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The occupation-number representation

Suppose each of our particles can sit in any one of $M$ states $\{\vert\phi_i\rangle \}$. These could be, for example Let's assume for convenience these states are orthogonal (though that is not essential, so long as they are linearly independent). Then we can construct a complete basis set for the $N$-particle system from the symmetrized products:
$\displaystyle \vert i_1,\ldots,i_N\rangle$ $\textstyle =$ $\displaystyle {1\over\sqrt{N!}}\det\left\vert\begin{array}{ccc}
\phi_{i_1}(r_1)...
..._{i_N}(r_1)&\ldots&\phi_{i_N}(r_N)
\end{array}\right\vert\quad\hbox{(fermions)}$ (55)
  $\textstyle =$ $\displaystyle {1\over\sqrt{N!}}{\rm perm}\left\vert\begin{array}{ccc}
\phi_{i_1...
...i_{i_N}(r_1)&\ldots&\phi_{i_N}(r_N)
\end{array}\right\vert\quad\hbox{(bosons)}.$ (56)

We can alternatively specify each $\vert i_1,\ldots,i_N\rangle $ by the occupation number $n_j$ of each state $\phi_j$: this is simply the number of times state $j$ sppears in the set $\{i_1,\ldots i_N\}$. For fermions the occupation number must be 0 or 1; for bosons it can be any integer. Let us divide the single-particle basis set into two subsets, corresponding to the system $S$ and the environment $E$. Now the problem has a direct product structure: to specify a configuration of the system (one of the basis kets $\{\vert i_1,\ldots,i_N\rangle \}$) we have to specify the values of the occupation numbers for both the system region and the environment region. If we know there are exactly $N$ particles in the system, there is a constraint on the occupation numbers that
\begin{displaymath}
\sum_{i\in S}n_i+\sum_{j\in E}n_j=N.
\end{displaymath} (57)

However, if the system is very large then it generally makes no difference whether we work at fixed $N$ or at fixed chemical potential; in that case, no constraint on the occupation numbers is necessary.
next up previous
Next: Handling direct-product systems Up: Descriptions of many-particle systems Previous: The many-particle wavefuncion
Andrew Fisher 2004-07-14