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Quantum energy doubt

  1. Nov 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, I have a problem related to some quantum concepts.

    One teacher of mine has tasked me with solving the next problem:

    Given a particle on its fundamental state in a potential pit of width a, V=0 between 0 and a and infinite at the rest of the space. Suddenly, the width changes to 2a, we proceed to measure the energy without changing the wave function.

    a) Which is the most likely value? And the probability of measuring that value?

    b) Which is the expected value of the energy? And its uncertainty?

    2. Relevant equations


    E=(P2/2m)+V(x)
    ^H=(-iħ*(∂/∂x))2/2m +V(x)
    E=∫dxψ*[^H]ψ+V(x), the integration limits are 0 and a.
    <E>=∫φEφE*, the integration limits are 0 and a.
    3. The attempt at a solution
    If we double the width of the pit , we double the integration limits; the eigenvalues of the function don't change if the function doesn't.
    I'n not sure how th keep with it, so any help would be appreciated.
    Thanks.
     
  2. jcsd
  3. Nov 23, 2014 #2

    Orodruin

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    If you explicitly measure the energy, you will measure an eigenvalue of the Hamiltonian. You need to find out which linear combinations of the new potential the eigenfunction of the old one corresponds to. The point is that the eigenfunction of the old potential is not an eigenfunction of the new one.
     
  4. Nov 23, 2014 #3
    Ψ=eigenfunction of the corresponding eighenvalue I assume that the particle can be considered free in the pit. Ψ=(1/√[2π])∫dk φ(k)Exp[kx-(hk2/2m*(t-t0))].
    Where φ=∫dx Exp[-ikx]Ψ(x, t0)
    I don't have the term Ψ(x, t0), so I can't calulate φ for both integration limits.
     
  5. Nov 23, 2014 #4

    Orodruin

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    So your first problem is then to figure out what the ground state wave function is by solving the Schrödinger equation. For later convenience, it will also be good to have the wave functions of the excited states.
     
  6. Nov 23, 2014 #5
    Which is :Eigenvalue= AExp[ik]+BExp[-ik], where k=√[2mE/ħ2]
    Am I correct?
    Because this doesn't solve the problem of my lack of initial condition
     
  7. Nov 23, 2014 #6

    Orodruin

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    Yes, but all k are not allowed. You need to make sure they fulfil the boundary conditions imposed by having infinite potential outside of the well. This is what will give you a discrete energy spectrum.
     
  8. Nov 23, 2014 #7
    If I write φ=Asin(kx)+Bcos(kx) and I say φ=0 at x=a, x=0, I can define energy as (nπħ)2/(2ma2)
    Then, I have the finite values of the energy. But I have an indetermination, of n.
    With this energy will I be able to solve φ for any x and then Ψ?
     
  9. Nov 23, 2014 #8

    Orodruin

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    So what n corresponds to the ground state of the first system, i.e., the one with the lowest energy? What is the k corresponding to the ground state in the original system?

    For the second system: How would you find the probability of a general wave function of being in the nth energy eigenstate?
     
  10. Nov 23, 2014 #9
    To the first question, with n=0 I have the lowest energy possible, so k would also be 0.
    To the second. I would integrate between 0 and a ∫|ΨΨ*|dx, for each x (the possible value); the inegral would be the probability to measure that x.
    Then I I have to find the right value of A so the inegral would have a value of 1
     
  11. Nov 23, 2014 #10

    Orodruin

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    No, this is not the case, n=0 results in a trivial solution, i.e., ##\psi = 0##.

    It is not about being in a particular position, the question is what is the probability of being in a particular energy eigenstate.
     
  12. Nov 23, 2014 #11
    Then n=1 and the chance of a certain eigenvalue is ∫|φφ*|2dx.
    Am I correct now?
     
  13. Nov 23, 2014 #12

    BvU

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    Not good. Re-read Oro's posts 2 and 4. By now you should have a (normalized!) function ##\Psi(x)## on [0,a] which is an eigenfunction of the old Hamiltonian. Extend it to [0,2a] with ##\Psi(x)=0## on [a,2a]. This ##\Psi## is not an eigenfunction of the new Hamiltonian, but a linear combination of these new eigenfunctions with each its different energy. Your task is to evaluate the coefficients.
     
  14. Nov 23, 2014 #13
    Allright, I gess I will figure how to do this with the information you have given me.
    Thank you very much both of you for your anwsers.
     
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