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.