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ph2b_hw1_soln

# ph2b_hw1_soln - PH2b Homework 1 solutions 1 Phillips 10.4...

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1 Phillips 10.4 Note first that, from equation (4.44), the given energy corresponds (up to exchange symme- try) to one of the particles in the ground state ( n x = n y = n z = 1), and one of the particles with one of the directions in the first excited state ( n x = 2, n y = n z = 1 for example). For the answers that follow, we arbitrarily choose the x direction to have the additional unit of energy – though one could equally have chosen the y or z direction. 1. Distinguishable implies no exchange symmetry, so we have something like Ψ = ψ 1 , 1 , 1 ( x p , y p , z p ) ψ 2 , 1 , 1 ( x q , y q , z q ) (1) or any of the other 5 choices of which subscript to set to “2”. 2. Identical bosons implies symmetric under exchange symmetry, so Ψ = 1 2 [ ψ 1 , 1 , 1 ( x p , y p , z p ) ψ 2 , 1 , 1 ( x q , y q , z q ) + ψ 1 , 1 , 1 ( x q , y q , z q ) ψ 2 , 1 , 1 ( x p , y p , z p )] (2) (note that p and q have been swapped between the terms). 3. Identical fermions implies antisymmetric under exchange. The spin state is symmetric, so the spatial wavefunction must be antisymmetric, Ψ = 1 2 [ ψ 1 , 1 , 1 ( x p , y

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ph2b_hw1_soln - PH2b Homework 1 solutions 1 Phillips 10.4...

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