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When studying the properties of solids we always say that electrons are in (stationary) energy eigenstates. The theory of conduction for example (with conduction bands and stuff) follows from the assumption that electrons are in energy eigenstates but why are they in such states in the first place? What prepared the electrons in energy eigenstates? And if they are not in energy eigenstates how is it possible that the assumption of them being in energy eigenstates works so well?

Example: I have a block of aluminum. It has never had the energy of its electrons measured by me nor anybody else: it just came out of production. If I use it as a conductor it has a resistance. Assuming its electrons are in energy eigenstates I can calculate its resistance. Why am I allowed to assume that its electrons are in energy eigenstates? If nothing ever measured the energy of the electrons of the aluminum they might be in a much different state.

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  • $\begingroup$ But is your block of aluminum well-isolated from the environment? One doesn't have to explicitly measure the energy of the electrons for the state to collapse. $\endgroup$ May 16 '20 at 23:37
  • $\begingroup$ @BioPhysicist okay, but why does it collapse in an energy eigenstate and not e. g. in a momentum eigenstate? $\endgroup$
    – Masterme
    May 16 '20 at 23:40
  • $\begingroup$ Questions: Why are energy eigenstates the main object of QM in general? Why is the Schrodinger equation about eigenstates of the Hamiltonian, and not some other operator? What is Hamilton's principle? $\endgroup$ May 17 '20 at 6:34
  • $\begingroup$ @MariusLadegårdMeyer Energy eigenstates are particularly useful because they make it easy to compute temporal evolution. Hamilton's principle is the least action principle. I don't see how this is relevant though $\endgroup$
    – Masterme
    May 17 '20 at 10:12
  • $\begingroup$ @MariusLadegårdMeyer The SE applies to all state functions. Not just energy eigenstates. $\endgroup$ May 17 '20 at 11:34
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I think you may be mistaken that the electrons are in energy eigenstates when looking at valence/conduction bands. The energy eigenstates are a complete basis, so we use them to express the state of the electrons, but that doesn't mean they're in an eigenstate. However, when talking about the valence and conduction bands, it's just easier to talk about the eigenstates rather than explicitly talk about the wavefunction of a single electron that is a superposition of different eigenstates. After all, we're interested in the energy needed for electron to jump bands (at which point it will have collapsed to a single energy band), and also the properties of the bulk material, not a single electron.

Also, in the comments you ask why not momentum eigenstates, but energy eigenstates. Here, these are the same thing. In fact quite often in QM they're the same since the Hamiltonian and momentum are simultaneously diagonalisable in many systems.

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  • $\begingroup$ The Hamiltonian is the sum of a kinetic part and a potential part. The kinetic part commutes with momentum but the potential part does not. How can they be simultaneously diagonalisable? $\endgroup$
    – Masterme
    May 17 '20 at 14:20
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    $\begingroup$ Because one of the assumptions we make in this model is that the crystalline structure is invariant under translations, i.e the Hamiltonian is invariant under translation (the potential is translation invariant). Since the momentum is the translation generator: $[H,e^{-iPa}]=0=H(1-iPa)-(1-iPa)H = -ia[H,P]$. So it follows that the Hamiltonian also commutes with the momentum. I'm guessing you've done Bloch's theorem? $\endgroup$
    – baker_man
    May 17 '20 at 15:11
  • $\begingroup$ Okay, that makes sense. Thanks. $\endgroup$
    – Masterme
    May 17 '20 at 15:21
  • $\begingroup$ You seem to be confusing momentum and quasimomentum. The former is not actually conserved in a crystal, unlike the latter. $\endgroup$
    – Ruslan
    May 18 '20 at 8:16
  • $\begingroup$ Ah thanks, and the quasimomentum is the conjugate momentum here? Sorry for causing any confusion $\endgroup$
    – baker_man
    May 19 '20 at 13:26
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I'm not sure if this explains it, but the electrons only fall into a particular eigenstate upon observation; before that, electrons exist in a state that may be expressed as a probabilistic superposition of all available eigenstates. Eigenstates exist due to the quantisation of angular momentum. (edit)

Also, possible duplicate: Why do electrons tend to be in energy eigenstates?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    May 17 '20 at 5:43
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I think it is not usually assumed that an electron (or system of electrons) is an energy eigenstate, but that at low temperatures the electron (or system of electrons) is in its ground state, which happens to be an energy eigenstate. In fact, for general temperatures a system at thermal equilibrium is generally assumed to be in thermal state, see e.g. http://militzer.berkeley.edu/diss/node13.html.

The point is that systems try to be in the state of lowest energy for low temperatures and most of the time low temperature is a good assumption.

This also why the Bohr model for the atom works so well. Here one essentially assumes that the atom is in its ground state. Most low energy phenomena can be described by valence electrons being excited etc and this is clear since energy eigenstates where the valence electrons are excited are the next to lowest energy eigenstates. Of course the atom, at moderate temperatures, can be considered to be in a superposition of the ground state and maybe some of the lowest excited states. So the dynamics usually happens "close to the ground state", unless of course we prepare the system in some way, by shooting a laser at it or something (I am very ignorant about applied physics). Of course this discussion applies equally well to solids or Fermi gas etc.

I want to also point out a confusion that I had with regards to this. The description of ground state, i.e. the state of lowest energy, applies to the lowest energy eigenstate of the single-particle system (e.g. lowest energy shell of atom) as well as to the lowest energy eigenstate of the $N$-particle system, i.e. the state where all fermions are "inside" the Fermi sphere. This should not be mixed up.

(Actually wanted to post this as a question, I am looking for agreement or disagreement for this take)

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  • $\begingroup$ If we assume so low temperatures that electrons are in their ground state, then we'll have no conduction in semiconductors — even doped, since the doping levels will be filled, Fermi-Dirac distribution will have no tails, so no charge carriers will be in the conduction & valence bands. $\endgroup$
    – Ruslan
    Jan 14 at 22:20
  • $\begingroup$ I agree. Clearly the system is not strictly in its ground state. I was just saying that the system is close to its ground state, in that the lower energy states tend to be occupied. I am mainly asking if it makes sense to argue this way to justify why such models work, and the system is not in an arbitrary superposition of energy states. $\endgroup$
    – jkb1603
    Jan 15 at 10:30

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