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Integral with symmetric infinitesimal bounds

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm reading something in my quantum physics book that says given a wavefunction ψ that is even, if we evaluate its integral from -ε to ε, the integral is 0. How can this be? I thought this is the property of odd functions.

    2. Relevant equations
    ψ=Aekx if x<0 and ψ=Be-kx if x>0, ε is infinitesimal change in x

    3. The attempt at a solution
    By boundary conditions, say at the origin, this will give A=B then the book says we can represent the wave function as ψ=Ae-k|x|. The wave function is even so the integral is 0 between -ε to ε.
  2. jcsd
  3. Oct 25, 2015 #2

    Ray Vickson

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    Homework Helper

    That is false. For very small, but positive ##\epsilon## the function ##\psi(x)## is very nearly constant (##=A##) over the interval ##-\epsilon \leq x \leq \epsilon##, so ##\int_{-\epsilon}^{\epsilon} \psi(x) \, dx \approx 2 A \epsilon##. Of course, if ##\epsilon## is infinitesimal, so is the integral, but infinitesimal does not mean zero.
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