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\bgroup\color{col1}$ N$\egroup-Boson systems

A basis for symmetric wave functions with \bgroup\color{col1}$ N$\egroup Bosons is constructed in the following way.

1. If we just have one possible state \bgroup\color{col1}$ \vert\nu_1\rangle$\egroup of the system, the symmetric state and the corresponding wave function is

\begin{displaymath}\begin{array}{cc}
\vert\underbrace{\nu_1,...,\nu_1}\rangle_S ...
...}(\xi_1)...\psi_{\nu_1}(\xi_N)\\
N {\rm times}&\\
\end{array}\end{displaymath}     (1.11)

This wave function is obviously symmetric.

2. If we have two particles ( \bgroup\color{col1}$ N=2$\egroup), the basis is constructed from the states \bgroup\color{col1}$ \vert\nu_1,\nu_2\rangle$\egroup with corresponding wave functions \bgroup\color{col1}$ \psi_{\nu_1}( \xi_1)\psi_{\nu_2}( \xi_2)$\egroup: this product is made symmetric,

$\displaystyle \vert\nu_1,\nu_2\rangle_S \leftrightarrow \langle\xi_1,\xi_2 \vert\nu_1,\nu_2\rangle_S$ $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{2}} \left[ \psi_{\nu_1}( \xi_1)\psi_{\nu_2}( \xi_2)
+ \psi_{\nu_1}( \xi_2)\psi_{\nu_2}( \xi_1) \right]$  
  $\displaystyle =$ $\displaystyle \hat{S} \psi_{\nu_1}( \xi_1)\psi_{\nu_2}( \xi_2).$ (1.12)

3. If we just have two possible state \bgroup\color{col1}$ \vert\nu_1\rangle$\egroup and \bgroup\color{col1}$ \vert\nu_2\rangle$\egroup for a system with \bgroup\color{col1}$ N$\egroup particles, \bgroup\color{col1}$ N_1$\egroup particles sit in \bgroup\color{col1}$ \vert\nu_1\rangle$\egroup and \bgroup\color{col1}$ N_2$\egroup particles sit in \bgroup\color{col1}$ \vert\nu_2\rangle$\egroup. We now have to symmetrize the states

    \begin{displaymath}\begin{array}{ccc}
\underbrace{\vert\nu_1,...,\nu_1},&\underb...
...xi_{N_2})
\\
N_1 {\rm times} & N_2 {\rm times}&\\
\end{array}\end{displaymath}  
    $\displaystyle N_1+N_2=N.$ (1.13)

If we apply the symmetrization operator \bgroup\color{col1}$ \hat{S}$\egroup to this product,
$\displaystyle \frac{1}{\sqrt{N!}}\sum_p \hat{\Pi}_p
\psi_{\nu_1}(\xi_1)...\psi_{\nu_1}(\xi_{N_1}) \psi_{\nu_2}(\xi_{N_1+1})...\psi_{\nu_2}(\xi_{N_2}),$     (1.14)

we get a sum of \bgroup\color{col1}$ N!$\egroup terms, each consisting of \bgroup\color{col1}$ N$\egroup products of wave functions. For example, for \bgroup\color{col1}$ N_1=1$\egroup and \bgroup\color{col1}$ N_2=2$\egroup we get
    $\displaystyle \frac{1}{\sqrt{3!}}\sum_p \hat{\Pi}_p \psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_3) =$ (1.15)
  $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{N!}}\Big[\psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_3) +
\psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_3) \psi_{\nu_2}(\xi_2)$ (1.16)
  $\displaystyle +$ $\displaystyle \psi_{\nu_1}(\xi_2) \psi_{\nu_2}(\xi_1) \psi_{\nu_2}(\xi_3) +
\psi_{\nu_1}(\xi_2) \psi_{\nu_2}(\xi_3) \psi_{\nu_2}(\xi_1)$ (1.17)
  $\displaystyle +$ $\displaystyle \psi_{\nu_1}(\xi_3) \psi_{\nu_2}(\xi_1) \psi_{\nu_2}(\xi_2) +
\psi_{\nu_1}(\xi_3) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_1)\Big],$ (1.18)

where in each line in the above equation we have \bgroup\color{col1}$ N_2!=2!$\egroup identical terms. Had we chosen an example with \bgroup\color{col1}$ N_1>1$\egroup and \bgroup\color{col1}$ N_2>1$\egroup, we would have got \bgroup\color{col1}$ N_1!N_2!$\egroup identical terms in each line of the above equation. The symmetrized wave function therefore looks as follows:
$\displaystyle \frac{1}{\sqrt{N!}} N_1! N_2! \left[ \mbox{\rm sum of }\frac{N!}{N_1!N_2!} \quad \mbox{\rm orthogonal wave functions} \right],$     (1.19)

which upon squaring and integrating would give
$\displaystyle \left[ \frac{1}{\sqrt{N!}} N_1! N_2!\right]^2 \frac{N!}{N_1!N_2!} = N_1!N_2!$     (1.20)

and not one! We therefore need to divide the whole wave function by \bgroup\color{col1}$ 1/\sqrt{N_1!N_2!}$\egroup in order to normalise it to one, and therefore the symmetric state with the corresponding normalised, symmetrical wave function is
    $\displaystyle \vert\nu_1,...,\nu_1,\nu_2,...,\nu_2\rangle_S$ (1.21)
  $\displaystyle \leftrightarrow$ $\displaystyle \frac{1}{\sqrt{N!}\sqrt{N_1!}\sqrt{N_2!}}\sum_p \hat{\Pi}_p
\psi_...
......\psi_{\nu_1}(\xi_{N_1}) \psi_{\nu_2}(\xi_{N_1+1})...\psi_{\nu_2}(\xi_{N_2}).$  

This is now easily generalised to the case where we have \bgroup\color{col1}$ N_1$\egroup particles in state \bgroup\color{col1}$ \nu_1$\egroup, \bgroup\color{col1}$ N_2$\egroup particles in state \bgroup\color{col1}$ \nu_2$\egroup,..., \bgroup\color{col1}$ N_r$\egroup particles in state \bgroup\color{col1}$ \nu_r$\egroup with
$\displaystyle \sum_{i=1}^r N_r =N.$     (1.22)

We then have
    $\displaystyle \vert\nu_1,...,\nu_1,\nu_2,...,\nu_2,...,\nu_r,...,\nu_r\rangle_S$ (1.23)
  $\displaystyle \leftrightarrow$ $\displaystyle \langle \xi_1 ,...,\xi_1,\xi_2,...,\xi_2,...,\xi_r,...,\xi_r
\vert\nu_1,...,\nu_1,\nu_2,...,\nu_2,...,\nu_r,...,\nu_r\rangle_S \equiv$  
  $\displaystyle \equiv$ $\displaystyle \frac{1}{\sqrt{N!}\sqrt{N_1!}\sqrt{N_2!}...\sqrt{N_r!}}\times$  
    $\displaystyle \sum_p \hat{\Pi}_p
\psi_{\nu_1}(\xi_1)...\psi_{\nu_1}(\xi_{N_1}) ...
...si_{\nu_2}(\xi_{N_2})
... \psi_{\nu_r}(\xi_{N-N_r+1})...\psi_{\nu_r}(\xi_{N})
.$  


next up previous contents index
Next: -Fermion systems Up: Basis vectors for Fermi Previous: Permutations   Contents   Index
Tobias Brandes 2005-04-26