Wave function of atoms and ground state

Question

There is something I really do not understand about quantum mechanics when we specifically solve, for example, the hydrogen atom. In every book about quantum mechanics we calculate the eigenfunctions by solving the time independent Schrödinger equation $$\hat{H} \Psi_{n}=E_{n}\Psi_{n} $$ But once we have done that the linearity of the Schrödinger differential equation allows us to find solutions of the form $$\Phi=\sum_{n} C_{n}\Psi_{n} $$ So my question is: Why do we always find the Hydrogen atom (after waiting some reasonable amount of time) in its ground state (if we don't apply any field) when the quantum theory allows us to have a superposition of states? My "naive" first thought would have been that we could find the Hydrogen atom in one eigenstate (with a certain probability), but we could also find it in another different eigenstate (with another probability), but the case is that we always find it in its ground state. Then I thought that maybe the initial state $\Phi(t=0)$ evolves with time making the ground state be the most likely after some time passed, but that doesn't make sense because $$\Phi(t)=\sum_{n} C_{n} e^{-it\hat{H}/\hbar} \Psi_{n}=\sum_{n} C_{n}e^{-itE_{n}/\hbar}\Psi_{n} $$ and that means that the probability of finding each state doesn't change with time. Of course I understand that there is also the point of view in which the wave function is "given" and we measure the probability of finding every eigenstate experimentally and we construct with these probabilities the wave function, but I thought that there would be a more "fundamental" physical reason for which the hydrogen (and I suppose any atom) prefer the ground state above any else.

I'm still an undergraduate student so maybe this question is very obvious for many of you.

Answer 1

Atom eventually comes in thermodynamic equilibrium with other atoms (via collisions), electromagnetic field (via emitting photons) and so on. Now one can take the Boltzmann distribution over different energy levels and calculate the probability for the atom not to be in the ground state, at a typical experimental temperature for spectroscopy. This could be rather different, however, if we discuss atoms in stars.

One could add that more generally the principle that the lowest energy states are filled first is known as electron shell filling order and Aufbau principle:

The filling of the shells and subshells with electrons proceeds from subshells of lower energy to subshells of higher energy. This follows the $n + ℓ$ rule which is also commonly known as the Madelung rule. Subshells with a lower $n + ℓ$ value are filled before those with higher $n + ℓ$ values. In the case of equal $n + ℓ$ values, the subshell with a lower n value is filled first.

Although the link with thermodynamics/statistical physics is rarely done explicitly in this context, it is taken for granted that the lowest energy states are filled first, which is just the expression of the principle of minimum energy, that is the second law of thermodynamics:

The principle of minimum energy is essentially a restatement of the second law of thermodynamics. It states that for a closed system, with constant external parameters and entropy, the internal energy will decrease and approach a minimum value at equilibrium.

The tricky part here is what we consider as a thermodynamic system. In spectroscopic context we usually talk about a large collection of atoms. Thus, even if we talk about shell filling in a single atom, a very large number of atoms is usually kept in mind.

In modern experiments, such as single-molecule spectroscopy, one can, in principle, "talk" to a single atom. This is usually found in its ground state, because the excited states have finite lifetime - they revert to the ground state via emission of photons. The opposite process is possible via absorption of a thermal photon, and its probability decreases with decreasing temperature, as dictated by the Planck's law for the black body radiation.

Answer 2

In introductory courses, the Hydrogen atom is typically solved as an isolated system subjected to classical fields. In more accurate quantum field theoretical treatments, the electromagnetic field itself is a quantum mechanical entity coupled to both the proton and electron in the Hydrogen atom: the proton may be treated as a classical "source current", and the electron as a Dirac particle (i.e. a Grassmann-number-valued, Fermionic quantum field with spin 1/2) that is induced near the proton in the ground state. This system will also include a certain number of excited bound states (i.e. with $\mathcal O(1)$ electrons localized near the proton), and transitions between these states will be determined by the photon-electron interaction (i.e. coupling between the quantized "electron field" and electromagnetic field): qualitatively, an excited bound-state electron coupled to a photon field in a mixed state can excite photons, but those photons will inevitably carry energy (and phase information) away from the electron until it settles into the ground state, which will tend to include a different mix of (virtual) electrons and positrons than the "free" particle state of an electron coupled to the electromagnetic field.

Answer 3

...the linearity of the Schrödinger differential equation allows us to find solutions of the form $$\Phi=\sum_{n} C_{n}\Psi_{n} $$

Not exactly, in general there is additional contribution due to integral of continuously indexed generalized eigenfunctions. This is important e.g. when describing scattering and ionization.

So my question is: Why do we always find the Hydrogen atom (after waiting some reasonable amount of time) in its ground state (if we don't apply any field) when the quantum theory allows us to have a superposition of states?

We do not really "find" the hydrogen atom in ground state. We often assume $\psi$ is the lowest energy eigenfunction, or in superposition of the lowest eigenfunctions, where the ground state is the major component, but this is valid only in some cases, e.g. when temperature is low and the hydrogen is not emitting radiation. In general, it is emitting some radiation, sometimes even visible (discharge lamp with hydrogen), and then the $\psi$ used to describe the atoms is of course not equal to the lowest energy eigenfunction, but it is in superposition. It is common to use density matrix instead to describe gas of known temperature, and there, unless the temperature is 0, one has multiple non-zero entries on the diagonal. When the system is radiating, even non-diagonal entries are non-zero.

Then I thought that maybe the initial state $\Phi(t=0)$ evolves with time making the ground state be the most likely after some time passed...

This is a correct idea, implied by the idea of spontaneous emission of radiation, but non-relativistic Schroedinger equation does not describe it, just as the Rutherford planetary model does not describe electron falling to lower orbits (in this model, the electron can circle on any orbit, it does not fall and does not emit radiation). We have to modify these models to describe emission of radiation and change of the orbit/psi function.

Mathematically, there are different ways to do it, e.g. we can put friction terms by hand into the Schr. equation. But if we want to describe it in agreement with the general idea of conservation of energy, we need to find the degrees of freedom which take away the energy. In physics, even with single atom, there are always at least EM field degrees of freedom, and this is due to the fact EM interaction is not actually Coulombic as in nonrel. Schr. equation, but it has finite speed of propagation $c$. Both in classical and quantum theory, taking into account theory of relativity (and thus introducing finite value of $c$) gives the EM field its own degrees of freedom, and the atom can lose energy to it.

...but I thought that there would be a more "fundamental" physical reason for which the hydrogen (and I suppose any atom) prefer the ground state above any else.

For single atom in vacuum (no radiation present), the fundamental reason is, there is a mechanism by which the atom loses energy to EM field. This could not happen if EM interaction was instantaneous Coulombic interaction, so we can say it is due to finite speed of EM interaction.

For single atom in an environment where other atoms and thermal radiation are present and act on the atom, there is another reason: the interaction with environment can also de-excite the atom. When this is due to external EM field, it is sometimes called stimulated emission (although the term really should refer to emission of radiation, the de-excitation of the atom is assumed to be linked to it).

Answer 4

It appears that the Schroedinger equation that you're working with is free from an interaction term that couples the atom from an electromagnetic field , which might include the vacuum state. Given that there is no such interaction present in the formulation, then all you've calculated was the states of an isolated atom. Using Boltzman distribution, or any type of assumption about theomodynamic equilibrium or upper-state lifetime depends crucially on an interaction term.