# Open problem? Square of the wave function $\Psi(x)_{x_o} = \delta(x-x_0)$ of a particle localized at a point $x_0$?

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
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

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).
@Martin-Why is that so? After position measurement on a system we do create these $\delta$-function states... do u mean they are unphysical? – SRS Feb 24 '14 at 12:17