If all microstates are equally probable, why do we see Hydrogen with its electron always in the ground state?

Question

I have learned in my Statistical Mechanics class that one of the fundamental assumptions of thermodynamics is that all microstates are equally probable. However, in the case of the Hydrogen atom, we know that we will most likely see the case where the electron is in the ground state. If all microstates are equally probable, we would see a variety of versions of hydrogen with the electron in an excited state. How can I reconcile the fact that 1) all microstates are equally probable 2) the ground state has the lowest energy, so the electron prefers to be in this state? I apologize if this question lacks rigor. Please let me know how I can improve it if so.

Answer 1

The statement isn't "all quantum states of a system are equally probable." That would be pretty ridiculous - every system would have infinite energy expectation value because it would share equal probability among infinitely many quantum states with ever increasing energy. Hydrogen atoms wouldn't just be in equal probabilities of each excited state, they would also have equal probabilities of existing in each ionized state up to the electron having infinite kinetic energy.

It's "every quantum state consistent with what we know about the system is equally probable. And usually what we know about the system is it has a well defined total energy that doesn't change with time."

This is also just the microcanonical ensemble (only one example of a set of assumptions that gives rise to statistical mechanics). The statement "at reasonable temperatures hydrogen is normally in the groundstate" probably comes from the canonical ensemble, where the probability of each state is proportional to $e^{-E_i/k_BT}$. So if $k_BT$ is small compared to $13\text{ eV}\sim 100,000\text{ K}$, then the groundstate will be astronomically more likely than excited states. Also note that because of the energy gap of hydrogen states, if the first electronically excited state is significantly populated, you probably also have a plasma, because the energy gap from 1S to 2S is bigger than the energy gap from 2S to ionization.

Consider also seeing my answer here where I attempt to explain where the microcanonical ensemble and the canonical ensemble come from. One could also just read the textbook explanations of these concepts in probably any stat mech textbook.

Answer 2

All microstates are equally probable in a thermodynamic system, i.e., in the limit of an infinite number of particles (while keeping some quantities constant, like density $N/V$ or the total energy.) A single hydrogen atom is not a thermodynamic system. However, if we had a system of $10^{24}$ (Avogadro constant) such atoms, we would have non-zero probability of finding some of them in excited states.

Furthermore, an atom is not isolated from the rest of the world - as a minimum it is coupled to the electromagnetic vacuum, which means that the excited states are among an infinite number of microstates of this atom+EM field system, where most of the microstates have atom in the ground state and a photon(s) present in any of the infinite number of the field modes.

Answer 3

I think your question deserves a slightly more complete answer.

The ``equal probability'' assumption is actually not so important.

What is better for students to keep in mind is that the teaching of physics theories involve a storyline that students need to follow and digest, and that in the course of the storyline, what is taken to be fundamental / important, will necessarily have to swap, even maybe a few times, in order to be understood by students. This is definitely one of the cases.

As mentioned by others, it is not that ``All microstates have the same probability.'' Instead, it is that "All microstates with the same total system energy have the same probability", and we have to add onto that, that we are only considering a single system energy, no changing from one energy to another. It is exceedingly obvious that this is kind of a silly assumption, because it is incredibly unphysical for a macroscopically large system, which necessarily must have a macroscopically large environment to interact with, to have a single total energy value. Indeed, this view will quickly be abandoned.

But one can see that there is a lot going for this. Any system has a phase space, where you can plot the Hamiltonian constant energy hypersurface contours. Imagine that at any time, a system is a point in this phase space, and the Hamiltonian sends it moving around the energy contour. The physical insight is that, if you have a whole ensemble of the system starting with their points in the same location in phase space, then by arbitrarily weak perturbations in energy by interaction with environment, these systems will be perturbed off the energy contour and thus move slight faster or slower than one that is fixed on the energy contour. If, later, another perturbation sends the system back onto the old energy contour, it would be slightly spread around. Thus, it makes sense for every point on the energy contour to have the same probability of actually being found as the microstate of a system. i.e. it is not total nonsense.

However, what this ``equal probability" assumption is really useful for, is to derive, by way of considering simple idealised combined systems and focusing only upon part of the systems, Boltzmann's probability distribution for different energies. This is the famous $e^{-E/k_B T}$ probability distribution. Once we derived in any one single simple case that this is a good, approximately true, probability distribution, the crucial step next is to turn around and postulate that this is true for all systems. That is, we turn around and swapped what is the fundamental assumption from "equal probability microstates" to "Boltzmann distributed energy macrostates". This has the physically much more sensible interpretation, because, as above, a macroscopically large system whereby thermodynamics is a good analysis tool, is also large enough to have an imprecise energy spread.

And similarly, later on, we would go on to talk about Gibbs' probability distribution, as even better than Boltzmann's.

Once we swap a fundamental assumption for another, the importance of the initial assumption is no longer as important as before. The viewpoint has completely changed. Students should reevaluate their entire understanding of the entire field of study in view of the new assumptions.

If you used good textbooks, they would talk about such things. e.g. Callen, and Ian Ford. The small disturbances of the energy contours is found in Feynman on statistical mechanics.