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

Question

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.

Answer 1

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.

Answer 2

I cannot understand Martin's answer, while I think that there is an excellent physical answer to the OP.

Most people forget that a Dirac's delta can be approximated by very many functions, as some parameter (say, $\sigma$) tends to zero. One such class of functions is of course $f(x) = (\sqrt{\pi}/\sigma) exp (-x²/\sigma²)$. Let us take this as an approximation to the probability distribution (not the wave function) of a quantum state representing a particle confined at the origin of coordinates. This function is smooth, integrable, normalized to 1.

Its associated wave-function, $\phi(x) = (\sqrt{\pi}/\sigma)^{1/2} exp (-x²/(2\sigma²))$ is also smooth and integrable over the same interval, and the value of this integral $\rightarrow 0$ as $\sigma \rightarrow 0$.

But we do not really care about the integrability of the wave-function per se, only that it yields meaningful results when we compute transition amplitudes. And for an arbitrary wave function $\psi(x)$ the transition amplitude is always proportional to

$\int \psi(x) \phi^*(x) dx$

which can be easily computed from the above in the limit $\sigma \rightarrow 0$: it always yields $0$, unless $\psi(x) = \phi(x)$, in which case it yields $1$. Does this make sense? Yes it does: whenever $\psi$ is not the wave function of a particle confined at the origin (i.e., when it represents a particle confined elsewhere, or when it represents a particle not confined at all) the two wave functions are orthogonal because they represent completely different physical states: the integral above (when squared) is the probability that an arbitrary quantum state be found at precisely the origin, which of course is zero for both smooth probability distributions and for particles confined at point which is not the origin.

Hence it makes perfect sense that the integral above vanishes, unless $\psi=\phi$ when it must yield certainty ($=1$).

To sum up, the wave function of a state representing a particle confined at the origin exists, is smooth and integrable so long as $\sigma > 0$, but we do not worry about that (at least from a physical point of view) because the wave function is not in itself an observable, because all we care about is that transition amplitudes exist, and these exist and make physical sense even in the limit $\sigma\rightarrow 0$.

The method of replacing a Dirac's delta with its approximants often leads to quite sensible answers.

Answer 3

Most scientists agree there are still some interpretational issues in QM, so it is hard to make unequivocal statements. IMO, point-like single quantum pure states (i.e., Hilbert space "rays") are not physically measurable or even physically realizable. They are idealizations of zero entropy and thus not realizable by the Nernst statement of the third law of thermodynamics. No physical object property can manifest existence in a spacetime interval of zero extent, e.g., no object can manifest a real-numbered position, not even its CG. This means every realizable quantum state is a mixed state of multiple concurrently existing incoherent pure states. (They are incoherent superpositions, otherwise they would be expressible as a coherent sum of pure states, i.e., a single component, zero entropy, pure state.) This, however, casts doubt on the probability interpretation of QM. The appropriate vertical metric that should be applied to the magnitude squared of a realizable quantum mixed state distribution must be a physical, or ontic, metric, not probability. So OP, real-numbered (point-like) eigenvalues are not physically realizable, all manifest physical property values are distributions, not real numbers.