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Ground state energy eigenvalue of particle in 1D potential

  1. Nov 9, 2015 #1
    1. The problem statement, all variables and given/known data
    a particle of mass m moves in 1D potential V(x),which vanishes at infinity.
    Ground state eigenfunction is ψ(x) = A sech(λx), A and λ are constants.
    find the ground state energy eigenvalue of this system.

    ans: -ħ^2*λ^2/2m

    2. Relevant equations
    <H> =E, H = Hamiltonian.
    p= i/ħ∂/∂x



    3. The attempt at a solution
    H = p^2/2m+ V(x)
    by normalization,

    |A| = λ
    i can take care of p^2/2m but what about V(x)
     
  2. jcsd
  3. Nov 9, 2015 #2

    mfb

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    It should be possible to find it based on the wave function (the Schroedinger equation is satisfied everywhere). There could be some solution that avoids this, not sure if the virial theorem or something similar can be used.
     
  4. Nov 9, 2015 #3
    solving the S-equation yields
    λ^2*ħ^2/2m(sech^2(λx)-tanh^2(λx))=E-V(x)
    or -λ^2ħ^2/2m= E-V(x)
    :headbang:
     
  5. Nov 9, 2015 #4

    mfb

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    If that is true for every x, evaluate it at x to infinity.
    It would need a constant potential, however, which doesn't fit to the given wave function.

    I don't understand where your equation comes from. The second derivative is not constant.
     
  6. Nov 9, 2015 #5
    sorry,
    used the wrong identity
    the final equation is,
    -λ^2*ħ^2/2m(1-2sech^2(λx))=E-V(x)
    it is a multiple choice question.
    since sech^2(λx) is 0 at infinity, i assume term 1 on lhs represents energy and term 2 represents potential.
     
  7. Nov 10, 2015 #6

    mfb

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    That gives a more reasonable potential, and it also gives E-V(x) at x to infinity, where you know the limit of V(x).
     
  8. Nov 10, 2015 #7

    BvU

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    Could you check this second derivative again ? Perhaps 1 - 2 tanh2 ?
     
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