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Does anybody know the status of the problem to define the wave function (non-relativistic Quantum Mechanics) of a particle localized at a definite point?

Landau-Lifshitz says in chapter 1 that this function is $\Psi(x)_{x_o} = \delta(x-x_0)$ and gives an explanation that it produces the correct probability density when it is used to span some other arbitrary wave function $\Psi(x)$. The problem is of course that the wave function given above squares to a non integrable function. As far as I know this problem is unsolved. My question is if anybody knows the status quo of this problem. I am sorry if this question may be duplicated, I could not find it amongst the answered questions.

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you should look into rigged Hilbert spaces, eg en.wikipedia.org/wiki/Rigged_Hilbert_space , physics.stackexchange.com/q/43515 , arxiv.org/abs/quant-ph/0502053 –  Christoph Sep 28 '13 at 10:27
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It should be stressed that the rigged Hilbert space formalism doesn't explain the meaning of the integral of the square of the Dirac distribution. –  Qmechanic Sep 28 '13 at 13:06

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Mathematically spoken, since you want your wave functions to be square integrable, your wave functions must be in $L^2$ or some subspace thereof. However, you won't find a function in this space that has a support on a countable set of points, since the Lebesgue integral cannot see countable sets (measure 0), hence there cannot be a function (i.e. no wave function) with support in a single point (incidentally, the delta function is not a "function" in a way for that reason).

This tells us that a wavefunction for a particle that is fully localized cannot be defined in the usual setting of square Lebesgue-integrable functions, which is not too tragic, because we don't really think it makes physical sense anyway.

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@Martin-Why is that so? After position measurement on a system we do create these $\delta$-function states... do u mean they are unphysical? –  Roopam Feb 24 at 12:17
    
@Roopam No, we don't create these delta-functions, because we can't accurately measure position. We cannot, in reality, measure position perfectly, because whatever instrument we can build, it will always measure position in some discrete set - and then the resulting projected wave function will be some wave-packet that is nicely square-integrable. So in this sense, delta functions are unphysical. The delta function is just a physical-mathematical approximation that makes our lives easier. Handling proper measurements of positions would otherwise require proper measure theory. –  Martin Feb 24 at 13:21
    
@Martin- But at least uncertainty principle does not forbid the precise position measurement when we do not care about its precise momentum simultaneously. So don't you think that in principle there is no problem with precise position measurement? –  Roopam Feb 24 at 13:46
    
As I see it, there is no mathematical reason to forbid us from precise position measurements other than that the resulting state does not lie in our state space - but that's just a problem of definition. We could work with rigged Hilbert spaces after all. But there is a precise PHYSICAL reason: We measure by measuring a number or something like this, however we may never measure an infinite amount of digits - there will always be some digitalization. –  Martin Feb 24 at 14:00
    
So, all in all, there is two things: a) our formalism says by definition that all physical objects are represented by wave-functions that are square integrable objects to have a probability interpretation. b) we encounter things that aren't square integrable - hence they have to be unphysical by definition so either we can find a physical reason, why they are unphysical, or we have to extend our formalism. Since I gave a purely physical reasoning (measuring at arbitrary but finite precision is possible, infinite precision is not) that's okay. –  Martin Feb 24 at 14:01

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