Graduate States & Observables: Are They Really Different?

[anuttarasammyak]

What would the spectral decomposition of this matrix be? I.e. Could you write it as S^=∑S|S⟩⟨S|?

Say entropy is an observable in QM, in general for the set of same states, observed value differs in each observation, so
\sigma_S^2=<S^2>-<S>^2 \neq 0
What are the eigenstates of entropy? I do not know whether e.g., superpositon of systems of T=0, T_1, T_2 are valid statement or not.

 

[Morbert]

Here's my crude attempt at framing Shannon entropy as an observable to be measured, motivated by this paper. Consider first an experiment to measure the observable $A$ of a system with state space $\mathcal{H}$ prepared in the state $\rho$. What we actually want to measure is not $A$, but the Shannon entropy of this experiment.

First we model multiple experimental runs by creating a new state space that is a product of $N$ copies of the original state space $$\mathcal{H'} = \mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\otimes\dots$$This allows us to frame the relative frequencies of multiple experimental runs as a single observable, such that a single outcome is an observed set of relative frequencies $\{f_i\}$ with probability $$p(\{f_i\}) = \mathrm{tr}\rho'\Pi_{\{f_i\}}$$ For a standard experiment, and for a large enough $N$, the probability for a particular set of relative frequencies should be close to 1. The relative frequencies are the observed frequencies, and will approximate the set of probabilities $\{p_i\}$ determined by $\rho$ and $A$. You can then compute Shannon entropy the usual way from these probabilities.

To emphasise: Shannon entropy seems to be more an objective property of the experiment than a property of the microscopic system to be measured.

 

[vanhees71]

Formally entropy is not an observable. This is clear from the formal definition a la von Neumann,
$$S=-\mathrm{Tr}(\hat{\rho} \ln \hat{\rho}),$$
i.e., it is a property of the state rather than a state-independent observable.

Also the above assertions concerning thermodynamic entropy don't show that you can "measure" entropy dircectly. What's quoted from Wikipedia above, i.e., measuring entropy differences by measuring $\mathrm{d} S=-\delta Q/T$, where $Q$ is the heat energy in the system, is indeed only indirect and only valid for the special case of equilibrium.