# Probability of having energy $E$ when $E$ is bounded

For a canonical ensemble the probability of a system to have energy $E$ is $P(E)=e^{-\beta E}$. For that we consider the that the system can have any energy between $0$ to $\infty$. What will be the probability if the system can have energy only within certain range say, $E_{min}\leq E \leq E_{max}$ ($E_{min},E_{max}>0$)?

Derivation for canonical ensemble

For a canonical ensemble I follow this derivation. Consider the total energy of system and reservoir ($E_1+E_2$) is constant and then optimize the total number of possible configuration of system and reservoir, i.e. $\frac{\partial}{\partial E_1} (\Omega_1 \Omega_1) =0$ (Statistical Mechanics, R K Pathria). This along with the relation $\frac{\partial E_2}{\partial E_1}=-1$ gives $\frac{\partial}{\partial E_1} \ln \Omega_1 = \frac{\partial}{\partial E_2} \ln \Omega_2 = \beta$. Assuming all the configurations are equally probable, the probability is given by $P(E_1)=\Omega_1^{-1} = e^{-\beta E_1}$.

But this doesn't imply $P(E)=0$ for $E>E_{max}~or~E<E_{min}$. The reason is probably that the boundedness condition ($E_{min}\leq E \leq E_{max}$) is not used in the derivation.

How to obtain the correct probability factor if the energy is bounded?

A similar problem ever puzzled me for a while. That was the spin-$\frac{1}{2}$ particle in magnetic field, where the energy spectrum is not only bounded but also discrete!

I persuade myself by a careful analysis of the derivation. Let's denote the heat bath by $H$, the system of interest by $S$ and the total system $T$, using them as subscripts. The number of states $\Omega$ has the following relation, $$\Omega_T(E) = \Omega_H(E_T-E) \Omega_S(E)$$ where $E$ is the energy of $S$.

Our fundamental postulate(equal a priori probability postulate) tells us the probability is $$P(E) = \frac{\Omega_T(E)}{\sum_{\text{all possible E}} \Omega_T(E) }$$ By definition of entropy in statistical mechanics, $$\Omega_T(E) = \Omega_S(E) e^{\ln \Omega_H(E_T-E)} = \Omega_S(E) e^{\frac{1}{k} S_H(E_T - E ) }$$ and the Taylor expansion we have, $$S_H(E_T - E ) = S_H(E_T) - E \frac{\partial S_H}{\partial E} + \frac{1}{2!} E^2 \frac{\partial^2 S_H}{\partial^2 E} + \cdots$$ The first order term is identified with the (inverse) temperature of the heat bath(also the temperature of the system by 0th law ). The second order term is the response function $$\frac{\partial S_H}{ \partial_E} = -\frac{1}{T_H^2} \frac{\partial T_H}{\partial E} = - \frac{1}{T_H^2 C_V}$$ the heat capacity for a heat bath should be divergent, after all that why it's called a heat bath! Presumably other terms are higher order response functions and should vanish for similar reasons. Eventually, we have the expansion $$S_H(E_T - E ) = S_H(E_T) - \frac{E}{T_H}$$

The complete expression for the probability is
$$P(E) = \frac{\Omega_S(E) e^{-\frac{E}{kT_H} } }{ e^{-S_H(E_T)/k}\sum_{\text{all possible E}} \Omega_T(E) } = \frac{\Omega_S(E) e^{-\frac{E}{kT_H} } }{Z} = \frac{g e^{-\frac{E}{kT_H} } }{Z}$$ notice that I have identified $\Omega_S(E)$ as the degeneracy of the states.

For bounded/discrete energy levels, those "missing" energy states simply have $$g = \Omega_s(E) = 0$$

We can use canonical ensemble even though we know very little about the system of interest, that's because we know it is in the equilibrium with the almost simplest(and fortunately the most common) thermodynamic environment - a heat bath.

• thanks @anecdote, and sorry for the delay. I was in vacation. I think then we need at least one more information to find the total probability $P(E)$ and that would be the the density of state ($g(E)$). As much I can remember, $P$ and $g$ is connected by a Laplace transformation with a kernel $E^{\beta E},~\beta=1/kT$ and you also have used the inverse transformation in the last definition of $P(E)$. So I think $g$ would be the probability at zero temperature and for finite temperature it has to be multiplied by proper factor depending on the statistics (MB,BE,FD) the system obeys. – Sumit Jan 4 '14 at 10:20
• @Sumit, Yes, statistics times $g(E)$ is the probability staying at this energy. Actually the density of states(degeneracy) is a trouble in practical calculation. – anecdote Jan 4 '14 at 14:31

The probability of state $i$ with energy $E_i$ is proportional to $e^{-\beta E_i}$ always, independently of the allowed interval of energies. The fact that system can have only energy inside finite interval of energies $\langle E_{min}, E_{max}\rangle$ implies that the system cannot be in state with energy that is not in this interval, hence these states are assigned probability 0. The rest of the states, with energy in the above interval, is assigned probability $p_k = Ne^{-\beta E_k}$ such that total probability is 1:

$$\sum_{k=k_{min}}^{k_{max}} Ne^{-\beta E_k} =1.$$

• the term N is I think is the partition function - that is just a scaling or better to say normalisation factor. With this factor the states with $E<E_{min}$ or $E>E_{max}$ have finite probability which conflicts with my initial assumption. – Sumit Dec 24 '13 at 20:13
• No, these states have prescribed probability 0, because they have energy outside the allowed interval. I've edited the answer. – Ján Lalinský Dec 24 '13 at 22:19