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Consider the quantum mechanical problem of a particle trapped in an infinite potential well. The solution to this problem using Schrodinger's equation gives us discrete energy eigenfunctions and eigenvalues. Suppose we already have a particle in some eigenstate, then its energy can be increased only if we impart an amount of energy equal to the difference between its present eigenstate and some other eigenstate (this is similar to an electrom in an atom).

The problem is how can this model be realised. I mean, if we already have the potential well and some particle with an energy not equal to some eigenvalue, then if try to trap this particle, will it instantaneously release energy? Is this a general feature of other quantum mechanical systems? I know that this problem is very ideal but I feel like some essential features like this can be better understood from this problem.

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    $\begingroup$ One of the simplest realizations is a molecule with aromatic ($\pi$-conjugated) system. In this system electron is delocalized but cannot escape molecule, therefore it is something like infinite potential. Modifying this simple molecule with help of appropriate additional fragments, you can explicitly see on experiment emission/absorption of photons with certain frequency, i.e. see that energy is quantized. The key point that electrons in a molucele are in a ground state. $\endgroup$ Jan 10 '20 at 22:20
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For a particle to have a definite energy, it must be in an energy eigenstate of the Hamiltonian that describes it. But this in no way means that our system always has to be in one of the energy eigenstates.

Take the simple case of particle in a box. The only condition on the wavefunctions is that it is $0$ at and outside the boundaries of the box and that it’s square integral inside is $1$. Any wavefunction that satisfies the above two conditions is a legitimate state that your system can be in. It is only that if you measure the energy, it’ll always be in one of the energy eigenstates.

For example, if I measure the position of a particle, the system will be in a position eigenstate that has no definite energy. The state is now a superposition of energy eigenstates after the measurement.

There is nothing sacred about energy eigenstates. We only care about them because the time evolution of states is trivial for energy eigenstates, ${\exp}\left(-\frac{iEt}{\hbar}\right)$.

As for why atoms are generally in ground state energy (an energy eigenstate) is that atoms are never isolated from perturbations. They have many decay mechanisms that lead them to lose energy. But this can only happen between energy eigenstates. Thus if you look for signs of these decays you’ll always find the atom to be in an energy eigenstate. Once it reaches the ground state, it can no longer decay and can only get excited if it absorbs an exact energy.

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  • $\begingroup$ Very clear answer! Guess your QM prof is better than OP's. $\endgroup$
    – Razor
    May 6 '20 at 17:05
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The problem is that you defined what your system is, concluded a discrete spectrum, then assume it has a definite energy prior to measurement, and on top of that it is not an allowed energy.

This (The system, not the error) can be realized with an analogy. A guitar string strapped on a guitar. Playing is a superposition of standing waves, and slightly and correctly touching a string and plucking would reveal an isolated pure note

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  • $\begingroup$ I get it, probably there is no point in talking about it before measurement. When I measure the system it is going to end up in some eigenstate. $\endgroup$ Jan 11 '20 at 8:58

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