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Is there an experimental system, or such that can be observed in nature where a particle's wave function assumes a form - $\psi(x)\propto \frac{1}{\sqrt{x^2+1}}$ such that $|\psi(x)|^2$ is Lorentzian?

An answer to this question Is there a condition of quantum mechanics that forbids Lorentzian distributions? claims that such wavefunctions are responsible for charge delocalization in molecules. I could not find a reference that discusses this.

My ultimate goal here is to understand whether quantities like $<\hat{X}>$ would assume finite values or indefinite values (since a Lorentzian distributions has no moments).

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$\psi(x) = \frac{1}{\sqrt{1+x^2}}$ is a perfectly valid wavefunction. However, one cannot find the expected value of $\hat X$ in this state, because $\psi$ is not in the domain of the $\hat X$ operator.

The appropriate domain on which $\hat X$ is self-adjoint is

$$D_X:= \left\{\psi \in L^2(\mathbb R) \ \left| \ \int_\mathbb R |x \psi(x)|^2 dx < \infty\right\}\right.$$

For all $\psi \in D_X$, we then have that $(\hat X \psi) (x) = x\psi(x)$. However, since you can see pretty quickly that $\int_\mathbb R |x\psi(x)|^2 dx \rightarrow \infty$ for $\psi(x) = 1/\sqrt{1+x^2}$, this wavefunction is not in the domain of the position operator, which makes

$$\langle \hat X\rangle_\psi \equiv \frac{\langle \psi,\hat X \psi\rangle}{\langle\psi,\psi\rangle}$$

undefined. There's nothing particularly weird about this. What it means is as follows: if you measure the position of each of $N$ identically prepared systems and denote the average as $\overline X_N$, then as $N\rightarrow \infty$, $\overline X_N$ will not converge to a finite value. If you wait long enough, you'll measure arbitrarily large positions, and this will occur sufficiently regularly that the overall average is not bounded as the number of trials gets larger.

Another way to think of it is that, given any $M>0$, you will measure your overall ensemble average position to be greater than $M$ (or less than $-M$) with probability 1 as the number of measurements tends to infinity.

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  • $\begingroup$ This is is a theoretical oriented answer, while I was asking an experimental question - where in nature do we find Lorentizan wavefunctions? I came across the claim that $<x>=\lim_{m \rightarrow \infty} = \intop_{-m}^{m} x|\psi(x)|^2dx=0$ because this is the only physically sensible way to take the limit in an actual physical system. So I want to know where do you encounter such situations in the lab and what was actually measured. $\endgroup$
    – proton
    Commented May 31, 2020 at 12:34
  • $\begingroup$ @proton You are asking if there is a situation in which the potential derived by Valter Moretti in this answer is a particularly useful model. I have never seen one, but that means fairly little. However, the "ultimate goal" part of your question is a theoretical one, not an experimental one, and is the primary target at which this answer is aimed. $\endgroup$
    – J. Murray
    Commented May 31, 2020 at 13:15
  • $\begingroup$ I am curious: how exactly does this fit in/jive with the mathematical formalism of the theory? The wave function clearly looks to inhabit an equivalence class in the relevant Hilbert space as it is square integrable, so the problem must be with the operator. Does this mean that these operators are not total functions on the state space? As I've always thought they were. (And that's why you don't play sloppy with math, kids!) $\endgroup$
    – The_Sympathizer
    Commented May 31, 2020 at 14:12
  • $\begingroup$ @J.Murray What I meant is what value of $<X>$ would I measure in practice for a system that is modeled in this way. In understand your mathematical derivation but the physical argument that I raised suggests that in practice things would behave differently. Is there anything wrong with it? $\endgroup$
    – proton
    Commented May 31, 2020 at 14:38
  • $\begingroup$ @proton The quantity $\langle X \rangle $ as referenced above is the expected average measured position in the limit as the number of measurements goes to infinity; in this case, the limit simply would not exist. In practice, if you measured 10 positions, you'd get some finite average. If you measured 100, you'd get another. If you measured 1000, you'd get yet another, and these averages would simply not converge to some limit as you kept going. $\endgroup$
    – J. Murray
    Commented May 31, 2020 at 15:59
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One potential example of an experimental system that has this type of Lorentzian "wavefunction" is cavity-enhanced parametric down conversion (PDC) photon sources. In this case it isn't the position wavefunction but the "wavefunction" in frequency space which is Lorentzian (more correctly the joint spectral intensity (JSI) of two photons is an Airy distribution but individual cavity modes are Lorentzian). This JSI is like the probability density function (PDFs) for two photons to be at frequencies $\omega_1$ and $\omega_2$ but the marginals describe single particle PDFs.

If you want to read more, I think this is a nice theory paper.

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In the post you mention there are some references. Also, they tackle your question about the interpretation of expected values. Nevertheless I think you are interestred in wavefunctions $\psi$ such that $\vert\psi\vert^2$ is a Lorentzian, which, you know, is not eactly the same!

Edit: Sorry to say that Loretzian wave-function is not square integrable.

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  • $\begingroup$ As far as I can tell none of these references over there say anything explicit about either the form of the wavefunction or the generating potential. Regarding my question about experimentally measurable quantities, I came across the claim that $<x>=0$ because the only sensible way to take the infinite limit is on a symmetric region. This contradict the answers in the link. $\endgroup$
    – proton
    Commented May 31, 2020 at 10:15
  • $\begingroup$ @proton The answer of Valter Moretti gives a wavefunction and shows how to construct the potential. Can you provide some reference for the claim $\langle x \rangle=0$? $\endgroup$
    – vin92
    Commented May 31, 2020 at 10:37
  • $\begingroup$ The face that you can construct the potential the way Valter Moretti did does not imply that there are experimental setting where this potential is realized. I do not have any reference to the claim I maid (it came from a discussion with a colleague), and I could not find any discussion on this issue in the literature. $\endgroup$
    – proton
    Commented May 31, 2020 at 10:56
  • $\begingroup$ @proton Okay now I understand the question. Unfortunately I'm a zero on the left in experimental physics. Sorry I can't help! $\endgroup$
    – vin92
    Commented May 31, 2020 at 11:08
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    $\begingroup$ Lorentzian wavefunctions are square-integrable. $\endgroup$
    – Emilio Pisanty
    Commented May 31, 2020 at 11:09

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