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1d potential V (-x)=-V (x) eigenfunctions.

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that for a 1d potential V (-x)=-V (x), the eigen functions of the Schrödinger equation are either symmetric/ anti-symmetric functions of x.


    2. Relevant equations


    3. The attempt at a solution
    I really don't know how to do it for odd potential.
    Let me show you how I am doing it for even potential.
    V (x)=V (-x)
    - (h2 / 2μ) [d2ψ(x)/dx2] + V (x) ψ(x)=E ψ (x).... (1)
    Making x to -x transformation we get
    - (h2 / 2μ) [d2ψ(-x)/dx2] + V (x) ψ(-x)=E ψ (-x)..(2)
    where i use V (-x)=V (x)
    Comparing (1) and (2) we see that ψ (x) and ψ(-x) eigenfunctions belongs to the same
    energy E.
    For a non degenerate state, ψ (-x ) must be a multiple of ψ (x):
    Ψ (-x) = λψ (x).
    Clearly ψ (x)= λψ (-x)=λ2ψ (x).
    λ2=1 or λ=+/- 1
    So ψ(-x)=+/- ψ(x)
    Now if i use odd potential in (2) eigenfunctions no longer belong to same energy E.
    The hamiltonian becomes weird for negative potential.
    Please help.


     
  2. jcsd
  3. Oct 5, 2014 #2

    BvU

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    Hello Sudip,
    I like your solution to the opposite problem (even potential), but I don't see how you have proven that there is no degeneracy.
    My book (Merzbacher, quantum mechanics, about 50 years old :-) ) only says that if ##\psi(x)## is a solution, then ##{1\over 2} [\psi(x)+\psi(-x)]## and ##{1\over 2} [\psi(x)-\psi(-x)]## are also solutions.

    If I imagine an odd-parity potential, e.g. a well from -a to 0 and a barrier from 0 to +a, I find it hard to believe the either/or statement can be proven. An even simpler V is a step from -V0 to + V0 at x = 0.

    I think any reasonable odd-parity V has to go to 0 when away from x=0, so there eigenfunctions are close to plane waves.

    Sorry I can't help you any better. Hope someone else can; if no: let us know if you find a sensible solution from teacher or by yourself.
     
  4. Oct 6, 2014 #3

    vela

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    I agree that the problem is flawed.
     
  5. Oct 6, 2014 #4
    How did I prove non degeneracy??
    Well , 1d bound states are non degenerate aren't they?
    I would like to quote somethings from a book by Dr.S.N.Ghosal(student of Emilio Segre) here :
    "The above classification of wavefunctions [λ=+/-1 so ψ(-x)=+/- ψ(x)] according to parity is possible only if Hamiltonian H^ remains invariant under parity operation i.e., if the interaction potential V (x) remain unchanged by parity operation, V (-x)=V(x)"
    Doesn't this statement raises question on the feasibility of the problem?
    I want to know in a simple yes or no is it possible to find symmetric as well antisymmetric functions for a potential well
    V(x)=-V0 for -a <x <0
    V (x)=0 elsewhere.
    Can't seem to find the above problem anywhere.A purely odd potential.
     
  6. Oct 6, 2014 #5

    vela

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    No, it's not possible. Your V(x) isn't odd either.
     
  7. Oct 6, 2014 #6
    Sorry.
    Will you say :
    V (x)= -V (a) for -a <x <0
    V (x)=0 elsewhere
    is odd??
     
  8. Oct 6, 2014 #7

    vela

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    No. An odd function satisfies V(-x) = -V(x) for all x. For your potential, V(a/2) = 0 while V(-a/2) = -V(a) ≠ 0 (I assume).

    BvU gave you an example of a simple odd potential: V(x) = V0 for x>0 and -V0 for x<0. Try solving the Schrodinger equation with this potential. You'll find the solutions aren't odd or even.
     
  9. Oct 6, 2014 #8

    BvU

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    I have to correct myself: after all the step function is a counter-example. V can't go to ##\infty##, but it doesn't have to go to 0 to be (mathematically - and thereby also physically) acceptable.

    I bring in another interesting potential function: a periodic square well . When x=0 is chosen such that V is even, eigenfunctions are even, but a shift of 1/4 period makes the potential odd, but not the eigenfunctions.
     
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