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Infinite Non-Square Well Potential

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Let ϕn(x) be the complete ortho-normal set of eigenstates of the Hamiltonian H = T +V
    and En, n = 0, 1, 2, are the corresponding eigenvalues. The E0 is the ground state energy.
    (a) If ϕ(x) is an arbitrary normalized state, show that E0 < = ∫dxϕ(x)Hϕ(x).

    (b) Consider one dimensional potential V0(x) = 0, abs|x| ≤ a/2 and = ∞ otherwise. The
    second one dimensional potential V1(x) = (4V|x|)/a - V_0, |x|≤ a/2, V > 0 and = ∞otherwise.
    If E0 and E1 are the ground state energies of the two potentials respectively. Find which
    one (1) E0 < E1 (2) E0 > E1 is correct and prove your judgement.

    2. Relevant equations

    Schrodinger's Equation

    3. The attempt at a solution

    I think I have gotten part A correct. However, I am so confused at a non-square well potential. I know that the infinite potential boundary conditions still apply here, but it throws me off with what is going on inside the well.

    Do I just do a normal solution of the T.I.S.E. and let my constant take care of the non-zero potential?

    Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 24, 2011 #2

    vela

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    Wouldn't the answer depend on the value of V0? You can shift the value of E1 around by changing V0, but E0 is fixed.
     
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