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Zero Point Energy and Bose einstein Condensates

  1. Dec 11, 2012 #1
    They say ZPE is the energy of the system at 0 K due to vibrations and even though the Schroedinger equation gives us the ground state minimum energy of the system (lets say by DFT/HF/QMC whatever), this energy is a little higher than that.

    My first question is, because I have learnt that SEqs is the most basic equation of QM, why are the atomic vibrations at 0 K not an out come of the SEqs. Because in reality now our ground state is redefined. If the system can NEVER have energy less than ZPE then isn't it the ground state energy. I know it might be very difficult to calculate, but apart from the electronic ground state shouldn't ZPE be also included in the ground state?? Or is this energy out of the purview of SEqns

    My second question is - if i am simulating (using DFT or other ab-initio method) the lattice system which has total integer spin (like lets say a collection of rubidium atoms) - then does my simulation actually give me a BOSE EINSTEIN condensate..??
  2. jcsd
  3. Dec 11, 2012 #2


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    When they say so they most probably refer to the solution of the electronic Schroedinger equation in Born Oppenheimer approximation.
    The vibrational energy has (and can) be calculated in a second step, so there is no problem with modern programs to calculate the ground state energy including the ZPE.
    To your second question: Forming a lattice and BEC exclude one another. So a lattice of Rb atoms cannot form a BEC.
  4. Dec 11, 2012 #3
    Thanks DrDu! I get it now. Programs like VASP can indeed calculate the ZPE. I still think though that it doesn't make much sense to use the ground state energy values not taking into account ZPE in the ground state energy, because including it gives us a better result.

    As for your answer to the second question:

    Of course BEC is not a lattice. I will try to reframe my question (which is inspired from a documentary which told that one of the first systems which was made into a BEC composed of rubidium atoms). Supposing I simply calculate for a given number of rubidium atoms using DFT at 0 K - why should the system not be a BEC? The only requirements for BEC are bosonic systems and extremely low temperatures which are both satisfied in this case I guess
  5. Dec 11, 2012 #4


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    There are several problems with this:
    A) DFT is a semi-empirical proceedure and functionals are neither designed nor sufficiently accurate for this kind of calculations.
    B) You would have to include the confining magnetic and electric potentials.
    C) That would be quite a formidable task for several thousand Rb atoms.
    D) It does not make much sense. You can obtain all the relevant parameters, like the scattering lengths from accurate calculations done on one or two atoms.
  6. Dec 11, 2012 #5
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