It is easy to understand the idea of macrostate and microstate in the context of classical statistical mechanics. The description of a microstate of a system requires the specification of position and momenta of each particle that comprises the system. It is a huge number of variables $\sim 10^{23}$ or more. Descrition of macrostate implies specifying fewer variables such as energy, temperature, pressure etc.

I understand the microstates are described by wavefunctions but I do not quite understand how a macrostate for a system is described in quantum statistics?

  • $\begingroup$ You have heard of density operators, haven't you? $\endgroup$ – user154997 Jul 6 '17 at 21:27
  • $\begingroup$ I have. But density operator $\hat{\rho}$ is the analog of classical phase space density $\rho(q,p,t)$. In classical statistical physics, macrostates are described by ensemble averages and not by $\rho(q,p,t)$. Isn't it? $\endgroup$ – SRS Jul 6 '17 at 21:29
  • $\begingroup$ I wrote an answer explaining the classical case a while ago (not trying to plug my stuff: there are other similar answers; just too lazy!). Now you just replace the integrals $\int d^3xd^3p$ by a trace and the phase space density with the density operator, and that's it. $\endgroup$ – user154997 Jul 6 '17 at 21:37
  • $\begingroup$ In both the classical and the quantum case, macrostates are described by probability distributions (formalized as phase space density in classical case, and density operator in quantum case) over the microstates of the system. The distribution that is appropriate in a given context can be obtained by maximizing the information entropy of the distribution subject to appropriate macroscopic constraints (e.g. fixing average energy for the canonical distribution). $\endgroup$ – joshphysics Jul 7 '17 at 6:53
  • $\begingroup$ @joshphysics Isn't a macrostate in the classical case described by macroscopic constraints? For example, $E,V,N$? Doesn't the specification of the macroscopic variables $E,V,N$ specify a macrostate? If yes, why do you say that the phase space density describes a macrostate? I'm confused. $\endgroup$ – SRS Jul 7 '17 at 9:57

Disclosure: The following is just mostly a close retelling of the "Precise expression" section of the Wikipedia article on the microcanonical ensemble.

A macrostate is a probability distribution over microstates. It is not, intrinsically, specified by "temperature" or "volume", those are very special kinds of macrostates:

A classical macrostate is given by a phase space distribution $\rho(p,q)$ and often given by macroscopic variables (e.g. $E,V,N$ in the microcanonical ensemble approach), where you implicitly assume it's an equilibrium macrostate where every microstate that has these same macroscopic parameters within a range is equally likely, i.e. the probability distribution here is $$ \rho(q,p) = \frac{1}{Z h^N C}f\left(H(q,p)-E,\omega\right),$$ for $h$ the coarse graining phase space volume, $C$ a coorection factor for indistinguishability, $Z$ a normalization factor and $\omega$ the range of energies we allow, where $f(x,\omega)$ is a function that peaks around $x=0$ with "width" $\omega$. Ideally, we'd take $\omega\to 0$, meaning $f$ is the Dirac delta.

In the same way, a quantum microstate is a density matrix, since a probability distribution over pure states is what a density matrix is. For the microcanonical ensemble, one would take energy eigenstates $\lvert E_i\rangle$ and write $$ \rho= \frac{1}{Z}\sum_i f(E_i - E,\omega) \lvert E_i\rangle\langle E_i\rvert,$$ but in this case the limit $\omega\to 0$ is somewhat dubious if the $E_i$ are discrete.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.