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Eigenfunction of the momentum "operator"

  1. Mar 6, 2006 #1
    which of the following functions is an eigenfunction of the momentum "operator"

    [tex]-i \hbar \frac{\partial}{\partial x}: [/tex]

    [tex]f_1 =cos(kx- \omega t)[/tex]
    [tex]f_2 =e^{a^2x}[/tex]
    [tex]f_3 =e^{-(\omega t+kx)}[/tex]

    for this question, i'm not sure what they are looking for...

    for f1
    [tex]i \hbar k sin(k x -\omega t) [/tex]
    for f2
    [tex]-i \hbar a^2 e^{a^2x} [/tex]
    for f3
    [tex]- \hbar ke^{-(\omega t+kx)}[/tex]

    the eigenvalues for f1 is -k^2
    the eigenvalues for f2 is a^2
    the eigenvalues for f3 is -ik

    how do I find the correct eigenfunction?
     
  2. jcsd
  3. Mar 6, 2006 #2
    careful with your assertions, remember an eigen function says,
    [tex] H \mid v \rangle = \lambda \mid v \rangle [/tex]
    in other words when you act on a vector, you get the function back times the eigenvalue right...check your calculations...hope this helps, sincerely,x
     
  4. Mar 6, 2006 #3

    Physics Monkey

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    Hi UrbanXrisis,

    Typically you say that f is an eigenfunction of an operator like [tex] \mathcal{O} = - i \hbar \frac{\partial}{\partial x} [/tex] if [tex] \mathcal{O} f = \lambda f[/tex] where [tex] \lambda [/tex] is just some number called the eigenvalue. In other words, if when taking the derivative of f you get the same function back up to an overall constant, then f is an eigenfunction of the momentum operator. There are issues associated with the ability to normalize such functions, but I don't think that's the point of this problem.
     
    Last edited: Mar 6, 2006
  5. Mar 6, 2006 #4

    AKG

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    Do you know what it means for a function f to be an eigenfunction of an operator T?
     
  6. Mar 6, 2006 #5
    ok, so taking [tex]f_2 =e^{a^2x}[/tex] as an example, the eigenfunction is [tex]e^{a^2x}[/tex] and the eigenvalue is [tex]a^2[/tex] because if I took the derivative of the function, it would keep get the eigenfunction multiplied by a^2 over and over again. what I dont understand is how to satisfy: [tex]-i \hbar \frac{\partial}{\partial x}[/tex]
     
  7. Mar 6, 2006 #6

    AKG

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    No, first look up the definition of an eigenfunction. There's no "satisfying" -ih(d/dx). -ih(d/dx) is an operator, you can apply it to any function. Consider the function f(x) = x². There's no "satisfying" f. You can plug in any number for x, and you get another number x². It's the same idea with operators. You plug in your function f into your operator, and you get another function.

    Plug the function f(x) = ea2x into -ih(d/dx), and you get:

    -ih(d/dx)(exp(a²x)) = -iha²exp(a²x) which is just another function, agree? It's g(x) = -iha²exp(a²x). Now the question is, is g(x) a scalar multiple of f(x)? Yes it is. What is that multiple (it's not a²).

    Try a new example, f(x) = cos(x). Plug it in:

    -ih(d/dx)(cos(x)) = ihsin(x)

    Again, this is just another function, g(x) = ihsin(x). Is this a scalar multiple of f(x)? That is, is there some complex constant z such that

    g(x) = zf(x) ?
    ihsin(x) = zcos(x) ?
    sin(x) = (z/ih)cos(x) ?

    No! Do you know why?
     
  8. Mar 7, 2006 #7
    ok, so the eigenvalue is really -iha² for ea2x?

    as for cos(x) the function changes to sin so it cant be an eigenfunction for -ih(d/dx)?
     
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