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.