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QM axioms¿always true?

  1. Sep 12, 2005 #1
    I have this question,when i was told QM they taught me some axioms:(1D)
    The wave function of the particle is given by the differential equation:

    [tex]i\hbar\frac{d\psi}{dt}=\frac{-\hbar^{2}}{2m}D^{2}\psi+V(x)\psi [/tex] with D=d/dx

    and that the eigenfunctions of [tex]H\phi=E_{n}\phi[/tex] are all orthogonal and are on L^2(R) function space....

    i have discussed in other forum a method to obtain RH by assuming that a Hamiltonian have its energies being the eigenvalues of a certain function f(x) my question in this case is if:

    a)does the Schroedinguer equation have always a solution independent of what the potential V is?.,let,s suppose that potential is dicontinous everywhere (impossible but mathematically true).

    b)are always the eigenfnction of the Hamiltonian on the space L^2(R)

    c)can we always say that exist a potential V so the energies of the system are the roots of a certain function f(x)?...

    i have some discussion with mathematicians there saying that ...
    Last edited: Sep 12, 2005
  2. jcsd
  3. Sep 12, 2005 #2


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    a) Functions that are continuous everywhere, but nowhere differentiable are also possible mathematically, but are physically meaningless (although some artificial potentials they can be used as excellent models). I`m sure that in all practical cases you will encounter a solution does exist.

    b) Are you asking whether all eigenfunctions of the Hamiltonian live in L^2(R)? That answer is no, so those functions are not physically realizable. They exist only as mathematical solutions to the D.E.
  4. Sep 12, 2005 #3


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    First thing's first:
    Yes, the axioms of quantum mechanics are ALWAYS TRUE.

    1.Technically, the SE is a second order PDE. Finding solutions to is not always simple. All solutions depend on the shape of the potential. In practice, the V functions is known (the methods to get it is beyond the preoccupation of the theorists) and the equations solutions are sought through various anaytical or perturbative methods.

    2. By the virtue of the Stone- von Neumann's theorem and that of Bargmann's transformation of the Fock representation of Born-Jordan CCR, we know for sure that [itex] L^{2} [/itex] is a "good" Hilbert space for wavefunctions. The trick appears when the operators (position, momentum and the Hamiltonian) are unbounded (they are in most cases), so that the domain of these operators are dense subspaces (called "nuclear subspaces") of [itex] L^{2} [/itex].

  5. Sep 12, 2005 #4


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    There is, to date, no reason to doubt the assuptions behind QM. And, of course, if somehow, somewhere these assuptions are not true, then sooner than later, some experiment will demonstrate the problem. And, actually, there are problems with the math of Quantum Field Theory, we simply do not know whether QFT is well-set.

    Most physicists are highly pragmatic, and tend not to worry about all the mathematial nicities -- hey, not to worry about unbounded operators.... And, as far as I know, when the mathematicians have caught up with us, they provide justification for our intuitive approach. All that really matters is that the Schrodinger Eq.has solutions for physically realistic potentials -- and it does -- talking here non-rel SE, and, I'm pretty sure such solutions also work in the case ofthe Dirac and Klein-Gordan equations.

    Reilly Atkinson
  6. Sep 12, 2005 #5
    the problem i have is this: in a math forum i provided a proof to RH that is that all the NOn-trivial zeros of the function [tex]\zeta(1/2+is)=0[/tex] to prove that i choose a Hamiltonian,whose eigenvalues are precisely the roots of this function and after that i prove ther are all real,i made the assumption that the potential is continous but some nasty mathematician keep saying that this is not true and that my affirmation is not valid (this of course without a proof), then this is my problem if given a nowhere-continous potential can we make the assumption that the SE will have Eigenvalues and eigenfunctions...
  7. Sep 12, 2005 #6


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    Your proof is way beyond my pay grade.

    As far as I know a requirement for solution of a Schrodinger (differential) equation is potentials continuous except at countable sets of measure zero, like a potential made from a countable number of square wells. What would be interesting, however, is to look at the electrical potential and/or fields of a classical particle undergoing nowhere-continuous motion.
    Tough stuff.
    Reilly Atkinson
  8. Sep 16, 2005 #7
    Also we can prove that for every set of energies [tex]E_{n}=g(n)[/tex] we can always find a potential in first order perturbation theory...

    [tex]E_{n}-E^{0}_{n}=<\phi|V|\phi>[/tex] where [tex]\phi E^{0}_{n} [/tex] are the eigenfucntions and eigenvalues of H0=p^{2}/2m
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