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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?:

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

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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.

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  • $\begingroup$ So the ground state would be lowest energy state? But the solution of the ground state of a one dimensional harmonic oscillator is $(n+ \frac{1}{2})\hbar \omega $. How we can put two electron in the ground state? $\endgroup$
    – user193422
    Aug 11 '19 at 17:00
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    $\begingroup$ the energies for one particle are $(n+1/2)\hbar \omega$ so the energies for two non-interacting particles are the sums of 1-particle energies, i.e. $(n_1+n_2+1)\hbar\omega$. The Pauli principle states that fermionic states must be antisymmetric so it is possible to have two fermions in (0,0) state, i.e. the lowest possible energy, if the spin part is antisymmetric. $\endgroup$ Aug 11 '19 at 17:25
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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.

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  • $\begingroup$ Why do you put similar spin in the same state? Is it not the violation of Pauli exclusion principle? Why do you write the energy for two dimensional harmonic oscillator. Elaborate your answer. $\endgroup$
    – user193422
    Aug 11 '19 at 6:56
  • $\begingroup$ @user193422 actually you are assuming that both electrons are in the same atom, that is only when we apply the Pauli's exclusion principle. Usually, when we talk about a two electron system, they are two free electrons, as in two independent electrons. $\endgroup$ Aug 11 '19 at 9:46
  • $\begingroup$ I am not putting the electrons in the same state. The two electrons in the $(0,0)$ state have quantum numbers $S_{z}=+\frac{1}{2}$ and $S_{z}=-\frac{1}{2}$. Note that $S_{z}$ commutes with the hamiltonian, so its eigenstates will have a $S_{z}$ label, which takes two values. So, the spin label doubles every original state present without spin. $\endgroup$
    – Mani Jha
    Aug 11 '19 at 11:02
  • $\begingroup$ As to why the energy is this way, since the hamiltonian of a system of non-interacting particles is the sum of the individual hamiltonians, the energy values get added too. So, $E=E_{1}+E_{2}$ where $E_{1}=(n_{1}+\frac{1}{2})\hbar\omega$ and similarly for $E_{2}$. $\endgroup$
    – Mani Jha
    Aug 11 '19 at 11:05
  • $\begingroup$ If anything the spectrum if $(n_1+n_2+1)$, not $1/2$. isn't the statement "if both spins are paired up, the ground state would be both electrons in (0,0) with opposite spins" self-contradicting? $\endgroup$ Aug 11 '19 at 13:59

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