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Problem: particle in infinitely deep potential

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data

    A square well: 0≤x≤L, V=0, for others x V=∞
    ψ=N[x(x-L)+Ax2(x-L)+Bx(x-L)2]
    H=-[STRIKE]h[/STRIKE]/2m*d2/dx2+V(x)

    a) Determine N
    b) Use the Rayleigh-Ritz variational method to obtain the <ψ|H|ψ> as a function of A and B, and minimize the energy with respect to A and B
    c) Discuss how many distinct values of the pair (A, B) that you obtain and do you get more than one energy? If yes, what are they?
    d) Compare you results to the exact particle in the box and discuss the error.
    a') Consider the same system, but now V(x)=x for 0≤x≤L
    Compute the first order perturbation correction to the ground state energy
    b') Compute the first order perturbation correction to the ground state wave function

    3. The attempt at a solution

    a) I found with SAGE mathematical software that N2=[itex]\frac{\sqrt{210}}{\sqrt{(2B2-3AB+2A2)L7+7(A-B)L6+7L5}}[/itex]
    b) I'm not sure in that. But I've calculated <E>=-[STRIKE]h[/STRIKE]/2m*∫ψ(x)*ψ''(x)*dx from 0 to L, it's another value depending from A, B and L.
    I don't understand, what does "minimize the energy with respect to A and B" means. It already depends only from A, B and L.
    c) It looks like that there must be some border values for energy, so it will be possible to determine A and B.
    d) Exact particle is E=[STRIKE]h[/STRIKE]2/8mL2, just need to know A and B in the value from b) to compare.

    a') <E>=-[STRIKE]h[/STRIKE]/2m*∫ψ(x)*(d2/dx2+x)[ψ(x)]*dx from 0 to L. Am I right in that solution? Or it is just -[STRIKE]h[/STRIKE]/2m*∫ψ(x)*(x)*ψ(x)*dx
    b') have no idea, how to calculate correction for the ground state of the wave function.
     
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 16, 2011 #2

    vela

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    The variational principle gives you an upper bound on the energy of the ground state; therefore, to get the best estimate, you want to find the least upper bound. In this case, your estimate will vary with A and B, so you want to find the value or values of A and B that minimize the function.
     
  4. Oct 16, 2011 #3
    Thank you.
    I've got 3 values of the <E>=5, 7 and 21 *([STRIKE]h[/STRIKE]2/mL2) after the finding derivatives of <E> with respect to A, B and AB.
    So, the minimum of <E> is 5([STRIKE]h[/STRIKE]2/mL2).

    But I'm still stuck with a') and b').
     
  5. Oct 17, 2011 #4

    vela

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  6. Oct 27, 2011 #5
    Vela, thanks for helping me!
     
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