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Sin(kx) is an eigenfunction of the KE operator?

  1. Apr 5, 2006 #1
    The function sin(kx) is an eigenfunction of the KE operator?

    my work:

    [tex]KE=\frac{- \hbar^2}{2m}\frac{\partial^2}{\partial x^2} sin(kx)[/tex]

    [tex]\frac{\hbar^2}{2m}\frac{sin(kx)}{k^2}[/tex]

    i'm not sure how to show that a function is an eigenfuntion. what other work do I need to do?
     
  2. jcsd
  3. Apr 5, 2006 #2

    Physics Monkey

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    Hi UrbanXrisis,

    You've calculated your derivative incorrectly. The definition of an eigenfunction [tex] f [/tex] with eigenvalue [tex] \lambda [/tex] of the kinetic energy operator is [tex] -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} f = \lambda f [/tex]. Your job is to see if your function satisfies this equation, and if it does, then you should find the eigenvalue [tex] \lambda [/tex].
     
    Last edited: Apr 5, 2006
  4. Apr 5, 2006 #3
    the eigen value is [tex]\frac{\hbar^2}{2mk^2}[/tex] isnt it? So to show that a function is an eigenfunction, it must satisfy [tex] -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} f = \lambda f [/tex] with [tex] \lambda [/tex] being a repeated term... aka the eigenvalue?

    That means the function sin(kx) is an eigenfunction of the KE operator right?
     
  5. Apr 5, 2006 #4

    Physics Monkey

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    Yes, you understand that it's an eigenfunction. However, you still have the eigenvalue wrong because you messed up the derivative.
     
  6. Apr 5, 2006 #5
    oh whoops.. i was thinking integral [tex]\frac{\hbar^2k^2}{2m}[/tex]

    so as long as the eigenvalue does not change, then a function is an eigenfunction?
     
  7. Apr 5, 2006 #6

    Physics Monkey

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    Yes, you have eigenvalue right now. Also, the eigenvalue doesn't change, it goes with the eigenvector. The key is that the operator acting on the eigenfunction gives you back the same function simply multiplied by a number, the eigenvalue.
     
  8. Apr 5, 2006 #7
    you said that lambda is a repeated term. what do you mean by "repeated term"?
     
  9. Apr 5, 2006 #8

    Physics Monkey

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    Actually, you used the phrase "repeated term", I never mentioned it. :smile: [itex] \lambda [/itex] is the eigenvalue, and it appears precisely as I described above.

    Hope this helps.
     
  10. Apr 6, 2006 #9

    Gokul43201

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    As a postscript to this discussion :

    Urbanxrisis, notice that the eigenvalue must have the dimensions of the operator. So, in this problem, your eigenvalue must have dimensions of energy. That should have alerted you to the error in calculating the derivatives.
     
  11. Apr 7, 2006 #10

    dextercioby

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    Actually, that operator has a purely continuous spectrum. Therefore, terms such as "eigenvalue", "eigenfunction", "eigenvector" are wrongly used in connection to that linear operator.

    Daniel.
     
  12. Apr 7, 2006 #11

    Physics Monkey

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    Thanks for the comment, dextercioby, but one must be a little careful here. The spectrum of that operator depends on its domain. For the whole real line, the operator turns out to be essentially self adjoint with continuous spectrum and hence technically no eigenvectors. On the half line, the operator has a deficiency index of (1,1) and thus has self adjoint extensions parameterized by a phase. Again, the spectrum is continuous, but depending on the extension one choooses, a discrete bound state can arise. Finally, on a finite interval, the deficiency index is (2,2) which means self adjoint extensions exist which are parameterized by [itex] U(2) [/itex]. In this case, each self adjoint extension has a discrete spectrum with eigenvalues and eigenvectors.

    In any case, the use of the eigen- words is commonplace amongst physicists even in situations where mathematicians have defined things differently.
     
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