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How do I find the energies of these states?

  1. Apr 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Here is the problem: http://imgur.com/XEqE4SY

    2. Relevant equations
    |psi_s_ms> = |s, ms> ⊗ Σ D_i_j |psi_i, psi_j>

    3. The attempt at a solution
    I know the singlet state in the |s, ms> basis is |0,0> = (1/sqrt(2))[ |up, down> - |down, up>] and that the hamiltonian for this system is H = (-hbar/2m)(∇1^2 + ∇2^2).

    How would you go about getting a value of energy. I feel like this is an easy question, I just don't know how to start. Is <0,0|H|0,0> the right approach? How would I even calculate that without an explicit expression for the state?

    My other approach is recognizing that [S, H] = 0 so spin eigenstates are simultaneous energy eigenstates. I don't know if this helps me?

    Can someone point me in the right direction?

    Thank you
  2. jcsd
  3. Apr 8, 2016 #2
    first you have to find out the solution for schrodinger equation invoking the boundary conditions for the two electron system and getting the energy eigen values and wave functions - naturally
    one gets energy levels which are degenerate and the splitting of levels will take place as states like s,p,d,f for various l values come up .then choose the lowest three states as asked by lookig at possible combinations of l and s values.
    if the states splits then the energy change has to be calculated.
  4. Apr 8, 2016 #3
  5. Apr 8, 2016 #4


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    Homework Helper

    Are you familiar with solving the eigenvalue problem for a single particle in an infinite potential well? Finding the eigenvalues (energy levels) for this problem is very similar to the one particle version as drvrm pointed out above, you just need to pair the wavefunctions from the two particles under the rule governed by Pauli principle.
    Yes, that's true. But for now, find first the general form of the eigenfunctions neglecting the spin and Pauli principle.
    Angular momentum operator is not a useful quantity in a 1D problem as it is equal to the zero operator.
    The inclusion of spins does not cause splitting as the Hamiltonian doesn't contain any of the spin operators.
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