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Verify whether eigenfunction or not?

  1. Dec 25, 2005 #1
    The wavefunction of a particle moving inside a one dimensional box of length L is non-zero only for 0<x<L.
    The normalised wavefunction is given by:
    [tex]\psi (x) = \sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L}[/tex]
    Is this wavefunction an eigenfunction of the x-component of the momentum operator [itex]\vec p = -i\hbar \vec \nabla[/itex]

    My work:
    I computed the partial derivative of [itex]\psi[/tex] with respect to 'x'. I got:
    [tex]\frac{\partial \psi}{\partial x} = \sqrt{\frac{2}{L}}\left(\frac{n\pi}{L}\right)\cos \frac{n\pi x}{L}[/tex]
    I don't think it is an eigenfunction of the operator but I don't know how to justify my answer. Help needed...
  2. jcsd
  3. Dec 25, 2005 #2


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    Well, what is the definition of an eigenfunction? Does this fit the definition? Since your operator involves an i you might want to put the function in complex form:
    [tex]sin(\frac{n\pic x}{L})= \frac{e^{\frac{n\pi x}{L}}- e^{-\frac{n\pi x}{L}}}{2}[/tex].
    Last edited: Dec 25, 2005
  4. Dec 25, 2005 #3
    The function should return from the operator exactly as is, except for a multiplicative scaling factor. It does not in this case.
  5. Dec 25, 2005 #4

    Tom Mattson

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    Well there you go, it's not an eigenfunction then. And you should not expect it to be an eigenfunction, because that wavefunction represents a particle which could either be traveling to the right or to the left.

    If you rewrite the sine function in terms of complex exponentials as HallsofIvy suggested, you will see explicitly that your wavefunction is a superposition of two momentum eigenstates.
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