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In the book, under the context of microcanonical ensemble, the author claims that the probability of a small system being in a particular state of energy $E$ is different from the probability of the small system having energy $E$, since the former probability is proportional to $e^{-E/k_BT}$ and the probability of the latter is the product of the Boltzmann probability and the number of states with that energy.

I wander how these two probabilities are different since both the probabilities need to calculate the number of states with energy $E$.

Moon Traveler
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  • It is something silly. The probability that the microstate is precisely (2,0) is less than when the macrostate is 2, because the macrostate is 2 can be (1,1) and (0,2) too. In particular, the former case is $p(\text{microstate})=e^{-E/k_BT}$ whereas the latter is $p(\text{macrostate})=g(E)e^{-E/k_BT}$, where $g(E)$ is just the multiplicity of the macrostate. – naturallyInconsistent May 11 '23 at 08:33
  • Thanks! It turns out that I didn't pay much attention to the word "particular". And It didn't occur to me the relationship that "system having energy $E$ is a macrostate" while "system in a particular state of energy E corresponding to a microstate." – Moon Traveler May 11 '23 at 10:59

2 Answers2

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Your statement, "I wonder how these probabilities are different since both the probabilities need to calculate the number of states with energy E" is incorrect, as, in your first case, the probability of finding the system in a particular state of energy E is given by the general formula:exp(-Ei/KT), where Ei is energy of i th energy eigenstate-whereas, in the second case, the entire system is taken to have energy E (to be interpreted as mean energy of the system), such that the sum of states having energy between E and E+dE must be multiplied by exp(-E/KT) (as all states are equally probable and hence characterized by the exponential factor exp(-E/kT)). Hence the probabilities are different in the two cases.

Puppeteer
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[...] the probability of a small system being in a particular state of energy $E$ is different from the probability of the small system having energy $E$, since the former probability is proportional to $e^{−E/k_BT}$ and the probability of the latter is the product of the Boltzmann probability and the number of states with that energy.

Let us consider a system of $N$ non-interacting spins (or DNA bonds), each of them having energy $\Delta$ when it is up, and energy $0$, when it is down. Suppose that the system has energy $m\Delta$, that is $m$ spins are up. There are ${N \choose m}$ ways to choose $m$ spins among $N$ to be pointed up. The probability of each such state (i.e., probability of a small system being in a particular state of energy E) is $$e^{-\frac{m\Delta}{k_BT}},$$ whereas the probability that system has energy $m\Delta$, that is the probability that the system is found in any of these states (i.e., probability of the small system having energy E), is $$ {N \choose m}e^{-\frac{m\Delta}{k_BT}}.$$

In other words: there are many microstates corresponding to the same macrostates. While every single microstate may have very low probability, the macrostate might have high probability due to many microstates realizing the same energy. Thus, in the example above, a state where all the spins point down ($m=0$), although having the lowest energy (if $\Delta>0$) is often less likely than the states where some spins are up.

In microcanonical ensemble Boltzmann entropy for the system discussed above is defined as $$S_B(m,N)=k_B\log {N \choose m}.$$

In canonical ensemble, where the energy is not fixed, we would minimize the free energy, which roughly corresponds to $$\frac{F}{k_BT}\sim \log\left[{N \choose m}e^{-\frac{m\Delta}{k_BT}}\right]=S_B(m,N)-\frac{m\Delta}{k_BT}.$$ (See this answer for a more general derivation.)

Roger V.
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    Very detailed answer, thanks! – Moon Traveler May 11 '23 at 11:00
  • Btw, is there any relationship between a microcanonical ensemble and a macrostate? I think that although a microcanonical ensemble is a set and a macrostate is just a state a macrostate can correspond to a microcanonical ensemble because a macrostate in the above context corresponds to microstates with the same energy $E$. And this idea probably only works for the above context since the definition of microstates is sometimes not based on energy and so do macrostates. – Moon Traveler May 12 '23 at 01:59
  • @MoonTraveler microstate is a particular configuration of the system - it is not specific to the microcanonical ensemble. Microcanonical is the one where the system is isolated from its surroundings, in canonical it can exchange energy with surroundings, in grand canonical it can also exchange particles. Energy is special, since it is an integral of motion. There are total 7 integrals of motion (energy + 3 components of momentum + 3 components of angular momentum), but in a center-of-mass coordinates and in non-rotating system, only energy matters. Other parameters do kot have the same status. – Roger V. May 12 '23 at 05:36
  • I see. So let's suppose that $A$ is the interested system and $A_t^i$ is the system in state $1$ at time $t$. The ensemble is the set of systems with all possible states(microstate), i.e. $\text{ensemble} = {A_t^1,A^t_2,....}$ and for the systems in the ensemble, they have the same macro quantities(macrostate), like volume, pressure. And the microcanonical ensemble requires each system of the ensemble to be isolated, that is, there is no interaction with the outside world – Moon Traveler May 12 '23 at 12:15
  • @MoonTraveler In microcanonical ensemble, all the systems are constrained to have the same energy. That is the ensemble is not all the microstates, but all the microstates of certain energy. When we say system is isolated we mean that it does not exchange energy or particles with surroundings = the energy and the particle number are fixed. – Roger V. May 12 '23 at 12:21
  • I suppose that energy can be regarded as a macro quantity, so the ensemble is just the set of systems with some certain macro quantities (energy, pressure etc.). Thus, do we must specify that all the systems are constrained to have the same energy or just need to state that all the systems are constrained to have the same macro quantity(probably not energy) when we are talking about the microcanonical ensemble? And what about other ensembles? – Moon Traveler May 12 '23 at 12:49
  • @MoonTraveler Energy is a function of state, which depends on parameters. If you want to see it as a constraint, you may want to check Jaynes' derivation of statistical mechanics from the maximum entropy principle. However, energy is special from the point of view of physics. – Roger V. May 12 '23 at 14:33