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As far as I know, quantum numbers range upto infinity, let it be a rigid rotator, the hydrogen atom, etc. I am thinking if there can be a system for which the quantum numbers of the wavefunctions, and hence the wavefunctions themselves, are finite. Mathematically, I am thinking about wavefunctions: $$ \Psi_{n_1, n_2,n_3,…,n_m} $$ Where, $$ n_1=1,2,3,…,N_1$$ $$ n_2=1,2,3,…,N_2$$ $$ ……….………$$ $$ n_m=1,2,3,…,N_m$$

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Considering the answer given to my I should like to extend my question: is there any finite dimensional alternative to the Hilbert space? This surely sounds like a joke but its rather more practical to have a finite world than an ideally infinite space.

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    $\begingroup$ "quantum numbers range from $1$ to $∞$" - you mean apart from (say) spin in the hydrogen atom? $\endgroup$
    – Emilio Pisanty
    Commented Mar 13, 2019 at 13:38
  • $\begingroup$ Of course yes, I am just talking about the principal quantum numbers. I hope wavefunctions without ANY quantum number ranging upto infinity. $\endgroup$
    – Awe Kumar Jha
    Commented Mar 13, 2019 at 13:43

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The Morse potential $$ V(r)=D_e(e^{-2a(r-r_c)}-2e^{-a(r-r_c)}) $$ is important as historically it is the first example of a continuous potential that supports finitely many bound states, i.e. there is an upper value of principal quantum number.

The energies are given by $$ \epsilon_n= -(\lambda-n-\textstyle\frac{1}{2})^2 $$ with $n=0,1,\ldots [\lambda-\frac{1}{2}]$ and $\lambda$ is the combination $$ \lambda=\frac{\sqrt{2mD_e}}{a\hbar}\, . $$ Basically the number of bound states is related to the depth $D_e$ of the potential and the length scale $a$ of this potential.

Before this example, only a finite well (which is discontinuous) was known to support finitely many bound states.

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  • $\begingroup$ I like this example, but doesn't any finite potential well support a finite number of (or zero) bound states? $\endgroup$
    – march
    Commented Mar 13, 2019 at 16:40
  • $\begingroup$ @march I would not bet on this being always true. Imagine for instance a Coulomb-like potential where the core is replaced by a uniformly charged sphere so the potential is linear in $r$ for some small $r/a$, but $1/r$ outside. Then by perturbation theory you'd still have infinitely many bound states but a technically finite potential at $r=0$. $\endgroup$
    – ZeroTheHero
    Commented Mar 13, 2019 at 17:09
  • $\begingroup$ That's interesting! If that turns out to be true, that's a serious violation of my intuition (which is fine, but still...). I mean: my reasoning would go something like this, thinking semi-classically. As the energy of the bound state increases, the number of nodes it has also increases, and hence the amount of kinetic energy it has increases. Eventually, the KE due to those extra nodes is larger than the difference between the bottom and top of the well, at which point the state is no longer bound. $\endgroup$
    – march
    Commented Mar 13, 2019 at 17:18
  • $\begingroup$ The Dirac delta well has a single bound state. Does that qualify? $\endgroup$
    – Bill N
    Commented Mar 13, 2019 at 17:33
  • $\begingroup$ However, maybe I'm using the finite square well in my thinking too much. There, you have to cram the nodes inside the well, and so the KE shoots up. For a potential that is less than 0 everywhere, perhaps the nodes just get pushed out farther and farther, making it so that the KE doesn't increase without bound (this seems to be what happens with hydrogenic wave functions, so...). $\endgroup$
    – march
    Commented Mar 13, 2019 at 17:46
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If we interpret "wavefunction" in the usual sense, as the space of states of a particle moving in $\mathbb{R}^n$, then this is not possible. This is simply because the Hilbert space is $L^2(\mathbb{R}^n)$, the space of square integrable functions, which is infinite dimensional; and every basis must have the same number of elements. You can have finite dimensional Hilbert spaces, of which spin is the most common example, but they're not really the wavefunctions of a particle moving in space.

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  • $\begingroup$ Or they can be - by using a suitable hamiltonian which restricts the dynamics to only a finite-dimensional subspace of an otherwise infinite-dimensional Hilbert space. $\endgroup$
    – Emilio Pisanty
    Commented Mar 13, 2019 at 15:14
  • $\begingroup$ @EmilioPisanty do you know of an example? I don't see how one can get around the fact that the Hilbert space is always $L^2$, a fact that is independent of the Hamiltonian. $\endgroup$
    – Javier
    Commented Mar 13, 2019 at 15:16
  • $\begingroup$ The full Hilbert space is always $L^2$, but the dynamics need not explore the full Hilbert space. A suitable example is a set of ions in an ion trap, addressed by lasers that are sharply resonant with only a few atomic and translational transitions: if the state starts off within that set of states, the dynamics will keep it there, and you can effectively forget about the rest of the Hilbert space. This is a foundational aspect to all of quantum information processing, be it on ions, cold atoms, superconducting circuits, photons, you name it. $\endgroup$
    – Emilio Pisanty
    Commented Mar 13, 2019 at 15:22
  • $\begingroup$ @EmilioPisanty. Or, really, any time there's some conserved quantity like angular momentum that would restrict the dynamics to a finite-dimensional subspace. $\endgroup$
    – march
    Commented Mar 13, 2019 at 19:03

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