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Delta Function Bound State

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle moves in one dimension in the delta function potential V= αδ(x). (where that is an 'alpha' ... not 'a')
    An initial wave function is given
    [tex] \Psi = A(a^2-x^2) [/tex] for x between -a and a and Psi=0 anywhere else
    What is the probability that an energy measurement will yield something other than:
    [tex] E=\frac{-m\alpha^2}{2\hbar^2} [/tex]

    note: when we solve the delta function potential problem, we find (for E<0) that there is only one bound state, and only one allowed energy: and that is exactly the one shown above.
    2. Relevant equations

    [tex] c_n = \int \Psi_n^* \Psi(x,0) dx[/tex]

    [tex] P(E_n) = abs(c_n)^2 [/tex]
    3. The attempt at a solution

    At first I was sure that since there is only one bound state in the delta function potential problem, and only one allowed energy, that the probability of measuring another energy should be zero. But when I calculate the coefficient c1, I do not get 1, i get something like .913 and then c1 squared is about .83 which would then be the probability of measuring the bound state energy.
    This is what is confusing me. If c1 is not equal to 1, then there must be some other c's with other eigenfunctions that must be combined linearly to produce the given initial wave function. But there is only one energy eigenfunction, so what are these other functions??
    And if there is only one energy eigenstate, is it possible to measure an energy other than the energy of the eigenstate?
    Thank you
     
  2. jcsd
  3. Oct 28, 2011 #2

    George Jones

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    Yes.
    Are you sure?
     
  4. Nov 1, 2011 #3
    We solved the schrodinger equation and found only one bound state with a corresponding energy. What we have done before is to expand a given state in terms of energy eigenstates, and the the constants in the summation give a probability density for the energy. But if we do such an expansion, since there is only one eigenstate, wouldn't its c value have to be 1? If not, then what other functions are we including in the linear combination?
     
  5. Nov 1, 2011 #4

    George Jones

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    Is the bound state the only energy eigenstate?
     
  6. Nov 2, 2011 #5
    i dont know of any others..
     
  7. Nov 2, 2011 #6

    vela

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    The key word is bound.
     
  8. Nov 3, 2011 #7
    If anybody has an explanation I would be happy to hear it. I have turned in the homework assignment already and decided it makes most sense to just calculate the c value for the bound state, and say

    [tex] P(E \neq E_1 ) = 1 - |c_1|^2 = .17 [/tex]

    What I am wondering about is how to express the state as a combination of eigenstates. Is it c1 times the bound state, and then plus some integral over scattering states?
     
  9. Nov 4, 2011 #8

    George Jones

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    Yes. From Quantum Mechanics by Robinettt
     
  10. Nov 4, 2011 #9
    Aha! Thank you
     
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