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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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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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  • $\begingroup$ @Martin-Why is that so? After position measurement on a system we do create these $\delta$-function states... do u mean they are unphysical? $\endgroup$
    – SRS
    Commented Feb 24, 2014 at 12:17
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    $\begingroup$ @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. $\endgroup$
    – Martin
    Commented Feb 24, 2014 at 13:21
  • $\begingroup$ @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? $\endgroup$
    – SRS
    Commented Feb 24, 2014 at 13:46
  • $\begingroup$ 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. $\endgroup$
    – Martin
    Commented Feb 24, 2014 at 14:00
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    $\begingroup$ @Roopam Yes, but that's besides the point. "Unphysical states" can't exist in nature (hence their name), so we don't have to worry about their normalization - that was what the original question was about. Of course we can't normalize them in the $L^2$-way, because they are not $L^2$. Besides, you are not obliged to work with "bases" like $\{|x\rangle\}$ - it's just very convenient to do so. $\endgroup$
    – Martin
    Commented Mar 8, 2014 at 11:10
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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.

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  • $\begingroup$ 1, There's a minor issue here that you seem to be taking a wave function $\phi_\sigma$ such that $\lim_{\sigma\to 0}\phi_\sigma\phi_\sigma^\ast = \delta$, instead of having the wave function itself be the nascent $\delta$. 2. There's a major issue here that it is not clear where you are taking the limits and what the limits are. Within the Hilbert space of $L^2$ functions, the sequence $\phi_\sigma$ converges to 0; one must carefully state in what space the limit has to be taken to yield a $\delta$. $\endgroup$
    – ACuriousMind ♦
    Commented Jan 17, 2017 at 16:30
  • $\begingroup$ @ACuriousMind I agree with point one: I am taking the probability distribution to be a Dirac's delta to represent the state of a particle confined at the origin. Physically, I cannot understand which kind of state is one with a wave function in the form of a Dirac's delta, can you? I disagree on point 2: the limit of the sequence of functions is what you state within $L^2$ indeed, but it is a Dirac's delta if we use the space of continuous functionals (i.e. distributions) over $L^1$ functions instead of $L^2$ . Look, this is quite standard theory of distribution by Laurent Schwartz. $\endgroup$
    – MariusMatutiae
    Commented Jan 17, 2017 at 16:57
  • $\begingroup$ @ACuriousMind Have you ever heard of the so-called delta-approximants? Look them up here, en.wikipedia.org/wiki/… $\endgroup$
    – MariusMatutiae
    Commented Jan 17, 2017 at 17:00
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    $\begingroup$ I know all that, but it is not clear to me why the weak convergence in the sense of distributions should be seen as the physically meaningful notion of convergence for states. The states are elements (or rays) of the Hilbert space, so it seems natural to me that the physically relevant notion of convergence is that of convergence in the Hilbert space norm. $\endgroup$
    – ACuriousMind ♦
    Commented Jan 17, 2017 at 17:06
  • $\begingroup$ @ACuriousMind You mean, apart from physical sense that says that the $\phi(x)$ above describes a very narrowly confined particle? Well, then you may see probability distributions as functionals acting on the space of dynamic quantities (Energy, position, linear and angular momentum, charge distribution and quadrupole moment,...) to yield scalars, i.e measurements. $\endgroup$
    – MariusMatutiae
    Commented Jan 17, 2017 at 17:32
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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.

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