# Physical Intuition for thermodynamic limit of the average occupancy, $\langle N\rangle$, of an orbital

Say your allowed occupancies for an orbital are $$0, 1, 2$$, and the values of energy associated with these occupancies are $$0, \epsilon, 2 \epsilon$$.

Give an expression for the ensemble average occupancy $$\langle N\rangle$$ when the system composed of this orbital is in thermal and diffusive contact with a reservoir at temperature $$\tau$$ and chemical potential $$\mu$$.

How does $$\langle N\rangle$$ behave in the limit:

(a) $$\tau \to \infty$$

(b) $$\tau \to 0$$ with $$\mu < \epsilon$$

(c) $$\tau \to 0$$ with $$\mu > \epsilon$$

I found $$\langle N\rangle$$ to be:

$$\langle N\rangle = \frac{\exp\left(\frac{\mu-\epsilon}{\tau}\right) + 2 \exp\left(\frac{2(\mu-\epsilon)}{\tau}\right)}{1 + \exp\left(\frac{\mu-\epsilon}{\tau}\right) + \exp\left(\frac{2(\mu-\epsilon)}{\tau}\right)}$$

And am reasonably confident it is correct. This means that

as $$\tau \to \infty$$, $$\space \langle N\rangle \to 1$$

as $$\tau \to 0$$, $$\space \langle N\rangle \to 0 \space$$ for $$\mu < \epsilon$$

as $$\tau \to 0$$, $$\space \langle N\rangle \to 2 \space$$ for $$\mu > \epsilon$$

But why should I expect these answers to be correct?

I believe that as $$\tau \to 0$$, the particles will enter the lowest energy state available to them, and as far as I understand chemical potential, it is sort of a measure for how much work must be done to remove a particle from a system.

There is not only thermal contact in this problem but also chemical. Hence the particles will enter the state with lowest $E - \mu N$ value as $\tau \to 0$.
• If $0 < \mu < \epsilon$, then $\epsilon - \mu < \epsilon$, so I would expect $\langle N\rangle \ge 1$, but this is not the case – infinitylord Dec 10 '17 at 15:58
• I don't understand how $\left< N\right> \geq 1$ follows from $\epsilon - \mu < \epsilon$. If $\epsilon - \mu > 0$, than addition of a particle increases value of $E-\mu N$, while this quantity should be minimal at equilibrium as $\tau \to 0$. – Gec Dec 10 '17 at 18:30