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Ground state in an infinite square well with length doubling

  1. Jun 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume a particle is in the ground state of an infinite square well of length L. If the walls of the well increase symmetrically such that the length of the well is now 2L WITHOUT disturbing the state of the system, what is the probability that a measurement would yield the particle in the ground state of the 2L-Well?

    2. Relevant equations
    Ground State of infinite well of length L: |ψ>= ∫√(2/L)cos(π/L*x)|x>dx from -L/2 to L/2

    3. The attempt at a solution
    Since the well has expanded, the current state of the system is no longer an eigenstate and needs to be renormalized because the range of integration has changed which yields
    ψ=√(1/L)cos(π/l*x).
    Now, the NEW ground state is just
    φ=∫√(1/L)cos(π/2L*x')|x'>dx'
    so the probability of finding the particle in the new ground state is just
    <φ|ψ>=∫∫<x'|x>(1/L)cos(π/2L*x')cos(π/L*x)dx'dx=
    ∫∫δ(x'-x)(1/L)cos(π/2L*x')cos(π/L*x)dx'dx=
    ∫(1/L)cos(π/2L*x)cos(π/L*x)dx from -L to L
    which I found to be equal to 4/3π, but the text claims the answer should be (8/3π)2! can someone point me to the mistake I've made somewhere?
     
  2. jcsd
  3. Jun 25, 2015 #2
    I've found my mistake! There is no need to renormalize the initial ground state because it is 0 outside the original limits! Also, The final integration should only be taken from -L/2 to L/2 for the same reason! but I still get an answer of 8/3π and NOT (8/3π)^2 like the book claims. Where does the square come from?
     
  4. Jun 25, 2015 #3

    blue_leaf77

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    The probability is interpreted as ##|\langle \phi | \psi \rangle|^2 ## because when you sum it up for all the expansion bases, you get unity.
     
  5. Jun 25, 2015 #4
    Thanks! It completely slipped my mind! LOL
     
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