5
$\begingroup$

Quantum numbers are supposed to denote every individual orbital. But if orbital shells are probability functions, then orbitals can't be definite, solid things. So in that case, there can be variation in the amount of energy given off when an electron drops between shells - it might, say, give off a tiny little bit more energy and drop to just below the orbital shell. Isn't this possible since orbitals are just probability functions - like "Here's where the electron probably is"? Not entirely sure where I was going with this, but I think the final question is, how come quantum numbers are only ever integers?

Edit: My question is about why quantum numbers as taught in schools are always integers. "Orbitals" as predicted by the Bohr model are in fact clouds of electrons, probability functions about where an electron probably is rather than a definite statement about where it definitely is. That means there's got to be wiggle room about how far an electron can be from the nucleus.

So does that mean that quantum numbers are an oversimplification, or just averages? Or am I just misunderstanding the whole "orbitals are just probability clouds" thing?

Edit: Ugh. Right. I'm an idiot. I forgot to mention that I'm only talking about the principal quantum number, n, the one telling which orbital the electron's in.

$\endgroup$

closed as unclear what you're asking by ACuriousMind, sammy gerbil, Michael Seifert, Qmechanic Jul 25 '16 at 18:21

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ Quantum numbers are not always integers: The energy of a free particle, for instance, is purely continuous. Whether the numbers are continuous or discrete depends on the specific system and observable you are considering. I'm not sure what kind of answer you expect to this - for instance, compact observables will always have discrete spectrum by the spectral theorem, but , for instance, position won't. $\endgroup$ – ACuriousMind Jul 25 '16 at 14:25
  • $\begingroup$ We can easily turn them into non-integers by using the natural logarithm of them. Not sure what your question is. If you want to know about the reason why there are energy levels instead of discrete spectra, that'd be a different question ... $\endgroup$ – Sanya Jul 25 '16 at 14:25
  • 1
    $\begingroup$ Spin can be a non-integer...1/2 $\endgroup$ – heather Jul 25 '16 at 14:27
  • 2
    $\begingroup$ You have a logic error - just because an orbital is a probability function does not then imply that they are not a definite thing. The orbital is well defined, with specific quantum numbers associated with it. $\endgroup$ – Jon Custer Jul 25 '16 at 14:50
  • $\begingroup$ You are confusing 2 different things : (1) a quantum number, which is the number of units of energy or momentum etc which an object has, and (2) the probability that the object has that amount of energy or momentum etc. $\endgroup$ – sammy gerbil Jul 27 '16 at 1:50
4
$\begingroup$

The orbitals that you are taught about at school are energy eigenstates i.e. eigenfunctions of the Hamiltonian. For a confined system like an electron in an atom the eigenstates of the Hamiltonian have discrete and precisely defined energies so the transition energies are all precisely defined.

In the real world electron states are only approximately eigenfunctions of the Hamiltonian because an eigenstate would be time independent i.e. have an infinite lifetime. Obviously in the real world this is never the case, and as a result the energies are not precisely defined and neither are the transition energies. This causes the broadening of spectral lines called lifetime broadening.

If the particle is not confined, e.g. a free electron, the Hamiltonian does not have discrete eigenstates and the energy can have any value. In this case there are no discrete energy states.

The reason the electron position is a probability distribution is because the eigenfunctions of the Hamiltonian are not eigenfunctions of the position operator. That is, a system cannot simultaneously have a precisely defined energy and a precisely defined position. You could regard this as another manifestation of the uncertainty principle.

$\endgroup$
2
$\begingroup$

Your confusion comes from unfortunate terminology

Quantum numbers are supposed to denote every individual orbital

quantum numbers denote the state of a quantum system as solution of the Schrödinger equation; in particular they often refer to the eigenvalues of a maximal set of operators used to describe the physics at hand. As an example the hydrogen atom can be described by the maximal set of operators $(H, L^2, L_z)$ with respective eigenvalues (quantum numbers) $(E_n, l, m_z)$ (therefore people use to say that the state of an electron is identified by its energy and its angular momentum).

But if orbital shells are probability functions

The probability functions you are mentioning are the wave functions associated to the state $|E_n, l, m_z\rangle$ defined as $\langle\,r\,|\,E_n, l, m_z\rangle$. You actually do not call this “orbital shell”, rather you call it “the wave function of the state”.

give off a tiny little bit more energy and drop to just below the orbital shell

Transition between states are well-defined and happen according to some well-determined rules (in particular see the Wigner-Eckart theorem).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.