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When we talk about the elementary problems in quantum mechanics like particle in a box, we first calculate the energy eigen-function. Then we say that the most general state is the linear combination or superposition of these basis eigen-functions. But when we go to the atoms, say hydrogen atom, we end up in calculating energy eigen-functions and say the electrons occupy these stationary states starting from least energy state (ground state). I have seen this in solid state physics too. For instance in nearly free electron model, we calculate energy eigen-functions with eigenvalues

$$E=\frac{h^2k^2}{2m}$$ where $k=n\pi/a$, $a$=length of the sample and assume that electrons are going to occupy these states. Here too there is no discussion of linear combination of these states.

So my question is why don't we talk about the state functions here that may be the linear combination of two or more than two stationary states in atoms or are the conditions under which electrons occupy only the stationary states ?

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    $\begingroup$ When you make a measurement the wavefunction collapses to an eigenstate, so the observables are only ever eigenvalues. $\endgroup$ – lemon Oct 5 '14 at 13:45
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    $\begingroup$ @lemon, it collapses to an eigenstate of an observable one measures, which is not necessarily the total energy. $\endgroup$ – Wildcat Oct 5 '14 at 14:02
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    $\begingroup$ Dissipation/decoherence is the crucial element here which everyone is ignoring. $\endgroup$ – DanielSank Oct 5 '14 at 15:45
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Why do electrons in an atom occupy only the stationary states?

This isn't true. An electron in an atom can be in any superposition of states. This is one of the basic postulates of quantum mechanics: linearity.

For example, say an atom has a ground state 1 and an excited state 2, and let's say we're able to prepare it in a pure state 2. It will decay electromagnetically to state 1. This decay is represented mathematically by a process in which the wavefunction becomes a mixture of states 1 and 2, with the amplitude of 2 decaying exponentially and the amplitude of 1 growing correspondingly.

Energy is special here only because many of the measuring devices we use to study atoms are energy-sensing devices. When we measure using one of these devices, we always get a definite energy. Take the two-state example again for simplicity. In the Copenhagen interpretation (CI), this is because of wavefunction collapse. In the many-worlds interpretation (MWI), the measuring device becomes a superposition, but it's a superposition of a state in which the device measured a single energy and another state in which the device measured the other energy. You can also discuss this in terms of decoherence.

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Depending on specific situations, I can think of two independent answers to the question "why we mostly only care about stationary states":

  1. decoherence by photon interaction (I won't focus on this aspect);

  2. all (effectively nondegenerate) systems becomes (improper) ensembles of stationary states IF we have uniform ignorance of time (I will focus on explaining this idea).

First, I assume you have basic knowledge about density matrix. (I think I have to do so, since otherwise my answer would be too long.)

Given an arbitrary initial density matrix $\rho_0$, we construct a time ensemble based on uniform distribution of time. Why do we want such time ensemble? Say we set the initial state, and then we wait while we don't record the time. Not recording time roughly means we have an uniform ignorance of time. Given that uniform ignorance of time, the density matrix describing the system is the time ensemble $\rho_{\text{time}}$. Say $U_t$ is the time evolution operator.

$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger$$

Say the initial state is in a pure state superposition $\rho_0=\left(\sum_m C_m \left|m \right>\right) \left( \sum_n C_n^\dagger \left< n \right|\right)$

$$\rho_{\text{time}} =\lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \ U_t\rho_0 U_t^\dagger = \lim_{t \rightarrow \infty} \frac{1}{t} \int^t_0 \mathrm{d}t \sum_{m,n} e^{-i (E_m-E_n)t} C_m C_n^\dagger \left| m \right> \, \left< n \right|$$

Taking the limit first gives the delta function $\delta_{m,n}$

$$\rho_{\text{time}}=\sum_{m,n} \delta_{m,n} C_m C_n^\dagger \left| m \right> \left< n \right|=\sum_{n} \left|C_n \right|^2 \left|n \right> \left< n \right|$$

Here we go! If we have an initial state and an uniform ignorance of time, all phase information is lost and the density matrix is diagonal in energy so it's an (improper) ensemble of energy states. For thermodynamic properties which are independent of time, such time ensemble capture all information about the system. Therefore, only energy states and the occupation probabilities of the states matter. The ideas of superpositon of energy states and interference between energy states become non important roughly speaking. (The above derivation requires non-degeneracy. The reason why it's often reasonable is that there is rarely perfect symmetry. I won't elaborate this more.)

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    $\begingroup$ This idea of uniform ignorance implies diagonalization and (improper) ensemble can also be applied to spacial ignorance. In that case, uniform ignorance of space implies the density matrix becomes a mixture of plane waves, which partially justifies most scattering theories. (The derivation is just to replace time evolution operator by translation operator, and integrate over all position of translation instead of all time of evolution) $\endgroup$ – Bohan Xu Apr 27 '18 at 13:39
  • $\begingroup$ Interesting point. Do you have some references about this (papers or textbooks)? $\endgroup$ – student Jun 19 '18 at 17:37
  • $\begingroup$ I'm not sure about your knowledge, so I start from the beginning. Pure state quantum mechanics: Sakurai Multiple particles & tensor product: Just google :D (You need to understand tensor product to understand density matrix and partial trace) For density matrix: any introduction part of open quantum system books, I used the book (not the overview) Decoherence QuantumToClassical Transition, Schlosshauer Given knowledge of density matrix, my post is already understandable. I'm not sure about further references. Proof of the Ergodic Theorem and the H-Theorem by Von Neumann 1929 might help $\endgroup$ – Bohan Xu Jun 21 '18 at 2:25
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One reason we focus on energy eigenstates is that atoms spend almost all of their time in an energy eigenstate, and their spectrum is a result of transitions between them.

Another reason is pedagogical: to peel back the onion one layer at a time. But before too long, many courses do include examples of systems that are not in an energy eigenstate. One popular example is the harmonic oscillator with the system in a Gaussian state that is displaced from the origin. All energy eigenstates are centered on the origin; the displaced Gaussian is a superposition of all energy eigenstates. When one calculates the time evolution one finds that the state oscillates back and forth about the origin ... simple harmonic motion.

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    $\begingroup$ This does not answer the question. You say that "atoms spend almost all of their time in an energy eigenstate", but OP knows that. He asks why do they spend almost all of their time in energy eigenstates? In other words, how do we know that? Does it comes from experimental observations (spectrum) or somewhere else? $\endgroup$ – Wildcat Oct 5 '14 at 14:55
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    $\begingroup$ @Wildcat No, the OP says "why don't we talk about the state functions here that may be the linear combination of two or more than two stationary states". He does not ask why they spend so much time there. Unless I really misread the question, I've addressed what he/she asked. $\endgroup$ – garyp Oct 5 '14 at 18:25
  • $\begingroup$ @garyp actually i wanted to ask why only the stationary states are preffered by the electrons or as you are saying "they spend most of their time in stationary states" but why? Why not the other states that may be the linear combination of the eigenstates.Is their any reason for this or is it just the experimental thing and we can only observe their behaviour? $\endgroup$ – Girish Pahwa Oct 6 '14 at 8:53
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This is an interesting question because you have to start dealing with the atom surrounded by its environment. Atoms with electrons in states other than the lowest possible filled-up states, have lots of opportunities. They can emit photons (and thereby go to a lower energy state) in many possible directions, and that photon can interact with any number of things in this world, such as photo-multiplier plates or retinas. We can say that the electron is more or less coupled to all the other charged particles in the universe. Given all the possible other configurations of the original electron and all the other charged particles, the system after some length of time will essentially have no probability of the electron being in the higher energy state.

To describe this process through time, one could describe the wavefunction of the electron as being in a superposition of states after some time, t1, and in a different superposition at a later time, t2. Eventually, the component at the higher energy has no amplitude.

If we try to measure the energy level, however, we can only interact with photons that match the gap between energy levels. Either the well-aimed photon gets absorbed, putting the energy level back up, or does not. If you look at it this way, the atom IS only in one eigenstate or the other.

To try to understand this further, you get into the realm of QM interpretation.

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  • $\begingroup$ I think almost all of think answer is irrelevant to the question. $\endgroup$ – Brandon Enright Oct 8 '14 at 22:55
  • $\begingroup$ The relation to the surrounding system is important. A state in superposition that can by some means be prevented from interacting with anything else will stay in superposition. $\endgroup$ – MoreThingsInHeaven Oct 9 '14 at 6:28
  • $\begingroup$ Personally I think this is the best answer so far, although far from perfect. I don't understand why it gets downvoted. Environment and decoherence are key to the question. $\endgroup$ – Chenfeng Oct 9 '14 at 19:29
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Edit after comments:

We have the experimental observation of fixed spectra coming from specific atoms. The same for nuclei, and both are stable in their ground state ( unless energetically disturbed or are unstable isotopes).

Quantum mechanical solutions reflect these experimental observations and the spectra of atoms and nuclei have been fitted with potential models, shell models.

when we talk about the elementary problems in quantum mechanics like particle in a box, we first calculate the energy eigenfunction.

This is a very specific problem and has set of eigenfunction. The particle can only be in one of these eigenfunctions , not in a superposition. The potential is very specific and the energy operator, the Hamiltonian, is known . If the electron, for example , were in a superposition of energy states about the hydrogen atom, it would have a probability to be in a higher than ground state without external energy input, and a probability then to decay to the ground state, violating energy conservation.

Then we say that the most general state is the linear combination or superposition of these basis eigenfunctions.

This is a general statement when the potentials are not specified, i.e., the Hamiltonian , is not specified. Then one has the possibility in order to describe the state of a particle to use the set of eigenfunctions coming from other specific quantum mechanical operators, for example solving for the eigenfunctions of the momentum operator .

Example:

Consider a free electron in one dimension that is described by the function

$$\Psi(x)= C_1\psi_1(x) +C_2\psi_2(x)$$

$$\psi _1(x) = \left ( \frac {1}{2L} \right )^{1/2} e^{ik_1x}$$

$$\psi _2(x) = \left ( \frac {1}{2L} \right )^{1/2} e^{ik_2x} \tag {5-23}$$

where k1 and k2 have different magnitudes. Although such a function is not an eigenfunction of the momentum operator or the Hamiltonian operator, we can calculate the average momentum and average energy of an electron in this state from the expectation value integral. (Note: in-this-state means described-by-this-function.)

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The function shown above belongs to a class of functions known as superposition functions, which are linear combinations of eigenfunctions. A linear combination of functions is a sum of functions, each multiplied by a weighting coefficient, which is a constant. The adjective linear is used because the coefficients are constants. The constants, e.g. $C_1$ and $C_2$ in the , give the weight of each component ($\psi_1$ and $\psi_2$) in the total wavefunction.

In the above example the $\psi_1(x)$ the $\psi_2(x)$ are an eigenfunction of the momentum operator and also the Hamiltonian operator ( since no potential in the problem) although the linear combination function $\Psi$ is not.

In general when the Hamiltonian is known the bound states are defined and the particles will fill up the energy levels starting from the lowest state sequentially. To get to a point where a superposition of eigenfunctions is necessary the system will not be a simple potential problem of a stable atom. A linear combination of energy eigenfunctions will be needed for the description of an ensemble of particles where some are in excited states and there are energy inputs by radiation, for example.

For your example after the edit: if the simple potential model is adequate in describing the system one avoids the complexity of linear combinations: the key is " nearly free", so there exists a solvable potential and that is used.

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    $\begingroup$ This answer is long, but the first paragraph is incorrect so I stopped reading at the error. A particle in a box can indeed be in a superposition state. For example, a state that localizes the particle. Such a state might have, for example, like a 3D bell-like shape. All eigenstates would be necessary to describe this state. $\endgroup$ – garyp Oct 5 '14 at 14:27
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    $\begingroup$ I second @garyp. This answer is incorrect. $\endgroup$ – DanielSank Oct 5 '14 at 15:09
  • $\begingroup$ @garyp and danielSank too, do you have a link for this state? I am always willing to learn . I am talking for something like the hydrogen atom, or the shell model of nuclear physics $\endgroup$ – anna v Oct 5 '14 at 15:09
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    $\begingroup$ Anna, you write "The particle can only be in one of these eigenfunctions, not in a superposition." - why do you think that is correct? That is exactly the assertion that the OP essentially asks about. $\endgroup$ – ACuriousMind Oct 5 '14 at 15:24
  • $\begingroup$ @ACuriousMind Nuclei are stable, atoms are stable, ( unless energetically disturbed) the limit on lifetime of a proton is a tiny experimental number. In a specific potential well, the electron will occupy the lowest empty level, maybe radiating a characteristic photon. It will not be dithering ( have a probability to be in another higher energy level) between energy levels, because we would be seeing several photons coming out and energy would not be conserved also. $\endgroup$ – anna v Oct 5 '14 at 15:41
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This answer is not in any way a contradiction to the previous answers; it is a slightly different perspective that may provide a more intuitive understanding.

The reason an electron stays in an eigenstate - in an orbital - is because it does not radiate when it is in that state.

When an electron is in a pure state, the effective charge distribution is static: no time dependence, no oscillation, and therefore no emission of radiation. (That's not 100% true, because when there is an empty orbital at a lower energy, the electron will eventually drop into that orbital-- but it will do so by emitting a photon whose energy matches the energy difference between the two orbitals, and whose frequency matches the difference between the Zwitterbewegung frequencies of the electron in each orbital.)

An oscillating charge density results in emission of an electromagnetic wave having the same frequency as the oscillation. When an electron in an atom is in a superposition of two states, interference between the two states results in what amounts to an oscillation of the charge density of the electron. That oscillation, it turns out, is precisely at the frequency of the photon that is emitted when the electron drops from the higher-energy state to the lower-energy state. (The temporal component of the interference is equal to the beat frequency between the zwitterbewegung frequencies of the electron in the two states.)

The following statement is not quite quantum mechanically correct, but it is close enough for most practical purposes:

Mixed eigenstates of different energy will always radiate. Because in an eigenstate the electron does not radiate, it stays in that state until something perturbs it.

Of course, the electron does radiate when there is an unoccupied lower-energy state. But the state transition half-life depends on perturbations. See for example (theory) and (experiment). In a laser, the active medium is selected to have a relatively long lifetime in a selected excited state. Perturbation due to a passing photon of the "correct" frequency triggers transition between the excited state and a particular lower-energy state, along with emission of a photon whose energy matches the energy difference between the two states - and whose frequency matches the difference between the Zitterbewegung frequencies of the electron in the two states.

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    $\begingroup$ This is completely wrong. Anything that's not the ground state will radiate; the differences you're trying to draw simply don't exist, and the reasons you give ("The reason an electron stays in an eigenstate - in an orbital - is because it does not radiate when it is in that state") are plain wrong. $\endgroup$ – Emilio Pisanty Apr 27 '18 at 8:02
  • $\begingroup$ No disagreement there: "when there is an empty orbital at a lower energy, the electron will eventually drop into that orbital--". Are you saying that the differences between the Zitterbewegung frequencies don't exist? Or are you saying that Zitterbewegung frequencies per se don't exist? $\endgroup$ – S. McGrew Apr 27 '18 at 12:45
  • $\begingroup$ I'm saying that if you prepare an electron in a 100% pure state in the $|\Psi\rangle = |2p_z\rangle$ state of hydrogen, then it will radiate, because Fermi's Golden Rule requires it to. Your answer claims the opposite, which is plain wrong. $\endgroup$ – Emilio Pisanty Apr 27 '18 at 12:54
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    $\begingroup$ Obviously it does radiate when there is an unoccupied lower-energy state. But the state transition lifetime depends on perturbations. See for example web2.ph.utexas.edu/~schwitte/PHY362L/QMnote.pdf (theory) and arxiv.org/pdf/1507.02427.pdf (experiment). In a laser, the active medium is selected to have a relatively long lifetime in the excited state. Perturbation due to a passing photon of the "correct" frequency triggers radiative transition between the excited state and a particular lower-energy state, the energy difference corresponding to (the same) photon frequency. $\endgroup$ – S. McGrew Apr 27 '18 at 13:22
  • $\begingroup$ I edited the answer to include the above references. $\endgroup$ – S. McGrew Apr 27 '18 at 13:34

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