When are all microstates equally probable? So I am taking my introductory statistical mechanics course, and this concept is something that I can not wrap my head around. My professor said that all microstates are equally probable and this is an axiom, always true.
However when I take a look at the canonical ensemble, I  find that the probability of a given microstate is $P_i=exp(-\beta E_i)$, which depends on $\beta$ and the energy of the given microstate. So if I have two microstates with different $E$ energies, then they will have different $P$ probabilities! They are NOT equally probable!
Another seemingly counter-example is the harmonic oscillator. It is known that a real (as in normal mass and energy scales) harmonic diatomic molecules have $\Delta E=h \nu$ energy differences between its vibration quant, and this results in that the vast majority of the molecules occupy the $\nu = 0$ state. How is this possible, when all states should be equally probable? By the way, is this example really different from the canonical ensemble, or is it concepually the same?
 A: In the microcanonical ensemble, all states with a fixed energy are equally probable. In classical mechanics, this follows essentially from ergodic theory (non-linear systems tend not to have 'many' conserved quantities). 
If you fix the energy of a 1D quantum harmonic oscillator, then you uniquely specify the state. Things are more interesting however if you have a collection of (weakly coupled) oscillators with fixed 'total energy' $E\approx E_1+E_2+\dots+E_N=\sum_{i=1}^N h\nu n_i$. (This formula ignores a contribution from interactions between oscillators, which we assume is found to be small empirically). 
In this case, for a given value of $E$ the number of states is given by $\sum_{\{n_i\}_{i=1}^N}\mathbb{1}(\sum_{i=1}^Nn_i=\frac{E}{h\nu})$. When $N$ is large, this is approximately the hyper-volume of an $N-1$ dimensional simplex. This lets you compute the entropy of a collection of the oscillators in the microcanonical ensemble, which then lets you compute the free energy per oscillator, which you can compare with the partition function of a single oscillator in the large $N$ limit. 
The probabilities you describe are derived from the canonical ensemble: the Boltzmann factors arise by assuming your system (a harmonic oscillator) is already weakly coupled to a thermal bath.  
A: Many microstates $N_i$ can have the same energy $E_i$, many can have another energy say $E_j$ ($N_j$). If $N_i$ is bigger than $N_j$ then $E_i$ is more probable than $E_j$ since all microstates are equally likely in equilibrium.  If its Max. Boltzman statistics then
$$\frac{P_i}{P_j} = \exp( (E_j-E_i)\cdot\beta)$$
at equilibrium.
A: I have been struggling with the same issue, this is what I have come to myself - it may or may not be correct.
Microstates are only equiprobable in an isolated system in thermal equilibrium. In the canonical ensemble we have a system and a reservoir, together forming a lager system we will call 'system + reservoir'. When distributing some energy E amongst the 'system + reservoir' we have far more microstates in which the reservoir has much more energy than the system. The 'system + reservoir' is isolated and in thermal equilibrium, so all micro states are equiprobable. We've established then that we are more likely to have a 'system + reservoir' in which the reservoir holds the bulk of the energy.
If, now, we examine the system, we see that its ‘system microstates’ are no longer equiprobable. Lets look at some of these ‘system microstates’ from the two possible extremes. (By ‘system microstates’ I am referring to a microstate only of the system, ‘system + reservoir microstates’ refer to the ‘true’ microstates.)
System Microstate A: The system has 0 energy.
This could also be considered a macrostate of the system - the macrostate of energy 0 has only one microstate - only one way to arrange 0 energy.
Macrostate B: A state in which the system has all the energy
There are many similar system microstates which contribute to the macrostate in which the system has all the energy of the 'system + reservoir' -  if you like, many different ways to distribute all the energy of the ‘system + reservoir’  among only the system.
However, we are still more likely to find the system in macrostate A, which has only one ‘system microstate’ associated with it, than B. This is because when looking not just at the system, but at the ‘system + reservoir’, we see that for each of the similar ‘system microstate that males up macrostate B, there is only one associated ‘system + reservoir microstate’– one way to distribute the remaining 0 energy within the reservoir for each of the different ways the energy is distributed within the system. For ‘system microstate’ A there are a vast number of ways the remaining (entirety of the) energy can be distributed amongst the reservoir – there are far more ‘system + reservoir’ microstates associated with the one ‘system microstate’ where the system has 0 energy in situation A, than there are different ‘system microstates’ with the same, maximal energy in situation B, and each of the latter has only one associated ‘system +reservoir’ microstate.
When viewing the system then, it appears that its microstates are not equiprobable when connected to the reservoir – ‘system microstates’ in which the system has more energy are  more numerous than states where the system has low energy, but they are far less probable, because they are associated with far less ‘system + reservoir’ microstates – it is these that dictate the likelihood of a macrostate. ‘system + reservoir microstates’ are true microstates in that they are equiprobable (since the ‘system + reservoir’ is an isolated system in thermal equilibrium).
Working through the maths reveals that individual ‘system microstates’ have probabilities given by the Boltzmann distribution. The key is that this unequal probability is just because less probable ‘system microstates’ have less associated ‘system+reservoir microsates’.
