Let's assume I have a one dimensional harmonic oscillator. The Eigen value 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?: 

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