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I had a question about the differences between a quantum state and a classical microstate.

Let's say we have an NVE ensemble, and we are trying to predict some property P of our ensemble. In our methods, we say that each microstate in out ensemble has some canonical coordinates $(\mathbf{r}^N, \mathbf{p}^N)$. The number of microstates in our ensembles is equal to $\Omega(N,V,E)$, and according to the a priori probability principle, the probability of each of these ensembles is $p=1/\Omega$.

Cool, that makes sense.

Now I have a quantum microcanonical ensemble, and I want to enumerate a macroscopic property P of my system. These are the notes that I was given:

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My question is, isn't $\mathcal{N} = \Omega$? I thought the definition of ensemble was that it was a list of boxes which had a certain configuration (or wavefunction), and each box had a given NVE. And by our definition of density of states, the number of boxes in our ensemble was equal to the density of states $\Omega$.

From my understanding, in classical ensembles, we have phase coordinates defining our system, in quantum systems, we have wave functions defining our system. Am I right in saying that?

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This notation seems a bit confusing to me and I would recommend reading a different text on the topic.

However, in the terms that you present, $\Omega(N,V,E)$ is the number of different (non-repeated) microstates compatible with the macrostate $(N,V,E)$. This is different from $\mathcal{N}$, which is the random collection of microstates compatibles with $(N,V,E)$. Think on an extreme example: you are interested in a macrostate $(N,V,E)$ which only have one compatible microstate, say $x$, then $P(x)=1\rightarrow\Omega=1$. However, the collection (or ensemble) of states have still $\mathcal{N}$ elements, in particular, these elements are all copies of the state $x$.

I don't understand the relationship between your doubt about $\mathcal{N}$ and the differences in quantum and classical states. Briefly, the statistics differ because of the intrinsic random nature of a quantum state (ket). While in classical systems the source of randomness is just due to the degeneracy of microstates compatible with a macrostate, a quantum microstate has also a stochastic nature that requires a different formalism (E.g. density matrix).

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This notation seems familiar. There is a small error here, if all the quantum states were equally likely, then degenerate eigenvalues would be more likely observed and $P_i \neq \dfrac{1}{\Omega}$. I have given the corrected statement here. Besides that, let me answer your questions.

My question is, isn't $\mathcal{N} = \Omega$?

No, $\mathcal{N}$ can be any very large number; it is the number of systems in the ensemble. $\Omega$ is number of the microstates of the system $\Omega = \sum_{i}\Omega (E_{i})$. The probability is given by $P_{i} = \dfrac{\Omega (E_{i})}{\Omega}$. Hope this also explains that $\Omega$ is not the density of states.

From my understanding, in classical ensembles, we have phase coordinates defining our system, in quantum systems, we have wave functions defining our system. Am I right in saying that?

The variables of state have no correlation with the likelihood of observing a system in a particular state. Phase coordinates in classical systems (defining the state) are not the same as a basis set chosen for the wavefunction (variables of state describing the system).

Correction: The correct statement in the book should be 'each eigenvalue is equally probable' or that $P_i = \dfrac{\text{deg}_{i}}{\Omega}$ where $\text{deg}_{i}$ is the the number of quantum system corresponding to the eigenvalue $E_{i}$, that is its degeneracy.

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