Clarification regarding the terminology of Microstates

Question

I would like to understand how microstates are defined or used in physics. Are microstates suppose to only mean eigenvalues of a given observable (or a generator of symmetry)? The reason for my confusion is the similarity between path integrals and partition function. The partition function is usually defined by

$$Z = \sum_{\{E_i\}} e^{-\beta E_i}$$

where $\{E_i\}$ is the entire collection of microstates. Now, consider the comparison with the path integral $$Z = \int [\mathcal{D}\phi] e^{-iS[\phi]}$$ for some field $\phi$. Now, by comparison, the path integral is equivalent to the summation over the entire microstate collection. So does that mean that I can call each field configuration $\phi(x)$ a microstate? This makes sense to me because in the momentum space $\tilde{\phi}(k)$ is indeed a microstate as $k^{\mu}$ is the eigenvalue of the generator of translation, so the original definition still holds. Is there any flaw in my understanding?

Answer 1

Generally, a microstate is just a state, i.e. a single, well-defined object that represents a particular configuration of a physical system, where by "configuration" we mean enough information to predict the time evolution of the system just from the configuration.

In contrast, a macrostate is a probability distribution over microstates that represents some more general notion of "state" - we define the possible microstates and their probability by some quantity like temperature and volume. A "partition function" is really not much more than the normalization factor of our probability distribution - if we know that our probability distribution is proportional to $f(x,p)$, then the normalization factor to get an actual probability distribution whose integral over all microstates is unity is $Z = \int f(x,p)\mathrm{d}^Nx\mathrm{d}^Np$.

Now, the quantum mechanical representation of a "probability distribution over states" is a density matrix, and the analog of the "integral over microstates" is the trace. The equation $$ Z = \sum_i \mathrm{e}^{-\beta E_i}$$ is not at all a definition of a partition function, it is the application of this general idea to the density matrix $\mathrm{e}^{-\beta H}$, where $H$ is the Hamiltonian of our system: This $Z$ is just $\mathrm{Tr}(\mathrm{e}^{-\beta H})$ expressed in the eigenbasis of $H$ (or, said another way: the trace is the sum of eigenvalues).

Now, the path integral is a tool that appears in quantum mechanics even when we're not doing statistical physics. It works even when we don't have a notion of "microstates" vs. "macrostates", just normal states. It turns out that for a quantum mechanical system, there is a rule to compute matrix elements of any observable $A(q)$ that is a function of our "fundamental variable" $q$ (no matter whether it's a position variable or a field) via $$ \langle x\vert A \vert y\rangle = \frac{\int_{q(0)=x}^{q(t_f)=y}A(q)\mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q(t)}{\int_{q(0)=x}^{q(t_f)=y}\mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q(t)}$$ i.e. the "partition function" $$ Z_{qm} = \int_{q(0)=x}^{q(t_f)=y}\mathrm{e}^{\mathrm{i}S[q]}\mathcal{D}q(t)$$ acts as a normalization factor for the amplitude/matrix element. This is crucially not the exact same notion as in statistical mechanics - we're normalizing some sort of probability (amplitude), yes, but this is not an expression for $\mathrm{Tr}(\mathrm{e}^{-\beta H}$), at least not obviously so.

It turns out that there is indeed a connection between the quantum statistical partition function and the quantum mechanical path integral - the quantum statistical partition function can be expressed as a path integral over the Euclidean ("Wick rotated") action and "paths periodic in imaginary time": $$ Z_{stat} = \int_{q(0) = q(\beta)} \mathrm{e}^{-S_E[q]}\mathcal{D}q(t)$$ This then leads to the slogan that quantum field theory in Minkowski space is "equivalent" to statistical mechanics in Euclidean space, though one has to be careful what exactly one means by equivalent there. See this question and its answers for more discussions of this connection