Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unbounded Hamiltonian leading to finite ground state

  1. Dec 28, 2013 #1
    If a Hamiltonian is unbounded from below, say the hydrogen atom where the Hamiltonian is -∞ at r=0, is there a way to tell if the ground state is bounded (e.g. hydrogen is -13.6 eV and not -∞ eV)?

    It seems if the potential is 1/r^2 or less, then the energy will be finite as:

    [tex]\int d^3 r (1/r^2) P(r) = \int r^2 dr (1/r^2) P(r)=1[/tex]

    where P(r) is the probability density.

    Or is this too naive and you have to work out P(r) which can help the integral converge or make it diverge?

    There are two things at play it seems, the potential and kinetic energies, and Heisenberg's uncertainty principle which keeps the hydrogen atom from falling into the nucleus.

    Is there a way to minimize H(p,r)=p^2/2m+V(r) with respect to r and p, using the constraint ΔrΔp<h/2, and seeing if the solution is finite?
     
  2. jcsd
  3. Jan 2, 2014 #2

    jfizzix

    User Avatar
    Science Advisor
    Gold Member

    I think I would look to the Virial theorem in quantum mechanics.

    For an interparticle potential [itex]V(r) = a r^{n}[/itex], and the system in an eigenstate of the Hamiltonian, the Virial theorem states that

    [itex]\langle 2 T \rangle = n \langle V \rangle[/itex], where the brackets denote expectation values. Expressing the Virial theorem in terms of the total energy [itex]E[/itex],

    [itex]\langle E \rangle = \frac{n+2}{n}\langle T\rangle = \frac{n+2}{2}\langle V\rangle[/itex].

    Since [itex]n[/itex] is a finite value, we at least know that if the expectation value of the potential energy is finite, the total energy must be finite as well.
     
  4. Jan 3, 2014 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Your question can be the subject of a book and it was Franz Rellich and later Tosio Kato who first gave the answer. First of all, the H-atom's Hamiltonian is bounded from below or semibounded according to the standard definition from functional analysis (you can simplify and discard the COM dynamics). Then one shows the existence of a minum value of the matrix elements for the KE+KP for the generic atomic Hamiltonian (again discarding COM dynamics) which is of course a finite sum of H-atom/ion-like Hamiltonians.

    There are a myriad of books on the mathematical treatment, you can first give it a try with Berezin and Shubin's "The Schrödinger Equation".
     
  5. Jan 3, 2014 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Being bounded below is an important difference between the classical and quantum Hamiltonians for hydorgen!

    Following on from dextercioby's post, a good, recent book that treats all this is "Quantum Theory for Mathematicians" by Hall,

    https://www.amazon.com/Quantum-Theo...uate-Mathematics/dp/146147115X/ref=pd_sim_b_1

    With respect to the hydrogen Hamiltonian ##H##, Hall writes
    The proof of 9.38 uses the Kato-Rellich Theorem, also proved in Hall.
     
    Last edited by a moderator: May 6, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Unbounded Hamiltonian leading to finite ground state
Loading...