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I Square integrable wave functions vanishing at infinity

  1. Oct 16, 2016 #1
    Hi!
    For the probability interpretation of wave functions to work, the latter have to be square integrable and therefore, they vanish at infinity. I'm reading Gasiorowicz's Quantum Physics and, as you can see in the attached image of the page, he works his way to find the momentum operator. My question is about what is inside the red box. SG says that "because the wave functions vanish at infinity, the first term (green box) does no contribute, and the integral gives..."
    If the waves functions vanish at infinity (and also their derivatives), why doesn't the second term (blue box) vanish aswell?
    Thank you for your time.
     

    Attached Files:

  2. jcsd
  3. Oct 16, 2016 #2

    blue_leaf77

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    Integrating the terms inside the green box you get
    $$
    \int_{-\infty} ^\infty d\left( \frac{d\psi^*}{dx}x\psi -x\psi^*\frac{d\psi}{dx}-|\psi|^2 \right) = \left( \frac{d\psi^*}{dx}x\psi -x\psi^*\frac{d\psi}{dx}-|\psi|^2 \right) \Big|_{-\infty} ^\infty = 0 - 0
    $$
    because ##\lim_{x\to \pm\infty}\psi(x) = \lim_{x\to \pm\infty}\psi^*(x) = \lim_{x\to \pm\infty}\frac{d\psi(x)}{dx} = \lim_{x\to \pm\infty}\frac{d\psi^*(x)}{dx} = 0##.
     
  4. Oct 16, 2016 #3

    stevendaryl

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    If you have an integral [itex]\int_{A}^{B} (\frac{d}{dx} Q(x)) dx[/itex], since integration is in some sense anti-differentiation, you can immediately evaluate the integral: [itex]\int_{A}^{B} (\frac{d}{dx} Q(x)) dx = Q(B) - Q(A)[/itex]. In the limit as [itex]A \rightarrow \infty[/itex] and [itex]B \rightarrow -\infty[/itex], you get zero, if [itex]Q(x)[/itex] goes to zero at [itex]\pm \infty[/itex].

    So if you can write the integrand (what's inside the integral) as a derivative of something, then its integral over all space will be zero.

    There is a collection of related facts about integrals that is called Stokes' theorem.

    [itex]\int_{A}^{B} \frac{dF}{dx} = F(B) - F(A)[/itex]: The integral of the derivative of a function over a region is equal to the difference of the value of the function on the endpoints.

    [itex]\int_{\mathcal{S}} \nabla \times \vec{F} \cdot \vec{dA} = \int_{\mathcal{C}} \vec{F} \cdot \vec{d\mathcal{l}}[/itex]: The integral of the curl of a function over a surface [itex]\mathcal{S}[/itex] is equal to the integral of the function over the curve [itex]\mathcal{C}[/itex] formed by the boundary of that surface.

    [itex]\int_V (\nabla \cdot \vec{F}) dV = \int_\mathcal{S} \vec{F} \cdot \vec{dA}[/itex]: The integral of the divergence of a function over a volume [itex]V[/itex] is equal to the integral of the function over the surface [itex]\mathcal{S}[/itex] formed by the boundary of [itex]V[/itex].

    In all three cases, if the function [itex]F[/itex] vanishes at infinity, then you get zero for all integrals, if the integral is over all the real line, or all the 2-D plane, or all of 3-D space.
     
  5. Oct 16, 2016 #4
    I've just realized this and "palmed" myself on the face really hard! Thank you for your time!
     
  6. Oct 17, 2016 #5

    bhobba

    Staff: Mentor

    Ahh grasshopper you might like to contact this guy and get a copy of his Phd theresis: R. de la Madrid
    https://arxiv.org/abs/quant-ph/0502053

    It's relaxation has led to many advances in applied math, especially probability theory:
    http://society.math.ntu.edu.tw/~journal/tjm/V7N4/0312_2.pdf

    The idea is they approximate what is physically achievable but are mathematically more tractable. For example in QM a particle with exact momentum is a wave-function to infinity. Physically impossible but an important insight nonetheless.

    Thanks
    Bill
     
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