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Calculating expansion coefficients for particle in 1-D box

  1. Oct 15, 2012 #1
    While reviewing for my midterm I came across an old problem that asked me to find the expansion coefficient for [itex]n=1[/itex] given an expression for the superposition wavefunction. I also know the expressions for the individual eigenstates because it is simply considering a particle in a 1-D box. I am asking regarding a possible misconception I have about how to go about solving this.

    Assume that the length of the one-dimensional box is [itex]a[/itex]; let [itex]\Psi(x)[/itex] be my superposition wavefunction, [itex]\phi_n(x)[/itex] my individual eigenstates, and [itex]c_n[/itex] my expansion coefficients. Then I have
    [tex]\int_0^a \phi_m^\star \Psi(x) \mathrm{d}x = \int_0^a \phi_m^\star \sum_{n=0}^\infty c_n\phi_n \mathrm{d}x[/tex]
    which trivially equals [itex]c_m[/itex] because the eigenstates are orthonormal, etc. However, I took a look at page 2 of
    http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2006/lecture-notes/lecture11.pdf [Broken]
    and the integration limits are from [itex]-\infty[/itex] to [itex]\infty[/itex]. What am I missing here? I don't understand why the integration limits would be like that instead of from [itex]0[/itex] to [itex]a[/itex], because that is the only range for which the wavefunction and eigenstates are physically relevant (since the particle is confined to this range). I would appreciate it if someone could clear up this point of confusion for me.

    The MIT lecture notes do change the limits of integration after the superposition wavefunction is explicitly written out as a summation of the eigenstates, which I don't particularly understand; since the two expressions are equivalent, I don't see why the limits of integration would have to be modified at all.

    (I hope I'm not missing something fundamental here.)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
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