1
$\begingroup$

Okay, I know you may say, there are infinite numbers of possible state of an electron between its ground state and ionic state, since energy levels get continuous as electron gets farther away from the nuclear. But that's not what I mean.

Here is my concern: Any linear combination of eigenstates is also a state for an electron in an atom. For example, an electron in 3d orbital (n=3, l=2) is thought to have only five possible states: dxy, dxz, dyz, dx2-y2, dz2, with literally no m eigenvalue. Yet, what about there linear combinations? Aren't they also possible states of an electron in 3d?

Thanks for any help!

$\endgroup$
  • $\begingroup$ You can't literally take a linear combination of states (or at least you can't define a linear structure on states that is useful for anything) but you can form a non-trivial linear combination of the vectors that represent various states, and thereby get a vector that represents some other state. So yes, as soon as you have $n$ distinct states, you have $n-1$ dimensions worth of additional states. It's not clear why this is bothering you, so in that sense it's not clear what you're asking. $\endgroup$ – WillO Aug 9 '18 at 5:41
  • 2
    $\begingroup$ Yes, there are an infinite number of bound states. When you talk of "five possible states" in your example, you really mean that there are five basis states that span that subspace of quantum states. The number of quantum states in that subspace, though, is infinite. As a simple example consider just a two state system with basis states |0> and |1>. The set of all possible quantum states is described by $\alpha$|0>+$\beta$|1> where the sum of the absolute squares of $\alpha$ and $\beta$ is 1. The number of quantum states in just this simple system is therefore infinite. $\endgroup$ – Samuel Weir Aug 9 '18 at 5:47
  • $\begingroup$ Samuel, so there are supposed to be infinite numbers of energy levels when the atom is in external fields, like crystal or something? Since there are infinite numbers of possible states and therefore they split into continuous energy spectrum? $\endgroup$ – JOE Aug 9 '18 at 6:07
  • 1
    $\begingroup$ A linear combination of the energy eigenstates is not an energy eigenstate because it isn't time independent. $\endgroup$ – John Rennie Aug 9 '18 at 6:51
3
$\begingroup$

An single-electron atom like hydrogen can be described by a single-particle wave function $\psi: \mathbb{R}^3 \to \mathbb{C}$ (neglecting spin here), an element of the Hilbert space $L^2(\mathbb{R}^3,\mathbb{C})$. This wave function describes the state of the electron -- so obviously, there are infinitely many possibilities. ($L^2$ is an infinite-dimensional Hilbert space, even.)

Now the bound states in an atom can be decomposed as a linear combination of the eigenstates of the Hamiltonian, i.e. the wave function can be decomposed in that basis. The point is that an electron can be in an arbitrary linear combination of eigenstates. But one often talks as if only those eigenstates "are there", because if you know how the eigenstates evolve, everything else is simple linear algebra.

$\endgroup$
  • $\begingroup$ If so, does it mean the energy levels become continuous when the atom is in external field like crystal field, rather than split in to finite and countable values? $\endgroup$ – JOE Aug 9 '18 at 11:30
  • $\begingroup$ @JOE The energy levels of the whole crystal are continuous, yes. It's not so useful to talk about the energy levels of a single atom when the atoms are strongly interacting, though, except in the mean-field approximation. In this approximation the bound-electron levels are still discrete, although countably infinite and slightly shifted due to the Stark effect. $\endgroup$ – tparker Aug 9 '18 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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