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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?

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The first step is to realize that the spatial wavefunctions for two non-interacting particles are 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.

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  • $\begingroup$ Why is it that the total wave function can be separated into a product of the spatial wave function and the spin wave function? $\endgroup$ – user148792 May 22 '18 at 12:24
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    $\begingroup$ @delickcrow123 By the method of separation of variables, wavefunctions for different, non-interacting degrees of freedom will be products. Just like the product wavefunctions for the spatial part. $\endgroup$ – ZeroTheHero May 22 '18 at 12:26
  • $\begingroup$ Can't the different degrees of freedom be entangled and, therefore, cannot be written simply as a product? $\endgroup$ – user148792 May 22 '18 at 12:36
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    $\begingroup$ @delickcrow123 that would be through linear combinations of product states, v.g. Bell states $\sim \vert +\rangle \vert -\rangle \pm \vert -\rangle \vert +\rangle $ $\endgroup$ – ZeroTheHero May 22 '18 at 13:03
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    $\begingroup$ @delickcrow123 I can search states however I want, but I know that the product states will be a basis and that therefore any state (entangled or not) will be a linear combination of product states. $\endgroup$ – ZeroTheHero May 22 '18 at 15:51

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