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Proof of the preservation of normalizability of the wavefunction

  1. Jan 16, 2015 #1
    p. 12 Introduction to Quantum Mechanics by Griffiths
    Equation 1.25: the differential operatot was factored. This to me seems like a mathematical trick or due to amazing foresight, but is there any underlying/guiding theory for this factorisation?
    Equation 1.27: the wavefunction was assumed to be zero at infinity which to me seems a bit weird, since by this assumption we enforced normalizability and got the answer we fabricated, so is my criticism valid? Is there a more rigorous proof?

    The image of the book is enclosed below.

    Thanks :)

    <a href='http://i.imgur.com/WSUwLhM' title=''><img src='http://i.imgur.com/WSUwLhM.jpg' alt='' title='Hosted by imgur.com' /></a>

    (I dont know why the image thumbnail isnt working)
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    [PLAIN]http://<a [Broken][/PLAIN] [Broken] href='http://i.imgur.com/WSUwLhM' [Broken][/PLAIN] [Broken] title=''><img src='http://i.imgur.com/WSUwLhM.jpg' alt='' title='Hosted by imgur.com' /></a>[/PLAIN] [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 16, 2015 #2

    stevendaryl

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    I'm not exactly sure what your question is. We assume that initially,

    [itex]\int |\psi|^2 dx = 1[/itex]

    So we assume that it starts off normalized, and that [itex]\psi \rightarrow 0[/itex] as [itex]x \rightarrow \pm \infty[/itex]

    Then we prove, based on the assumption that it is true at time [itex]t=0[/itex], that it will be true for all time.
     
  4. Jan 16, 2015 #3

    vanhees71

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    The question is, how to prove that the wave function is normalized to 1 for all times, if it was normalized to 1 at the initial time ##t=0##.

    The answer is: That follows from the Schrödinger equation and the self-adjointness of the Hamiltonian. Working in position representation ("wave mechanics") it reads
    $$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$
    Here and in the following I use "natural units", where ##\hbar=1##.

    The self-adjointness of the Hamiltonian means that for all square-integrable wave functions
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi_1(\vec{x})^* \hat{H} \psi_2(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} [\hat{H} \psi_1(\vec{x})]^* \psi_2(\vec{x}).$$
    Now you can prove that the norm of the wave function is conserved under the time evolution following from the Schrödinger equation:
    $$\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(t,\vec{x}) \psi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \left [\partial_t \psi^*(t,\vec{x}) \psi(t,\vec{x}) + \psi^*(t,\vec{x}) \partial_t \psi(t,\vec{x}) \right ].$$
    Now from the Schrödinger equation and its complex conjugate this gives
    $$\frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(t,\vec{x}) \psi(t,\vec{x}) = \mathrm{i} \int_{\mathbb{R}^3} \left [(\hat{H} \psi)^* \psi-\psi^* \hat{H} \psi \right ]=0.$$
    Thus the norm is preserved under time evolution and thus, if the wave function is normalized to 1 at any time it stays so. This must be so from the physical point of view, because the probality to find the particle somewhere in the whole universe is always 1 (because in non-relativistic physics particles usually are not destroyed in interactions in contrast to relativistic physics, where the creation and annihilation of particle-antiparticle pairs and similar processes are the usual business :-)).
     
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