Boltzmann Distribution With Constraints

Question

I have a problem with showing the existence of Boltzmann distribution given some constraints.

Consider $p_1,...,p_n$ a Boltzmann distibution, where $p_i=\frac{\epsilon^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{-\beta \cdot E_j}}$.

Define the expected value $E = \sum_{i}^{} p_i \cdot E_i$.

Whether Boltzmann distribution exists in the three following cases?

1. $E_1<E<\frac{\sum_{i}{} E_i}{n}$

2. $E>\frac{\sum_{i}{} E_i}{n}$

3. $E=\frac{\sum_{i}{} E_i}{n}$

Let's consider the first case, when $E_1<E<\frac{\sum_{i}{} E_i}{n}$.

$E= \sum_{i}^{} p_i \cdot E_i = \sum_{i}^{} \frac{e^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{- \beta \cdot E_j}}$.

Whether $E_1<\sum_{i}^{} \frac{e^{-\beta \cdot E_i}}{\sum_{j}^{} \epsilon^{- \beta \cdot E_j}}<\frac{\sum_{i}^{} E_i}{n}$?

The problem is so far I don't have an idea how to approach the solution.

Would appreciate any help.

Addendum: I would like to thank joriki for his great answer.

Let's try to make rigorous proof.

When $E_1<E_2$, by definition of Boltzmann probability $p_1>p_2$. Show that $E<\frac{E_1+E_2}{2}$. When $p_1>p_2$ and $p_1+p_2=1$ then $p_1>0.5$ and $p_2<0.5$.

$E_1p_1 + E_2p_2 $? $0.5E_1 + 0.5E_2$

$E_1p_1 + E_2(1-p_1) $? $0.5E_1 + 0.5E_2$

$E_1(p_1-0.5) + E_2(0.5-p_1) $? $0$

Where $p_1-0.5>0$ and $0.5-p_1<0$, because $p_1>0.5$.

I am really stuck here.

At least let's show that $E_1<E$

$E_1p_1 + E_2p_2$?$E_1$

$E_1(p_1-1) + E_2p_2$ ? 0

$-E_1p_2 + E_2p_2$ ?0

$p_2(E_2-E_1)>0$

Let's try to it show for the general case $\sum_{i}^{} E_ip_i > E_1$

[sorry for the last attempt]

$\sum_{i}^{} E_ip_i $?$ E_1$

$E_1p_1 + \sum_{i=2}^{} E_ip_i$?$ E_1$

$E_1(p_1-1) + \sum_{i=2}^{} E_ip_i$?$0$

$p_1+...+p_n =1$

$E_1(-\sum_{i=2}^{} p_i) + \sum_{i=2}^{} E_ip_i$?$0$

$\sum_{i=2}^{} p_i (\sum_{i=2}^{} E_i - E_1) > 0$

In addition, finally I get the idea in 2) the trick with derivative, unfortunately I cannot get it equal to negative variance, but I am keep trying, the computations get very long. But one minor point, by showing that derivative is negative we conclude that $E$ is a concave function with maximum at $\beta = 0$, which is exactly like in the 3) case where the distribution exists, so why Boltzmann distribution is impossible in 2)?

Answer 1

1. Take $E_1\lt E_2$; then $p_1\gt p_2$ and thus $E_1\lt E\lt(E_1+E_2)/2$.

2. This is impossible for non-negative $\beta$. To see this, form $\partial E/\partial\beta$, show that it's the negative variance of $E$ and thus non-positive, and infer that $E$ is never greater than its value at $\beta=0$, which is the mean.

3. This occurs either if all $E_i$ are equal, or for $\beta=0$, corresponding to $T\to\infty$.

[Edit in response to addendum:]

For 1., you started out on the right path; then you can write

\begin{align} p_1E_1+p_2E_2 &= p_1\left(\frac{E_1+E_2}2+\frac{E_1-E_2}2\right)+p_2\left(\frac{E_1+E_2}2-\frac{E_1-E_2}2\right) \\ &= (p_1+p_2)\frac{E_1+E_2}2+(p_1-p_2)\frac{E_1-E_2}2 \\ &= \frac{E_1+E_2}2+(p_1-p_2)\frac{E_1-E_2}2 \\ &\lt \frac{E_1+E_2}2\;. \end{align}

I'm not sure what you're trying to do to show that the mean is $\gt E_1$. The mean of elements that are not all equal is greater than the least of the elements; I doubt that you're expected to prove that ab initio.