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Quantum mechanics ground state

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Why is the ground state always symmetric and first excited state anti-symmetric?
    OR Why does the ground state always have no node and first excited state has one node?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 19, 2010 #2

    Matterwave

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    This is not true in general. The ground state is symmetric only for symmetric potentials.

    As to why exactly this is the case, I've seen some sketchy proofs, but I cannot offer anything deep. Perhaps the next poster will have a better response.
     
  4. Apr 20, 2010 #3
    I think you should consider a more specific example .. because the situation would be different (i.e) for different boundaries ..

    say you have a particle in a box, and the length is from x = 0 to x = a .. when you solve the Schrodinger equation you will end up with a solution that is neither even nor odd ..

    but if the length is from x = -a/2 to x = a/2 .. then for n=1,3,5.. it is even (symmetric) and for n=2,4,6... is it odd (antisymmetric)
     
  5. Apr 21, 2010 #4
    But generally speaking, is it true that ground state must have no node? Why?
     
  6. Apr 21, 2010 #5
    The greater the curvature of a wave function, the more kinetic energy that it has. Wave functions with nodes have more curvature than nodeless wave functions. Since we are trying to minimize the energy, it makes sense to me that wave functions with no nodes would be good candidates to be the ground state so long as the potential energy doesn't make them have greater total energy than wave functions with nodes. This certainly isn't a proof, but hopefully it helps.
     
  7. Apr 22, 2010 #6
    Anyone knows any simple proof?
     
  8. Apr 23, 2010 #7
    im not sure if that would help..

    a symmetric function is such a function which has the property f(x)=f(-x) .. and the antisymmertric f(x) = - f(-x) ..
     
  9. Apr 23, 2010 #8
    I think that second way that jasony phrased his question is really more of what he is aiming for: why does the ground state have no nodes? Correct me if I'm wrong jasony.
     
  10. Apr 25, 2010 #9
    yes why does the ground state has no node? Can anyone provide a simple maths proof?
     
  11. Aug 20, 2010 #10
    Hi, I know this is an old thread, but I have a simple Hamiltonian whose ground state is anti-symmetric.

    [tex]H=\left[\begin{array}{cc}0 & \Delta\\\Delta & 0\end{array}\right]\,.[/tex]

    Eigenstates and eigenenergies are:

    [tex]\psi_{\eta}=\left[\begin{array}{c}1\\ \eta\end{array}\right]\,,\quad\varepsilon_{\eta}=\eta\Delta\,,[/tex]

    where [itex]\eta=\pm 1[/itex]. The ground state corresponds to [itex]\eta=-1[/itex], which is anti-symmetric.
     
  12. Aug 20, 2010 #11

    kuruman

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    Badphysicist is on the right track.

    The Schrodinger Equation in one dimension can be written as

    [tex]\frac{d^2 \psi}{dx^2}+k^2 \psi=0[/tex]

    where

    [tex]k^2=\frac{2m(E-V)}{\hbar^2}[/tex]

    For the ground state, lowest energy means lowest possible value of k.
    Lowest possible value of k means longest possible value of the sinusoidal wavelength inside the well.
    A wavefunction with a node has shorter wavelength than a wavelength with no nodes.
    That's why.
     
    Last edited: Aug 20, 2010
  13. Aug 20, 2010 #12
    Well, that demonstrates that the ground-state for the 1D Schroedinger equation in a piece-wise constant potential contains no nodes. If that's the OP's question then you can simply solve the SE and verify that the ground state indeed contains no nodes.

    If you want to prove that it's true in the case of, say [itex]H=\sum_i\left\{-\nabla^2_i+V(\vec{r_i})\right\} + \sum_{i\neq j}U(\vec{r_i}, \vec{r_j})[/itex] (or any Hamiltonian you can write down) then it's not so easy to show that the ground state really contains no nodes as far as I'm aware - although it is true.

    You can look at appendix A of this thesis http://thesis.library.caltech.edu/1007/ for some hints of how this might be proven in the case of a single particle.

    Also try looking for the derivations of the equations used in DFT. I'm not sure if that's so useful, but they relate the ground state energy of a system its particle density, so you might find from these equations that for some reason the ground state never contains a nodal point.
     
  14. Aug 24, 2010 #13
    its the how of the question. not the why?
     
  15. Aug 24, 2010 #14
    i think answer is simple. ground state is n=0 that is even. first excited stat is n=1 odd.


    by the way you can use ladder operators.
     
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