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Energy eigenstates provide a convenient basis for solving quantum mechanics problems, but they are by no means the only allowable states. Yet it seems to me that particles/systems are assumed to be in energy eigenstates "in nature".

Some examples of what I mean:

  • Solving the Schrodinger equation for the Hydrogen atom gives the standard $|n,l,m \rangle$ basis of energy/angular momentum eigenstates. But we speak of "filling up" these orbitals with electrons, or transitions between energy levels. Why should we expect to find the electrons only in such energy eigenstates, as opposed to say, some arbitrary superposition?
  • In quantum statistical mechanics we have Bose-Einstein and Fermi Dirac distributions that give the number particles in a state of energy $\epsilon$, but why must a particle be found in a state of definite energy to begin with?
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    $\begingroup$ You see, energy eigenstates are stationary states - they don't evolve with time. If we look at the time evolution, $|\Psi(t)> = e^{\frac{-iHt}{\hbar}}|\Psi(0)>$. This is exponential of an operator. However, in the energy basis, it becomes just the exponential of a number! This convenience gets us around the otherwise difficult task of computing the exponential of an operator. So our solution would take the form, $|\Psi(t)> = \sum_{n}e^{\frac{-iE_n t}{\hbar}}|\Psi_n>$. So solving the problem reduces to finding the energy eigenvectors. $\endgroup$
    – guru
    Commented Jun 30, 2013 at 1:28
  • $\begingroup$ sorry, i think i misunderstood the question, you are asking about why the electron in the hydrogen atom is assumed to be in a energy eigenstate. this is a good question and i dont really know the answer to that. explanations for line spectra of hydrogen in many textbooks assume that the electron is originally in an energy eigenstate and transits to another eigen state emitting a photon. what is the basis for this assumption that it exists only in energy eigenstates? i dont think any of the answers below have really answered this question! $\endgroup$
    – guru
    Commented Jul 2, 2013 at 4:27
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    $\begingroup$ @guru please do not set equations in comments on separate lines. Comments are not meant to be fully expressive and if you must have that formatting then you should probably write an answer. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 4, 2013 at 3:59

4 Answers 4

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If we take a system, and let it evolve for some indefinite amount of time, it will be in an incoherent mixture of energy eigenstates. Many systems we encounter in nature have been sitting for some time, and not interacting with the environment (much). These can be considered to be in energy eigenstates.

For example, suppose we consider an atom in a gas. Let's assume that the last interaction it had with the environment was a collision with another atom, which put it in some state which we can consider as a superposition of energy eigenstates. Now, let's look at it. Even if we know everything about the collision except how long ago it collided, we can't determine the phase of this superposition, so we might as well think about it as being in a probabilistic mixture of energy eigenstates.

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  • $\begingroup$ So energy eigenstates are 'attractor states'? This reminds me of an intuitive explanation someone gave once for the central limit theorem (CLT happens because gaussians transform as gaussians so are 'attractor states' in probability space). $\endgroup$
    – Matt
    Commented Sep 2, 2021 at 2:38
  • $\begingroup$ I wonder if there's an explanation from the schrodinger equation as to why a noninteracting atom in a pure state would be robust to small perturbations in its state vector towards an incoherent mixture. $\endgroup$
    – Matt
    Commented Sep 2, 2021 at 14:28
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Is there a more sophisticated answer than "This is done because isolated low-temperature systems tend to give up energy more often than they receive it, and thus, gravitate toward their ground states. Thus, even if they start in a mixed energy state, the system will radiate away energy until it reaches its ground state, which is obviously an energy eignenstate."?

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    $\begingroup$ Well, I thought that the Hamiltonian being the generator of time translations/the time evolution operator would be relevant. Eigenfunctions of the Hamiltonian propagate in time, with only an oscillatory phase. They remain eigenstates of anything that commutes with the Hamiltonian, i.e. any good symmetries. EDIT: Just reread the Q. Maybe not totally relevant after all! $\endgroup$
    – innisfree
    Commented Jun 29, 2013 at 20:25
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    $\begingroup$ @innisfree: you could say the same thing about a mixed energy eigenstate, though--it would just oscillate with two frequencies. $\endgroup$
    – Zo the Relativist
    Commented Jun 29, 2013 at 20:31
  • $\begingroup$ I would tend to agree that there is probably not much more to it. $\endgroup$
    – joshphysics
    Commented Jun 29, 2013 at 20:32
  • $\begingroup$ @JerrySchirmer you're quite right. $\endgroup$
    – innisfree
    Commented Jun 29, 2013 at 20:44
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    $\begingroup$ @yuval, you've got the cart before the horse in a way. It isn't that not being in a ground state causes the system to radiate, it's that the system cannot radiate in the ground state, because, by the very meaning of ground state, there is no lower energy state to fall to via radiation. $\endgroup$
    – Alfred Centauri
    Commented Jun 29, 2013 at 23:53
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  1. Let me start with a "sociological" remark: it seems to me that many people do believe (erroneously) that a system can only be in an eigenstate of energy. I guess this may have some historical roots - IMHO, some of the great founders of quantum mechanics did emphasize "quantum jumps" more than they should have.

  2. I would say "a convenient basis for solving quantum mechanics problems" contains not just "energy eigenstates", those states are typically also eigenstates of the components of total momentum of the system.

  3. innisfree mentions in his/her comment that the Hamiltonian is "the generator of time translations". So maybe the reason that eigenstates of total energy and momentum form a convenient basis is the homogeneity of space and time. A similar argument is valid for angular momentum as well, but different components of angular momentum do not commute.

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Heisenberg discovers Quantum mechanics (Matrix mechanics), by considering transitions between different energy levels of. In fact, the position $X(t)$ is no more a simple quantity, but has to be written $X_{ij}(t)$, because it is a transition between 2 energy levels $i$ and $j$, so $X(t)$ is a matrix (in fact, it is an infinite matrix, so it is called an operator). For instance, for the harmonic oscillator, we may write :

$$X(t) \sim ae^{-i\omega t} + a^+e^{i\omega t}$$ with $[a,a^+]=1$

Here $a$ and $a^+$ are operators which make the transition between energy levels of index $n$ and $n+1$

So the origin of Quantum Mechanics is clearly based on energy levels, and the position operator $X(t)$ is clearly made of transitions between energy levels. So it is not surprising that the energy basis is more fundamental in Quantum mechanics.

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    $\begingroup$ We discovered energy eigenstates first because they are in some sense more fundamental; it's not that they are more fundamental because we discovered them first. $\endgroup$
    – Peter Shor
    Commented Jun 30, 2013 at 14:41
  • $\begingroup$ @PeterShor : In this case, the two (fundamental/discovery) match. In Quantum Field Theory , the fields are no more than transitions between "momentum/energy" levels, too. So, from my point of view, it is a general quantum pattern. $\endgroup$
    – Trimok
    Commented Jun 30, 2013 at 16:49

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