$N$-Boson systems

A basis for symmetric wave functions with $N$ Bosons is constructed in the following way.

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

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

This wave function is obviously symmetric.

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

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

$$= \hat{S} \psi_{\nu_1}( \xi_1)\psi_{\nu_2}( \xi_2).$$ (1.12)

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

$$\begin{array}{ccc} \underbrace{\vert\nu_1,...,\nu_1},&\underb... ...xi_{N_2}) \\ N_1 {\rm times} & N_2 {\rm times}&\\ \end{array}$$

$$N_1+N_2=N.$$ (1.13)

If we apply the symmetrization operator $\hat{S}$ to this product,

$$\frac{1}{\sqrt{N!}}\sum_p \hat{\Pi}_p \psi_{\nu_1}(\xi_1)...\psi_{\nu_1}(\xi_{N_1}) \psi_{\nu_2}(\xi_{N_1+1})...\psi_{\nu_2}(\xi_{N_2}),$$ (1.14)

we get a sum of $N!$ terms, each consisting of $N$ products of wave functions. For example, for $N_1=1$ and $N_2=2$ we get

$$\frac{1}{\sqrt{3!}}\sum_p \hat{\Pi}_p \psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_3) =$$ (1.15)

$$= \frac{1}{\sqrt{N!}}\Big[\psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_3) + \psi_{\nu_1}(\xi_1) \psi_{\nu_2}(\xi_3) \psi_{\nu_2}(\xi_2)$$ (1.16)

$$+ \psi_{\nu_1}(\xi_2) \psi_{\nu_2}(\xi_1) \psi_{\nu_2}(\xi_3) + \psi_{\nu_1}(\xi_2) \psi_{\nu_2}(\xi_3) \psi_{\nu_2}(\xi_1)$$ (1.17)

$$+ \psi_{\nu_1}(\xi_3) \psi_{\nu_2}(\xi_1) \psi_{\nu_2}(\xi_2) + \psi_{\nu_1}(\xi_3) \psi_{\nu_2}(\xi_2) \psi_{\nu_2}(\xi_1)\Big],$$ (1.18)

where in each line in the above equation we have $N_2!=2!$ identical terms. Had we chosen an example with $N_1>1$ and $N_2>1$, we would have got $N_1!N_2!$ identical terms in each line of the above equation. The symmetrized wave function therefore looks as follows:

$$\frac{1}{\sqrt{N!}} N_1! N_2! \left[ \mbox{\rm sum of }\frac{N!}{N_1!N_2!} \quad \mbox{\rm orthogonal wave functions} \right],$$ (1.19)

which upon squaring and integrating would give

$$\left[ \frac{1}{\sqrt{N!}} N_1! N_2!\right]^2 \frac{N!}{N_1!N_2!} = N_1!N_2!$$ (1.20)

and not one! We therefore need to divide the whole wave function by $1/\sqrt{N_1!N_2!}$ in order to normalise it to one, and therefore the symmetric state with the corresponding normalised, symmetrical wave function is

$$\vert\nu_1,...,\nu_1,\nu_2,...,\nu_2\rangle_S$$ (1.21)

$$\leftrightarrow \frac{1}{\sqrt{N!}\sqrt{N_1!}\sqrt{N_2!}}\sum_p \hat{\Pi}_p \psi_... ......\psi_{\nu_1}(\xi_{N_1}) \psi_{\nu_2}(\xi_{N_1+1})...\psi_{\nu_2}(\xi_{N_2}).$$

This is now easily generalised to the case where we have $N_1$ particles in state $\nu_1$, $N_2$ particles in state $\nu_2$,..., $N_r$ particles in state $\nu_r$ with

$$\sum_{i=1}^r N_r =N.$$ (1.22)

We then have

$$\vert\nu_1,...,\nu_1,\nu_2,...,\nu_2,...,\nu_r,...,\nu_r\rangle_S$$ (1.23)

$$\leftrightarrow \langle \xi_1 ,...,\xi_1,\xi_2,...,\xi_2,...,\xi_r,...,\xi_r \vert\nu_1,...,\nu_1,\nu_2,...,\nu_2,...,\nu_r,...,\nu_r\rangle_S \equiv$$

$$\equiv \frac{1}{\sqrt{N!}\sqrt{N_1!}\sqrt{N_2!}...\sqrt{N_r!}}\times$$

$$\sum_p \hat{\Pi}_p \psi_{\nu_1}(\xi_1)...\psi_{\nu_1}(\xi_{N_1}) ... ...si_{\nu_2}(\xi_{N_2}) ... \psi_{\nu_r}(\xi_{N-N_r+1})...\psi_{\nu_r}(\xi_{N}) .$$