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Reading the quantum mechanics textbook we are told the wave function for a definite position at $a$ is $\psi(x)=\delta(x-a)$. Yet, also we are told that the probability must be $\int|\psi(x)|^2 dx$=1. But $\int_{-\infty}^{\infty} |\delta(x-a)|^2 dx = \delta(0) = \infty$. (In fact squaring a delta function is not allowed as it is a "distribution").

Hence it seems like we really want $\psi(x)=\delta(x-a)^{1/2}$ or something similar? Perhaps (formerly) $\psi(x) = \frac{\delta(x-a)}{\sqrt{\delta(0)}}$. Or restricting $x$ to a periodic dimension?

In fact how can we even use the Dirac Delta function at all as a wave function when it gives a ridiculous probability! (And QM is after all based on probabilities.)

Is there a mathematically rigorous way out of this conundrum? I express the flaw as follows:

  1. We assume $\psi$ is a delta function when position is known 100%
  2. To get probabilities we take absolute square of wave function
  3. Distributions such as delta function can't be squared.
  4. Number 3 contradicts with number 2.

To my eyes, it seems like quantum mechanics would require a mathematics of square-roots of distributions. The same problem occurs in QFT if we want the wave function of a definite valued field $\Psi[\phi]$.

Qmechanic
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    I'm a bit rusty in functional analysis, so I'll let someone else fill in the details, but the problem is that the position operator has no eigenvectors. We only pretend it has because it makes calculations easier, but the appropriate manipulation would be to deal always with wavepackets. Also, I think no text will say the position operator normalizes to $1$, but rather to a Dirac delta. – Níckolas Alves Jun 10 '22 at 22:37
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    Related/Possible duplicates: https://physics.stackexchange.com/q/43515/50583, https://physics.stackexchange.com/q/351658/50583 and their linked questions [simple answer: There are no actual wavefunctions of "definite position"] – ACuriousMind Jun 10 '22 at 23:23

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