Why are accessible states taken as eigenstates in statistical physics? Is the resolution via decoherence?

Question

See for example this question or the answers to this question, neither of which seem to say exactly why the system Hamiltonian eigenstates (and not some other complete basis) are used. It is presumably something that baffles every person learning elementary statistical physics for the first time: why are the "accessible states" always (approximate) eigenstates of the system Hamiltonian? Why are superpositions of these eigenstates not admissible, or other bases entirely?

I cannot seem to find the passage now, but in reading Schlosshauer's textbook on Decoherence I seem to remember him discussing how, in the thermodynamic limit, decoherence's resolution of the problem of the preferred basis (that basis which does not change during entanglement with the environment) implies as a theorem that system eigenstates are selected. Is this generally the underlying reason that, in statistical physics, we only consider system eigenstates? If not, what is said justification or where can I find it?

Answer 1

I'll present a complement to AXensen's very good answer.

Statistical physics of quantum systems can be based on the general description of the state of a quantum system in terms of the density operator $\hat \rho$ (a positive semi-definite, trace one, hermitian operator, on the Hilbert space of the system). The matrix elements of $\hat \rho$ define the so-called statistical density matrix.

The density matrix, or density operator, is the most general description of the system's quantum state of interest, including the case of local measurements of entangled states. Every set of basis vectors of the Hilbert space can be used to get a matrix representation of $\hat \rho$, for equilibrium and non-equilibrium states.

However, the time evolution of $\hat \rho$ is controlled by the Liouville–von Neumann equation: $$ i \hbar \frac{\partial \hat \rho}{\partial t}= [\hat H,\hat \rho], $$ where $\hat H$ is the Hamiltonian operator of the system. Thus, in the case of equilibrium systems, where $\hat \rho$ must be time-independent to ensure time independent averages of any time-independent observable $\hat O$, we have that the density operator commutes with the Hamiltonian, thus admitting a basis of common eigenvectors. Such a possibility ensures a special place for the Hamiltonian eigenstates in equilibrium statistical mechanics. Notice that this is the quantum counterpart of a similar consequence of the classical Liouville's theorem in classical statistical mechanics, stating that equilibrium ensembles ar characterized by a probability density function of the phase space poin through the Hamiltonian ($\rho(q,p)=f(H(q,p))$).

Answer 2

This answer is long, and this is a subject that can be approached in many different ways. And I'm not an expert in this - just a guy who took a well-taught statistical mechanics course in grad school. So credit to Prof. Dung-Hai Lee; I also used his notes as reference for this. Feel free to vote in a way that reflects how helpful and how correct you found this to be. If you find part of this argument to be incomplete or suspect, I'd probably agree with you - leave a comment.

Very handwavy answer:

In quantum mechanics, energy is the only special operator, because the operator that tells you how a state changes in time is given by: $$ \hat{U}(t)=e^{-i\hat{H}t/\hbar} $$ So if you're going to do statistical mechanics with the eigenstates of any operator, it makes most sense to do it with the Hamiltonian eigenstates.

Better answer:

The question you are asking requires starting the explanation of statistical mechanics from scratch. There are several possible starting points for statistical mechanics.

The first approach I'll discuss is called the microcanonical ensemble. The argument for this approach goes as follows: An isolated system which is in equilibrium can be said to have a certain energy, and perhaps a small error on that energy $\Delta E\ll E$, because the energy of a system doesn't change with time or as it comes to equilibrium. Other than that, we have no idea which quantum mechanical state it's in - so we assume it could be in any of the quantum mechanical states with energy $E\pm\Delta E$ with equal probability.

For this ensemble, and for all others, I should make a clarification - we don't actually believe that it's more likely for our system to be in a pure energy eigenstate as opposed to a linear combination of several energy eigenstates within that range. We're only interested in how many linearly independent states there are - could be any basis at this point. We want to know the dimensionality of the subsbace of the system's entire hilbertspace that is consistent with our only knowledge - the energy. For example, if a hydrogen atom is known to have energy $-13.6\pm0.1\text{ eV}/4$, there are 8 linearly independent states the electron could be in (two spin states times [1 l=0 state+3 l=1 states]).

The reason for this "principle of equal probability" assumption is that it maximizes entropy. The Von Neumann entropy of a system is the sum of the negative logs of the probability of each state. this is maximized when every state has equal probability. And Von Neumann entropy has been shown to give the same phenomenology as any other definition of entropy, which we know increases over time. So if the system has had a long time to come to equilibrium, it's entropy is maximized given the constraint that it could not change energy. I believe better derivations of this principle are an active area of research. Under the principle of equal probability, the Von Neumann entropy turns into $S=k_B\ln W$, where $W$ is the number of states consistent with your known energy. $k_B$ is just some arbitrary constant that brings our quantum definition of entropy in line with the much older understanding of entropy from before quantum mechanics was known.

So what am I saying, am I saying a state is always after long enough in an equal-weights superposition of all the energy eigenstates with that same energy? Hell no. I'm saying that if I prepared the same system identically many times, each time I would have a different linear combination of those energy eigenstates, and any linear combination would be equally likely.

But this discussion would be woefully incomplete without also discussing the canonical ensemble. The canonical ensemble makes a different assumption that the system has a well defined temperature. Temperature tells us which direction heat flows. When two systems are brought into thermal contact, heat flows from the system of higher temperature to the system of lower temperature. Many systems we encounter in everyday life have a well defined temperature, which means there's a bigger system (whose energy, not temperature is well defined, so it can be treated with the microcanoical ensemble) with that temperature that our smaller system is in thermal contact with. Our smaller system is actually not equally likely to be in any energy state. Instead, it has a probability to be in a state $i$ with energy $E_i$ of: $$ P_i\propto e^{-E_i/k_BT} $$ Say system 1 is the big system and system 2 is the small system. Now let $W_n(E)$ be the number of states of system $n$ with energy $E$ in system $n$. Then the probability of system 2 being in a state having energy $E_2$ is: $$ \propto W_1(E-E_2)=\exp\left[\frac{S_1(E-E_2)}{k_B}\right]=\exp\left[\frac{S_1(E)-(dS_1(E)/dE)E_2}{k_B}\right]$$ $$\propto\exp\left[-\frac{E_2}{k_BT}\right] $$ Where $S_1$ is the entropy of system 1. Note that the definition of temperature (almost because of this argument) is $(dS_1(E)/dE)=1/T$.

So in summary, we talk about energy eigenstates because a system's energy is the thing that's most often well defined about a system we're going to do statistical mechanics on. We make no other assumptions and we assign equal probability - partly out of ignorance, and partly because this assumption maximizes Von Neumann entropy. And finally, when we arrive at the canonical ensemble which describes a system with a temperature, this fact too can be derived from the previous microcanonical ensemble.

Answer 3

The usual argument in the classical statistical mechanics is that macroscopic properties of a system should be expressible in terms of its integrals of motion (otherwise, they depend on the particular microscopic state, and the whole concept of statistical physics fails.) There are seven of these:

• energy: $E$
• three components of momentum: $P_x,P_y,P_z$
• three components of the angular momentum: $L_x,L_y,L_z$

Working in the center of mass system removes the three components of momentum. Then the statistical physics textbooks also limit themselves to discussing the systems that do not rotate (although this is an interesting and relevant case, e.g., when discussing stars.) This leaves us with the energy - all the macroscopic properties of a system in thermodynamic equilibrium should be expressible in terms of its energy. Microcanonical distribution and the rest immediately follow.

Same logic applies to quantum mechanics, even though it may seem less obvious at first: Hamiltonian corresponds to the only conserved macroscopic quantity.

Broading and smearing of states are indeed used as euphemisms for decoherence in statistical physics. Essentially, this implies vanishing of non-diagonal elements of the density matrix - this could be even taken as the definition of decoherence.