Timeline for Quantum ideal gas - Canonical ensemble - Occupation number summation notation (Huang)
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2015 at 13:40 | vote | accept | xihiro | ||
| Jan 22, 2015 at 13:39 | comment | added | xihiro | I was thinking of identical particles. That means the degeneracy of $E_n$ is determined by the number of ways we can distribute the same $p_i$s momenta among the identical particles (thus resulting in the same configuration). Thanks! | |
| Jan 22, 2015 at 10:48 | comment | added | glS | @nvon yes I agree with what you wrote (note however that in your examples the particles are not identical, otherwise you should have for example also the state $E_{1_c} = |\epsilon_2,\epsilon_1,\epsilon_2 \rangle$, but that is not relevant to your problem I think). | |
| Jan 21, 2015 at 21:41 | comment | added | xihiro | I think we were saying the same thing, but the repeated $j$ index got me confused. I edited the post (EDIT) by adding an example - is that in agreement to what you wrote? | |
| Jan 21, 2015 at 16:46 | comment | added | glS | @nvon I edited the post. Tell me if that helps | |
| Jan 21, 2015 at 16:45 | history | edited | glS | CC BY-SA 3.0 |
added 771 characters in body
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| Jan 21, 2015 at 16:34 | comment | added | glS | @nvon in the case of free particles you can think of the index $j$ as labeling the momenta, i.e. as being your $\textbf p$. Given that the energy dispersion relation is $E_{\textbf p} = \textbf p^2/2m$ there is a one-to-one corrispondence between the two. So call it $n_j$ or $n_{\textbf p}$, the key point is that you are labeling the single particle states. | |
| Jan 21, 2015 at 16:22 | comment | added | glS | The index $j$ labels the one particles states, meaning that each particle is in some state which we label $j=1,2,...$. With $\{n_j\}$ we mean the $N$-particles state with $n_1$ particles in the state $j=1$, $n_2$ particles in the state $j=2$ and so on. The condition $$ \sum_j n_j = N,$$ where $N$ is the total number of particles is implicit in these calculations (and in the sum (2)). With (2) we mean that we can have $n_1$ of the $N$ particles in the state $j=1$, $n_2$ in the state $j=2$ and so on. Each of these states is equally valid, and so we sum over all of them. | |
| Jan 21, 2015 at 16:17 | comment | added | xihiro | I am not sure, the $j$ index you used does not make much sense to me - isn't the occupation number the number of particles occupying a given state? As far as I could understand, what you wrote suggests that, if I consider (*) (assuming the addendum up to this part is correct) I may not only have several particles with the same energy, $\epsilon_i$, corresponding to $E_j$, but there may also be particles with energy $\epsilon_i$ corresponding to another eigenvalue $E_k$. The total number of particles with energy $\epsilon_i$ will then be $n_\vec{p}$ (like in the book), right? | |
| Jan 21, 2015 at 13:23 | history | answered | glS | CC BY-SA 3.0 |