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Proving an equation is an eigenstate of the momentum operator.

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data
    A free particle (de Broglie wave) may be represented by the wave-function

    [itex]\psi[/itex](x)=Aeikx

    Show that this is an eigenstate of the momentum operator [itex]\hat{p}[/itex]=-[itex]\hbar[/itex][itex]\frac{\delta}{\deltax}[/itex]


    2. Relevant equations
    [itex]\hat{p}[/itex]un(x)=anun(x)
    an is the eigenvalue
    un(x) is the corresponding eigenfunction

    3. The attempt at a solution
    Ok so first I have
    [itex]\hat{p}[/itex]=-[itex]\hbar[/itex][itex]\frac{\delta}{\deltax}[/itex]​
    which has the boundary condition that un is periodic in the range L. Then the eigenvalue equation is
    -i[itex]\hbar\frac{\delta}{\deltax}[/itex][itex]\psi[/itex](x)=an[itex]\psi[/itex](x)​
    where [itex]\psi[/itex](x) is the eigenfunction un(x)
    The solution of the equation is then
    [itex]\psi[/itex](x)=[itex]\hbar[/itex]eianx
    =[itex]\psi[/itex](x)=Aeikx
    where k is the eigenvalue an and A is [itex]\hbar[/itex] (Plancks Constant)

    Ok so what I would like to know is:
    Is the above correct?
    If so why is it correct?
    If its wrong why is it wrong?
    What exactly is an eigenfunction and eigenvalue?
    What is this range L?

    As you can probably guess from the above questions I know absolutely nothing on this topic. I have a sort of understanding on what an eigenfunction and eigenvalue are but after going through revision notes on this stuff numerous times I still can't get my head round the mathematics of it, how everything changes to the way it does. The only reason why I have the above working out is because the revision notes on the topic is very similar to the question but is quite vague on how one thing transitions to the next.
     
    Last edited: Sep 30, 2011
  2. jcsd
  3. Sep 30, 2011 #2
    When an operator acts on a function, the function may change. But when an operator acts on one of the eigenfunctions of this same operator, the function does not change
     
  4. Sep 30, 2011 #3
    So are you saying that -iℏ(δ/deltax) shouldnt change to ℏeianx since its acting on the eigenfunction of the same operator? Sorry if this is wrong just trying to understand what you said.
     
  5. Sep 30, 2011 #4
    When an operator acts on one of its own eigenfuctions, it only multiplies the given fuction simply by a number but it does not result in a different function.
     
  6. Sep 30, 2011 #5

    vela

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    You have[tex]\hat{p}=-i\hbar\frac{\partial}{\partial x}[/tex]and [itex]\psi(x) = Ae^{ikx}[/itex], so you can show that[tex]\hat{p}\psi(x) = -i\hbar\frac{\partial}{\partial x}(Ae^{ikx}) = \hbar k \,\psi(x)[/tex]I'll leave it to you to fill in the steps. Applying [itex]\hat{p}[/itex] to this wave function results in a multiple of the wave function, so this wave function is called an eigenfunction or eigenstate of [itex]\hat{p}[/itex], and the multiple, [itex]\hbar k[/itex], is called the eigenvalue.

    If you want to go with the approach you originally took, you have[tex]-i\hbar\frac{d\psi}{dx} = a_n \psi[/tex]
    (Since [itex]\psi(x)[/itex] only depends on x, we can replace the partial derivative by a regular derivative.) That equation is separable, so you get
    [tex]-i\hbar\frac{d\psi}{\psi} = a_n\,dx[/tex]
    Now you'd want to integrate both sides and solve for [itex]\psi[/itex]. The constant of integration eventually turns into A. By comparing your solution to the form of [itex]\psi[/itex] given, you can determine what the eigenvalue [itex]a_n[/itex] is equal to in terms of k.
     
  7. Oct 1, 2011 #6
    So going from my original approach if i continue from:

    -i[itex]\hbar[/itex][itex]\frac{\delta\psi}{\psi}[/itex]=andx

    then integrate through I get:

    -i[itex]\hbar[/itex]ln[itex]\psi[/itex]+constant=anx+constant

    divide through by -i[itex]\hbar[/itex] and move the constant from the LHS to the RHS giving:

    ln[itex]\psi[/itex]=[itex]\frac{anx}{-i\hbar}[/itex]+constant

    convert into exponential form:

    [itex]\psi[/itex]=Ae[itex]\frac{anx}{-i\hbar}[/itex]

    where k=[itex]\frac{an}{\hbar}[/itex]

    Is this correct?

    Also I'm not sure if this happens on any of your computers/laptops but on mine some of the equations are not displaying properly. If they aren't showing up right what has made it go like that?
     
  8. Oct 1, 2011 #7

    vela

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    Yes, that looks correct.

    Use LaTeX to write the entire expression rather than individual symbols. Don't use sub and sup tags inside itex tags. Use _ for subscripts and ^ for superscripts, e.g. a_n and e^{ikx}.
     
  9. Oct 1, 2011 #8
    Ok thanks for that, the equation issue was bugging me for ages haha. Also thanks for the help vela and grzz I think I understand how operators work now.
     
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