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?:

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

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?:

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