Why do electrons in an atom occupy only the stationary states?

Question

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 ?

Answer 1

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.

Answer 2

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 "slow environment" (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.)

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.