General solution of a specific eigenvalue problem.












2












$begingroup$


Consider a Schroedinger-like equation with a generalized harmonic potential:



$$
left(sum_imu_i^2frac{partial^2}{partial x_i^2}-sum_{ij}Omega_{ij}x_ix_j+Eright)Psi=0,
$$
where indices run from 1 to $N$, $x_i$ are real coordinates, $mu_i$ are positive real numbers, and $Omega$ is a positive-definite symmetric real matrix.



Is the following statement valid:



Any eigenvalue of the problem is determined by a $N$-tuple of non-negative integer numbers $(n_1,n_2,dots,n_N)$ as
$$
E_{n_1n_2dots n_N}=sum_{i=1}^Nleft(1+2n_iright)omega_i,
$$
where $omega$'s are the square roots of the eigenvalues of the matrix $tildeOmega$:
$$tildeOmega_{ij}=mu_imu_jOmega_{ij}$$?



If the general solution of the problem is known I would be thankful for corresponding hints and/or references.






pde eigenvalues-eigenvectors quantum-mechanics sturm-liouville linear-pde







share|cite|improve this question











$endgroup$












  • $begingroup$
    I made a small edit to your post and added a few tags. Hope this is OK.
    $endgroup$
    – Robert Lewis
    Jun 13 '17 at 18:34






  • 1




    $begingroup$
    I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
    $endgroup$
    – user
    Jun 14 '17 at 7:52






  • 1




    $begingroup$
    The proof of the positive definiteness has appeared to be trivial.
    $endgroup$
    – user
    Jun 14 '17 at 13:10
















2












$begingroup$


Consider a Schroedinger-like equation with a generalized harmonic potential:



$$
left(sum_imu_i^2frac{partial^2}{partial x_i^2}-sum_{ij}Omega_{ij}x_ix_j+Eright)Psi=0,
$$
where indices run from 1 to $N$, $x_i$ are real coordinates, $mu_i$ are positive real numbers, and $Omega$ is a positive-definite symmetric real matrix.



Is the following statement valid:



Any eigenvalue of the problem is determined by a $N$-tuple of non-negative integer numbers $(n_1,n_2,dots,n_N)$ as
$$
E_{n_1n_2dots n_N}=sum_{i=1}^Nleft(1+2n_iright)omega_i,
$$
where $omega$'s are the square roots of the eigenvalues of the matrix $tildeOmega$:
$$tildeOmega_{ij}=mu_imu_jOmega_{ij}$$?



If the general solution of the problem is known I would be thankful for corresponding hints and/or references.






pde eigenvalues-eigenvectors quantum-mechanics sturm-liouville linear-pde







share|cite|improve this question











$endgroup$












  • $begingroup$
    I made a small edit to your post and added a few tags. Hope this is OK.
    $endgroup$
    – Robert Lewis
    Jun 13 '17 at 18:34






  • 1




    $begingroup$
    I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
    $endgroup$
    – user
    Jun 14 '17 at 7:52






  • 1




    $begingroup$
    The proof of the positive definiteness has appeared to be trivial.
    $endgroup$
    – user
    Jun 14 '17 at 13:10














2












2








2


2



$begingroup$


Consider a Schroedinger-like equation with a generalized harmonic potential:



$$
left(sum_imu_i^2frac{partial^2}{partial x_i^2}-sum_{ij}Omega_{ij}x_ix_j+Eright)Psi=0,
$$
where indices run from 1 to $N$, $x_i$ are real coordinates, $mu_i$ are positive real numbers, and $Omega$ is a positive-definite symmetric real matrix.



Is the following statement valid:



Any eigenvalue of the problem is determined by a $N$-tuple of non-negative integer numbers $(n_1,n_2,dots,n_N)$ as
$$
E_{n_1n_2dots n_N}=sum_{i=1}^Nleft(1+2n_iright)omega_i,
$$
where $omega$'s are the square roots of the eigenvalues of the matrix $tildeOmega$:
$$tildeOmega_{ij}=mu_imu_jOmega_{ij}$$?



If the general solution of the problem is known I would be thankful for corresponding hints and/or references.






pde eigenvalues-eigenvectors quantum-mechanics sturm-liouville linear-pde







share|cite|improve this question











$endgroup$




Consider a Schroedinger-like equation with a generalized harmonic potential:



$$
left(sum_imu_i^2frac{partial^2}{partial x_i^2}-sum_{ij}Omega_{ij}x_ix_j+Eright)Psi=0,
$$
where indices run from 1 to $N$, $x_i$ are real coordinates, $mu_i$ are positive real numbers, and $Omega$ is a positive-definite symmetric real matrix.



Is the following statement valid:



Any eigenvalue of the problem is determined by a $N$-tuple of non-negative integer numbers $(n_1,n_2,dots,n_N)$ as
$$
E_{n_1n_2dots n_N}=sum_{i=1}^Nleft(1+2n_iright)omega_i,
$$
where $omega$'s are the square roots of the eigenvalues of the matrix $tildeOmega$:
$$tildeOmega_{ij}=mu_imu_jOmega_{ij}$$?



If the general solution of the problem is known I would be thankful for corresponding hints and/or references.






pde eigenvalues-eigenvectors quantum-mechanics sturm-liouville linear-pde




pde eigenvalues-eigenvectors quantum-mechanics sturm-liouville linear-pde






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 13 '17 at 18:30









Robert Lewis

48.1k23167




48.1k23167










asked Jun 13 '17 at 16:48









useruser

5,45411030




5,45411030












  • $begingroup$
    I made a small edit to your post and added a few tags. Hope this is OK.
    $endgroup$
    – Robert Lewis
    Jun 13 '17 at 18:34






  • 1




    $begingroup$
    I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
    $endgroup$
    – user
    Jun 14 '17 at 7:52






  • 1




    $begingroup$
    The proof of the positive definiteness has appeared to be trivial.
    $endgroup$
    – user
    Jun 14 '17 at 13:10


















  • $begingroup$
    I made a small edit to your post and added a few tags. Hope this is OK.
    $endgroup$
    – Robert Lewis
    Jun 13 '17 at 18:34






  • 1




    $begingroup$
    I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
    $endgroup$
    – user
    Jun 14 '17 at 7:52






  • 1




    $begingroup$
    The proof of the positive definiteness has appeared to be trivial.
    $endgroup$
    – user
    Jun 14 '17 at 13:10
















$begingroup$
I made a small edit to your post and added a few tags. Hope this is OK.
$endgroup$
– Robert Lewis
Jun 13 '17 at 18:34




$begingroup$
I made a small edit to your post and added a few tags. Hope this is OK.
$endgroup$
– Robert Lewis
Jun 13 '17 at 18:34




1




1




$begingroup$
I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
$endgroup$
– user
Jun 14 '17 at 7:52




$begingroup$
I think I was able to prove the statement, except for showing that $tildeOmega$ is positive-definite. I would be thankful for any hint concerning a proof of the latter property.
$endgroup$
– user
Jun 14 '17 at 7:52




1




1




$begingroup$
The proof of the positive definiteness has appeared to be trivial.
$endgroup$
– user
Jun 14 '17 at 13:10




$begingroup$
The proof of the positive definiteness has appeared to be trivial.
$endgroup$
– user
Jun 14 '17 at 13:10










1 Answer
1






active

oldest

votes


















1












$begingroup$

This is just a bunch of N linear oscillators, a system readily diagonalizable, upon defining real coordinates $y_iequiv x_i/mu_i$, so you have
$$
left(sum_i frac{partial^2}{partial y_i^2}-sum_{ij}tilde Omega_{ij}y_iy_j+Eright)Psi=0,
$$
with $tilde Omega$ real, symmetric (symmetrize by the ys) and positive.



As such, it is orthogonally diagonalizable to its real positive eigenvalues' matrix $tilde { tilde Omega}_{ij}=delta_{ij}omega^2_i$, with eigenvectors $z_i$, to wit
$$
left(sum_i ( frac{partial^2}{partial z_i^2}- omega_{i}^2 z_i^2)+Eright)Psi=0,
$$
N decoupled harmonic oscillators.






share|cite|improve this answer









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    $begingroup$

    This is just a bunch of N linear oscillators, a system readily diagonalizable, upon defining real coordinates $y_iequiv x_i/mu_i$, so you have
    $$
    left(sum_i frac{partial^2}{partial y_i^2}-sum_{ij}tilde Omega_{ij}y_iy_j+Eright)Psi=0,
    $$
    with $tilde Omega$ real, symmetric (symmetrize by the ys) and positive.



    As such, it is orthogonally diagonalizable to its real positive eigenvalues' matrix $tilde { tilde Omega}_{ij}=delta_{ij}omega^2_i$, with eigenvectors $z_i$, to wit
    $$
    left(sum_i ( frac{partial^2}{partial z_i^2}- omega_{i}^2 z_i^2)+Eright)Psi=0,
    $$
    N decoupled harmonic oscillators.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      This is just a bunch of N linear oscillators, a system readily diagonalizable, upon defining real coordinates $y_iequiv x_i/mu_i$, so you have
      $$
      left(sum_i frac{partial^2}{partial y_i^2}-sum_{ij}tilde Omega_{ij}y_iy_j+Eright)Psi=0,
      $$
      with $tilde Omega$ real, symmetric (symmetrize by the ys) and positive.



      As such, it is orthogonally diagonalizable to its real positive eigenvalues' matrix $tilde { tilde Omega}_{ij}=delta_{ij}omega^2_i$, with eigenvectors $z_i$, to wit
      $$
      left(sum_i ( frac{partial^2}{partial z_i^2}- omega_{i}^2 z_i^2)+Eright)Psi=0,
      $$
      N decoupled harmonic oscillators.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        This is just a bunch of N linear oscillators, a system readily diagonalizable, upon defining real coordinates $y_iequiv x_i/mu_i$, so you have
        $$
        left(sum_i frac{partial^2}{partial y_i^2}-sum_{ij}tilde Omega_{ij}y_iy_j+Eright)Psi=0,
        $$
        with $tilde Omega$ real, symmetric (symmetrize by the ys) and positive.



        As such, it is orthogonally diagonalizable to its real positive eigenvalues' matrix $tilde { tilde Omega}_{ij}=delta_{ij}omega^2_i$, with eigenvectors $z_i$, to wit
        $$
        left(sum_i ( frac{partial^2}{partial z_i^2}- omega_{i}^2 z_i^2)+Eright)Psi=0,
        $$
        N decoupled harmonic oscillators.






        share|cite|improve this answer









        $endgroup$



        This is just a bunch of N linear oscillators, a system readily diagonalizable, upon defining real coordinates $y_iequiv x_i/mu_i$, so you have
        $$
        left(sum_i frac{partial^2}{partial y_i^2}-sum_{ij}tilde Omega_{ij}y_iy_j+Eright)Psi=0,
        $$
        with $tilde Omega$ real, symmetric (symmetrize by the ys) and positive.



        As such, it is orthogonally diagonalizable to its real positive eigenvalues' matrix $tilde { tilde Omega}_{ij}=delta_{ij}omega^2_i$, with eigenvectors $z_i$, to wit
        $$
        left(sum_i ( frac{partial^2}{partial z_i^2}- omega_{i}^2 z_i^2)+Eright)Psi=0,
        $$
        N decoupled harmonic oscillators.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 16:25









        Cosmas ZachosCosmas Zachos

        1,810522




        1,810522






























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