1. The problem statement, all variables and given/known data 2. Relevant equatio

Question

1. The problem statement, all variables and given/known data


2. Relevant equations

H=p12/2m+p22/2m

3. The attempt at a solution

Since they are electrons, they are spin 1/2 particles and arefermions. Thus, their wavefunction must be antisymmetric withrespect to the exchange of particle labels. This means thetwo particles have a total spin of 0 or 1, so there are twowavefunctions for the ground state?

S=0: |0,0,s=0,m=0> and S=1: |0,1,s=0,m=0>. Am I on the right track here?

Explanation / Answer

Label the box energy levels by|1;n>=n(r1) as the spacialwavefunction for particle 1 in eigenstate n of the box. Label thespin for particle 1 having up as |1+> NOTE: I haven't normalized the total wavefunctions below. The lowest energy is found by populating the n=0 orbital twice. Theonly way to do this is if one spin is up and the other in down, bythe Pauli principle. So the ground state is 0 = |1;0>|1+>|2;0>|2-> -|1;0>|1->|2;0>|2+> 0 = |1;0>|2;0>(|1+>|2-> -|1->|2+>) Here the spacial part is symmetric, while the spin part isantisymmetry under exchange of 1 and 2. There is only one spin combination that is anti-symmetric. So thereis only 1 possibility for the ground state wavefunction. The next lowest energy is when one electron is in the n=0 and theother is in n=1. Lets make the singlet first 1 = (|1;0>|2;1> +|1;1>|2;0>)(|1+>|2-> - |1->|2+>) and the triplet states will be 1 = (|1;0>|2;1> -|1;1>|2;0>)(|1+>|2-> + |1->|2+>) 1 = (|1;0>|2;1> -|1;1>|2;0>)|1+>|2;+> 1 = (|1;0>|2;1> -|1;1>|2;0>)|1->|2-> I hope this helps, my notation is very bad.

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