Why exactly cant two thumbprints be alike

None of those are really "good" answers. I will concur. Moreover, it is really more a question that ends up touching onto meta-mathematics, than physics.

I am not really sure one can levy a specific reason, other than the very general reason that suitably complex equations cannot be solved exactly as a general principle. But I believe the explanation below, while not a solid or firm answer, is nonetheless likely to suffice.

One way you can think of it is that there are only a finite number of operations we have for expressing solutions to equations, but an infinite number of possible actual solution functions - although I don't think that quite gets at it, either because you can combine the "expressing" functions in an infinite number of ways. Perhaps a better way to say it is that when we go to solve equations "exactly" what we are talking about is to express them in terms of the composition of some limited number of "base" functions we consider "acceptable" - almost always this includes and is at least the elementary functions, which are those composed of the constant functions, the four arithmetical operations (+, -, *, /), and the exponential ($\exp$) and logarithmic ($\log$) functions over the complex field ($\mathbb{C}$), and the closure of this set is taken under functional composition ($\circ$). We may also include a few non-elementary "special" functions as well like the gamma function ($\Gamma$) or error function ($\mathrm{erf}$, the integral of the Gaussian).

And effectively what happens is that these do not provide us with enough "degrees of freedom", so to speak, to express all functions of suitable generality. While I'm not sure of a result that proves this for the general composition operator, a very similar result along this line is the fact that the space of all differentiable (say) functions - like those that come out as solutions to a differential equation like Schrodinger - is algebraically infinite-dimensional, that is, the cardinal of any Hamel basis is infinite, meaning that if we want to express any differentiable function as a linear combination

$$f(x) = a_0 f_0(x) + a_1 f_1(x) + \cdots + a_N f_N(x)$$

of basis functions $f_j$, then we actually need a basis set with infinitely many independent functions we can pull the $f_j$ from (i.e. they cannot be expressed in terms of each other) to be able to have enough expressing power to express every differentiable function $f$ in this fashion. (Such a set is called Hamel basis.) For the quantum theory of atoms, this is even more direct, it is the infinite-dimensionality of the Hilbert space.

Likewise, I'd strongly suspect something similar is happening when we allow general composition as well - it doesn't provide us with enough power, but I do not know of a proof.

However, this doesn't really answer the question as more specifically to why that a more restricted set of functions, like the solutions of a specific kind of differential equation - and in particular its eigenvectors, cannot be expressed with a suitable composition of a "nice" set of base functions. After all, linear constant-coefficient differential equations are famously known to have a completely-expressible solution set. Many other very specific differential equations have similar sets. Although, a key point may be that the Schrodinger equation offers effectively infinite freedom through the choice of the potential function $U(\mathbf{r})$, and more generally its Hamiltonian operator $\hat{H}$.

That said, with a very restricted parameter set, say atoms very specifically, it may indeed be there is a finite or easily-described infinite set of functions that can provide a suitable basis set. We just don't, as of yet, know what the "atom functions" are or how to describe them.

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answered Sep 9 '18 at 14:43

The_SympathizerThe_Sympathizer