In quantum mechanics, if the wavefunction is normalizable, then it would represent a particle. Why does it not represent a particle when it is not normalizable?


By definition, the probability to detect a particle with normalized wavefunction $\psi(x)$ in an interval $[x_1,x_2]$ is $$ P(x_1,x_2) = \int_{x_1}^{x_2}\lvert\psi(x)\rvert^2\mathrm{d}x.$$ If the wavefunction is not normalized, but normalizable, i.e. the integral $C := \int_{-\infty}^\infty\lvert\psi(x)\rvert^2\mathrm{d}x$ is finite, then we can still define this probability as $$ P(x_1,x_2) = \frac{1}{C}\int_{x_1}^{x_2}\lvert\psi(x)\rvert^2\mathrm{d}x.$$ This is essentially just one of the basic postulates of quantum mechanics - states are not vectors in Hilbert space, but rays (or elements of the projective Hilbert space), and it does not matter whether you choose a normalized or an unnormalized representant of a ray to compute physical quantities.

However, an unnormalizable wavefunction is not a member of any ray - it does not lie in the Hilbert space, usually $L^2(\mathbb{R})$, on which quantum mechanics takes place, because the elements of $L^2(\mathbb{R})$ by definition have finite integrals, i.e. are normalizable. In particular, there is no way to give a prescription how to compute $P(x_1,x_2)$ from it. Therefore, an unnormalizable wavefunction is not a state in the sense of quantum mechanics, it does not represent a physically meaningful or accessible state (although it may be an idealization of one, like the states $\lvert x\rangle$).

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    $\begingroup$ Note: normalisation of states has nothing to do with vector vs ray. Both may be normalised just the same. Rays play no role here, as far as I can see. Thoughts? $\endgroup$ – AccidentalFourierTransform Jul 28 at 20:33
  • $\begingroup$ @AccidentalFourierTransform (Ah, my favourite, trying to figure out what I was thinking in an answer from years ago) I think what I wanted to stress here is that the "real" state is the ray, not any particular vector, and so the freedom to normalize (or not) a normalizable wavefunction corresponds to our freedom of picking an arbitrary representant for the ray. For a non-normalizable wavefunction, it's not an element of the Hilbert space to begin with, and hence also not a representant of any ray. I don't disagree that one could have written an answer without mentioning rays at all, though. $\endgroup$ – ACuriousMind Jul 28 at 21:15

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