Quantum Entropy and Energy States Part II

 In this note we wish to examine three ideas. First we consider the case of W(x,t)=Sum over n an exp(-iEn t) Wn(x) ((1)) (for n=0,1 as an example) and note that the von Neuman entropy is 0 for such a state. This is pointed out in (1) together with the fact that there exists a von Neumann density operator exp(-H/T) (H=Hamiltonian operator) which yields the same result as Shannon’s entropy S using P(ei) if one uses  dQ/T=Sum over n  en dP(en)/T = dS. It seems then that there is a contradiction if the usual von Neumann entropy is 0 yet using a density operator = exp(-H/T) yields nonzero results. A resolution is suggested in (2) which deals with information theory and we see if it applies to ((1)).

    Secondly, we consider the Shannon’s form  P(e)ln(P(e)) in more detail. In Part I and in previous notes we argued that ln ensures that S=S1+S2 if one has two systems. We also argued that dE= Sum over n  en dP(en) = TdS if dWork =0. In this note, we argue that a stronger idea may be at work, namely the idea that  Eave= Sum over n  en P(en/T)  and TS are essentially the same function except for a function of T. This approach may be used to show that ln is the function associated with entropy.

   Finally, we note as in Part I that when using either the von Neumann approach or Shannon’s entropy one must know which probabilities to use. For a system with energy levels en, P(n)=P(en/T) is an immediate candidate, but there may be more probabilities in the system related to a particle jumping from one ei to another. In such a case, if one uses these probabilities and then sums over some of them leaving only P(en/T), the form of entropy changes from that of Shannon’s form. (Originally, however, Shannon’s form is used before summing.)