Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditions in Wave Function, and Integration

  1. Oct 11, 2015 #1
    Untitled.png
    I want to know how this integral will equal zero?

    I know that Ψ will fall to zero as x goes to infinity
    and i know that Ψ must fall to zero very quickly , Ψ must fall to zero faster than 1/√|x|

    all of this will help evaluating this integral

    i tried to solve it as follows

    Untitled2.png

    The first term will vanish due to the previous reasons
    but how the second term will vanish?
     
  2. jcsd
  3. Oct 11, 2015 #2

    ShayanJ

    User Avatar
    Gold Member

    If ## \psi ## is vanishingly small at infinity, how can it have finite differences at infinity?
     
  4. Oct 11, 2015 #3

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##\psi(x) = \frac{sin(x^2)}{x}##

    Or at least a function based on this for large ##|x|##
     
  5. Oct 11, 2015 #4

    bhobba

    Staff: Mentor

    The full elucidation of the question you are asking needs Rigged Hilbert Spaces which requires considerable background in functional analysis. It most definitely is not recommended for the beginner.

    As a halfway stop, and an extremely important part of the tool kit of anyone into applied math, not just physics, is what's called Distribution theory. I think everyone should have a copy of the following wonderful book on it:
    https://www.amazon.com/Theory-Distr...52137149X/ref=mt_hardcover?_encoding=UTF8&me=

    It will repay its study many times over. Its worth it for its treatment of the Fourier Transform alone. Conventional treatments become bogged down with difficult issues of convergence etc - but the Distribution theory approach bypasses all of that with great elegance.

    To get back to your original query at the start in QM simply view all the functions you deal with as zero outside some finite region - such are called functions of compact support and you will see why that view is reasonable when you study Distribution theory. You will occasionally encounter functions that most definitely are not like that but in the sense of Distribution theory can be viewed as approximations to functions of compact support. Simply replace it by a function of compact support and things like the second term that worries you vanishes. Its not valid math but to start with you wont run into any issues.

    Later you can learn the technicalities of functional analysis and Rigged Hilbert Spaces and see how it resolves these issues with full rigour, but to start with bite your tongue and simply think of the functions you deal with being of compact support.

    As an aside when I first learned QM I became absorbed with issues like that and did a long sojourn into the exotica that resolves such things. I know from experience worrying about it now is counter productive - you can get to the bottom of it later.

    Thanks
    Bill
     
    Last edited: Oct 11, 2015
  6. Oct 12, 2015 #5

    samalkhaiat

    User Avatar
    Science Advisor

    If [itex]\frac{\partial^{n}\psi}{\partial x^{n}}[/itex] does not approach zero as [itex]x \to \pm \infty[/itex], the operator [itex](\frac{\hbar}{i} \frac{\partial}{\partial x})^{n+1}[/itex] may not be Hermitean. For this reason, In QM, it is assumed that the wavefunction and its partial derivatives to finite order vanish as [itex]x \to \pm \infty[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook