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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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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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