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My understanding from reading about quantum mechanics is that the state of a particle such as an electron can be kept in a superposition of energy states for an extended period of time, when it is not interacting with other particles, but that as soon as atoms interact with each other, their electrons will tend to an energy eigenstate.

But energy eigenstates are just one particular basis for representing quantum states of an electron. Why do particles tend to eigenstates of the hamiltonian, as opposed to elements of some random other basis?

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    $\begingroup$ Does this answer your question? Why is it often assumed that particles are found in energy eigenstates? $\endgroup$ Commented May 14 at 13:58
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    $\begingroup$ The best answer in the link above is the first comment. "You see, energy eigenstates are stationary states - they don't evolve with time." $\endgroup$
    – mmesser314
    Commented May 14 at 14:22
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    $\begingroup$ @mmesser314, I ofcourse know that they are stationary states, but this is a handwavy answer. Why should they move to stationary states? $\endgroup$
    – user56834
    Commented May 14 at 14:53
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    $\begingroup$ @RileyScottJacob, I don't think so... (I saw this question before posting, by the way). I don't see how those answers clarify why it is energy eigenstates specifically, rather than some random other basis that they tend to. $\endgroup$
    – user56834
    Commented May 14 at 14:55
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    $\begingroup$ "I don't see how those answer clarify why it is energy eigenstates specifically..." It is not energy eigenstates specifically. You are asserting, without proof, that "their electrons will tend to an energy eigenstate," but this is not necessarily true. The only way this might be true is if you are talking about the ground state and if the temperature is low compared to other energy scales. $\endgroup$
    – hft
    Commented May 14 at 17:14

4 Answers 4

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But energy eigenstates are just one particular basis for representing quantum states of an electron. Why do particles tend to eigenstates of the hamiltonian,

States don't "tend to" do anything other than evolve according to the well-known principles of Quantum Mechanics.

Setting aside measurement, states evolve in time according to the time-dependent Schrodinger equation.

When measured, states collapse to the eigenstate(s) of the measured observable, regardless of if the observable is energy, or position, or whatever.


The only energy eigenstate that you might reasonably say states "tend to" is the ground state. If a state has some small coupling to the outside world, considered as a heat bath, then at low enough temperature you would expect the state to be the ground state.

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  • $\begingroup$ I don't think this answer the question: 1) when we talk about atom orbitals, we never mention (at least not explicitly) that we measure their energy. We do as if these are the states in which atoms are found "in the wild" without any observer. 2) the ground state that you mention is the ground state w.r.t. energy. Why energy? Why not minimize something else. I do not see how this answer addresses the distinguished nature of energy. $\endgroup$
    – M. Winter
    Commented May 26 at 9:14
  • $\begingroup$ I answered the question as I understood it. As you can probably see, OP has selected a different answer as "correct." But anyways, the orbitals that "we" talk about are just the single-particle solutions to some single particle (possibly self-consistent) Hamiltonian. Those orbitals, or their Slater determinants, are not the solutions to the many-body Hamiltonian that describes the atoms "in the wild," since we can not solve for their wave functions exactly for anything but Hydrogen. The question of "why energy" is easy. The energy is an integral of the motion. $\endgroup$
    – hft
    Commented May 26 at 18:34
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If a system is undergoing interference and information is copied out of that system that tends to suppress the interference: this process is called decoherence. For macroscopic objects this is usually a result of interaction with the environment and it selects a set of states that are robust under interaction with the environment that are highly localised in position and momentum and act approximately classical and such states aren't necessarily energy eigenstates: see Sections IV and V of Zurek's review Decoherence, einselection, and the quantum origins of the classical, see also

https://arxiv.org/abs/1111.2189

If interaction with the environment is relatively weak and is dominated by the self Hamiltonian this tends to prefer energy eigenstates.

The set of preferred states can be explained by decoherence, so collapse is unnecessary for explaining the states we see when we do observations. In addition you can only tell what states are preferred by working out the actual consequences of quantum equations of motion not by ad hoc, unclear and often unstated modifications of those equations of motion.

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    $\begingroup$ I love how often people vote to close a question or say the question is wrong right before an amazing answer like this appears. (Actually I don't love it but rather dislike it, but you get what I mean). $\endgroup$
    – user56834
    Commented May 15 at 12:19
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    $\begingroup$ no I genuinely meant that I think it's an amazing answer. It's just that people voted to close the question (and if that had happened, I wouldn't have gotten your answer). $\endgroup$
    – user56834
    Commented May 15 at 13:14
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    $\begingroup$ @user56834 Thinking and answering is difficult, voting to close is easy and cheap. $\endgroup$
    – user402514
    Commented May 15 at 23:15
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    $\begingroup$ That is exactly what I intended with my comment. I think most other answers really misunderstood the (very good and reasonable) question. +1 $\endgroup$ Commented May 16 at 9:08
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    $\begingroup$ @alanf I think "actually I don't love it" was meant as clarification of the sarcasm in the first sentence $\endgroup$
    – MJD
    Commented May 17 at 11:49
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Sometimes other basis states are interesting. Neutrino flavor eigenstates are not energy eigenstates. So netrinos oscillate, changing from one to the next.

FermiLab's Even Bananas series of video has a lot on neutrinos. In particular, How do neutrino oscillations work?

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    $\begingroup$ Your answer is also a illustration of the traveling "neutrino" not being a collapsed energy eigenstate, but, instead, a superposition of three such, for all of its "long" life from birth to observation. $\endgroup$ Commented May 14 at 15:13
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The fundamental aspect to consider is what is the operator you are dealing with. Eigenvalues and Eigenstates are associated to specific operators; in the case of an energy operator, for example, such as the Hamiltonian describing the energy of a basic non-relativistic quantum system, the eigenvalues will be energy eigenvalues, and this can be proven by looking at the dimension of those eigenvalues $\lambda$ that satisfy the defining equation in operator form $H\psi=\lambda\psi$, where $H$ is the Hamiltonian of the system. Similarly, you have momentum eigenvalues for the momentum operator etc. Crucially, the eigenvalues and eigenstates are what completely describes the observables, i.e. the quantity than can/could be detected experimentally, so their determination is very informative. The determination of such operators that are physically meaningful is an important process in the construction of quantum theories. The construction of spin-operators, for example, are considerably less straightforward.

Note: I have not addressed the (considerably) trickier case of Lagrangian-based quantum systems but I believe this answer helps to shed light on OP's question.

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