# Probability in canonical ensemble

For a system in thermal contact with reservoir (bath) at constant temperature, why in text books like Modern Statistical Mechanics (Chandler), they use the notion that the probability of the system being in a state with energy E is directly proportional to the number of microstates of the reservoir (bath) ?

I think you mean "The probability of having a certain energy is proportional to the number of microstates with that energy if it is at constant $T$".

Yes, that's true, see: the probability (in equilibrium) of the system being in a certain microstate is

$$p(i)= \frac{1}{Z} \cdot e^{-\beta E(i)}$$

The probability of measuring an energy $E_0$ is the sum of probabilities of all microstates with that energy.

In other words, the probability of $E=E_0$ is the sum of having $E_0$ via microstate 1 + having it via microstate 2 +...

Mathematically:

$$p(E_0) = \sum_{i \ | \ E(i)=E_0} p(i)$$

And, since $p(i)$ depends only on $Z$, $E$ and $T$, and all those are constant, then you have

$$p(E_0) = p(i) \cdot\sum_{i \ | \ E(i)=E_0} 1 = p(i)\cdot N$$

Where $N$ is the number of microstates with $E(i)=E_0$.

Hence you have

$$p(E_0)=\frac{N(E_0)}{Z} e^{-\beta E_0}\quad \propto N(E_0)$$

Pathria's "Statistical mechanics" equivalent chapter on the canonical ensemble has a better explanation on why this is true (pg.44). Their argument goes as follows:

1. Fix the state of the system and its energy $E_{v}$ (although that is somewhat counterintuitive in the canonical ensemble stage, you can still imagine doing it)

2. The microcanonical ensemble applies to the isolated system+bath and assume system is small so $E=E_v+E_B$, $E_v\ll E_B$.

3. Now the available states to the system+bath are only the ones of the bath since we fixed the energy of the system! So one has to define a new microstate density related to the bath $\Omega_{B}(E_{B})$ since the sybsystem under consideration is the bath and it has energy $E_B$.

4. By expanding $\ln\Omega_{B}$ around $E$ the rest of the argument that proves the Boltzmann distribution goes through since the bath and the system+bath have the same temperature, which implies that:

$\frac{\partial \ln\Omega_B}{\partial E'}\Big|_{E'=E}=\frac{\partial \ln\Omega}{\partial E'}\Big|_{E'=E}$

where in the last equality corrections to $\mathcal{O}(E_v)$ have been neglected.

I hope this breakdown helps.