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In normalising the complete wavefunction of a free partcile; $V(x)=0$

we arrive at

$\int_{-\infty}^{\infty}\Psi_{k} ^\dagger\Psi_{k}dx=|A|^{2}\int_{-\infty}^{\infty}dx=|A|^{2}\left(\infty\right)$

Which implies that this wave function is not normalisable.

Mathematically, it is not normalisable but that's as far as my understanding goes. I would like to see a further and more related explanation for which a non-normalisable wave function implies that a free particle cannot exists in a stationary state and also that such a free particle does not have a definite energy.

Physkid
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  • These wave functions are perfectly normalizable, you are simply using the wrong math to do it. This has nothing to do with the question whether such a particle is stationary, or not, or whether it has a definite energy. A classical body doesn't have a definite energy, either. One can always add an arbitrary constant to a scalar potential without changing the physics and the kinetic energy is observer dependent. – CuriousOne Jan 02 '16 at 05:30
  • @CuriousOne What do you mean by 'wrong' math? Am I to use Fourier transform? – Physkid Jan 02 '16 at 05:50
  • The proper mathematical tools that can do what you want can be found in "functional analysis" and Fourier transforms do fall under that discipline as a special application, but one can do much broader things with it, which is why physicists usually don't bother. We wing all of this by introducing a crude form of distribution with Dirac's delta function. – CuriousOne Jan 02 '16 at 05:54
  • There are many existing questions that deal with this e.g. Normalizing the solution to free particle Schrödinger equation. Also see this search, and similar searches using variations on normalise, normalize, normalisation etc. – John Rennie Jan 02 '16 at 06:22

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