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QM eigenfns and momentum op

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    which of the following functions are eigenfunctions of the momentum op? what are the eigenvalues in these cases?

    [tex]\psi[/tex](x)=Asin(kx)
    [tex]\psi[/tex](x)=Asin(kx)-Acos(kx)
    [tex]\psi[/tex](x)=Acos(kx)+iAsin(kx)
    [tex]\psi[/tex](x)=Aexp((ik(x-a))


    2. Relevant equations
    Pop=hbar/i(d/dx)


    3. The attempt at a solution
    ok now my understanding of an eigenfn is that if u multiply the operator and function, ad the result is the function*constant, then the that function is an eigenfn, im not quite sure though what the eigen value is... i think its the value of the constant. so i did the math(its probably wrong)

    1.(hbar/i)(Akcos(kx))[tex]\neq[/tex]const*fn
    2.(hbar/i)(Akcos(kx)+Aksin(kx))[tex]\neq[/tex]const*fn
    3.(hbar/i)(Aksin(kx)+iAkcos(kx))=(hbar/i)(k)*fn, thus eigen value=(hbar*k/i)???
    4.(hbar/i)(ikAexp(ik(x-a)))=(hbar*k)*fn, thus eigen value=hbar*k

    plz can some help me if im wrong... im just a bit unsure, i it looks ok, but can anyone confirm, TY
     
  2. jcsd
  3. Apr 9, 2009 #2
    The eigenvalue problem of an operator is:

    [tex]A\psi=\lambda \psi[/tex]

    A is the operator, [tex]\lambda[/tex] is the eigenvalue, and [tex]\psi[/tex] is the eigenfunction.

    your logic was correct, and you calculated the first two correctly and the last one correctly, but you didnt differentiate the third one correctly ;)

    You could even guess the eigenvalues of the momentum operator, before calculating the above. What does the eigenvalue of these operators mean in QM ;))
     
  4. Apr 9, 2009 #3
    ok thankyou very much, ok then to fix up the 3rd one is it...
    (hbar/i)(-Aksin(kx)+iAkcos(kx))=(-hbar/i)(k)*fn thus eigen value is (-hbar*k/i)

    does that seem better.. always forget bout negative... lol, is an eigen value allowed to be complex??? a complex number is still constant right?
     
  5. Apr 9, 2009 #4
    Thats still not right :(

    According to what you wrote:

    [tex]-A\sin(kx)+iA\cos(kx)[/tex]

    is the original function, but that is :[tex]A\cos(kx)+iA\sin(kx)[/tex]

    So you should write:

    [tex]-A\sin(kx)+iA\cos(kx)=i\cdot i A\sin(kx)+iA\cos(kx)=i(A\cos(kx)+iA\sin(kx))[/tex]

    Hence you have to multiply the number you got by i to get the eigenvalue.

    And the momentum operator, and all of the operators representing phyisical quantities in QM are hermitian or self adjoint, hence the always have real eigenvalues, so if you get a complex eigenvalue, then you screwed something :))
     
  6. Apr 9, 2009 #5
    ahh ok yes i thought so, to confirm again the correct answer should be
    (hbar/i)(k)(i(Acos(kx)+iAsin(kx)))=hbar*k*fn, so hbar*k is eigenfn... that makes sense because its the same fn as the 4th one... i think.

    can i also ask, how do the type ur calculations using latex... i have no idea how, and it would help a lot = )
     
  7. Apr 9, 2009 #6

    Pengwuino

    User Avatar
    Gold Member

    Clicking on the actual formulas will pop-up the LaTeX coding associated with them.
     
  8. Apr 9, 2009 #7
    ya i see, thankyou very much, you have been a great help = ) much appreciated
     
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