILLATOR PART I.- In this case the hamiltonian for the two dimensional isotropic harmonic oscillator is Hˆ = 1 p2 + p2 + mω2 (ˆ2 + y 2 ), the previous hamiltonian can be divided in a part ˆx ˆy x ˆ 2m 2 that correspond with the x direction and another in the y direction, both hamiltonians are independent, for the x direction I can choose ax = mω x + mω px and for the y direction ˆ 2¯ h ˆ ı ˆ ay = mω y + mω py . ˆ 2¯ h ˆ ı ˆ √ √ I find ax |nx , ny = nx |nx − 1, ny , a† |nx , ny = nx + 1|nx + 1, ny , ay |nx , ny = ˆ ˆx ˆ √ † √ ny |nx , ny − 1 , ay |nx , ny = ny + 1|nx , ny + 1 , where nx = 0, 1, 2, . . ., ny = 0, 1, 2, . . . ˆ and N ˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , in such a way that the hamiltonian ˆ ˆ ˆ ˆ can be expressed as H = Nx + Ny + 1 hω, and the spectrum for this case is Enx ,ny = ¯ (nx + ny + 1) hω, where nx = 0, 1, 2, . . ., ny = 0, 1, 2, . . ., the vacuum state is obtained as ¯ mx my ax ay |0, 0 = 0 and a general state is given by |mx , my = √ 1 ˆ ˆ a† ˆ a† ˆ |0, 0 . (mx !)(my !) mω ı PART II.- If I choose, ax ˆ = 2¯ h x+ ˆ p ˆ mω x and for the y direction ay = ˆ √ √ − mω 2¯ h y− ˆ ı p ˆ mω y , I find ax |nx , ny = ˆ nx |nx − 1, ny , a† |nx , ny = nx + 1|nx + 1, ny , ˆx ay |nx , ny = ˆ |ny ||nx , ny + 1 , a† |nx , ny = |ny − 1||nx , ny − 1 , where nx = 0, 1, 2, . . ., ˆy ny = . . . , −2, −1, 0 and N ˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny . ˆ ˆ ˆ ˆ ¯ In this case the hamiltonian will be H = Nx + Ny hω, and the spectrum for this case is Enx ,ny = (nx + ny ) hω, where nx = 0, 1, 2, . . ., ny = . . . , −2, −1, 0, ¯ the vacuum state is obtained as ax ay |0, 0 = 0 and a general state is |mx , −|my | = ˆ ˆ mx |my | √ 1 a† ˆ a† ˆ |0, 0 . (mx !)(|my |!)
  • 3. 3 mω ı PART III.- I can choose, ax = − ˆ 2¯ h x− ˆ p ˆ mω x and for the y direction ay = ˆ mω 2¯ h y+ ˆ ı p ˆ mω y , in such a way that, ax |nx , ny = ˆ |nx ||nx + 1, ny , a† |nx , ny = ˆx √ √ |nx − 1||nx − 1, ny , ay |nx , ny = ny |nx , ny − 1 , a† |nx , ny = ny + 1|nx , ny + 1 , where ˆ ˆy nx = . . . , −2, −1, 0, ny = 0, 1, 2, . . . and Nˆx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , the ˆ ˆ ˆ ˆ ¯ hamiltonian can be expressed as H = Nx + Ny hω and the spectrum for this case is Enx ,ny = (nx + ny ) hω, where nx = . . . , −2, −1, 0, ny = 0, 1, 2, . . ., the vacuum state is obtained as ¯ |mx | my ax ay |0, 0 = 0 and a general state is given by |−|mx |, my = √ 1 ˆ ˆ a† ˆ a† ˆ |0, 0 . (|mx |!)(my !) PART IV.- The hamiltonian can be divided in a part corresponding with the x direction and another in the y direction, both hamiltonians are independent, for the x direction I can choose ax = − mω x − mω px and for the y direction ay = − mω y − mω py . ˆ 2¯ h ˆ ı ˆ ˆ 2¯ h ˆ ı ˆ I find ax |nx , ny = ˆ |nx ||nx + 1, ny , a† |nx , ny = ˆx |nx − 1||nx − 1, ny , ay |nx , ny = ˆ |ny ||nx , ny + 1 , a† |nx , ny ˆy = |ny − 1||nx , ny − 1 , where nx = . . . , −2, −1, 0, ny = ˆ ˆ . . . , −2, −1, 0 and Nx |nx , ny = nx |nx , ny , Ny |nx , ny = ny |nx , ny , in such a way that ˆ ˆ ˆ the hamiltonian can be written as H = Nx + Ny − 1 hω, and the spectrum for this case is ¯ Enx ,ny = (nx + ny − 1) hω, where nx = . . . , −2, −1, 0, ny = . . . , −2, −1, 0, the vacuum state ¯ is obtained as ax ay |0, 0 = 0, with the final expression for a general state | − |mx |, −|my | = ˆ ˆ |mx | |my | √ 1 a† ˆ a† ˆ |0, 0 . (|mx |!)(|my |!) CONCLUSIONS AND PERSPECTIVES I have deduced that two different classical systems have more than one quantum spectra. I have verified all these mathematical results and are correct. My personal point of view is that there are more systems with this characteristic and in fact, using similar arguments for the case of n-dimensional isotropic oscillators the number of different spectra is undeniable not unique. The reader can study in a similar way the Kepler problem, a charged particle in a constant gravitational field, problems in quantum field theory and so on. The hamiltonians are quadratic and people can think that the spectra must be positive, I show the opposite, you can study quadratic systems in the mathematical theory of differential equations and probably you could find new spectra. REFERENCES [1] V. I. Arnold, Mathematical methods of classical mechanics, Moscow, Springer-Verlag, New York, (1980). [2] W. Heisenberg, The Physical Principles of the Quantum Theory, University of Chicago Press, Chicago, (1930). [3] P. A. M. Dirac, The Principles of Quantum Mechanics, Oxford University Press, Oxford, (1981). [4] J. J. Sakurai, Modern Quantum Mechanics, Late, University of California, Los Angeles, Addison-Wesley Publishing Company, Inc. (1994).