Ground state of two electrons in one dimensional S.H.O

Question

Let's assume I have a one dimensional harmonic oscillator. The eigenvalue of the oscillator would be $E= (n+ \frac{1}{2}) \hbar \omega$.

Now I have two electrons (their spins are identical, I mean either both are spin up or spin down) and I want to find the ground spin state of the oscillator.

If I want to look at the triplet of the two electron system I can have two of the similar spin directions which are $$|{\uparrow \uparrow}\rangle$$ $$|{\downarrow \downarrow}\rangle.$$

Here is how I understand it:

Since both electrons spins are are identical, we can not put them in the same quantum number. Like if we put first electron in the state $n=0$, next one has to be in the first excited state (n=1).

Do you think I can write the spin state of similar spins for the lowest ground state like this?:

$$ \alpha |{\uparrow_0 \uparrow_1}\rangle + \beta |{\downarrow_0 \downarrow_1}\rangle$$

Answer 1

This depends on the statistics of your particles and with electrons (fermions) the total (i.e. spin plus spatial parts) wave function must be antisymmetric. Since the spectrum is $(n_1+n_2+1)\hbar\omega$, the lowest energy state is one where $n_1=n_2=0$ and the spatial part of the wavefunction is thus $$ \psi_{00}(x_1,x_2)= \psi_0(x_1)\psi_0(x_2) $$ which is symmetric under permutation of the particle numbers. Thus, the spin part of the wavefunction must be antisymmetric, i.e. must be $\chi_1(\uparrow)\chi_2(\downarrow)-\chi_1(\downarrow)\chi_2(\uparrow)$, where $\chi(\uparrow)$ is a spin up state, etc. This is the singlet state, not a state in the triplet. Hence the total wavefunction (up to normalization) is given by $$ \Psi(x_1,x_2)=\psi_0(x_1)\psi_0(x_2)\left[\chi_1(\uparrow)\chi_2(\downarrow)-\chi_1(\downarrow)\chi_2(\uparrow)\right]. $$ The three (unnormalized) spin states of a triplet states are \begin{align} \vert 11\rangle &= \chi_1(\uparrow)\chi_2(\uparrow)\, , \\ \vert 10\rangle &=\chi_1(\uparrow)\chi_2(\downarrow)+\chi_1(\downarrow)\chi_2(\uparrow)\, , \\ \vert 1,-1\rangle &=\chi_1(\downarrow)\chi_2(\downarrow) \end{align} and these states are symmetric. Any linear combo of these states is also symmetric so the lowest possible energy state for this must be spatially antisymmetric, i.e. must be of the form (again unnormalized). $$ \phi(x_1,x_2)=\psi_0(x_1)\psi_1(x_2)-\psi_1(x_1)\psi_0(x_2) $$ with total state $$\Psi_{01}=\left(\psi_0(x_1)\psi_1(x_2)-\psi_1(x_1)\psi_0(x_2)\right)\left[a\vert 11\rangle +b \vert 10\rangle+c\vert 1,-1\rangle \right]\, , $$ still antisymmetric under exchange of particle labels.

Answer 2

First note the spectrum. The energy actually is $(n_{1}+n_{2}+1)\hbar\omega$. So, the ground state actually is $(0,0)$, and there are two degenerate first excited states $(1,0)$ and $(0,1)$. The spin label doubles the number of states for each set of quantum numbers.

So, if both spins are paired up, the ground state would be both electrons in $(0,0)$ with opposite spins. Otherwise, one spin would be promoted to one of the next levels. In that case, your expression for the state seems to be correct.