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Eigenfunction of an Operator

  1. Sep 3, 2007 #1
    How can i prove that [tex]u(x)=exp(-x^{2}/2)[/tex] is the eigenfunction of [tex]\hat{A} = \frac{d^{2}}{dx^{2}}-x^2 [/tex].(if i don't know the eigenfunction how can i find it from expression of A operator)
     
  2. jcsd
  3. Sep 3, 2007 #2
    By substitution- how else?
     
  4. Sep 3, 2007 #3
    Like This ?

    Is this true? [tex]\frac{d^{2}u(x)}{dx^2}-x^{2}u(x) = a u(x)[/tex] u(x) is the eigenfunction and a is eigenvalue for this function
     
    Last edited: Sep 3, 2007
  5. Sep 3, 2007 #4
    By definition, yes, it's true. You can check for something being an eigenfunction by applying the operator to the function, and seeing if it does indeed just scale it. You find eigenfunctions by solving the (differential) equation Au = au.
     
  6. Sep 3, 2007 #5

    HallsofIvy

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    Notice that you are not required to find an eigenfunction- you are already given it. You only need to show that this function is an eigenfunction by applying the operator to it and showing that you get a multiple of the function.
     
  7. Sep 3, 2007 #6
    Yes ,But?

    Main question is, if i don't know the eigenfunction in the begining ,how can i find the eigenfunction of operator A ? i can't solve the de. [tex]\frac{d^{2}u(x)}{dx^2}-x^{2}u(x) = a u(x)[/tex]
    help me ?
     
    Last edited: Sep 3, 2007
  8. Sep 3, 2007 #7
    Then you need practise with solving differential equations. However, the equations are in general difficult. In this case, you'll need to have learnt how to do series solutions of differential equations. The procedure for this equation is something like (I haven't done it by hand for years -- it's a standard enough that I can remember the form of the solution):

    1. Assume that u=P(x) exp(-x^2), because the for the solution to be physical, it must equal zero at infinity (in general be L^2 integrable).
    2. Get a differential equation for P, and solve by series method. This will show that the only solutions which are physical (by the same requirements again) are finite order polynomials -- which are called the Hermite polynomials. They are a well-known polynomial series.
    3. Normalise, if necessary.

    The steps are not difficult if you're already acquainted, but tedious.
     
  9. Sep 3, 2007 #8
    Thanks sir.

    Thanks sir ,I agree with you.This DE. can be solved by using power series solution method.
     
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