One of the common derivations of the canonical ensemble goes as follows:

Assume there is a system in the contact with heat reservoir which together form an isolated system. Heat can be exchanged between the system and reservoir until thermal equilibrium is established and both are at temperature $T$.

Suppose system is in microstate $i$ and that it has energy $E_{i}$, energy of the reservoir is than defined as $E_R = E - E_i$. In the last equation $E$ is the energy of the total isolated system which must be constant.

We want to determine the number of microstates of the total isolated system in which the system is in microstate $i$ or the probability that isolated system will be in such a state. Number of such microstates is denoted as $\Omega_i(E)$ and is equal to the number of microstates of the reservoir $\Omega_R(E - E_i)$ since the system is in particular microstate $i$.

Total number of microstates of the isolated system is defined as: $$ \Omega (E) = \sum_{i=1} \Omega_R(E - E_i) $$

that is, it is equal to sum of the number of the reservoir microstates over every microstate of the system $i$.

Not going further into the derivation, number of microstates and probability mentioned previously are given by: $$\Omega_R(E - E_i) = \Omega_R(E)e^{- \beta E_i}$$ $$ P_i = \frac {e^{-\beta E_i}} {\sum_{j=1}^N e^{-\beta E_j}} $$

where the denominator of the previous equation is the canonical partition function $Z$ and $N$ is the number of the microstates in which the system can be.

Now, we come to my question.

According to the notation given, microstate of the system $i$ is determined by the system's energy $E_i$.

However, we know that the task of statistical thermodynamics is to connect macrostate of the system with microstates in which it can be found. Energy of the system is its macrostate and therefore such a macrostate is compatible with many different microstates in which system may be.

Because of this, energy of the system $E_i$ can't depend only on particular microstate $i$ since the same microstate can be compatible with more than one energy of the system.

  • 1
    $\begingroup$ When you write "Energy of the system is its macrostate", you refer to what energy and what system? $\endgroup$ Jul 4, 2022 at 22:03
  • $\begingroup$ In canonical ensemble, $E_i$ is just the energy of the microstate $i$. $\endgroup$
    – Meng Cheng
    Jul 5, 2022 at 0:12
  • $\begingroup$ @GiorgioP In the derivation of canonical ensemble, this is internal energy and the system is the one in contact with the reservoir. System is closed as energy can be transfered, but not particles. $\endgroup$ Jul 5, 2022 at 8:11
  • $\begingroup$ @MengCheng Yes, this is what I don't understand. Energy of the system should be its macrostate which means it can't be only determined by the microstate $i$ since the same microstate can be compatible with more than one energy of the system. $\endgroup$ Jul 5, 2022 at 8:14
  • $\begingroup$ @DarioMirić > "the same microstate can be compatible with more than one energy of the system." --- Why do you think this? This is not how the standard statistical physics works. One microstate means one energy, usually value of the system Hamiltonian at that microstate. $\endgroup$ Jul 5, 2022 at 12:14

1 Answer 1


$P_i$ here is the probability of a microstate. When calcilating the value of any macroscopic variable we will deal with a sum like $$ \langle A\rangle =\sum_i A_iP_i, $$ which is then converted into sum over energies using the density-of-states $$ g(E)=\sum_i\delta(E-E_i), $$ that is $$ \langle A\rangle =\sum_i A_iP_i=\int dE g(E)P(E)=Z^{-1}\int dE g(E)e^{-\beta E}. $$ We thus express a macroscopic variable in terms of the properties of the microstates. What is essential for this to work is that this macroscopic variable has the same value in all these microstates - which is another way of saying that all these microstates correspond to the same macrostate.

Remark: the question does not seem to be specific to the canonical ensemble - the only difference between the ensembles is the expression for probability $P_i$ (even if microcanonical ensemble might be sometimes unwieldy).

  • $\begingroup$ Yes, I agree. In equations I've written summing is done over all possible microstates ($N$ is number of all microstates). However what I am not sure about is that particular microstate $i$ can be compatible with more than one energy of the system $E_i$. This means that for any $i$ there may be more than one term in the partition function. However, equation form of the partition function doesn't allow more than one term for each microstate. $\endgroup$ Jul 5, 2022 at 9:21
  • $\begingroup$ @DarioMirić there is some ambiguity in your question about about what you call system: there is the sub-system or system of interest that we are interested in, and the system that comprises the subsystem and the reservoir. Anyway, microstate implies knowing positions and momenta of all the particles - it cannot correspond to more than one energy. $\endgroup$ Jul 5, 2022 at 9:29
  • $\begingroup$ To be clear, I used terms $isolated$ $system$ and $system$. The latter is what you call $system$ $of$ $interest$. $\endgroup$ Jul 5, 2022 at 9:34
  • $\begingroup$ However, isn't the microstate just an instanteneous configuration of particles over available energy states? What this means that system goes to different instanteneous configurations, but if there is no energy exchanged with the surroundings, all configurations (microstates) must have the same total energy. This argument is used in deriving Boltzmann distribution which we get by maximizing weight of the configuration with constraints that total energy and number of particles is conserved. $\endgroup$ Jul 5, 2022 at 9:42
  • $\begingroup$ @DarioMirić this is energy conservation - why does it surprise you? $\endgroup$ Jul 5, 2022 at 9:44

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