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Why is the solution of the schrödinger equation always symmetric or antisymmetric?

  1. Apr 18, 2007 #1

    I read that in an 1D potential, the solution for the Schrödinger equation is always either symmetric or antisymmetric if the potential is a symmetric function: V(x) = V(-x).

    How can I proof this?

    Thanks, for ur answers!
  2. jcsd
  3. Apr 18, 2007 #2
    The proof for this is typically done using parity operators. If your hamiltonian is given by
    [tex]\mathcal{H} = \hat{p}^2/2m + V(\hat{x})[/tex]
    Write down the time-independent Schrodinger equation, then flip the signs on all the x-coordinates and see what this imposes on the wave function.

    If you want a better discussion of this, check out Chapter 4 of Sakurai.
  4. Apr 19, 2007 #3
    Great! Thanks for this answer!
  5. Apr 19, 2007 #4
    I have another question to the same problem: In which situation can the eigenstates be degenerated?
  6. Apr 19, 2007 #5
    I'm not familiar with the word "degenerated". If you mean "having degenerate eigenstates", degeneracies usually arise when you have another observable [tex]\mathcal{O}[/tex] such that [tex]\left [ \mathcal{H}, \mathcal{O} \right ] - 0[/tex]. This implies that an eigenstate of the hamiltonian is also an eigenstate of your new observable (I leave it to you to figure out why).

    What frequently happens in this case is that there are multiple values of [tex]\mathcal{O}[/tex] for a given energy eigenvalue, and so you end up with degeneracies in the energy spectrum.

    Of course, sometimes things end up being more degenerate than they should be. For example, in hydrogen, the energy levels don't depend on the [tex]\ell[/tex] quantum number, although in general a spherically symmetric potential leads to an [tex]\ell[/tex] dependent energy spectrum. This is called an "accidental degeneracy". In the case of the hydrogen atom, the degeneracy arises because the angular momentum operators aren't the only ones that commute with the hamiltonian, and the underlying group symmetry of the hydrogen hamiltonian is SO(4).
  7. Apr 20, 2007 #6


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    A superposition of solutions is also a solution. A superposition of a symmetric and an antisymmetric solution is neither symmetric nor antisymmetric. Therefore, the solution does NOT need to be either symmetric or antisymmetric.
  8. Apr 20, 2007 #7

    The solution is not always symmetric or antisymmetric !!!
    It depends on the symmetries of the forces in the system.
    If the potential is symmetric (V(-x)=V(x)), the the hamiltonian cumutes with the inversion operator (P), and the operators H and P share a common basis of eigenvectors.

    Read about the consequence of [A,B]=0, apply that to your case here: [H,P]=0 .
  9. Apr 20, 2007 #8


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    Are we talking about solutions of the Schrodinger equation, or about eigenstates of the Hamiltonian operator? A superposition of solutions is a solution, whereas a superposition of eigenstates is NOT an eigenstate.
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