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Unformatted text preview: PH2b: Homework 1 solutions
January 10, 2007 1 Phillips 10.4 Note ﬁrst that, from equation (4.44), the given energy corresponds (up to exchange symmetry) to one of the particles in the ground state (nx = ny = nz = 1), and one of the particles with one of the directions in the ﬁrst excited state (nx = 2, ny = nz = 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 = ψ1,1,1 (xp , yp , zp )ψ2,1,1 (xq , yq , zq ) or any of the other 5 choices of which subscript to set to “2”. 2. Identical bosons implies symmetric under exchange symmetry, so 1 Ψ = √ [ψ1,1,1 (xp , yp , zp )ψ2,1,1 (xq , yq , zq ) + ψ1,1,1 (xq , yq , zq )ψ2,1,1 (xp , yp , zp )] 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, 1 Ψ = √ [ψ1,1,1 (xp , yp , zp )ψ2,1,1 (xq , yq , zq ) − ψ1,1,1 (xq , yq , zq )ψ2,1,1 (xp , yp , zp )] (3) 2 4. Again, identical fermions implies antisymmetric under exchange. However, the spin state is antisymmetric, so the spatial wavefunction must be symmetric for the overall wavefunction to be antisymmetric. thus 1 Ψ = √ [ψ1,1,1 (xp , yp , zp )ψ2,1,1 (xq , yq , zq ) + ψ1,1,1 (xq , yq , zq )ψ2,1,1 (xp , yp , zp )] (4) 2 (Alternative answers that are acceptable: You could have included all choices of which direction to use for the additional unit of energy, weighted by the appropriate prefactor; you could have explicitly shown the spin part of the wavefunctions for parts (c) and (d); and you h could have included the time dependence as factors of the form eiEt/¯ ). 1 (1) (2) ...
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This note was uploaded on 01/08/2011 for the course PH 2b taught by Professor Martin during the Winter '08 term at Caltech.
 Winter '08
 Martin
 Physics, Energy, Work

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