How to understand "microstate" quantum mechanically in the definition of microcanonical ensemble?

Question

There exists some discussion on this topic like

Microstates of the system in microcanonical ensemble

Definition of quantum microcanonical ensemble in Landau & Lifshitz

and others but I am still confused of it.

What is a microstate exactly? Microcanonical ensemble assumes (like pathria's book,chapter 5) in energy eigenbasis the density matrix takes the form (in my notation degeneracy contributes a larger $\Gamma$)$$\rho_{mn}=\frac{1}{\Gamma}\delta_{mn}$$

So the first question is: it seems that the natural way to understand a microstate is just a solution of Schrodinger's equation, whose energy expectation value is equal to the given $E_0$.

And the second one: The usual definition is (more or less like) all accessible microstates have the same probability. So why energy is superior? Say, why we don't take the definition:

In eigenbasis of operator $O$, the density matrix takes the form $$\rho_{mn}=\frac{1}{\Gamma}\delta_{mn}$$

Is this because we can't do this, or we can in principle define something like this, energy basis is just some convention, and for each observable operator $O$ like $P$ or $S$, we can define a $(N,V,O)$ ensemble?

Answer 1

So the first question is: it seems that the natural way to understand a microstate is just a solution of Schrodinger's equation, whose energy expectation value is equal to the given E0.

We are dealing here with a many-particle physics ($N_A\sim 10^{23}$), which means that the eigenstates of the Hamiltonian describing such a system for a given energy are potentially highly degenerate (unless we talk specifically about the ground state.) The definition of the density matrix is then that the system can be with equal probability in any of these degenerate eigenstates corresponding to the given energy. This is just a quantum equivalent of the microcanonical ensmeble, where all the combinations of particle velocities adding up to the same energy are equally probable: $$ \rho(E)=\frac{1}{\Gamma} \delta\left(\sum_{i=1}^{N}\frac{p_i^2}{2m}-E\right), $$ where the partition function is $$ \Gamma(E)=\int dp_1\int dp_2...\int dp_N\delta\left(\sum_{i=1}^{N}\frac{p_i^2}{2m}-E\right)$$ (possibly with factors taking into account particle permutations.)

If we consider now an $N$-particle quantum state parametrized by the particle momenta eigenvalues, $p_i=\hbar k_i$, so that $n=(k_1,k_2,...,k_N)$, then the density matrix is just $$ \rho_{n,m}=\rho_{k_1,k_2,...,k_N; k_1',k_2',...,k_N'}= \begin{cases} \delta_{k_1,k_1'}\delta_{k_2,k_2'}...\delta_{k_N,k_NN'}\Gamma^{-1}, \text{ if } \sum_{i=1}^{N}\frac{\hbar^2k_i^2}{2m}=E,\\ 0\text{, otherwise} \end{cases} $$ (with the appropriate quantum caveats for the particle permutations.)

And the second one: The usual definition is (more or less like) all accessible microstates have the same probability. So why energy is superior?

Statistical mechanics aims at describing complex multi-particle systems in terms of macroscopic quantities. The most natural of these quantities characterizing the system are its integrals of motion, that is three components of the momentum of the center-of-mass, three components of the angular momentum, and energy. Provided that the system is not rotating and we are working in its center-of-mass reference frame, the energy is the only thing that remains. This is a key point for statistical mechanics in general, not only quantum statistical mechanics.

Answer 2

1. Considering a system with Hamilton operator $H$, the so-called "microstates" $|n\rangle$ are just energy eigenstates obeying the eigenvalue equation $H |n \rangle =E_n |n\rangle$. It is possible to choose an orthonormal basis $\{|n \rangle\}$ of the corresponding Hilbert space $\mathcal{H}$ with the normalization conditions $\langle m|n\rangle=\delta_{mn}$ and the completness relation $\sum\limits_n |n \rangle \langle n|=\mathbb{I}_{\mathcal H}$.

2. The density operator of a microcanonical ensemble $$ \rho_m := \frac{\chi_{[E-\Delta,E]}(H)}{{\rm Tr} \,\chi_{[E-\Delta,E]}(H)}=\frac{1}{\Gamma} \sum\limits_{E-\Delta \le E_n\le E}\! |n \rangle \langle n |, \qquad \Gamma= \sum\limits_{E-\Delta \le E_n \le E} 1 \tag{1} \label{1}, $$ ($\chi_I$ denotes the characteristic function of the interval $I$) fulfils the required conditions $$\rho_m =\rho_m^\dagger \ge 0, \quad {\rm Tr} \, \rho_m=1 \tag{2} \label{2}$$ for a (mixed) state. The specific form of $\rho_m$ in \eqref{1} is based on the plausible assumption that an isolated system in thermodynamic equilibrium can be characterized by equal probabilities $1/\Gamma$ for the accessible microstates $|n\rangle$ subject to the condition $E-\Delta \le E_n \le E$.

3. The time evolution of a density operator $\rho(t)$ in the Schrödinger picture is given by $$ \rho(t)= e^{-iHt/\hbar}\rho(0) e^{i Ht/\hbar} \tag{3} \label{3}$$ or, equivalently, by the von Neumann equation $$ \dot{\rho}(t)= -\frac{i}{\hbar }[H, \rho(t)]. \tag{4} \label{4}$$ As $\rho_m$, defined in \eqref{1}, commutes with the Hamilton operator $H$, $$\rho_m(t)= \rho_m(0) \tag{5} \label{5}$$ is time-independent, as to be expected for the desription of a state in thermodynamic equilibrium. Thus time-independence of the density operator requires the condition $$ [H, \rho]=0 \tag{6} \label{6}$$ being obviously fulfilled if $\rho$ is chosen as a function of $H$ itself, like $\rho_m$ in \eqref{1}. The density operator of the canonical ensemble, $$ \rho_c:= \frac{e^{-H/kT}}{{\rm Tr} \, e^{-H/kT}}, \tag{7} \label{7}$$ may serve as a further example. On the other hand, choosing $\rho=f(O)$ with an operator $[H, O] \ne 0$, (implying in general also $[H , f(O)]\ne 0$) would not be wise in view of the condition \eqref{6}.

Answer 3

Since you are doing Pathria and using density operators, I will assume that you are asking about the quantum case.

The microstate that goes as ingredient into the microcanonical ensemble, is the maximally specified Hamiltonian eigenstate. That is, find a CSCO for your system under consideration, such that all the degeneracies are counted once and only once. Then each unique allowed state in such an enumeration is a microstate.

And the third one:

I've answered your first question; where is the second question?

So why energy is superior?

That comes from the fact that you are doing (statistical) thermodynamics. If you understood any bit from your undergraduate thermodynamics and undergraduate statistical thermodynamics, you should have understood what Boltzmann and friends were saying, and that leads to the special-ness of the Hamiltonian basis.

I mean, you can really only deal with equilibria, at least at the level you are discussing. Equilibrium states are, by definition, time translation invariant, at least on humanly accessible timescales. By Noether's theorem, that privileges the Hamiltonian basis.

Note that this is not something you can side-step. If something is out-of-equilibrium, we have the fact that Hamiltonian flows conserve entropy whereas non-equilibrium systems must generate entropy. This means that the very assumption of using Hamiltonians (of just the system itself) is wrong in that case. You can, of course, purify your system and use a larger Hamiltonian that you later trace away the environment degrees of freedom and thus obtain the generation of entropy, but almost nobody (outside of studying decoherence) does that.

In practice, it is important to first fully understand equilibrium (statistical) thermodynamics before going onto any topic more complicated than that.