Find the wave functions of 2 identical particles in infinite square well given their spin I want to find the wave functions for the three first energy states  of  2 identical particles in an infinite square well for two cases of spin 
1) both have zero spin
2) both have 1/2 spin
I know that $$\psi_n =\psi_{\mathrm{space}} \psi_{\mathrm{spin}}$$ but I have a problem finding both $\psi_{\mathrm{space}}$ and $\psi_{\mathrm{spin}}$
For the wave function of space, I don't know what does it mean for 2 identical particles to be in a specific energy state, what should $n_1$, $n_2$ be.  For 1 particle it is $n=1$ for the ground state $n=2$ for the first excited state and so on, but what about 2 particles?
And for the wave function of spin,  I can't find a formula or method to find it, can you explain the process I must follow?
 A: The first step is to realize that the spatial wavefunctions for two non-interacting particles are sums of the form $\psi_{n_1}(x_1)\psi_{n_2}(x_2)$ and so have energy $E=E_{n_1}+E_{n_2}\sim n_1^2+n_2^2$.  Thus the lowest energy states have the lowest sums $n_1^2+n_2^2$.
For the spin-1/2 case, the total wavefunction must be fully antisymmetric since spin-1/2 particles are fermions.  There are four possible types of spin wavefunctions for two spin-1/2 particles:
\begin{align}
\chi^1_1&= \chi^{1/2}_+(1)\chi^{1/2}_+(2)\, ,\\
\chi^1_{-1}&= \chi^{1/2}_- (1)\chi^{1/2}_-(2)\, ,\\
\chi^1_0&= \chi^{1/2}_+(1)\chi_-^{1/2}(2)+\chi^{1/2}_-(1)\chi^{1/2}_+(2)\, ,\\
\chi^0_0&=\chi^{1/2}_+(1)\chi_-^{1/2}(2)-\chi^{1/2}_-(1)\chi^{1/2}_+(2)
\end{align}
The first three are symmetric, while the last is antisymmetric under interchange of the particle label $1\leftrightarrow 2$.  Thus, the first three must be combined with a spatially antisymmetric wavefunction, and the last to a spatially symmetric wavefunction, to produce a full antisymmetric total wavefunction.
There are three possible types of spatial wavefunctions. The first two are symmetric wavefunctions  of the form
$$
\psi^+(x_1,x_2)=\left\{\begin{array}{lol}
\psi_{n_1}(x_1)\psi_{n_2}(x_2)+\psi_{n_2}(x_1)\psi_{n_1}(x_2)& 
\hbox{for } n_1\ne n_2\\
\psi_{n}(x_1)\psi_{n}(x_2)&\hbox{for } n_1=n_2=n
\end{array}
\right.
$$
There is also an antisymmetric combination for $n_1\ne n_2$:
$$
\psi^-(x_1,x_2)=\psi_{n_1}(x_1)\psi_{n_2}(x_2)-\psi_{n_1}(x_2)\psi_{n_2}(x_1)\, .
$$
which can occur when $n_1\ne n_2$.  The permutation symmetry here is on the spatial position $x_1\leftrightarrow x_2$, which results from the interchange particle labels $1\leftrightarrow 2$.
The lowest energy wave function is $\psi_{1}(x_1)\psi_1(x_2)$ and is space-symmetric.  Thus, it must be combined with $\chi^0_0$ to produce the fully antisymmetric state with lowest energy.
The next lowest are $\psi_1(x_1)\psi_2(x_2)\pm \psi_1(x_2)\psi_2(x_1)$.  The symmetric combination must be multiplied by $\chi^0_0$ to make it fully antisymmetric.  The antisymmetric combination can be multiplied by any one of the $\chi^1_m$ to make it fully antisymmetric.  This should be enough to produce what you need for the spin-1/2 case.
For the spin-0 case there is only one single particle wavefunction and thus there is only one product spin wavefunction:
$\chi^0_0(1)\chi^0_0(2)$; it is fully symmetric under the exchange of particle labels $1\leftrightarrow 2$.  It must therefore always be multiplied by a spatially symmetric wavefunction since spin-0 particles are bosons.  You can work out the details on this using the three types of spatial wavefunctions given above.
