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Quantum 1D box obtain an expression for the normalization constant

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    An electron in a one-dimensional box with walls at x =(o,a) is in the quantum state
    psi = A o<x<a/2
    psi = -A a/2<x<a
    A) obtain an expression for the normalization constant, A.
    B) What is the lowest energy of the electron that will be measured in this state?

    2. Relevant equations
    Not given anything. But its a chapter on Hermitian operators, and Hamiltonian.

    3. The attempt at a solution
    So for part a I think I am just supposed to normalize, so 1=integral of A*A ... and I get A=(a)^(-1/2) which i think is my normalization constant.

    But is it asking not for the constant but to work back and find an equation?

    For part b,,, I just operated on psi with the Hamiltonian and because all I had were constants, I got zero... which is boring if true, but I think isn't the answer I am looking for.

    If anyone could help out, many thanks.
     
  2. jcsd
  3. Nov 23, 2015 #2

    DrClaude

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    Staff: Mentor

    It is only asking for A, in order to normalize the wave function, such that you can do part B.

    That is indeed not the correct approach. What you have to do is decode what
    actually means. What possible results can you get if you measure the energy of the electron?
     
  4. Nov 24, 2015 #3


    Don't you find the energy by using the Hamiltonian operator? If this isn't I have no idea what to do with this.
     
  5. Nov 24, 2015 #4

    DrClaude

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    What do the postulates of QM say about measurements?
     
  6. Nov 24, 2015 #5
    The only measurable results are the eigenvalues associated with the operator... but I don't know how that helps because zero is a valid result.
     
  7. Nov 24, 2015 #6

    DrClaude

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    Yes, and given the wave function, what is the probability of finding a given eigenvalue?
     
  8. Nov 24, 2015 #7
    Do I find the expansion coefficent?
     
  9. Nov 24, 2015 #8

    DrClaude

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    Yes. You basically need to calculate the coefficients, starting from the ground state and going up in energy, to find the first one that is not 0. Then you can get the lowest possible measured energy.
     
  10. Nov 24, 2015 #9
    So

    OK so I haven't done the work yet. But just intuition it's something like the second state?
     
  11. Nov 24, 2015 #10

    DrClaude

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    Staff: Mentor

    Good intuition. The ground state is even with respect to the center of the box, while ψ is odd, so the integral will be zero. You should get a non-zero coefficient with the first excited state (n = 2).
     
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