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shootmeproton
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Homework Statement
The spin 1/2 electrons are placed in a one-dimensional harmonic oscillator potential of angular frequency ω. If a measurement of $$S_z$$ of the system returns $$\hbar$$. What is the smallest possible energy of the system?
Homework Equations
$$\hbar\omega(n+1/2)|n>$$
The Attempt at a Solution
By computing the total spin, we get $$\hbar$$. This indicates that the electrons are in one of the triplet state, meaning it is symmetric.
We can rule out the ground state because the ground state has antisymmetric spatial wavefunction, thus not in the triplet.
Now, one of those three triplets have the next lowest energy after ground state but I do not know which state would have the next lowest energy...
The triplets are:
$$|\uparrow>|\uparrow>$$
$$\frac{1}{\sqrt{2}}(|\downarrow>|\uparrow>+|\uparrow>|\downarrow>)$$
$$|\downarrow>|\downarrow>$$
Any suggestion as to how to find the next lowest energy level?
In addition, I do not know how to compute the energy level solely based on spin of the electrons. Is there a relevant equation that describes the relation between spin and energy level of the system?