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Let us suppose that we have a system $(A,B)$ consisting of two independent subsystems $A$ and $B$. Suppose that $A$ has $m$ states having energies $E_1^{(A)}, \dots, E_m^{(A)}$ with probabilities $p_1, \dots, p_m$ respectively and system $B$ has $n$ states having energies $E_1^{(B)}, \dots, E_n^{(B)}$ with probabilities $q_1, \dots, q_n$ respectively.

Suppose that the average energy of system $A$ is defined as $E^{(A)}=\sum_i p_i' E_i^{(A)}$, where $p_i'=\frac{\phi(p_i)}{\sum_j \phi(p_j)}$ for some nice function $\phi$ defined on $[0,\infty)$, $\phi(0)=0$.

Similarly for system $B$ its given by $E^{(B)}=\sum_j q_j' E_j^{(B)}$.

Suppose that the combined system has energies, $$E_{ij}^{(A,B)}=E_i^{(A)}+E_j^{(B)} \text{ with probabilities } r_{i,j}=p_i\cdot q_j$$ for $i=1,\dots, m, j=1, \dots, n$ and $E^{(A,B)}=\sum_{i,j} r_{i,j}'E_{ij}^{(A,B)}$.

The question is , for what functions $\phi$, we have $E^{(A,B)}=E^{(A)}+E^{(B)}$?

Obviously for $\phi(x)=x^s, s>0$, it holds. My intuition suggests that in order for the additivity to hold the function should factor like $\phi(xy)=\phi(x)\phi(y)$. So the only functions which satisfy this Cauchy functional equation are $\phi(x)=x^s, s>0$. I know its not a rigorous enough proof. Can someone prove or disprove this?

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That can't be true, since you can scale $\phi$ by a constant without changing anything. Also, I don't see why $\phi(x)=x^s$ should lead to additivity. –  joriki Nov 23 '11 at 10:01
Thats not a big deal. Its fine if one can show $\phi(x)=k\cdot x^s$. –  Ashok Nov 23 '11 at 10:05
joriki, a computation would show $\phi(x)=x^s$ leads to additivity. –  Ashok Nov 23 '11 at 10:07
I see now -- I suggest to delete "Obviously" in the question :-) –  joriki Nov 23 '11 at 10:33

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