I am a bit confused about the difference between macrostate and microstate in the microcanonical ensemble. So I have read that for the microcanonical ensemble, the probabilities of each microstate are equal $$ p = 1/\Omega $$ where $\Omega$ is the number of microstates. For a given number of particles $N$ the number of microstates of particles distributed over discreet energy levels is given by

$$\Omega = \frac{N!}{\prod_i n_i} $$

where $n_i$ is the number of particles in the $i$th energy level. Maximising $\Omega$ given the constraint that the particle number is constant $\sum_i n_i = N $ and the energy is constant $\sum_i \varepsilon_i n_i = E$ gives

$$ p_i \propto e^{-\varepsilon_i/kT}$$

This gives a probability for a particle to be in the $i$th energy level. I thought if we were in the microcanonical ensemble all probabilities are equal?


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    $\begingroup$ What is equal is the probability of each microstate, that is, a given distribution of particles with total energi E. Each microstate will have particles with different energies with some distribution. Then, after averaging across all posible microstates, you get the probability $p_i$ $\endgroup$
    – user126422
    Commented Apr 21, 2017 at 1:32
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    $\begingroup$ Firstly , you lost the factorial in the formula for $\Omega$: $n_i!$. Secondly , $\ p_i$ is not the probability of a microstate of the microcanonical ensemble $\endgroup$ Commented Jun 27, 2017 at 9:33
  • $\begingroup$ You are getting the concepts of micro canonical and canonical ensembles wrong. In microcanonical ensemble all the accessible states of a member system is characterised by fixed energy $E$. So essentially, if you arbitrarily pick up any system and measure energy you are ascertained to get $E$. So the probability density function is not a function of energy. Here probability is $\dfrac{1}{\Omega}= \dfrac{\Pi_i n_i}{N!}$. $\endgroup$ Commented May 13, 2021 at 4:50

1 Answer 1


Here is how to derive the microcanonical and the canonical distributions. In all cases $$ \Omega(\{n_i\}) = \frac{N!}{\prod_i n_i!} $$


All $n_i$ in all distributions of the ensemble have the same energy: $E_i=E$ for all $i$. The problem is to find the distribution that maximizes $\Omega$ under the sole constraint $$ \sum_i n_i = N $$ The solution is $$ \frac{n_i^*}{N} = \frac{1}{\Omega} \doteq p_i $$ is the multiplicity of distirbution $\{n_i\}$.


We don't know the energy if microstate $i$ but we know that the average energy per microstate is the same in all distributions of the ensemble: $E_\text{tot}/N = \bar E$ for all distributions $\{n_i\}$. The problem now is to find $\{n_i\}$ that maximizes $\Omega$ under the following two conditions: $$ \sum_i n_i = N,\quad \sum_i n_i E_i = N \bar E $$ The solution now is $$ \frac{n_i^*}{N} = \frac{e^{-\beta E_i}}{Q} \doteq p_i, $$

The key difference is that all microstates in the microcanonical ensemble have the same energy while in the canonical ensemble the have the same energy only on average.


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