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2 non-interacting fermions in 1D SHO

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Two identical non-interacting spin 1/2 particles are in the one-dimensional simple harmonic oscillator potential V(x) = kx2/2. The particles are in the lowest-energy triplet state.
    a. Write down the normalized space part of the wave function.
    b. Calculate the energy of this state.
    c. If the position of both particles are measured, what is the probability that both particles will be located on the right-hand side of the minimum in the potential?


    2. Relevant equations



    3. The attempt at a solution
    Since they are fermions, the wave equation is anti-symmetric. If X(x1, x2) is the spin state and Y(x1, x2) is the position state, then Psi = YX. Since the spin state is symmetric in the lowest energy triplet state, the position state must be anti-symmetric.

    So, Y(x1, x2) = A[Ya(x1)Yb(x2) - Yb(x1)Ya(x2)]

    Here is my problem: in the lowest energy triplet state, both particles are in ground state. So the only difference in state is from the spin. If this assumption is correct, then a = b and Y = 0, which would make Psi = 0, so cannot be right. I think I am misunderstanding something fundamental here. I hope someone can clarify this situation for me.
     
  2. jcsd
  3. Feb 9, 2010 #2

    vela

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    You're assuming that in the lowest-energy triplet state, both particles are in the single-particle ground state. The reason you're getting the confusing result is because the assumption is wrong.

    There are many states where the spins of the electron are in the triplet state. One, and possibly more if there is degeneracy, will have the lowest energy among them. That's the state the particles are in, and it's not necessarily the ground state of the system.
     
  4. Feb 9, 2010 #3
    Thanks for the response, but I'm afraid I still don't understand. If I don't know the spin state, how do I know whether to use a symmetric or anti-symmetric spatial equation? Do I have to use them all, multiplying by appropriate spatial equation to make the product anti-symmetric, and then sum the products together?
     
  5. Feb 9, 2010 #4

    vela

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    You have the spin state. What you've shown so far is that with that spin state, you can't be in the system's ground state. The lowest-energy triplet state and lowest-energy state are not the same state.
     
  6. Feb 10, 2010 #5
    It would seem that you are correct. Since the spatial equation cancels to zero at the ground state, thus making psi forbidden at that state, the lowest energy is with one fermion in n = 1 and the other in n = 2. Thanks for your cryptic yet insightful response.
     
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