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SU(3) symmetry and simple zeros of w. f.

  1. Jan 29, 2008 #1
    Can someone help me? It is correct at all to make the conection between the SU(3) symmetry and zeros of (radial) wave function? To make more clear: can I say that the fact that radial wave function has only the simple zeros automatically excludes the existence of SU(3) symmetry for given quantum system? Please, answer me soon, it is very important to me.
    Thanks in advance
    stel
     
  2. jcsd
  3. Jan 29, 2008 #2

    Tom Mattson

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    SU(3) of what, exactly? Usually when talking about symmetries we say something like "SU(3) of color" or "SU(3) of flavor". SU(3) by itself is just a mathematical group. It's only when you use it to describe something physical that a quantum system can be said to "have" that symmetry.
     
  4. Feb 3, 2008 #3
    Obviously I had to be more specific. Well, I will write something about the ‘history’ of the problem that delivers my question. You probably heard of the ‘particle in the box’ model or ‘spatially confined system’ model with applications in many diverse fields of physics: atomic and molecular, solid state, astrophysics … An example is the hydrogen atom placed in the center of the sphere with impenetrable walls (also called ‘compressed atom’ model and introduced in the physics to explain the pressure effect on the H atom spectrum). Those systems have interesting eigenspectrum properties. It is obviously that the degeneracy of their energy levels will be changed in comparison with the analogous unconfined (or free, or, as someone says, standard) quantum systems as a consequence of the different symmetry their Hamiltonians posses (or ‘have’ as you say). Thus, the symmetry group of the confined H atom is SO(3) group in contrast to SO(4) group of the unconfined H atom, and Coulomb degeneracy, characteristic of that last system energy levels, is removed. But, when the radius of the confining sphere has a specific value (l+1)(l+2) (here l is orbital quantum number), hidden or conditional symmetry appears (there is a couple references about it where the symmetry group of that system is not explicitly named) and, of course, specific degeneracy (states with orbital quantum numbers l and l+2 have the same energy). What is going to happen when, instead of H atom, three-dimensional isotropic harmonic oscillator (3D IHO) is centrally enclosed in the sphere with impenetrable walls? My intention was to see how the spatial confinement change the degeneracy of the energy levels of the 3D IHO, i. e. to establish if confined 3D harmonic oscillator has some kind of symmetry, higher than obvious SO(3) symmetry and that is the moment when one of the colleagues enters the story. He claims that I cannot get anything and prove anything applying the generators of the SU(3) group, symmetry group for unconfined 3D IHO (similar procedure was done for confined H atom with the generators of SO(4) group in the references I have already mentioned) on the wave function of the confined 3D harmonic oscillator state ‘due to the well-known fact that the zeros of Schroedinger wave functions are simple’ (means, of the radial wave functions as the functions in radial coordinate). I could not agree with him and accept that usefulness of the SU(3) group is connected with the fact that the zeros of the radial wave functions are not simple. By the way, I do not know if there is some quantum system with that property. Moreover, the radial wave functions of the unconfined 3D IHO states have the simple zeros (except the zero in the origin, but the same is valid for confined HO) and this system have SU(3) symmetry and corresponding degeneracy of its energy levels! In my opinion symmetry of the system is determined by the existence of the operators commuting with Hamiltonian and the commutator values among those operators specify the symmetry group. But me and the colleague still stand on our standpoints each. Who is right?

    stel
     
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