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Relation between harmonic oscillator potential and spin

  1. Aug 20, 2014 #1
    1. The problem statement, all variables and given/known data

    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?


    2. Relevant equations

    $$\hbar\omega(n+1/2)|n>$$

    3. 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?
     
  2. jcsd
  3. Aug 25, 2014 #2

    nrqed

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    Gold Member

    Watch out. They say that the [itex] z \, component [/itex] is measured to be ## \hbar##. They are not talking about the total spin here, but the z component. This means that both their spin must be up. Therefore their spin function is symmetric. From this what can we say about their energy?
     
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