1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof that d/dx is anti-hermitian by integration by parts

  1. Jul 7, 2011 #1
    The attempt at a solution

    [tex]
    \begin{equation*}
    \begin{split}
    \ -\ i\int\psi^* \frac{\partial{\psi}}{\partial x}= \\ -i(\psi^*\psi\ - \int\psi \frac{\partial\psi^*}{\partial x})\space\ (?)
    \end{split}
    \end{equation*}
    [/tex]

    I thought [itex] \psi^*\psi\ = \ constant\neq\ 0.[/itex] However, it vanishes in the correct proof.
     
  2. jcsd
  3. Jul 7, 2011 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi HotMintea! :smile:

    Not sure what this is :redface:, but doesn't [constant]ab = 0 ?
     
  4. Jul 7, 2011 #3

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    As tiny-tim pointed out, you need limits of integration.

    However, psi*psi is not a constant. It is only a constant when integrated over all space, but it is not integrated in this case.

    The real reason the first term vanishes is that for limits at plus or minus infinity, the wave-function must decay to 0 to be normalizable. For periodic boundary conditions, the two limits are the same point so they must subtract to 0.
     
  5. Jul 7, 2011 #4
    Hi tiny-tim and Matterwave! :smile:
    For a decaying wave function, is the following correct?:
    [tex]
    \begin{equation*}
    \begin{split}
    \ -\ i\int_{-\infty}^{\infty} \psi^*\frac{\partial \psi}{\partial x}\ =\\-\ i\ (\psi^*\psi\ |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \psi\frac{\partial \psi^*}{\partial x})\ = \\ i\int_{-\infty}^{\infty}\psi\frac{\partial\psi^*}{\partial x}
    \end{split}
    \end{equation*}
    [/tex]
    Thus, [itex] \psi\ = \ 0 \ at \ x\ = -\infty\ \ and \ \infty [/itex]?
    For periodic boundary conditions, [itex]\ -\ i\int_{a}^{a} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{a}\psi\frac{\partial\psi^*}{\partial x}\ = \ 0 [/itex] ?
    For [itex] \psi\ = \ Ae^{i(k x-\omega t)} [/itex] where A is an arbitrary constant, [itex] \psi^* \psi\ = \ A^2 \ = \ constant. [/itex] Is [itex] \psi^* \psi\ = \ constant [/itex] not true in general? (Example taken from: http://en.wikipedia.org/wiki/Normalizable_wave_function#Example_of_normalization)
     
    Last edited: Jul 8, 2011
  6. Jul 8, 2011 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi HotMintea! :smile:
    Yes.
    No, [itex] \psi[/itex] is the same (at each boundary), so the [] is zero, not the ∫.
    That's a featureless plane wave, so of course [itex] \psi^* \psi[/itex] is constant.

    Not all waves are so uninteresting. :wink:
     
  7. Jul 8, 2011 #6

    :wink:!
    For the periodic boundary conditions, is the following correct?:
    [tex]
    \begin{equation*}
    \begin{split}
    \ -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = -\ i\ (\psi^*\psi\ |_{a}^{b} - \int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x})\
    \end{split}
    \end{equation*}
    [/tex]
    However, [itex] \psi(a)\ = \psi(b).[/itex] Moreover, conjugates [itex] \psi^*(a)\ = \psi^*(b). [/itex] Thus, [itex] \psi^*\psi\ |_{a}^{b}\ = \ 0.[/itex]

    Therefore,
    [tex]\begin{equation*}\begin{split} -\ i\int_{a}^{b} \psi^*\frac{\partial \psi}{\partial x}\ = \ i\int_{a}^{b}\psi\frac{\partial\psi^*}{\partial x} \end{split}\end{equation*}[/tex]
     
    Last edited: Jul 8, 2011
  8. Jul 8, 2011 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

  9. Jul 8, 2011 #8
    Thanks for your help! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof that d/dx is anti-hermitian by integration by parts
Loading...