What is the smallest energy of a system of 2 spin 1/2 electrons? This question comes directly from "Conquering the Physics GRE" by Kahn and Anderson.

Two spin-1/2 electrons are placed in a one-dimensional harmonic oscillator potential of angular frequency $\omega$. If a measurement of $S_z$ of the system returns $\hbar$, which of the following is the smallest possible energy of the system?

(The answer is $2\hbar\omega$.)
There is a solution in the book but I can't follow it. Can someone hopefully take the time to walk me through this?
Edit: Book solution
Here is the book solution:

Total $S_z=\hbar$ means the electrons must be in the triplet state, which is symmetric. For a totally antisymmetric wavefunction, the spatial wavefunction must be antisymmetric. This knocks out the ground state, where both electrons are in the $n=0$ state of the harmonic oscillator, since after antisymmetrization this vanishes identically. So the next available state is an antisymmetrized version having $n=0$ and $n=1$:
  $$\psi_{spatial}=\frac{1}{\sqrt{2}}(|0\rangle_1 |1\rangle_2-|1\rangle_1 |0\rangle_2).$$
  This is an energy eigenstate with energy $\hbar\omega/2+3\hbar\omega/2$.

My updated questions:


*

*Why do the states need to be antisymmetrized?

*How does the equation for $\psi_{spatial}$ give the energy of $\hbar\omega /2+3\hbar\omega /2$?


I hope that clarifies my question.
 A: The fact that the spin of the system is unit means that the system is in the "triplet state." Equally, the net spin of the system is spin 1, you can see following the standard prescription of "addition of angular momentum" for two spin half particles that there are three possible spin states, all of which are symmetric with respect to interchange of the fermions.
$$ \psi_1 = |+\rangle |+\rangle \\
\psi_2 = |-\rangle |-\rangle \\
\psi_3 = |+\rangle |-\rangle + |-\rangle |+\rangle
$$
Fermion statistics require that the net state of an $N$ fermion system be antisymmetric under interchange of any two fermions. Recall that the energy levels of the harmonic oscillator follow 
$$ E_n = \hbar \omega \left(n+\frac{1}{2}\right) $$
for the $n$'th harmonic. The specific form of the wavefunction is not important, but you can look them up on wikipedia here. What is important is that any state of the system can be written as 
$$ \psi = \sum \limits_i c_i \psi_i(1)\psi_i(2) $$
where $c_i$ is the amplitude for the state and $\psi_i (1,2)$ are the one particle wavefunctions for particle 1 and 2 respectively (written in the "eigenbasis" of some operator with eigenvalues $i$, in this case we take the Hamiltonian so that we have states of definite energy). 
We know we can write $\psi$ as the product of the spin part and the spacial part. From above, we can conclude that the spin part of the state is symmetric, therefore, the spacial part must be antisymmetric. The lowest possible energy spacial wavefunction is then (disregarding normalization) 
$$
\psi_\text{spacial} = \psi_0(1) \psi_1(2) - \psi_1(1)\psi_0(2)
$$
Here we have set $c_i$ to 0 for all $i$ except this combination since any other combination must have larger values for $n$ or not be antisymmetric. This is the only way to get an antisymmetric combination of $n=0$ and $n=1$, and that's the lowest energy level since trying to write an antisymmetric state with both $n = 0$ would cause a cancellation and result in $\psi = 0$.
Then the total energy of such a state is the sum of the energies of the individual states, namely, $E_\text{net} = E_0 + E_1$, which gives you the answer of $E = 2 \hbar \omega$.
A: Fermions can't be in the same state according to the Pauli exclusion principle. That's why two electrons with equal spins should occupy two different energy levels. The lowest combination is E0+E1
